Signals and Systems: Lecture Notes
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6.003: Signals and Systems Signals and Systems
September 8, 2011 1
6.003: Signals and Systems Today’s handouts: Single package containing • Slides for Lecture 1 • Subject Information & Calendar Lecturer: Denny Freeman Instructors: Elfar Adalsteinsson Russ Tedrake TAs: Phillip Nadeau Wenbang Xu Website: mit.edu/6.003 Text: Signals and Systems – Oppenheim and Willsky
2
6.003: Homework Doing the homework is essential for understanding the content. • •
where subject matter is/isn’t learned equivalent to “practice” in sports or music
Weekly Homework Assignments • •
Conventional Homework Problems plus Engineering Design Problems (Python/Matlab)
Open Office Hours ! • •
Stata Basement Mondays and Tuesdays, afternoons and early evenings
3
6.003: Signals and Systems Collaboration Policy • •
Discussion of concepts in homework is encouraged Sharing of homework or code is not permitted and will be re ported to the COD
Firm Deadlines • • •
Homework must be submitted by the published due date Each student can submit one late homework assignment without penalty. Grades on other late assignments will be multiplied by 0.5 (unless excused by an Instructor, Dean, or Medical Official).
4
6.003 At-A-Glance Thursday
Friday
L1: Signals and Systems
R2: Difference Equations
HW1 R3: Feedback, due Cycles, and Modes
L3: Feedback, Cycles, and Modes
R4: CT Systems
L4: CT Operator Sep 20 Representations
HW2 Student Holiday: due No Recitation
L5: Laplace Transforms
R5: Laplace Transforms
Sep 27 L6: Z Transforms
HW3 R6: Z Transforms due
L7: Transform Properties
R7: Transform Properties
L9: Frequency Response
R8: Convolution and Freq. Resp.
Tuesday
Wednesday
Sep 6
Registration Day: No Classes
R1: Continuous & Discrete Systems
Sep 13
L2: Discrete-Time Systems
Oct 4
L8: Convolution; Impulse Response
EX4
Oct 11
Columbus Day: No Lecture
HW5 L10: Bode R9: Bode Diagrams Diagrams due
R10: Feedback and Control
Oct 18
L11: DT Feedback and Control
HW6 R11: CT Feedback due and Control
L12: CT Feedback and Control
R12: CT Feedback and Control
Exam 2 HW7 No Recitation
L14: CT Fourier Series
R13: CT Fourier Series
L13: CT Feedback Oct 25 and Control
Exam 1 No Recitation
Nov 1
L15: CT Fourier Series
EX8 due
R14: CT Fourier Series
L16: CT Fourier Transform
R15: CT Fourier Transform
Nov 8
L17: CT Fourier Transform
HW9 R16: DT Fourier due Transform
L18: DT Fourier Transform
Veterans Day: No Recitation
Nov 15
L19: DT Fourier Transform
L20: Fourier Relations
R17: Fourier Relations
HW10
Exam 3 No Recitation
Nov 22 L21: Sampling
EX11 R18: Fourier due Transforms
Thanksgiving: No Lecture
Thanksgiving: No-Recitation
Nov 29 L22: Sampling
HW12 R19: Modulation due
L23: Modulation
R20: Modulation Study Period Final Exams: No-Recitation
EX13 R21: Review
L25: Applications of 6.003
Breakfast with Dec 13 Staff
EX13 R22: Review
Study Period: No Lecture
Dec 20
Final finals Examinations: finals No Classes finals
Dec 6
L24: Modulation
finals
5
finals
6.003: Signals and Systems Weekly meetings with class representatives • •
help staff understand student perspective learn about teaching
Tentatively meet on Thursday afternoon Interested? ...
6
The Signals and Systems Abstraction Describe a system (physical, mathematical, or computational) by the way it transforms an input signal into an output signal.
signal in
system
7
signal out
Example: Mass and Spring x(t)
y(t)
x(t)
y(t)
t
mass & spring system
8
t
Example: Tanks r0 (t)
h1 (t) r1 (t)
h2 (t) r2 (t) r0 (t)
r2 (t) t
tank system
9
t
Example: Cell Phone System
sound out
sound in
sound in
sound out
t
cell phone system
10
t
Signals and Systems: Widely Applicable The Signals and Systems approach has broad application: electrical, mechanical, optical, acoustic, biological, financial, ... x(t)
y(t) mass & spring system
t
t
r0 (t) h1 (t) r1 (t)
r0 (t)
r2 (t)
h2 (t)
tank system
t
t
r2 (t) sound in
sound out
t
11
cell phone system
t
Signals and Systems: Modular The representation does not depend upon the physical substrate.
sound out
sound in
E/M cell sound optic sound cell E/M tower tower in phone out fiber phone
focuses on the flow of information, abstracts away everything else 12
Signals and Systems: Hierarchical Representations of component systems are easily combined. Example: cascade of component systems
sound in
E/M cell optic cell E/M tower tower fiber phone phone
sound out
Composite system
sound in
cell phone system
sound out
Component and composite systems have the same form, and are analyzed with same methods. 13
Signals and Systems Signals are mathematical functions. • •
independent variable = time dependent variable = voltage, flow rate, sound pressure x(t)
y(t) mass & spring system
t
r0 (t)
t
r2 (t) tank system
t
sound in
t
sound out
t
cell phone system
14
t
Signals and Systems continuous “time” (CT) and discrete “time” (DT)
x[n]
x(t)
n
t 0
2
4
6
8
10
0
2
4
6
8
10
Signals from physical systems often functions of continuous time. • mass and spring • leaky tank Signals from computation systems often functions of discrete time. • state machines: given the current input and current state, what is the next output and next state. 15
Signals and Systems Sampling: converting CT signals to DT
x(t)
x[n] = x(nT )
n
t 0T 2T 4T 6T 8T 10T
0
2
4
6
8
10
T = sampling interval
Important for computational manipulation of physical data. • •
digital representations of audio signals (e.g., MP3) digital representations of images (e.g., JPEG) 16
Signals and Systems Reconstruction: converting DT signals to CT zero-order hold
x(t)
x[n]
n 0
2
4
6
8
t 0
10
2T 4T 6T 8T 10T
T = sampling interval
commonly used in audio output devices such as CD players 17
Signals and Systems Reconstruction: converting DT signals to CT piecewise linear
x(t)
x[n]
n 0
2
4
6
8
t 0
10
2T 4T 6T 8T 10T
T = sampling interval
commonly used in rendering images 18
Check Yourself Computer generated speech (by Robert Donovan) f (t)
t
Listen to the following four manipulated signals: f1 (t), f2 (t), f3 (t), f4 (t). How many of the following relations are true? • • • •
f1 (t) = f (2t) f2 (t) = −f (t) f3 (t) = f (2t) f4 (t) = 13 f (t) 19
Check Yourself Computer generated speech (by Robert Donovan) f (t)
t
Listen to the following four manipulated signals: f1 (t), f2 (t), f3 (t), f4 (t). How many of the following relations are true? 2 • • • •
√ f1 (t) = f (2t) f2 (t) = −f (t) X f3 (t) = f (2t) X √ f4 (t) = 13 f (t) 20
Check Yourself f (x, y)
−250
0
250
y
−250
250
x
−250
0
250
f1 (x, y) = f (2x, y) ?
x
250 0 −250
0 −250
−250
0
250
How many images match the expressions beneath them? y y
250
y
0
−250
0
250
x
f2 (x, y) = f (2x−250, y) ? 21
−250
0
250
x
f3 (x, y) = f (−x−250, y) ?
Check Yourself
0
f (x, y)
250
x
−250
0
250
x
f1 (x, y) = f (2x, y) ?
250 0 −250
0 −250
0 −250
0 −250 −250
y
250
y
250
y
250
y
−250
0
250
x
−250
0
250
x
f2 (x, y) = f (2x−250, y) ? f3 (x, y) = f (−x−250, y) ?
√ x=0 → f1 (0, y) = f (0, y) x = 250 → f1 (250, y) = f (500, y) x=0 → f2 (0, y) = f (−250, y) x = 250 → f2 (250, y) = f (250, y) x=0 → f3 (0, y) = f (−250, y) x = 250 → f3 (250, y) = f (−500, y)
22
X √ √ X X
Check Yourself f (x, y)
−250
0
250
y
−250
250
x
−250
0
250
f1 (x, y) = f (2x, y) ?
x
250 0 −250
0 −250
−250
0
250
How many images match the expressions beneath them? y y
250
y
0
−250
0
250
x
f2 (x, y) = f (2x−250, y) ? 23
−250
0
250
x
f3 (x, y) = f (−x−250, y) ?
The Signals and Systems Abstraction Describe a system (physical, mathematical, or computational) by the way it transforms an input signal into an output signal.
signal in
system
24
signal out
Example System: Leaky Tank Formulate a mathematical description of this system.
r0 (t)
h1 (t)
r1 (t)
What determines the leak rate?
25
Check Yourself
The holes in each of the following tanks have equal size. Which tank has the largest leak rate r1 (t)?
1. 2.
3.
4.
26
Check Yourself
The holes in each of the following tanks have equal size. Which tank has the largest leak rate r1 (t)? 2
1. 2.
3.
4.
27
Example System: Leaky Tank Formulate a mathematical description of this system.
r0 (t)
h1 (t)
Assume linear leaking:
r1 (t)
r1 (t) ∝ h1 (t)
What determines the height h1 (t)?
28
Example System: Leaky Tank Formulate a mathematical description of this system.
r0 (t)
h1 (t)
r1 (t)
Assume linear leaking:
r1 (t) ∝ h1 (t)
Assume water is conserved:
dh1 (t) ∝ r0 (t) − r1 (t) dt
Solve:
dr1 (t) ∝ r0 (t) − r1 (t) dt 29
Check Yourself
What are the dimensions of constant of proportionality C?
� � dr1 (t) = C r0 (t) − r1 (t) dt
30
Check Yourself
What are the dimensions of constant of proportionality C? inverse time (to match dimensions of dt)
� � dr1 (t) = C r0 (t) − r1 (t) dt
31
Analysis of the Leaky Tank Call the constant of proportionality 1/τ . Then τ is called the time constant of the system. dr1 (t) r0 (t) r1 (t) = − dt τ τ
32
Check Yourself
Which tank has the largest time constant τ ?
1. 2.
3.
4.
33
Check Yourself
Which tank has the largest time constant τ ?
1. 2.
3.
4.
34
4
Analysis of the Leaky Tank Call the constant of proportionality 1/τ . Then τ is called the time constant of the system. dr1 (t) r0 (t) r1 (t) = − dt τ τ Assume that the tank is initially empty, and then water enters at a constant rate r0 (t) = 1. Determine the output rate r1 (t).
r1 (t)
time (seconds) 1
2
3
Explain the shape of this curve mathematically. Explain the shape of this curve physically. 35
Leaky Tanks and Capacitors Although derived for a leaky tank, this sort of model can be used to represent a variety of physical systems. Water accumulates in a leaky tank.
r0 (t)
h1 (t)
r1 (t)
Charge accumulates in a capacitor.
ii
io + v −
C
i − io dv = i ∝ i i − io dt C
analogous to 36
dh ∝ r0 − r1 dt
MIT OpenCourseWare http://ocw.mit.edu
6.003 Signals and Systems Fall 2011
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
6.003: Signals and Systems Discrete-Time Systems
September 13, 2011
1
Homework
Doing the homework is essential to understanding the content. Weekly Homework Assigments • •
tutor (exam-type) problems: answers are automatically checked to provide quick feedback engineering design (real-world) problems: graded by a human
Learning doesn’t end when you have submitted your work! • • • • •
solutions will be posted on Wednesdays at 5pm read solutions to find errors and to see alternative approaches mark the errors in your previously submitted work submit the markup by Friday at 5pm identify ALL errors and get back half of the points you lost!
2
Discrete-Time Systems
We start with discrete-time (DT) systems because they • • •
are conceptually simpler than continuous-time systems illustrate same important modes of thinking as continuous-time are increasingly important (digital electronics and computation)
3
Multiple Representations of Discrete-Time Systems
Systems can be represented in different ways to more easily address different types of issues. Verbal description: ‘To reduce the number of bits needed to store a sequence of large numbers that are nearly equal, record the first number, and then record successive differences.’ Difference equation: y[n] = x[n] − x[n − 1] Block diagram: x[n]
+ −1
y[n]
Delay
We will exploit particular strengths of each of these representations.
4
Difference Equations
Difference equations are mathematically precise and compact. Example: y[n] = x[n] − x[n − 1] Let x[n] equal the “unit sample” signal δ[n], δ[n] =
1, 0,
if n = 0; otherwise. x[n] = δ[n]
n −1 0 1 2 3 4
We will use the unit sample as a “primitive” (building-block signal) to construct more complex signals.
5
Check Yourself
Solve y[n] = x[n] − x[n − 1] given x[n] = δ[n] How many of the following are true? 1. 2. 3. 4. 5.
y[2] > y[1] y[3] > y[2] y[2] = 0 y[n] − y[n−1] = x[n] − 2x[n−1] + x[n−2] y[119] = 0
6
Step-By-Step Solutions
Difference equations are convenient for step-by-step analysis.
Find y[n] given x[n] = δ[n]:
y[n] = x[n] − x[n − 1] y[−1] = x[−1] − x[−2]
=0−0=0
y[0] = x[0] − x[−1]
=1−0=1
y[1] = x[1] − x[0]
= 0 − 1 = −1
y[2] = x[2] − x[1]
=0−0=0
y[3] = x[3] − x[2] ...
=0−0=0
x[n] = δ[n]
y[n]
n
n
−1 0 1 2 3 4
−1 0 1 2 3 4 7
Check Yourself
Solve y[n] = x[n] − x[n − 1] given x[n] = δ[n] How many of the following are true?
4 √
1. 2. 3. 4. 5.
y[2] > y[1] y[3] > y[2] y[2] = 0 y[n] − y[n−1] = x[n] − 2x[n−1] + x[n−2] y[119] = 0
8
X √ √ √
Step-By-Step Solutions
Block diagrams are also useful for step-by-step analysis.
Represent y[n] = x[n] − x[n − 1] with a block diagram: start “at rest”
x[n]
y[n]
+ −1
Delay
x[n] = δ[n]
0
y[n]
n
n
−1 0 1 2 3 4
−1 0 1 2 3 4 9
Step-By-Step Solutions
Block diagrams are also useful for step-by-step analysis.
Represent y[n] = x[n] − x[n − 1] with a block diagram: start “at rest”
+
1 −1
Delay
1 0
−1
x[n] = δ[n]
y[n]
n
n
−1 0 1 2 3 4
−1 0 1 2 3 4 10
Step-By-Step Solutions
Block diagrams are also useful for step-by-step analysis.
Represent y[n] = x[n] − x[n − 1] with a block diagram: start “at rest”
+
1→0 −1
Delay
0 → −1
−1
x[n] = δ[n]
y[n]
n
n
−1 0 1 2 3 4
−1 0 1 2 3 4 11
Step-By-Step Solutions
Block diagrams are also useful for step-by-step analysis.
Represent y[n] = x[n] − x[n − 1] with a block diagram: start “at rest”
+
1→0 −1
Delay
−1 0 → −1
0
x[n] = δ[n]
y[n]
n
n
−1 0 1 2 3 4
−1 0 1 2 3 4 12
Step-By-Step Solutions
Block diagrams are also useful for step-by-step analysis.
Represent y[n] = x[n] − x[n − 1] with a block diagram: start “at rest”
−1
+
0 −1
Delay
−1
0
x[n] = δ[n]
y[n]
n
n
−1 0 1 2 3 4
−1 0 1 2 3 4 13
Step-By-Step Solutions
Block diagrams are also useful for step-by-step analysis.
Represent y[n] = x[n] − x[n − 1] with a block diagram: start “at rest”
+
0 −1
Delay
−1 −1 → 0
0
x[n] = δ[n]
y[n]
n
n
−1 0 1 2 3 4
−1 0 1 2 3 4 14
Step-By-Step Solutions
Block diagrams are also useful for step-by-step analysis.
Represent y[n] = x[n] − x[n − 1] with a block diagram: start “at rest”
+
0 −1
Delay
0 −1 → 0
0
x[n] = δ[n]
y[n]
n
n
−1 0 1 2 3 4
−1 0 1 2 3 4 15
Step-By-Step Solutions
Block diagrams are also useful for step-by-step analysis.
Represent y[n] = x[n] − x[n − 1] with a block diagram: start “at rest”
+
0 −1
Delay
0 0
0
x[n] = δ[n]
y[n]
n
n
−1 0 1 2 3 4
−1 0 1 2 3 4 16
Step-By-Step Solutions
Block diagrams are also useful for step-by-step analysis.
Represent y[n] = x[n] − x[n − 1] with a block diagram: start “at rest”
+
0 −1
Delay
0 0
0
x[n] = δ[n]
y[n]
n
n
−1 0 1 2 3 4
−1 0 1 2 3 4 17
Check Yourself
DT systems can be described by difference equations and/or block diagrams. Difference equation: y[n] = x[n] − x[n − 1] Block diagram: x[n]
+ −1
y[n]
Delay
In what ways are these representations different?
18
Check Yourself
In what ways are difference equations different from block diagrams? Difference equation: y[n] = x[n] − x[n − 1] Difference equations are “declarative.” They tell you rules that the system obeys. Block diagram: x[n]
+ −1
y[n]
Delay
Block diagrams are “imperative.” They tell you what to do. Block diagrams contain more information than the corresponding difference equation (e.g., what is the input? what is the output?) 19
From Samples to Signals
Lumping all of the (possibly infinite) samples into a single object — the signal — simplifies its manipulation. This lumping is an abstraction that is analogous to • representing coordinates in three-space as points • representing lists of numbers as vectors in linear algebra • creating an object in Python
20
From Samples to Signals
Operators manipulate signals rather than individual samples. x[n]
+ −1
y[n]
Delay
Nodes represent whole signals (e.g., X and Y ). The boxes operate on those signals: • • •
Delay = shift whole signal to right 1 time step Add = sum two signals −1: multiply by −1
Signals are the primitives.
Operators are the means of combination.
21
Operator Notation
Symbols can now compactly represent diagrams. Let R represent the right-shift operator: Y = R{X} ≡ RX where X represents the whole input signal (x[n] for all n) and Y represents the whole output signal (y[n] for all n) Representing the difference machine x[n]
+ −1
Delay
with R leads to the equivalent representation Y = X − RX = (1 − R) X
22
y[n]
Operator Notation: Check Yourself
Let Y = RX. Which of the following is/are true: 1. 2. 3. 4. 5.
y[n] = x[n] for all n y[n + 1] = x[n] for all n y[n] = x[n + 1] for all n y[n − 1] = x[n] for all n none of the above
23
Check Yourself Consider a simple signal: X
n −1 0 1 2 3 4
Then
Y = RX
n −1 0 1 2 3 4
Clearly y[1] = x[0]. Equivalently, if n = 0, then y[n + 1] = x[n]. The same sort of argument works for all other n.
24
Operator Notation: Check Yourself
Let Y = RX. Which of the following is/are true: 1. 2. 3. 4. 5.
y[n] = x[n] for all n y[n + 1] = x[n] for all n y[n] = x[n + 1] for all n y[n − 1] = x[n] for all n none of the above
25
Operator Representation of a Cascaded System
System operations have simple operator representations. Cascade systems → multiply operator expressions. Y1
+
X −1
+ −1
Delay
Using operator notation: Y1 = (1 − R) X Y2 = (1 − R) Y1 Substituting for Y1 : Y2 = (1 − R)(1 − R) X
26
Delay
Y2
Operator Algebra
Operator expressions can be manipulated as polynomials. Y1
+
X −1
+ −1
Delay
Using difference equations: y2 [n] = y1 [n] − y1 [n − 1] = (x[n] − x[n − 1]) − (x[n − 1] − x[n − 2]) = x[n] − 2x[n − 1] + x[n − 2] Using operator notation: Y2 = (1 − R) Y1 = (1 − R)(1 − R) X
= (1 − R)2 X
= (1 − 2R + R2 ) X
27
Delay
Y2
Operator Approach
Applies your existing expertise with polynomials to understand block diagrams, and thereby understand systems.
28
Operator Algebra
Operator notation facilitates seeing relations among systems.
“Equivalent” block diagrams (assuming both initially at rest): Y1 + + Y2 X −1
−1
Delay
+
X
Delay
Y
Delay
−2 Delay
Equivalent operator expressions: (1 − R)(1 − R) = 1 − 2R + R2 The operator equivalence is much easier to see. 29
Check Yourself
Operator expressions for these “equivalent” systems (if started “at rest”) obey what mathematical property? +
X −1
Delay
Y
+
Y
Delay Delay
X −1
Delay
Delay
1. commutate 2. associative 3. distributive 4. transitive 5. none of the above
30
Check Yourself
+
X −1
Delay
Y
+
Y
Delay
Y = R(1 − R)X
Delay
X −1
Delay
Delay
Y = (R − R2 )X
Multiplication by R distributes over addition.
31
Check Yourself
Operator expressions for these “equivalent” systems (if started “at rest”) obey what mathematical property? 3 +
X −1
Delay
Y
+
Y
Delay Delay
X −1
Delay
Delay
1. commutate 2. associative 3. distributive 4. transitive 5. none of the above
32
Check Yourself
How many of the following systems are equivalent to Y = (4R2 + 4R + 1) X ?
Delay
X
Delay
X
+
2
+
Delay
Delay
+
2
+
4
Y
Delay
X
Delay
+
4
33
+
Y
Y
Check Yourself Delay
X
+
2
Delay
+
2
Y
Y = (2R + 1)(2R + 1) X —————————————————————————– Delay
X
+
Delay
+
4
Y
Y = (4R2 + 4R + 1) X —————————————————————————– Delay X
Delay
+
4
+
Y
Y = (4R2 + 4R + 1) X —————————————————————————–
All implement Y = (4R2 + 4R + 1) X 34
Check Yourself
How many of the following systems are equivalent to Y = (4R2 + 4R + 1) X ? 3
Delay
X
Delay
X
+
2
+
Delay
Delay
+
2
+
4
Y
Delay
X
Delay
+
4
35
+
Y
Y
Operator Algebra: Explicit and Implicit Rules
Recipes versus constraints.
Recipe: subtract a right-shifted version of the input signal from a copy of the input signal. +
X −1
Y Y = (1 − R) X
Delay
Constraint: the difference between Y and RY is X. X
+
Y Y = RY + X Delay
(1 − R) Y = X
But how does one solve such a constraint? 36
Example: Accumulator
Try step-by-step analysis: it always works. Start “at rest.” + x[n] y[n] Delay Find y[n] given x[n] = δ[n]:
y[n] = x[n] + y[n − 1] y[0] = x[0] + y[−1] = 1 + 0 = 1
x[n] = δ[n]
y[1] = x[1] + y[0]
=0+1=1
y[2] = x[2] + y[1] ... y[n]
=0+1=1
n
n
−1 0 1 2 3 4
−1 0 1 2 3 4
37
Example: Accumulator
Try step-by-step analysis: it always works. Start “at rest.” + x[n] y[n] Delay Find y[n] given x[n] = δ[n]:
y[n] = x[n] + y[n − 1] y[0] = x[0] + y[−1] = 1 + 0 = 1
x[n] = δ[n]
y[1] = x[1] + y[0]
=0+1=1
y[2] = x[2] + y[1] ... y[n]
=0+1=1
n
n
−1 0 1 2 3 4
−1 0 1 2 3 4
Persistent response to a transient input! 38
Example: Accumulator
The response of the accumulator system could also be generated by a system with infinitely many paths from input to output, each with one unit of delay more than the previous. +
X Delay Delay
Delay
Delay
Delay
Delay
...
...
Y = (1 + R + R2 + R3 + · · ·) X 39
Y
Example: Accumulator
These systems are equivalent in the sense that if each is initially at rest, they will produce identical outputs from the same input. (1 − R) Y1 = X1
⇔?
Y2 = (1 + R + R2 + R3 + · · ·) X2
Proof: Assume X2 = X1 : Y2 = (1 + R + R2 + R3 + · · ·) X2
= (1 + R + R2 + R3 + · · ·) X1
= (1 + R + R2 + R3 + · · ·) (1 − R) Y1
= ((1 + R + R2 + R3 + · · ·) − (R + R2 + R3 + · · ·)) Y1
= Y1
It follows that Y2 = Y1 . It also follows that (1 − R) and (1 + R + R2 + R3 + · · ·) are reciprocals. 40
Example: Accumulator
The reciprocal of 1−R can also be evaluated using synthetic division. 1 +R +R2 +R3 + · · · 1−R 1 1 −R R R −R2 R2 R2 −R3 R3 R3 −R4 ··· Therefore 1 = 1 + R + R2 + R3 + R4 + · · · 1−R 41
Feedback
Systems with signals that depend on previous values of the same signal are said to have feedback.
Example: The accumulator system has feedback.
X
+
Y Delay
By contrast, the difference machine does not have feedback. x[n]
+ −1
Delay
42
y[n]
Cyclic Signal Paths, Feedback, and Modes
Block diagrams help visualize feedback.
Feedback occurs when there is a cyclic signal flow path. +
X
Y
R X
−2
+
Y
R Delay
acyclic
cyclic
Acyclic: all paths through system go from input to output with no cycles. Cyclic: at least one cycle. 43
Feedback, Cyclic Signal Paths, and Modes
The effect of feedback can be visualized by tracing each cycle through the cyclic signal paths. X
+
Y p0
Delay
x[n] = δ[n]
y[n]
n
n
−1 0 1 2 3 4
−1 0 1 2 3 4
Each cycle creates another sample in the output.
44
Feedback, Cyclic Signal Paths, and Modes
The effect of feedback can be visualized by tracing each cycle through the cyclic signal paths. X
+
Y p0
Delay
x[n] = δ[n]
y[n]
n
n
−1 0 1 2 3 4
−1 0 1 2 3 4
Each cycle creates another sample in the output.
45
Feedback, Cyclic Signal Paths, and Modes
The effect of feedback can be visualized by tracing each cycle through the cyclic signal paths. X
+
Y p0
Delay
x[n] = δ[n]
y[n]
n
n
−1 0 1 2 3 4
−1 0 1 2 3 4
Each cycle creates another sample in the output.
46
Feedback, Cyclic Signal Paths, and Modes
The effect of feedback can be visualized by tracing each cycle through the cyclic signal paths. X
+
Y p0
Delay
x[n] = δ[n]
y[n]
n
n
−1 0 1 2 3 4
−1 0 1 2 3 4
Each cycle creates another sample in the output. The response will persist even though the input is transient.
47
Check Yourself
How many of the following systems have cyclic signal paths? +
X
+
R
Y
+
X R
R
X
+
+
+
Y
R
+
X
+ R
R
48
Y
Y R
Check Yourself
How many of the following systems have cyclic signal paths? 3 +
X
+
R
Y
+
X R
R
X
+
+
+
Y
R
+
X
+ R
R
49
Y
Y R
Finite and Infinite Impulse Responses
The impulse response of an acyclic system has finite duration, while that of a cyclic system can have infinite duration.
+
X −1
Y
X
+
Y Delay
Delay
n
n
−1 0 1 2 3 4
−1 0 1 2 3 4
50
Analysis of Cyclic Systems: Geometric Growth
If traversing the cycle decreases or increases the magnitude of the signal, then the fundamental mode will decay or grow, respectively. If the response decays toward zero, then we say that it converges. Otherwise, we it diverges.
51
Check Yourself
How many of these systems have divergent unit-sample responses?
X
+
Y Delay
0.5 X
+
Y Delay
1.2 X
+
Delay
1.2 Delay
0.5
52
Y
Check Yourself y[n] X
+
Y 0.5
Delay
n −1 0 1 2 3 4 y[n] X
+
Y 1.2
Delay
n −1 0 1 2 3 4 y[n] X
+
Delay
0.5
1.2
Y
Delay
n −1 0 1 2 3 4 53
Check Yourself y[n] X
+
Y 0.5
X
Delay
n −1 0 1 2 3 4 y[n] X
+
Y 1.2
√
Delay
n −1 0 1 2 3 4 y[n] X
+
Delay
1.2
Y X
0.5
Delay
n −1 0 1 2 3 4 54
Check Yourself
How many of these systems have divergent unit-sample responses? 1
X
+
Y X 0.5
X
+
Y 1.2
X
Delay
+
√
Delay
Delay
1.2
Y X
0.5
Delay
55
Cyclic Systems: Geometric Growth
If traversing the cycle decreases or increases the magnitude of the signal, then the fundamental mode will decay or grow, respectively. X
+
Y 0.5
X
Delay
+
Y 1.2
y[n]
Delay
y[n]
n
n
−1 0 1 2 3 4
−1 0 1 2 3 4
These are geometric sequences: y[n] = (0.5)n and (1.2)n for n ≥ 0. These geometric sequences are called fundamental modes. 56
Multiple Representations of Discrete-Time Systems
Now you know four representations of discrete-time systems. Verbal descriptions: preserve the rationale. “To reduce the number of bits needed to store a sequence of large numbers that are nearly equal, record the first number, and then record successive differences.” Difference equations: mathematically compact. y[n] = x[n] − x[n − 1] Block diagrams: illustrate signal flow paths. + x[n] −1
y[n]
Delay
Operator representations: analyze systems as polynomials. Y = (1 − R) X
57
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6.003: Signals and Systems Feedback, Poles, and Fundamental Modes
September 15, 2011
1
Homework
Doing the homework is essential to understanding the content. Weekly Homework Assigments • •
tutor (exam-type) problems: answers are automatically checked to provide quick feedback engineering design (real-world) problems: graded by a human
Learning doesn’t end when you have submitted your work! • • • • •
solutions will be posted on Wednesdays at 5pm read solutions to find errors and to see alternative approaches mark the errors in your previously submitted work submit the markup by Friday at 5pm identify ALL errors and get back half of the points you lost!
2
Last Time: Multiple Representations of DT Systems
Verbal descriptions: preserve the rationale. “To reduce the number of bits needed to store a sequence of large numbers that are nearly equal, record the first number, and then record successive differences.” Difference equations: mathematically compact. y[n] = x[n] − x[n − 1] Block diagrams: illustrate signal flow paths. + x[n] −1
y[n]
Delay
Operator representations: analyze systems as polynomials. Y = (1 − R) X
3
Last Time: Feedback, Cyclic Signal Paths, and Modes
Systems with signals that depend on previous values of the same signal are said to have feedback.
Example: The accumulator system has feedback. X
+
Y Delay
By contrast, the difference machine does not have feedback. +
X −1
Delay
4
Y
Last Time: Feedback, Cyclic Signal Paths, and Modes
The effect of feedback can be visualized by tracing each cycle through the cyclic signal paths. X
+
Y p0
Delay
x[n] = δ[n]
y[n]
n
n
−1 0 1 2 3 4
−1 0 1 2 3 4
Each cycle creates another sample in the output.
5
Last Time: Feedback, Cyclic Signal Paths, and Modes
The effect of feedback can be visualized by tracing each cycle through the cyclic signal paths. X
+
Y p0
Delay
x[n] = δ[n]
y[n]
n
n
−1 0 1 2 3 4
−1 0 1 2 3 4
Each cycle creates another sample in the output.
6
Last Time: Feedback, Cyclic Signal Paths, and Modes
The effect of feedback can be visualized by tracing each cycle through the cyclic signal paths. X
+
Y p0
Delay
x[n] = δ[n]
y[n]
n
n
−1 0 1 2 3 4
−1 0 1 2 3 4
Each cycle creates another sample in the output.
7
Last Time: Feedback, Cyclic Signal Paths, and Modes
The effect of feedback can be visualized by tracing each cycle through the cyclic signal paths. X
+
Y p0
Delay
x[n] = δ[n]
y[n]
n
n
−1 0 1 2 3 4
−1 0 1 2 3 4
Each cycle creates another sample in the output. The response will persist even though the input is transient.
8
Geometric Growth: Poles
These unit-sample responses can be characterized by a single number — the pole — which is the base of the geometric sequence. X
+
Y p0
y[n] =
y[n]
Delay
pn 0, 0,
if n >= 0; otherwise.
y[n]
y[n]
n −1 0 1 2 3 4 p0 = 0.5
n −1 0 1 2 3 4 p0 = 1
n −1 0 1 2 3 4 p0 = 1.2
9
Check Yourself
How many of the following unit-sample responses can be represented by a single pole?
n
n
n
n
n
10
Check Yourself
How many of the following unit-sample responses can be represented by a single pole? 3
n
n
n
n
n
11
Geometric Growth
The value of p0 determines the rate of growth.
y[n]
y[n]
y[n]
y[n]
z −1
p0 < −1: −1 < p0 < 0: 0 < p0 < 1: p0 > 1:
magnitude magnitude magnitude magnitude
0
1
diverges, alternating sign converges, alternating sign
converges monotonically
diverges monotonically
12
Second-Order Systems
The unit-sample responses of more complicated cyclic systems are more complicated. X
+
Y R 1.6 R −0.63
y[n]
n −1 0 1 2 3 4 5 6 7 8 Not geometric. This response grows then decays. 13
Factoring Second-Order Systems
Factor the operator expression to break the system into two simpler systems (divide and conquer). X
+
Y R 1.6 R −0.63
Y = X + 1.6RY − 0.63R2 Y (1 − 1.6R + 0.63R2 ) Y = X (1 − 0.7R)(1 − 0.9R) Y = X
14
Factoring Second-Order Systems
The factored form corresponds to a cascade of simpler systems. (1 − 0.7R)(1 − 0.9R) Y = X X
Y2
+ 0.7
+
R
0.9
(1 − 0.7R) Y2 = X
X
0.9
R
(1 − 0.9R) Y = Y2
Y1
+
Y
R
+
Y 0.7
(1 − 0.9R) Y1 = X
R
(1 − 0.7R) Y = Y1
The order doesn’t matter (if systems are initially at rest).
15
Factoring Second-Order Systems
The unit-sample response of the cascaded system can be found by multiplying the polynomial representations of the subsystems. Y 1 1 1 = = × X (1 − 0.7R)(1 − 0.9R) (1 − 0.7R) (1 − 0.9R) _ _ _ _ = (1 + 0.7R + 0.72 R2 + 0.73 R3 + · · ·) × (1 + 0.9R + 0.92 R2 + 0.93 R3 + · · ·) Multiply, then collect terms of equal order: Y = 1 + (0.7 + 0.9)R + (0.72 + 0.7 × 0.9 + 0.92 )R2 X + (0.73 + 0.72 × 0.9 + 0.7 × 0.92 + 0.93 )R3 + · · ·
16
Multiplying Polynomial
Graphical representation of polynomial multiplication. Y = (1 + aR + a2 R2 + a3 R3 + · · ·) × (1 + bR + b2 R2 + b3 R3 + · · ·) X 1
1 a X
...
R
a2
R2
a3
R3
+
...
...
b
R
b2
R2
b3
R3
+
Y
...
Collect terms of equal order: Y = 1 + (a + b)R + (a2 + ab + b2 )R2 + (a3 + a2 b + ab2 + b3 )R3 + · · · X 17
Multiplying Polynomials
Tabular representation of polynomial multiplication. (1 + aR + a2 R2 + a3 R3 + · · ·) × (1 + bR + b2 R2 + b3 R3 + · · ·)
1 aR 2 a R2 a 3 R3 ···
1
bR
b2 R2
b3 R3
···
1 aR a2 R 2 a3 R 3 ···
bR abR2 a2 bR3 a3 bR4 ···
b2 R2 ab2 R3 a2 b2 R4 a3 b2 R5 ···
b3 R3 ab3 R4 a2 b3 R5 a3 b3 R6 ···
··· ··· ··· ··· ···
Group same powers of R by following reverse diagonals: Y = 1 + (a + b)R + (a2 + ab + b2 )R2 + (a3 + a2 b + ab2 + b3 )R3 + · · · X y[n]
n −1 0 1 2 3 4 5 6 7 8 18
Partial Fractions
Use partial fractions to rewrite as a sum of simpler parts.
X
+
Y R 1.6 R −0.63
Y 1 1 4.5 3.5 = = = − 2 (1 − 0.9R)(1 − 0.7R) 1 − 0.9R 1 − 0.7R X 1 − 1.6R + 0.63R
19
Second-Order Systems: Equivalent Forms
The sum of simpler parts suggests a parallel implementation.
4.5 3.5 Y = − X 1 − 0.9R 1 − 0.7R
X
Y1
+ 0.9
+
R Y2
+ 0.7
4.5
−3.5
R
If x[n] = δ[n] then y1 [n] = 0.9n and y2 [n] = 0.7n for n ≥ 0. Thus, y[n] = 4.5(0.9)n − 3.5(0.7)n for n ≥ 0. 20
Y
Partial Fractions
Graphical representation of the sum of geometric sequences. y1 [n] = 0.9n for n ≥ 0 n −1 0 1 2 3 4 5 6 7 8 y2 [n] = 0.7n for n ≥ 0 n −1 0 1 2 3 4 5 6 7 8 y[n] = 4.5(0.9)n − 3.5(0.7)n for n ≥ 0
n −1 0 1 2 3 4 5 6 7 8 21
Partial Fractions Partial fractions provides a remarkable equivalence. + X Y R 1.6 R −0.63
X
Y1
+ 0.9
+
Y
R Y2
+ 0.7
4.5
−3.5
R
→ follows from thinking about system as polynomial (factoring). 22
Poles
The key to simplifying a higher-order system is identifying its poles. Poles are the roots of the denominator of the system functional when R → z1 . Start with system functional: Y 1 1 1 = = = 2 (1−p0 R)(1−p1 R) (1−0.7R) (1−0.9R) X 1 − 1.6R+0.63R | {z } | {z } p0 =0.7
1 and find roots of denominator: z Y 1 z2 z2 = = 2 = 1.6 0.63 (z−0.7) (z−0.9) X z −1.6z+0.63 1− + 2 | {z } | {z } z z z =0.7 z =0.9
Substitute R →
0
The poles are at 0.7 and 0.9. 23
1
p1 =0.9
Check Yourself
Consider the system described by 1 1 1 y[n] = − y[n − 1] + y[n − 2] + x[n − 1] − x[n − 2] 4 8 2 How many of the following are true? 1. 2. 3. 4. 5.
The unit sample response converges to zero. There are poles at z = 21 and z = 14 . There is a pole at z = 12 . There are two poles. None of the above
24
Check Yourself
1 y[n] = − y[n − 1] + 4 � � 1 1 2 1+ R− R Y 4 8 H(R) =
=
1. 2. 3. 4. 5.
1 1
y[n − 2] + x[n − 1] − x[n − 2] 8 2 � � 1
= R − R2 X
2
R − 12 R2 Y = X 1 + 14 R − 18 R2 1
1 1
z − 2 z 2
1 + 14 z1 − 18 1 2 z
=
z − 12
z − 12
�
��
�
=
z 2 + 14 z − 18
z + 12 z − 14
The unit sample response converges to zero.
There are poles at z = 12 and z = 14 .
There is a pole at z = 12 .
There are two poles. None of the above 25
Check Yourself
1 y[n] = − y[n − 1] + 4 � � 1 1 2 1+ R− R Y 4 8 H(R) =
=
1 1
y[n − 2] + x[n − 1] − x[n − 2] 8 2 � � 1
= R − R2 X
2
R − 12 R2 Y = X 1 + 14 R − 18 R2 1
1 1
z − 2 z 2
1 + 14 z1 − 18 1 2 z
=
z − 12
z − 12
�� � � =
z 2 + 14 z − 18
z + 12 z − 14
√
1. 2. 3. 4. 5.
The unit sample response converges to zero. There are poles at z = 12 and z = 14 . X
1 There is a pole at z = 2 . √ X
There are two poles. None of the above X
26
Check Yourself
Consider the system described by 1 1 1 y[n] = − y[n − 1] + y[n − 2] + x[n − 1] − x[n − 2] 4 8 2 How many of the following are true? 1. 2. 3. 4. 5.
2
The unit sample response converges to zero. There are poles at z = 21 and z = 14 . There is a pole at z = 12 . There are two poles. None of the above
27
Population Growth
28
Population Growth
29
Population Growth
30
Population Growth
31
Population Growth
32
Check Yourself
What are the pole(s) of the Fibonacci system? 1. 2. 3. 4. 5.
1 1 and −1 −1 and −2 1.618 . . . and −0.618 . . . none of the above
33
Check Yourself
What are the pole(s) of the Fibonacci system? Difference equation for Fibonacci system: y[n] = x[n] + y[n − 1] + y[n − 2] System functional: Y 1 H= = X 1 − R − R2 Denominator is second order → 2 poles.
34
Check Yourself
Find the poles by substituting R → 1/z in system functional. Y 1 1 z2 → H= = = X z2 − z − 1 1 − R − R2 1 − z1 − 12 z
Poles are at √ 1± 5 z= = 2
φ or
−
1 φ
where φ represents the “golden ratio” √ 1+ 5 φ= ≈ 1.618 2 The two poles are at 1 z0 = φ ≈ 1.618 and z1 = − ≈ −0.618 φ
35
Check Yourself
What are the pole(s) of the Fibonacci system? 1. 2. 3. 4. 5.
1 1 and −1 −1 and −2 1.618 . . . and −0.618 . . . none of the above
36
4
Example: Fibonacci’s Bunnies
Each pole corresponds to a fundamental mode.
φ ≈ 1.618
and
−
1 ≈ −0.618 φ
1 n − φ
φn
n
n
−1 0 1 2 3 4
−1 0 1 2 3 4
One mode diverges, one mode oscillates!
37
Example: Fibonacci’s Bunnies
The unit-sample response of the Fibonacci system can be written as a weighted sum of fundamental modes. φ
1
√ 5
1R φ
√ Y 1 φ 5 H= = = + 2 1 − φR 1 + X 1−R−R
φ 1 h[n] = √ φn + √ (−φ)−n ; 5 φ 5
n≥0
But we already know that h[n] is the Fibonacci sequence f : f : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, . . . Therefore we can calculate f [n] without knowing f [n − 1] or f [n − 2] !
38
Complex Poles
What if a pole has a non-zero imaginary part?
Example:
Y 1 = X 1 − R + R2 1 z2 = = z2 − z + 1 1 − z1 + 12 z
√
Poles are z = 12 ± 23 j = e±jπ/3 .
What are the implications of complex poles?
39
Complex Poles
Partial fractions work even when the poles are complex. Y 1 1 1 = × = √ jπ/3 −jπ/3 X j 3 1−e R 1−e R
�
e jπ/3 e−jπ/3 − 1 − e jπ/3 R 1 − e−jπ/3 R
�
There are two fundamental modes (both geometric sequences): e jnπ/3 = cos(nπ/3) + j sin(nπ/3) and e−jnπ/3 = cos(nπ/3) − j sin(nπ/3)
n
n
40
Complex Poles
Complex modes are easier to visualize in the complex plane. e jnπ/3 = cos(nπ/3) + j sin(nπ/3) Im e j1π/3 e j2π/3 e j3π/3
e j4π/3 e j4π/3 e j3π/3
e j2π/3
e j0π/3 Re
n
e j5π/3 e−jnπ/3 = cos(nπ/3) − j sin(nπ/3) Im e j5π/3 e j0π/3 Re
n
e j1π/3 41
Complex Poles
The output of a “real” system has real values. y[n] = x[n] + y[n − 1] − y[n − 2]
Y 1
H= = X 1 − R + R2 1 1 × = jπ/3 R 1 − e� 1 − e−jπ/3 R � ! 1 e jπ/3 e−jπ/3 − = √ j 3 1 − e jπ/3 R 1 − e−jπ/3 R 1 2 (n + 1)π h[n] = √ e j(n+1)π/3 − e−j(n+1)π/3 = √ sin 3 3 j 3 h[n] 1 n −1
42
Check Yourself
Unit-sample response of a system with poles at z = re±jΩ .
n
Which of the following is/are true? 1. 2. 3. 4. 5.
r < 0.5 and Ω ≈ 0.5 0.5 < r < 1 and Ω ≈ 0.5 r < 0.5 and Ω ≈ 0.08 0.5 < r < 1 and Ω ≈ 0.08 none of the above
43
Check Yourself
Unit-sample response of a system with poles at z = re±jΩ .
n
Which of the following is/are true? 1. 2. 3. 4. 5.
r < 0.5 and Ω ≈ 0.5 0.5 < r < 1 and Ω ≈ 0.5 r < 0.5 and Ω ≈ 0.08 0.5 < r < 1 and Ω ≈ 0.08 none of the above
44
2
Check Yourself X
+
R
R
R
Y
How many of the following statements are true? 1. This system has 3 fundamental modes. 2. All of the fundamental modes can be written as geometrics. 3. Unit-sample response is y[n] : 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1 . . . 4. Unit-sample response is y[n] : 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1 . . . 5. One of the fundamental modes of this system is the unit step.
45
Check Yourself X
+
R
R
R
Y
How many of the following statements are true? 4 1. This system has 3 fundamental modes. 2. All of the fundamental modes can be written as geometrics. 3. Unit-sample response is y[n] : 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1 . . . 4. Unit-sample response is y[n] : 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1 . . . 5. One of the fundamental modes of this system is the unit step.
46
Summary
Systems composed of adders, gains, and delays can be characterized by their poles. The poles of a system determine its fundamental modes. The unit-sample response of a system can be expressed as a weighted sum of fundamental modes. These properties follow from a polynomial interpretation of the sys tem functional.
47
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6.003: Signals and Systems Continuous-Time Systems
September 20, 2011
1
Multiple Representations of Discrete-Time Systems
Discrete-Time (DT) systems can be represented in different ways to more easily address different types of issues. Verbal descriptions: preserve the rationale. “Next year, your account will contain p times your balance from this year plus the money that you added this year.” Difference equations: mathematically compact. y[n + 1] = x[n] + py[n] Block diagrams: illustrate signal flow paths. x[n]
+
Delay
y[n]
p Operator representations: analyze systems as polynomials. (1 − pR) Y = RX
2
Multiple Representations of Continuous-Time Systems Similar representations for Continuous-Time (CT) systems. Verbal descriptions: preserve the rationale. “Your account will grow in proportion to your balance plus the rate at which you deposit.” Differential equations: mathematically compact. dy(t) = x(t) + py(t) dt Block diagrams: illustrate signal flow paths. Z t x(t)
( · ) dt
+
y(t)
−∞
p Operator representations: analyze systems as polynomials. (1 − pA)Y = AX
3
Differential Equations
Differential equations are mathematically precise and compact. r0 (t)
h1 (t) r1 (t) We can represent the tank system with a differential equation. dr1 (t) r0 (t) − r1 (t) = dt τ You already know lots of methods to solve differential equations: • general methods (separation of variables; integrating factors) • homogeneous and particular solutions • inspection Today: new methods based on block diagrams and operators, which provide new ways to think about systems’ behaviors. 4
Block Diagrams
Block diagrams illustrate signal flow paths.
DT: adders, scalers, and delays – represent systems described by linear difference equations with constant coefficents. x[n]
+
Delay
y[n]
p CT: adders, scalers, and integrators – represent systems described by a linear differential equations with constant coefficients. Z t x(t)
( · ) dt
+
−∞
p Delays in DT are replaced by integrators in CT. 5
y(t)
Operator Representation
CT Block diagrams are concisely represented with the A operator. Applying A to a CT signal generates a new signal that is equal to the integral of the first signal at all points in time. Y = AX is equivalent to t
y(t) =
x(τ ) dτ −∞
for all time t.
6
Check Yourself +
X
A
y(t) ˙ = x(t) ˙ + py(t)
Y
p X
X
+
p
A
˙ = x(t) + py(t) y(t)
Y
+
˙ = px(t) + py(t) y(t)
Y p
A
Which block diagrams correspond to which equations?
1.
2.
3.
4. 7
1
5. none
Check Yourself +
X
A
y(t) ˙ = x(t) ˙ + py(t)
Y
p X
X
+
p
A
˙ = x(t) + py(t) y(t)
Y
+
˙ = px(t) + py(t) y(t)
Y p
A
Which block diagrams correspond to which equations?
1.
2.
3.
4. 8
1
5. none
Evaluating Operator Expressions
As with R, A expressions can be manipulated as polynomials.
Example: W
+
X A
+
Y
A
Z t w(t) = x(t) +
x(τ )dτ −∞ Z t
y(t) = w(t) +
w(τ )dτ −∞ Z t
y(t) = x(t) +
Z t x(τ )dτ +
−∞
Z t
Z τ2
x(τ )dτ + −∞
−∞
−∞
W = (1 + A) X Y = (1 + A) W = (1 + A)(1 + A) X = (1 + 2A + A2 ) X 9
x(τ1 )dτ1 dτ2
Evaluating Operator Expressions
Expressions in A can be manipulated using rules for polynomials. • Commutativity: A(1 − A)X = (1 − A)AX • Distributivity: A(1 − A)X = (A − A2 )X • Associativity:
�
� � � (1 − A)A (2 − A)X = (1 − A) A(2 − A) X
10
Check Yourself
Determine k1 so that these systems are “equivalent.” +
X
+
A
A
−0.7 X
+
Y
−0.9 A
A
Y
k1 k2 1. 0.7
2. 0.9
3. 1.6
4. 0.63
11
5. none of these
Check Yourself
Write operator expressions for each system. +
X
W
A
+
A
−0.7
Y
−0.9
(1+0.7A)(1+0.9A)Y = A2 X (1+0.7A)W = AX W = A(X −0.7W ) → → (1+0.9A)Y = AW Y = A(W −0.9Y ) (1+1.6A+0.63A2 )Y = A2 X X
+
W
A
A
Y
k1 k2 W = A(X +k1 W +k2 Y ) Y = AW k1 = −1.6
→ 12
Y = A2 X +k1 AY +k2 A2 Y (1−k1 A−k2 A2 )Y = A2 X
Check Yourself
Determine k1 so that these systems are “equivalent.” +
X
+
A
A
−0.7 X
+
Y
−0.9 A
A
Y
k1 k2 1. 0.7
2. 0.9
3. 1.6
4. 0.63
13
5. none of these
Elementary Building-Block Signals
Elementary DT signal: δ[n].
δ[n] =
1, 0,
if n = 0; otherwise δ[n] 1 n 0
• • •
simplest non-trivial signal (only one non-zero value) shortest possible duration (most “transient”) useful for constructing more complex signals
What CT signal serves the same purpose? 14
Elementary CT Building-Block Signal
Consider the analogous CT signal: w(t) is non-zero only at t = 0. ⎧ ⎨0 w(t) = 1 ⎩ 0
t0 w(t) 1 t 0
Is this a good choice as a building-block signal? Z t ( · ) dt
w(t)
−∞
The integral of w(t) is zero! 15
0
No
Unit-Impulse Signal
The unit-impulse signal acts as a pulse with unit area but zero width. p (t) 1 2
δ(t) = lim p (t)
unit area
→0
−
p1/4 (t)
p1/2 (t)
t
p1/8 (t) 4 2
1 1 − 2
1 2
t
1 − 4
1 4 16
t
1 1 − 8 8
t
Unit-Impulse Signal
The unit-impulse function is represented by an arrow with the num ber 1, which represents its area or “weight.” δ(t) 1 t It has two seemingly contradictory properties: • it is nonzero only at t = 0, and • its definite integral (−∞, ∞) is one ! Both of these properties follow from thinking about δ(t) as a limit:
p (t) 1 2
δ(t) = lim p (t)
unit area
→0
17
−
t
Unit-Impulse and Unit-Step Signals
The indefinite integral of the unit-impulse is the unit-step. Z t u(t) =
δ(λ) dλ =
−∞
1; 0;
t≥0 otherwise u(t) 1 t
Equivalently δ(t)
A
18
u(t)
Impulse Response of Acyclic CT System
If the block diagram of a CT system has no feedback (i.e., no cycles), then the corresponding operator expression is “imperative.”
+
X
+
A
A
Y = (1 + A)(1 + A) X = (1 + 2A + A2 ) X If x(t) = δ(t) then y(t) = (1 + 2A + A2 ) δ(t) = δ(t) + 2u(t) + tu(t)
19
Y
CT Feedback
Find the impulse response of this CT system with feedback. x(t)
+
A
y(t)
p Method 1: find differential equation and solve it. y˙(t) = x(t) + py(t) Linear, first-order difference equation with constant coefficients.
Try y(t) = Ceαt u(t).
Then y˙(t) = αCeαt u(t) + Ceαt δ(t) = αCeαt u(t) + Cδ(t).
Substituting, we find that αCeαt u(t) + Cδ(t) = δ(t) + pCeαt u(t).
Therefore α = p and C = 1
→
y(t) = ept u(t).
20
CT Feedback
Find the impulse response of this CT system with feedback. x(t)
+
A
y(t)
p Method 2: use operators. Y = A (X + pY ) Y A = X 1 − pA Now expand in ascending series in A: Y = A(1 + pA + p2 A2 + p3 A3 + · · ·) X If x(t) = δ(t) then y(t) = A(1 + pA + p2 A2 + p3 A3 + · · ·) δ(t) 1 1 = (1 + pt + p2 t2 + p3 t3 + · · ·) u(t) = ept u(t) . 2 6 21
CT Feedback
We can visualize the feedback by tracing each cycle through the cyclic signal path. x(t)
+
y(t)
A p
y(t) = (A + pA2 + p2 A3 + p3 A4 + · · ·) δ(t)
1 1
= (1 + pt + p2 t2 + p3 t3 + · · ·) u(t) 2 6 y(t)
1 t
0 22
CT Feedback
We can visualize the feedback by tracing each cycle through the cyclic signal path. x(t)
+
y(t)
A p
y(t) = (A + pA2 + p2 A3 + p3 A4 + · · ·) δ(t)
1 1
= (1 + pt + p2 t2 + p3 t3 + · · ·) u(t) 2 6 y(t)
1 t
0 23
CT Feedback
We can visualize the feedback by tracing each cycle through the cyclic signal path. x(t)
+
y(t)
A p
y(t) = (A + pA2 + p2 A3 + p3 A4 + · · ·) δ(t)
1 1
= (1 + pt + p2 t2 + p3 t3 + · · ·) u(t) 2 6 y(t)
1 t
0 24
CT Feedback
We can visualize the feedback by tracing each cycle through the cyclic signal path. x(t)
+
y(t)
A p
y(t) = (A + pA2 + p2 A3 + p3 A4 + · · ·) δ(t)
1 1
= (1 + pt + p2 t2 + p3 t3 + · · ·) u(t) 2 6 y(t)
1 t
0 25
CT Feedback
We can visualize the feedback by tracing each cycle through the cyclic signal path. x(t)
+
y(t)
A p
y(t) = (A + pA2 + p2 A3 + p3 A4 + · · ·) δ(t)
1 1
= (1 + pt + p2 t2 + p3 t3 + · · ·) u(t) = ept u(t) 2 6 y(t)
1 t
0 26
CT Feedback
Making p negative makes the output converge (instead of diverge). x(t)
+
A −p
y(t) = (A − pA2 + p2 A3 − p3 A4 + · · ·) δ(t)
1 1 = (1 − pt + p2 t2 − p3 t3 + · · ·) u(t) 2 6
27
y(t)
CT Feedback
Making p negative makes the output converge (instead of diverge). x(t)
+
y(t)
A −p
y(t) = (A − pA2 + p2 A3 − p3 A4 + · · ·) δ(t)
1
1
= (1 − pt + p2 t2 − p3 t3 + · · ·) u(t) 2 6 y(t)
1 t
0 28
CT Feedback
Making p negative makes the output converge (instead of diverge). x(t)
+
y(t)
A −p
y(t) = (A − pA2 + p2 A3 − p3 A4 + · · ·) δ(t)
1
1
= (1 − pt + p2 t2 − p3 t3 + · · ·) u(t) 2 6 y(t)
1 t
0 29
CT Feedback
Making p negative makes the output converge (instead of diverge). x(t)
+
y(t)
A −p
y(t) = (A − pA2 + p2 A3 − p3 A4 + · · ·) δ(t)
1
1
= (1 − pt + p2 t2 − p3 t3 + · · ·) u(t) 2 6 y(t)
1 t
0 30
CT Feedback
Making p negative makes the output converge (instead of diverge). x(t)
+
y(t)
A −p
y(t) = (A − pA2 + p2 A3 − p3 A4 + · · ·) δ(t)
1
1
= (1 − pt + p2 t2 − p3 t3 + · · ·) u(t) 2 6 y(t)
1 t
0 31
CT Feedback
Making p negative makes the output converge (instead of diverge). x(t)
+
y(t)
A −p
y(t) = (A − pA2 + p2 A3 − p3 A4 + · · ·) δ(t)
1
1
= (1 − pt + p2 t2 − p3 t3 + · · ·) u(t) = e−pt u(t) 6 2 y(t)
1 t
0 32
Convergent and Divergent Poles
The fundamental mode associated with p converges if p < 0 and
diverges if p > 0. X
+
A
Y
p
p0
y(t)
y(t)
1
1 0
t
0
33
t
Convergent and Divergent Poles
The fundamental mode associated with p converges if p < 0 and
diverges if p > 0. X
+
A
Y
p Im(p)
Convergent
Divergent
Re(p)
Re(p)
34
CT Feedback
In CT, each cycle adds a new integration. x(t)
+
y(t)
A p
y(t) = (A + pA2 + p2 A3 + p3 A4 + · · ·) δ(t) 1
1
= (1 + pt + p2 t2 + p3 t3 + · · ·) u(t) = ept u(t) 2 6 y(t)
1 t
0 35
DT Feedback
In DT, each cycle creates another sample in the output. X
+
Y p0
Delay
y[n] = (1 + pR + p2 R2 + p3 R3 + p4 R4 + · · ·) δ[n] = δ[n] + pδ[n − 1] + p2 δ[n − 2] + p3 δ[n − 3] + p4 δ[n − 4] + · · · y[n]
n −1 0 1 2 3 4
36
Summary: CT and DT representations
Many similarities and important differences. y[n] = x[n] + py[n − 1]
y˙(t) = x(t) + py(t) X
+
A
Y
X
+
Y
p
p
Delay
A 1 − pA
1 1 − pR
e pt u(t)
pn u[n]
37
Check Yourself
Which functionals represent convergent systems? 1 1−
1
1 R2 4
1 − 14 A2
1 1 + 2R + 43 R2 √ x √ 1. x
√ √ 2. x x
1 1 + 2A + 43 A2 √ √ 3. √ √
√
x 4. x √
38
5. none of these
Check Yourself
1 1−
1 R2 4
1
=
1 (1 −
1 R)(1 + 1 R) 2 2
(1 −
1 A)(1 + 1 A) 2 2
1
√ both inside unit circle
left & right half-planes
X
1 1 = 3 1 2 (1 + 2 R)(1 + 32 R) 1 + 2R + 4 R
inside & outside unit circle
X
1 1 = 3 1 2 (1 + 2 A)(1 + 32 A) 1 + 2A + 4 A
both left half plane
1−
1 A2 4
=
√
39
Check Yourself
Which functionals represent convergent systems? 1 1−
1
1 R2 4
1 − 14 A2
1 1 + 2R + 43 R2 √ x √ 1. x
√ √ 2. x x
4
1 1 + 2A + 43 A2 √ √ 3. √ √
√
x 4. x √
40
5. none of these
Mass and Spring System
Use the A operator to solve the mass and spring system. x(t) F = K x(t) − y(t) = M y¨(t) y(t)
x(t)
+
K M
y¨(t)
A
−1 K A2 Y M = K A2 X 1 + M 41
˙ y(t)
A
y(t)
Mass and Spring System
Factor system functional to find the poles.
K A2 K 2 Y
MA = M K =
(1 − p0 A)(1 − p1 A) X 1 + M A2
1+
K 2 A = 1 − (p0 + p1 )A + p0 p1 A2 M
The sum of the poles must be zero. The product of the poles must be K/M . p0 = j
K M
p1 = −j
K M
42
Mass and Spring System
Alternatively, find the poles by substituting A → 1
s .
The poles are then the roots of the denominator.
K A2 Y = MK X 1 + M A2
Substitute A → 1s : K Y M = K X s2 + M r K s = ±j M
43
Mass and Spring System
The poles are complex conjugates. Im s s-plane
q
K M
≡ ω0
Re s
q K ≡ −ω − M 0
The corresponding fundamental modes have complex values. fundamental mode 1: ejω0 t = cos ω0 t + j sin ω0 t fundamental mode 2: e−jω0 t = cos ω0 t − j sin ω0 t 44
Mass and Spring System
Real-valued inputs always excite combinations of these modes so that the imaginary parts cancel. Example: find the impulse response. K A2 K Y
A A
M M =
=
−
X 1 + K A2 p 0 − p1 1 − p 0 A 1 − p 1 A M ω02 A A − = 2jω0 1 − jω0 A 1 + jω0 A A ω0 A
ω0 − = 2j 1 + jω0 A
2j 1 − jω0 A ' v " ' v " makes mode 1
makes mode 2
The modes themselves are complex conjugates, and their coefficients are also complex conjugates. So the sum is a sum of something and its complex conjugate, which is real.
45
Mass and Spring System
The impulse response is therefore real.
Y ω0 A ω0 A = − X 2j 1 − jω0 A 2j 1 + jω0 A The impulse response is ω0 jω0 t ω0 −jω0 t e = ω0 sin ω0 t ; h(t) = e − 2j 2j
t>0
y(t)
t
0
46
Mass and Spring System
Alternatively, find impulse response by expanding system functional.
x(t)
+
ω02
y¨(t)
A
˙ y(t)
−1 ω02 A2 Y = = ω02 A2 − ω04 A4 + ω06 A6 − + · · · X 1 + ω02 A2 If x(t) = δ(t) then
3
5
2 4t 6t − + · · · , t ≥ 0
y(t) = ω 0
t − ω 0
+ ω0
5! 3!
47
A
y(t)
Mass and Spring System
Look at successive approximations to this infinite series.
∞ � �l 0 ω02 A2 Y 2 2 2 2 = = ω A −ω A 0 0 X 1 + ω02 A2 l=0
If x(t) = δ(t) then ∞ � �l 0 y(t) = ω02 −ω02 A2l+2 δ(t) l=0
= ω02 t − ω04
t5 t7 t9 t3 + ω06 − ω08 + ω010 − + · · · = ω0 sin ω0 t 3! 5! 7! 9! y(t)
t
0
48
Summary: CT and DT representations
Many similarities and important differences. y[n] = x[n] + py[n − 1]
y˙(t) = x(t) + py(t) X
+
A
Y
X
+
Y
p
p
Delay
A 1 − pA
1 1 − pR
e pt u(t)
pn u[n]
49
MIT OpenCourseWare http://ocw.mit.edu
6.003 Signals and Systems Fall 2011
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
6.003: Signals and Systems Z Transform
September 22, 2011
1
Concept Map: Discrete-Time Systems
Multiple representations of DT systems.
Delay → R
Block Diagram X
+
+ Delay
System Functional Y
1 Y = H(R) = X 1 − R − R2
Delay
Unit-Sample Response index shift
h[n] : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .
Difference Equation
System Function z2 Y (z) = 2 H(z) = X(z) z −z−1
y[n] = x[n] + y[n−1] + y[n−2] 2
Concept Map: Discrete-Time Systems
Relations among representations.
Delay → R
Block Diagram X
+
+ Delay
System Functional Y
1 Y = H(R) = X 1 − R − R2
Delay
Unit-Sample Response index shift
h[n] : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .
Difference Equation
System Function z2 Y (z) = 2 H(z) = X(z) z −z−1
y[n] = x[n] + y[n−1] + y[n−2] 3
Concept Map: Discrete-Time Systems
Two interpretations of “Delay.”
Delay → R
Block Diagram X
+
+ Delay
System Functional Y
1 Y = H(R) = X 1 − R − R2
Delay
Unit-Sample Response index shift
h[n] : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .
Difference Equation
System Function z2 Y (z) = 2 H(z) = X(z) z −z−1
y[n] = x[n] + y[n−1] + y[n−2] 4
Concept Map: Discrete-Time Systems
Relation between System Functional and System Function.
Delay → R
Block Diagram X
+
+ Delay
System Functional Y
1 Y = H(R) = X 1 − R − R2
Delay
Unit-Sample Response index shift
h[n] : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .
Difference Equation
R → z1
System Function z2 Y (z) = 2 H(z) = X(z) z −z−1
y[n] = x[n] + y[n−1] + y[n−2] 5
Check Yourself
What is relation of System Functional to Unit-Sample Response
Delay → R
Block Diagram X
+
+ Delay
System Functional Y
1 Y = H(R) = X 1 − R − R2
Delay
Unit-Sample Response index shift
h[n] : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .
Difference Equation
System Function z2 Y (z) = 2 H(z) = X(z) z −z−1
y[n] = x[n] + y[n−1] + y[n−2] 6
Check Yourself
Expand functional in a series: Y 1
= H(R) = X 1 − R − R2 1 +R +2R2 +3R3 +5R4 +8R5 + · · · 1 − R − R2 1 1 −R −R2 R +R2 R −R2 −R3 2R2 +R3 2R2 −2R3 −2R4 3R3 +2R4 3R3 −3R4 −3R5 ···
H(R) =
1 = 1 + R + 2R2 + 3R3 + 5R4 + 8R5 + 13R6 + · · · 1 − R − R2 7
Check Yourself
Coefficients of series representation of H(R) 1 = 1 + R + 2R2 + 3R3 + 5R4 + 8R5 + 13R6 + · · · H(R) = 1 − R − R2 are the successive samples in the unit-sample response! h[n] : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . . If a system is composed of (only) adders, delays, and gains, then H(R) = h[0] + h[1]R + h[2]R2 + h[3]R3 + h[4]R4 + · · · =
�
h[n]Rn
n
We can write the system function in terms of unit-sample response!
8
Check Yourself
What is relation of System Functional to Unit-Sample Response?
Delay → R
Block Diagram X
+
+ Delay
System Functional Y
1 Y = H(R) = X 1 − R − R2
Delay
H(R) =
P
h[n]Rn
Unit-Sample Response index shift
h[n] : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .
Difference Equation
System Function z2 Y (z) = 2 H(z) = X(z) z −z−1
y[n] = x[n] + y[n−1] + y[n−2] 9
Check Yourself
What is relation of System Function to Unit-Sample Response?
Delay → R
Block Diagram X
+
+ Delay
System Functional Y
1 Y = H(R) = X 1 − R − R2
Delay
H(R) =
P
h[n]Rn
Unit-Sample Response index shift
h[n] : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .
Difference Equation
System Function z2 Y (z) = 2 H(z) = X(z) z −z−1
y[n] = x[n] + y[n−1] + y[n−2] 10
Check Yourself
Start with the series expansion of system functional:
H(R) =
� X
h[n]Rn
n
Substitute R → H(z) =
� X
1 : z
h[n]z −n
n
11
Check Yourself
What is relation of System Function to Unit-Sample Response?
Delay → R
Block Diagram X
+
+ Delay
System Functional Y
1 Y = H(R) = X 1 − R − R2
Delay
H(R) =
P
h[n]Rn
Unit-Sample Response index shift
h[n] : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .
H(z) = Difference Equation
P
h[n]z −n
System Function z2 Y (z) = 2 H(z) = X(z) z −z−1
y[n] = x[n] + y[n−1] + y[n−2] 12
Check Yourself
Start with the series expansion of system functional: H(R) =
� X
h[n]Rn
n
Substitute R → H(z) =
� X
1 : z
h[n]z −n
n
Today: thinking about a system as a mathematical function H(z) rather than as an operator.
13
Z Transform
We call the relation between H(z) and h[n] the Z transform. H(z) =
X �
h[n]z −n
n
Z transform maps a function of discrete time n to a function of z. Although motivated by system functions, we can define a Z trans form for any signal. X(z) =
∞ X �
x[n]z −n
n=−∞
Notice that we include n < 0 as well as n > 0 → bilateral Z transform (there is also a unilateral Z transform with similar but not identical properties). 14
Simple Z transforms
Find the Z transform of the unit-sample signal.
δ[n]
n x[n] = δ[n] X(z) =
∞ � X
x[n]z −n = x[0]z 0 = 1
n=−∞
15
Simple Z transforms
Find the Z transform of a delayed unit-sample signal.
x[n]
n x[n] = δ[n − 1] X(z) =
∞ � X
x[n]z −n = x[1]z −1 = z −1
n=−∞
16
Check Yourself
What is the Z transform of the following signal.
n x[n] = 78 u[n]
n −4−3−2−1 0 1 2 3 4
1.
1 1 − 78 z
2.
1 1 − 78 z −1
3.
z 1 − 78 z
17
4.
z −1 1 − 78 z −1
5. none
Check Yourself
What is the Z transform of the following signal. n x[n] = 78 u[n]
n −4−3−2−1 0 1 2 3 4 X(z) =
∞ � X
n=−∞
∞ X � 7 n −n z u[n] = 8
n=0
18
7 n −n 1 z = 8 1 − 78 z −1
Check Yourself
What is the Z transform of the following signal.
2
n x[n] = 78 u[n]
n −4−3−2−1 0 1 2 3 4
1.
1 1 − 78 z
2.
1 1 − 78 z −1
3.
z 1 − 78 z
19
4.
z −1 1 − 78 z −1
5. none
Z Transform Pairs
The signal x[n], which is a function of time n, maps to a Z transform X(z), which is a function of z.
n 7 x[n] = u[n] 8
↔
1
X(z) =
1 − 78 z −1
For what values of z does X(z) make sense? The Z transform is only defined for values of z for which the defining sum converges. ∞ n ∞ n � X X � 7 7 1 −n X(z) = z u[n] = z −n = 8 8 1 − 78 z −1 n=−∞ n=0
7
7
Therefore z −1 < 1, i.e., |z| > .
8 8 20
Regions of Convergence
The Z transform X(z) is a function of z defined for all z inside a Region of Convergence (ROC). x[n] =
n 7 u[n] 8
ROC: |z| >
↔
X(z) =
1 1−
7 8
21
7 z −1 8
;
|z| >
7 8
Z Transform Mathematics
Based on properties of the Z transform. Linearity: if x1 [n] ↔ X1 (z) for z in ROC1 and x2 [n] ↔ X2 (z) for z in ROC2 then x1 [n] + x2 [n] ↔ X1 (z) + X2 (z) for z in (ROC1 ∩ ROC2 ). Let y[n] = x1 [n] + x2 [n] then ∞
X � Y (z) = y[n]z −n
=
=
n=−∞
∞
� X
(x1 [n] + x2 [n])z −n
n=−∞
∞ � X
x1 [n]z
−n
n=−∞
+
∞
� X
x2 [n]z −n
n=−∞
= X1 (z) + X2 (z) 22
Delay Property
If x[n] ↔ X(z) for z in ROC then x[n − 1] ↔ z −1 X(z) for z in ROC. We have already seen an example of this property. δ[n] ↔ 1 δ[n − 1]
↔
z −1
More generally, ∞
� X X(z) = x[n]z −n
n=−∞
Let y[n] = x[n − 1] then ∞ ∞ X � � −n Y (z) = y[n]z = x[n − 1]z −n n=−∞
n=−∞
Substitute m = n − 1 ∞
X � Y (z) = x[m]z −m−1 = z −1 X(z)
m=−∞
23
Rational Polynomials
A system that can be described by a linear difference equation with constant coefficients can also be described by a Z transform that is a ratio of polynomials in z. b0 y[n] + b1 y[n − 1] + b2 y[n − 2] + · · · = a0 x[n] + a1 x[n − 1] + a2 x[n − 2] + · · · Taking the Z transform of both sides, and applying the delay property b0 Y (z)+b1 z −1 Y (z)+b2 z −2 Y (z)+· · · = a0 X(z)+a1 z −1 X(z)+a2 z −2 X(z)+· · · H(z) =
Y (z) a0 + a1 z −1 + a2 z −2 + · · · = X(z) b0 + b1 z −1 + b2 z −2 + · · ·
=
a0 z k + a1 z k−1 + a2 z k−2 + · · ·
b0 z k + b1 z k−1 + b2 z k−2 + · · ·
24
Rational Polynomials
Applying the fundamental theorem of algebra and the factor theo rem, we can express the polynomials as a product of factors. H(z) =
=
a0 z k + a1 z k−1 + a2 z k−2 + · · · b0 z k + b1 z k−1 + b2 z k−2 + · · ·
(z − z0 ) (z − z1 ) · · · (z − zk )
(z − p0 ) (z − p1 ) · · · (z − pk )
where the roots are called poles and zeros.
25
Rational Polynomials
Regions of convergence for Z transform are delimited by circles in the Z-plane. The edges of the circles are at the poles. Example: x[n] = αn u[n] X(z) =
∞ � X
αn u[n]z −n =
n=−∞
∞
� X
αn z −n
n=0
1 −1 ; = αz |α| = z−α ROC
x[n] = αn u[n]
z z−α n
−4−3−2−1 0 1 2 3 4 26
z -plane
α
Check Yourself
What DT signal has the following Z transform?
z -plane z 7 ; |z| < 7 8 z−8
ROC 7 8
27
Check Yourself
Recall that we already know a function whose Z transform is the
outer region. ROC
7 n
x[n] = 8
u[n]
z z − 78 n
−4−3−2−1 0 1 2 3 4 What changes if the region changes? The original sum X(z) =
∞ n � X 7
n=0
8
z −n
does not converge if |z| < 87 . 28
z -plane
7 8
Check Yourself
The functional form is still the same,
H(z) =
Y (z) z .
= X(z) z − 78
Therefore, the difference equation for this system is the same, 7 y[n + 1] − y[n] = x[n + 1] . 8 Convergence inside |z| = 78 corresponds to a left-sided (non-causal)
response. Solve by iterating backwards in time:
8 y[n] = (y[n + 1] − x[n + 1]) 7
29
Check Yourself
Solve by iterating backwards in time: 8 y[n] = (y[n + 1] − x[n + 1]) 7 Start “at rest”: n x[n] y[n] >0 0 0 0 1 0
−1 0 − 87
2 −2 0 − 87
3 −3 0 − 87 ··· · · ·
−n n − 87
−n 8 y[n] = − ; 7
n
7 n 1.
Check Yourself
Find the inverse transform of Y (z) =
2 z ; z−1
|z| > 1.
y[n] corresponds to unit-sample response of the right-sided system 2 2 2 Y z 1 1 = = = X z−1 1−R 1 − z −1 = 1 + R + R2 + R3 + · · · × 1 + R + R2 + R3 + · · · 1
R2
R
R3
R2 R3 R3 R4 R4 R5 R5 R6 ··· ··· ∞ X � Y (n + 1)Rn = 1 + 2R + 3R2 + 4R3 + · · · = X 1 R R2 R3 ···
1 R R2 R3 ···
R R2 R3 R4 ···
n=0
y[n] = h[n] = (n + 1)u[n]
47
··· ··· ··· ··· ··· ···
Check Yourself
Table lookup method. 2 z Y (z) = ↔ z−1 z ↔ z−1
y[n] =? u[n]
48
Properties of Z Transforms
The use of Z Transforms to solve differential equations depends on several important properties. Property
x[n]
Linearity
ax1 [n] + bx2 [n]
Multiply by n
z −1 X(z)
R
dX(z) dz
R
−z
nx[n] ∞ � X
ROC
aX1 (z) + bX2 (z) ⊃ (R1 ∩ R2 )
x[n − 1]
Delay
Convolve in n
X(z)
x1 [m]x2 [n − m]
m=−∞
49
X1 (z)X2 (z)
⊃ (R1 ∩ R2 )
Check Yourself
Table lookup method.
2 z Y (z) = z−1 z z−1 2 1 d z =z −z z−1 dz z − 1 2 z d z = z × −z z−1 dz z − 1
50
↔
y[n] =?
↔
u[n]
↔
nu[n]
↔
(n + 1)u[n + 1] = (n + 1)u[n]
Concept Map: Discrete-Time Systems
Relations among representations.
Delay → R
Block Diagram X
+
+ Delay
System Functional Y
1 Y = H(R) = X 1 − R − R2
Delay
Unit-Sample Response index shift
h[n] : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . . Z transform
Difference Equation
System Function z2 Y (z) = 2 H(z) = X(z) z −z−1
y[n] = x[n] + y[n−1] + y[n−2] 51
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6.003: Signals and Systems Laplace Transform
September 27, 2011 1
Mid-term Examination #1 Wednesday, October 5, 7:30-9:30pm, No recitations on the day of the exam. Coverage:
CT and DT Systems, Z and Laplace Transforms Lectures 1–7 Recitations 1–7 Homeworks 1–4
Homework 4 will not collected or graded. Solutions will be posted. Closed book: 1 page of notes (8 21 × 11 inches; front and back). Designed as 1-hour exam; two hours to complete. Review sessions during open office hours. Conflict? Contact before Friday, Sept. 30, 5pm. Prior term midterm exams have been posted on the 6.003 website. 2
Concept Map for Discrete-Time Systems Last time: relations among representations of DT systems. Delay → R
Block Diagram X
+
+ Delay
System Functional Y
1 Y = H(R) = X 1 − R − R2
Delay
Unit-Sample Response index shift
h[n] : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .
Difference Equation
System Function z2 Y (z) = 2 H(z) = X(z) z −z−1
y[n] = x[n] + y[n−1] + y[n−2] 3
Concept Map for Discrete-Time Systems Most important new concept from last time was the Z transform. Delay → R
Block Diagram X
+
+ Delay
System Functional Y
1 Y = H(R) = X 1 − R − R2
Delay
Unit-Sample Response index shift
h[n] : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . . Z transform
Difference Equation
System Function z2 Y (z) = 2 H(z) = X(z) z −z−1
y[n] = x[n] + y[n−1] + y[n−2] 4
Concept Map for Continuous-Time Systems Today: similar relations among representations of CT systems. Delay → R
Block Diagram +
X
R −
+
R −
System Functional
2A2 Y = X 2 + 3A + A2
Y
1 2
1
Impulse Response ˙ x(t)
R
x(t)
h(t) = 2(e−t/2 − e−t ) u(t)
Differential Equation
System Function Y (s) 2 = 2 X(s) 2s + 3s + 1
˙ + y(t) = 2x(t) 2¨ y (t) + 3y(t) 5
Concept Map for Continuous-Time Systems Corresponding concept for CT is the Laplace Transform. Delay → R
Block Diagram +
X
R −
+
R −
System Functional
2A2 Y = X 2 + 3A + A2
Y
1 2
1
Impulse Response ˙ x(t)
R
x(t)
h(t) = 2(e−t/2 − e−t ) u(t) Laplace transform
Differential Equation
System Function Y (s) 2 = 2 X(s) 2s + 3s + 1
˙ + y(t) = 2x(t) 2¨ y (t) + 3y(t) 6
Laplace Transform: Definition Laplace transform maps a function of time t to a function of s. Z X(s) = x(t)e−st dt
There are two important variants: Unilateral (18.03) Z ∞ X(s) = x(t)e−st dt 0
Bilateral (6.003) Z ∞ X(s) = x(t)e−st dt −∞
Both share important properties. We will focus on bilateral version, and discuss differences later. 7
Laplace Transforms Example: Find the Laplace transform of x1 (t):
x1 (t) =
e−t 0
x1 (t) 1
if t ≥ 0 otherwise
t ∞ Z ∞ Z ∞ 1 e−(s+1)t −t −st −st e e dt = X1 (s) = x1 (t)e dt = = s+1 −(s + 1) 0 −∞ 0
0
provided Re(s + 1) > 0 which implies that Re(s) > −1. s-plane ROC
1 ; Re(s) > −1 s+1
−1
8
Check Yourself x2 (t) x2 (t) =
e−t − e−2t 0
if t ≥ 0 otherwise
0 Which of the following is the Laplace transform of x2 (t)? 1 1. X2 (s) = (s+1)(s+2) ; Re(s) > −1 1 2. X2 (s) = (s+1)(s+2) ; Re(s) > −2 s ; Re(s) > −1 3. X2 (s) = (s+1)(s+2) s 4. X2 (s) = (s+1)(s+2) ; Re(s) > −2
5. none of the above
9
t
Check Yourself Z ∞ X2 (s) =
(e−t − e−2t )e−st dt
0
Z ∞ e
= 0
=
−t −st
e
Z ∞ dt −
e−2t e−st dt
0
1 1 (s + 2) − (s + 1) 1 − = = s+1 s+2 (s + 1)(s + 2) (s + 1)(s + 2)
These equations converge if Re(s + 1) > 0 and Re(s + 2) > 0, thus Re(s) > −1. s-plane 1 ; Re(s) > −1 (s + 1)(s + 2)
10
ROC
−2 −1
Check Yourself x2 (t) x2 (t) =
e−t − e−2t 0
if t ≥ 0 otherwise
0
t
Which of the following is the Laplace transform of x2 (t)? 1 1 1. X2 (s) = (s+1)(s+2) ; Re(s) > −1 1 2. X2 (s) = (s+1)(s+2) ; Re(s) > −2 s ; Re(s) > −1 3. X2 (s) = (s+1)(s+2) s 4. X2 (s) = (s+1)(s+2) ; Re(s) > −2
5. none of the above
11
Regions of Convergence Left-sided signals have left-sided Laplace transforms (bilateral only). Example: x3 (t) x3 (t) =
−e−t 0
if t ≤ 0 otherwise
t −1
0 −(s+1)t −e X3 (s) = x3 (t)e−st dt = −e−t e−st dt = −(s + 1) −∞ −∞ Z ∞
Z 0
−∞
ROC
provided Re(s + 1) < 0 which implies that Re(s) < −1. s-plane 1 ; Re(s) < −1 s+1
−1
12
=
1 s+1
Left- and Right-Sided ROCs Laplace transforms of left- and right-sided exponentials have the same form (except −); with left- and right-sided ROCs, respectively. Laplace transform
time function
e−t u(t)
s-plane
1
0
−e−t u(−t) 1 s+1
t −1
13
ROC
−1
ROC
1 s+1
t
−1
s-plane
Check Yourself
Find the Laplace transform of x4 (t). x4 (t) x4 (t) = e−|t| 0 1. X4 (s) = 2. X4 (s) = 3. X4 (s) = 4. X4 (s) =
2 1−s2 2 1−s2 2 1+s2 2 1+s2
;
−∞ < Re(s) < ∞
;
−1 < Re(s) < 1
;
−∞ < Re(s) < ∞
;
−1 < Re(s) < 1
5. none of the above
14
t
Check Yourself Z ∞ X4 (s) =
e−|t| e−st dt
−∞
Z 0 =
e(1−s)t dt +
Z ∞
−∞
0
0 e(1−s)t = (1 − s)
−∞
=
∞ e−(1+s)t + −(1 + s) 0
1 1 + − s} + s} |1 {z |1 {z
Re(s)−1
1+s+1−s 2 = ; (1 − s)(1 + s) 1 − s2
−1 < Re(s) < 1
The ROC is the intersection of Re(s) < 1 and Re(s) > −1.
15
Check Yourself Laplace transform of a signal that is both-sided is a vertical strip. x4 (t) x4 (t) = e−|t| t
0 s-plane X4 (s) =
2 1 − s2
ROC
−1
−1 < Re(s) < 1
16
1
Check Yourself
Find the Laplace transform of x4 (t).
2
x4 (t) x4 (t) = e−|t| 0 1. X4 (s) = 2. X4 (s) = 3. X4 (s) = 4. X4 (s) =
2 1−s2 2 1−s2 2 1+s2 2 1+s2
;
−∞ < Re(s) < ∞
;
−1 < Re(s) < 1
;
−∞ < Re(s) < ∞
;
−1 < Re(s) < 1
5. none of the above
17
t
Time-Domain Interpretation of ROC Z ∞ X(s) =
x(t) e−st dt
−∞
x1 (t)
s-plane t −1
x2 (t)
s-plane t −2 −1
x3 (t)
−1
s-plane t −1
x4 (t)
s-plane t −1 18
1
Time-Domain Interpretation of ROC Z ∞ X(s) =
x(t) e−st dt
−∞
x1 (t)
s-plane t −1
x2 (t)
s-plane t −2 −1
x3 (t)
−1
s-plane t −1
x4 (t)
s-plane t −1 19
1
Time-Domain Interpretation of ROC Z ∞ X(s) =
x(t) e−st dt
−∞
x1 (t)
s-plane t −1
x2 (t)
s-plane t −2 −1
x3 (t)
−1
s-plane t −1
x4 (t)
s-plane t −1 20
1
Time-Domain Interpretation of ROC Z ∞ X(s) =
x(t) e−st dt
−∞
x1 (t)
s-plane t −1
x2 (t)
s-plane t −2 −1
x3 (t)
−1
s-plane t −1
x4 (t)
s-plane t −1 21
1
Check Yourself
2s The Laplace transform 2 corresponds to how many of s −4 the following signals?
1. e−2t u(t) + e2t u(t) 2. e−2t u(t) − e2t u(−t) 3. −e−2t u(−t) + e2t u(t) 4. −e−2t u(−t) − e2t u(−t)
22
Check Yourself Expand with partial fractions: 2s 1 1 = + 2 + 2} − 2} s −4 |s {z |s {z pole at −2 pole at 2 pole −2 2
function e−2t e2t
right-sided; ROC e−2t u(t); Re(s) > −2 e2t u(t); Re(s) > 2
left-sided (ROC) Re(s) < −2 2 t −e u(−t); Re(s) < 2
−e−2t u(−t);
1. e−2t u(t) + e2t u(t)
Re(s) > −2 ∩ Re(s) > 2
Re(s) > 2
2. e−2t u(t) − e2t u(−t)
Re(s) > −2 ∩ Re(s) < 2
−2 < Re(s) < 2
3. −e−2t u(−t)+e2t u(t)
Re(s) < −2 ∩ Re(s) > 2
none
4. −e−2t u(−t) − e2t u(−t)
Re(s) < −2 ∩ Re(s) < 2
Re(s) < −2
23
Check Yourself
2s The Laplace transform 2 corresponds to how many of s −4 the following signals? 3
1. e−2t u(t) + e2t u(t) 2. e−2t u(t) − e2t u(−t) 3. −e−2t u(−t) + e2t u(t) 4. −e−2t u(−t) − e2t u(−t)
24
Solving Differential Equations with Laplace Transforms Solve the following differential equation: y˙(t) + y(t) = δ(t) Take the Laplace transform of this equation. L {y˙(t) + y(t)} = L {δ(t)} The Laplace transform of a sum is the sum of the Laplace transforms (prove this as an exercise). L {y˙(t)} + L {y(t)} = L {δ(t)} What’s the Laplace transform of a derivative?
25
Laplace Transform of a Derivative Assume that Z ∞X(s) is the Laplace transform of x(t): X(s) = x(t)e−st dt −∞
Find the Laplace transform of y(t) = x˙ (t). Z ∞ Z ∞ −st st Y (s) = y(t)e dt = x(t) ˙ e|− {z} dt −∞ −∞ |{z} u
v˙
∞ Z ∞ −st x(t)(− = x(t) e| {z } − | se {z })dt |{z} |{z} −∞ −∞ −st
v
u
u˙
v
The first term Z ∞ must be zero since X (s) converged. Thus Y (s) = s x(t)e−st dt = sX(s) −∞
26
Solving Differential Equations with Laplace Transforms Back to the previous problem: L {y˙(t)} + L {y(t)} = L {δ(t)} Let Y (s) represent the Laplace transform of y(t). Then sY (s) is the Laplace transform of y˙(t). sY (s) + Y (s) = L {δ(t)} What’s the Laplace transform of the impulse function?
27
Laplace Transform of the Impulse Function Let x(t) = δ(t). Z ∞ X(s) = Z−∞ ∞ = Z−∞ ∞
δ(t)e−st dt δ(t) e−st t=0 dt δ(t) 1 dt
= −∞
=1 Sifting property: Multiplying f (t) by δ(t) and integrating over t sifts out f (0).
28
Solving Differential Equations with Laplace Transforms Back to the previous problem: sY (s) + Y (s) = L {δ(t)} = 1 This is a simple algebraic expression. Laplace transform converts a differential equation y˙(t) + y(t) = δ(t) to an equivalent algebraic equation. sY + Y = 1
29
Solving Differential Equations with Laplace Transforms Back to the previous problem: sY (s) + Y (s) = L {δ(t)} = 1 This is a simple algebraic expression. Solve for Y (s): 1 Y (s) = s+1 We’ve seen this Laplace transform previously. y(t) = e−t u(t)
(why not y(t) = −e−t u(−t) ?)
Notice that we solved the differential equation y˙(t)+y(t) = δ(t) without computing homogeneous and particular solutions.
30
Solving Differential Equations with Laplace Transforms Summary of method. Start with differential equation: y˙(t) + y(t) = δ(t) Take the Laplace transform of this equation: sY (s) + Y (s) = 1 Solve for Y (s): 1 Y (s) = s+1 Take inverse Laplace transform (by recognizing form of transform): y(t) = e−t u(t)
31
Solving Differential Equations with Laplace Transforms Recognizing the form ... Is there a more systematic way to take an inverse Laplace transform? Yes ... and no. Formally, Z σ+j ∞ 1 X(s)est ds 2πj σ−j∞ but this integral is not generally easy to compute. x(t) =
This equation can be useful to prove theorems. We will find better ways (e.g., partial fractions) to compute inverse transforms for common systems.
32
Solving Differential Equations with Laplace Transforms Example 2: y¨(t) + 3y˙(t) + 2y(t) = δ(t) Laplace transform: s2 Y (s) + 3sY (s) + 2Y (s) = 1 Solve: Y (s) =
1 1 1 = − (s + 1)(s + 2) s+1 s+2
Inverse Laplace transform: y(t) = e−t − e−2t u(t) These forward and inverse Laplace transforms are easy if • •
differential equation is linear with constant coefficients, and the input signal is an impulse function.
33
Properties of Laplace Transforms Usefulness of Laplace transforms derives from its many properties. Property
x(t)
X(s)
ROC
Linearity
ax1 (t) + bx2 (t)
aX1 (s) + bX2 (s)
⊃ (R1 ∩ R2 )
x(t − T )
X(s)e−sT
R
Delay by T
dX(s) ds X(s + α)
shift R by −α
sX(s)
⊃R
x(τ ) dτ
X(s) s
⊃ R ∩ Re(s) > 0
x1 (τ )x2 (t − τ ) dτ
X1 (s)X2 (s)
⊃ (R1 ∩ R2 )
Multiply by t
−
tx(t)
Multiply by e−αt
x(t)e−αt
Differentiate in t
dx(t) dt Z t
Integrate in t −∞
R
Z ∞ Convolve in t −∞
and many others! 34
Initial Value Theorem If x(t) = 0 for t < 0 and x(t) contains no impulses or higher-order singularities at t = 0 then x(0+ ) = lim sX(s) . s→∞
Z ∞ Consider lim sX(s) = lim s s→∞
s→∞
As s → ∞ the function
x(t)e
−st
Z ∞ dt = lim
s→∞ 0
−∞
e−st
x(t) se−st dt.
shrinks towards 0.
e−st s=1 s=5
s = 25
t
1 Area under e−st is → area under se−st is 1 → lim se−st = δ(t) ! s→∞ sZ Z ∞ ∞ −st lim sX(s) = lim x(t)se dt → x(t)δ(t)dt = x(0+ ) s→∞
s→∞ 0
0
(the 0+ arises because the limit is from the right side.) 35
Final Value Theorem If x(t) = 0 for t < 0 and x(t) has a finite limit as t → ∞ x(∞) = lim sX(s) . s→0
Z ∞ Consider lim sX(s) = lim s s→0
s→0
x(t)e
−st
−∞
Z ∞ dt = lim
s→0 0
x(t) se−st dt.
As s → 0 the function e−st flattens out. But again, the area under se−st is always 1. e−st
s = 25
s=1 s=5
x(∞) t
As s → 0, area under se−st monotonically shifts to higher values of t (e.g., the average value of se−st is 1s which grows as s → 0). In the limit, lim sX(s) → x(∞) . s→0
36
Summary: Relations among CT representations Delay → R
Block Diagram +
X
R −
+
R −
System Functional
2A2 Y = X 2 + 3A + A2
Y
1 2
1
Impulse Response ˙ x(t)
R
x(t)
h(t) = 2(e−t/2 − e−t ) u(t) Laplace transform
Differential Equation
System Function Y (s) 2 = 2 X(s) 2s + 3s + 1
˙ + y(t) = 2x(t) 2¨ y (t) + 3y(t)
Many others: e.g., Laplace transform of a circuit (see HW4)! 37
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6.003: Signals and Systems Discrete Approximation of Continuous-Time Systems
September 29, 2011
1
Mid-term Examination #1
Wednesday, October 5, 7:30-9:30pm, No recitations on the day of the exam. Coverage:
CT and DT Systems, Z and Laplace Transforms Lectures 1–7 Recitations 1–7 Homeworks 1–4
Homework 4 will not collected or graded. Solutions will be posted. Closed book: 1 page of notes (8 12 × 11 inches; front and back). No calculators, computers, cell phones, music players, or other aids. Designed as 1-hour exam; two hours to complete. Review sessions during open office hours. Conflict? Contact before Friday, Sept. 30, 5pm. Prior term midterm exams have been posted on the 6.003 website. 2
Concept Map
Today we will look at relations between CT and DT representations. Delay → R
Block Diagram X
+
+ Delay
DT
System Functional Y
Y 1 = H(R) = X 1 − R − R2
Delay
Unit-Sample Response index shift
h[n] : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .
Difference Equation
CT
System Function Y (z) z2 = 2 X(z) z −z−1
y[n] = x[n] + y[n−1] + y[n−2]
H(z) =
Delay → R
Block Diagram +
X
R −
+
R −
System Functional
1 2
1
T D
Y 2A2 = X 2 + 3A + A2
Y
Impulse Response x(t) ˙
R
x(t)
h(t) = 2(e−t/2 − e−t ) u(t)
CT Differential Equation
2¨ y (t) + 3y(t) ˙ + y(t) = 2x(t)
System Function Y (s) 2 = 2 X(s) 2s + 3s + 1
3
Discrete Approximation of CT Systems Example: leaky tank r0 (t)
h1 (t) r1 (t) R
X
AX
Block Diagram +
X
−
1 τ
System Functional
R
A Y = X A+τ
Y
Impulse Response x(t) ˙
R
x(t)
h(t) = τ1 e−t/τ u(t)
Differential Equation
System Function Y (s) 1 = X(s) 1 + τs
τ r˙1 (t) = r0 (t) − r1 (t)
H(s) =
Today: compare step responses of leaky tank and DT approximation. 4
Check Yourself (Practice for Exam)
What is the “step response” of the leaky tank system?
u(t)
s(t) =?
Leaky Tank
1
1
1.
2.
τ
t
τ
1
t
1
3.
4.
τ
t
τ
5. none of the above 5
t
Check Yourself
What is the “step response” of the leaky tank system? de:
τ r˙1 (t) = u(t) − r1 (t)
t < 0: r1 (t) = 0 t > 0: r1 (t) = c1 + c2 e−t/τ r˙1 (t) = − cτ2 e−t/τ Substitute into de: τ − cτ2 e−t/τ = 1 − c1 − c2 e−t/τ
→
c1 = 1
Combine t < 0 and t > 0: r1 (t) = u(t) + c2 e−t/τ u(t) c2 r˙1 (t) = δ(t) + c2 δ(t) − e−t/τ u(t) τ Substitute into de: c2 τ (1 + c2 )δ(t) − τ e−t/τ u(t) = u(t) − u(t) − c2 e−t/τ u(t) τ
r1 (t) = (1 − e−t/τ )u(t)
6
→
c2 = −1
Check Yourself
Alternatively, reason with systems!
δ(t) δ(t)
δ(t)
A A+τ
h(t) = τ1 e−t/τ u(t)
u(t)
A A+τ
s(t) =?
A A+τ
s(t) =?
u(t) A A A+τ
Z t
h(t) A
s(t) = −∞
t 1 I 1 −tI /τ ' ' u(t )dt = e e−t /τ dt' = (1 − e−t/τ ) u(t) −∞ τ 0 τ t
s(t) =
h(t0 )dt0
7
Check Yourself
What is the “step response” of the leaky tank system?
u(t)
s(t) =?
Leaky Tank
1
2
1
1.
2.
τ
t
τ
1
t
1
3.
4.
τ
t
τ
5. none of the above 8
t
Forward Euler Approximation
Approximate leaky-tank system using forward Euler approach. Approximate continuous signals by discrete signals: xd [n] = xc (nT ) yd [n] = yc (nT ) Approximate derivative at t = nT by looking forward in time: y [n+1] − yd [n] y˙ c (nT ) = d T
yc (t) yd [n+1]
yd [n] nT
(n+1) T 9
t
Forward Euler Approximation
Approximate leaky-tank system using forward Euler approach. Substitute xd [n] = xc (nT )
yd [n] = yc (nT )
yc (n + 1)T − yc nT y [n + 1] − yd [n]
y˙ c (nT ) ≈ = d T T into the differential equation τ y˙ c (t) = xc (t) − yc (t) to obtain τ y [n + 1] − yd [n] = xd [n] − yd [n] . T d Solve: T T yd [n + 1] − 1 − yd [n] = xd [n] τ τ 10
Forward Euler Approximation Plot.
1 T τ
= 0.1
T τ
= 0.3
t
1 t
1 T τ
=1
t
1 T τ
= 1.5
t
1 T τ
τ
=2
t
Why is this approximation badly behaved for large Tτ ? 11
Check Yourself
DT approximation: T T yd [n] = xd [n] yd [n + 1] − 1 − τ τ Find the DT pole. T τ τ 3. z = T
2. z = 1 −
1. z =
4. z = − 5. z =
1 1 + Tτ
12
τ T
T τ
Check Yourself
DT approximation:
T T yd [n] = xd [n] yd [n + 1] − 1 − τ τ Take the Z transform: T T Yd (z) = Xd (z) zYd (z) − 1 − τ τ Solve for the system function: T Y (z) τ H(z) = d = Xd (z) z− 1− T τ
Pole at z = 1 −
T . τ
13
Check Yourself
DT approximation: T T yd [n] = xd [n] yd [n + 1] − 1 − τ τ Find the DT pole.
2
T τ τ 3. z = T
2. z = 1 −
1. z =
4. z = − 5. z =
1 1 + Tτ
14
τ T
T τ
Dependence of DT pole on Stepsize z 1 t
T τ
= 0.1
T τ
= 0.3
z
1 t
z
1 T τ
t
=1 z
1 T τ
t
= 1.5 z
1 τ
T τ
t
=2
The CT pole was fixed (s = − τ1 ). Why is the DT pole changing? 15
Dependence of DT pole on Stepsize
Dependence of DT pole on T is generic property of forward Euler. Approach: make a systems model of forward Euler method. CT block diagrams: adders, gains, and integrators:
X
A
Y
y˙(t) = x(t)
Forward Euler approximation: y[n + 1] − y[n] = x[n] T Equivalent system:
X
T
+
R
Y
Forward Euler: substitute equivalent system for all integrators. 16
Example: leaky tank system Started with leaky tank system:
+
X
−
1 τ
R
Y
Replace integrator with forward Euler rule:
X
+
−
1 τ
+
T
R
Write system functional: T R T R
T R
Y τ τ = τ 1−R =
=
TR T R
X 1 + T R 1 − R +
1 − 1 −
τ 1−R τ τ Equivalent to system we previously developed: T T yd [n] = xd [n] yd [n + 1] − 1 − τ τ 17
Y
Model of Forward Euler Method
Replace every integrator in the CT system
X
A
Y
with the forward Euler model:
X
T
+
Substitute the DT operator for A: T 1 TR T A= → = z 1 = z−1 s 1−R 1− z z−1 . T Or equivalently: z = 1 + sT .
Forward Euler maps s →
18
R
Y
Dependence of DT pole on Stepsize Pole at z = 1 − Tτ = 1 + sT .
z
1 t
T τ
= 0.1
T τ
= 0.3
z
1 t
z
1 T τ
t
=1 z
1 T τ
t
= 1.5 z
1 τ
T τ
t 19
=2
Forward Euler: Mapping CT poles to DT poles
Forward Euler Map: →
s 0
1
− T1
0
− T2
−1
s
1 T
− T1
− T2
z = 1 + sT
z z → 1 + sT
−1
1
DT stability: CT pole must be inside circle of radius T1 at s = − T1 . −
2 1 3).
1
H(s) = 1+
1 s + Q ω0
s plane ω0
s 2 ω0 log |H(jω)|
r
1 1 − 2Q
2
0 −1
−1 −
1 2Q r − 1−
1 2Q
−2 2
log
−2
48
−1
0
1
2
ω ω0
Check Yourself
Find dependence of peak magnitude on Q (assume Q > 3).
Analyze with vectors.
−1
low frequencies
high frequencies
ω/ω0
ω/ω0
1 − 2Q
σ/ω0
−1
1 − 2Q
σ/ω0
1 1 ×2= 2Q Q
1×1=1
Peak magnitude increases with Q ! 49
Frequency Response of a High-Q System
As Q increases, the width of the peak narrows.
1
H(s) =
1 s + 1+ Q ω0 s plane ω0
s 2 ω0 log |H(jω)|
r
2 1 1 − 2Q
0 −1
−1 −
1 2Q
2−2 1 − 1 − 2Q −2 r
50
log −1
0
1
2
ω ω0
Frequency Response of a High-Q System
As Q increases, the width of the peak narrows.
1
H(s) =
1 s + 1+ Q ω0 s plane ω0
s 2 ω0 log |H(jω)|
r
1 1 − 2Q
2
0 −1
−1 −
1 2Q r − 1−
1 2Q
−2 2
log
−2
51
−1
0
1
2
ω ω0
Frequency Response of a High-Q System
As Q increases, the width of the peak narrows.
1
H(s) =
1 s + 1+ Q ω0 s plane ω0
s 2 ω0 log |H(jω)|
r
1 1 − 2Q
2
0 −1
−1 −
1 2Q r − 1−
1 2Q
−2 2
log
−2
52
−1
0
1
2
ω ω0
Frequency Response of a High-Q System
As Q increases, the width of the peak narrows.
1
H(s) =
1 s + 1+ Q ω0 s plane ω0
s 2 ω0 log |H(jω)|
r
1 1 − 2Q
2 0 −1
−1 −
1 2Q r − 1−
1 2Q
−2 2
log
−2
53
−1
0
1
2
ω ω0
Frequency Response of a High-Q System
As Q increases, the width of the peak narrows.
1
H(s) =
1 s + 1+ Q ω0 s plane ω0
s 2 ω0 log |H(jω)|
r
1 1 − 2Q
2 0 −1
−1 −
1 2Q r − 1−
1 2Q
−2 2
log
−2
54
−1
0
1
2
ω ω0
Check Yourself
Estimate the “3dB bandwidth” of the peak (assume Q > 3). Let ωl (or ωh ) represent the lowest (or highest) frequency for which the magnitude is greater than the peak value divided by √ 2. The 3dB bandwidth is then ωh − ωl .
s plane ω0
log |H(jω)| r
1 1 − 2Q
2
0 −1
−1 −
1 2Q r − 1−
1 2Q
−2 2
log
−2
−1
55
0
1
2
ω ω0
Check Yourself
Estimate the “3dB bandwidth” of the peak (assume Q > 3).
Analyze with vectors. low frequencies
high frequencies
ω/ω0
ω/ω0 1+
1−
−1
1 − 2Q
1 2Q
1 2Q σ/ω0
−1
√ √ 1 2 2Q × 2 = Q2
Bandwidth approximately
1 − 2Q
σ/ω0
√ √ 1 2 2Q × 2 = Q2
1 Q 56
Frequency Response of a High-Q System
As Q increases, the phase changes more abruptly with ω.
1
H(s) =
1 s + 1+ Q ω0 s plane ω0
s 2 ω0 ∠H(jω) 0 −π/2
−1 −π
log
−2
57
−1
0
1
2
ω ω0
Frequency Response of a High-Q System
As Q increases, the phase changes more abruptly with ω.
1
H(s) =
1 s + 1+ Q ω0 s plane ω0
s 2 ω0 ∠H(jω)
r
2 1 1 − 2Q
0
−π/2 −1 −
1 2Q
2−π 1 − 1 − 2Q −2 r
58
log −1
0
1
2
ω ω0
Frequency Response of a High-Q System
As Q increases, the phase changes more abruptly with ω.
1
H(s) =
1 s + 1+ Q ω0 s plane ω0
s 2 ω0 ∠H(jω)
r
1 1 − 2Q
2
0
−π/2 −1 −
1 2Q r − 1−
1 2Q
−π 2
log
−2
59
−1
0
1
2
ω ω0
Frequency Response of a High-Q System
As Q increases, the phase changes more abruptly with ω.
1
H(s) =
1 s + 1+ Q ω0 s plane ω0
s 2 ω0 ∠H(jω)
r
1 1 − 2Q
2
0
−π/2 −1 −
1 2Q r − 1−
1 2Q
−π 2
log
−2
60
−1
0
1
2
ω ω0
Frequency Response of a High-Q System
As Q increases, the phase changes more abruptly with ω.
1
H(s) =
1 s + 1+ Q ω0 s plane ω0
s 2 ω0 ∠H(jω)
r
1 1 − 2Q
2 0 −π/2
−1 −
1 2Q r − 1−
1 2Q
−π 2
log
−2
61
−1
0
1
2
ω ω0
Check Yourself Estimate change in phase that occurs over the 3dB bandwidth.
1
H(s) =
1 s + 1+ Q ω0 s plane ω0
s 2 ω0 ∠H(jω)
r
2 1 1 − 2Q
0
−π/2 −1 −
1 2Q
2−π 1 − 1 − 2Q −2 r
62
log −1
0
1
2
ω ω0
Check Yourself
Estimate change in phase that occurs over the 3dB bandwidth.
Analyze with vectors. low frequencies
high frequencies
ω/ω0
ω/ω0 1+
1−
−1
1 − 2Q
1 2Q
1 2Q σ/ω0
−1
π π π − = 2 4 4 Change in phase approximately
1 − 2Q
σ/ω0
π π 3π + = 2 4 4 π . 2
63
Summary
The frequency response of a system can be quickly determined using Bode plots. Bode plots are constructed from sections that correspond to single poles and single zeros. Responses for each section simply sum when plotted on logarithmic coordinates.
64
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6.003 Signals and Systems Fall 2011
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6.003: Signals and Systems CT Feedback and Control
October 20, 2011
1
Mid-term Examination #2
Wednesday, October 26, 7:30-9:30pm, No recitations on the day of the exam. Coverage:
Lectures 1–12 Recitations 1–12 Homeworks 1–7
Homework 7 will not be collected or graded. Solutions will be posted. Closed book: 2 pages of notes (8 12 × 11 inches; front and back). No calculators, computers, cell phones, music players, or other aids. Designed as 1-hour exam; two hours to complete. Review sessions during open office hours. Conflict? Contact before Friday, Oct. 21, 5pm. 2
Feedback and Control
Feedback: simple, elegant, and robust framework for control.
X
+
E
C
Y
− controller
plant
S sensor
Last time: robotic driving.
di = desiredFront do = distanceFront 3
Feedback and Control
Using feedback to enhance performance. Examples: • • • • •
increasing speed and bandwidth controlling position instead of speed reducing sensitivity to parameter variation reducing distortion stabilizing unstable systems − magnetic levitation − inverted pendulum
4
Op-amps
An “ideal” op-amp has many desireable characteristics.
V+
Vo = K (V+ − V− )
K V−
• • • • •
high speed large bandwidth high input impedance low output impedance ...
It is difficult to build a circuit with all of these features.
5
Op-Amp
∠K(jω)|
|K(jω)| [log scale]
The gain of an op-amp depends on frequency.
105 104 103 102
ω [log scale] 1
10
102
103
104
105
106
10
102
103
104
105
106
0 − π2
ω [log scale] 1
Frequency dependence of LM741 op-amp.
6
Op-Amp
∠K(jω)|
|K(jω)| [log scale]
Low-gain at high frequencies limits applications. audio frequencies
105 104 103 102
ω [log scale] 1
10
102
103
104
105
106
10
102
103
104
105
106
0 − π2
ω [log scale] 1
Unacceptable frequency response for an audio amplifier.
7
Op-Amp
An ideal op-amp has fast time response.
Vi Vo
K
Step response:
Vo (t) = s(t)
Vi (t) = u(t) 1
A t
t
8
Check Yourself
s(t) A
τ
t [seconds]
∠K(jω)| |K(jω)| [log scale]
Determine τ for the unit-step response s(t) of an LM741. 105 104 103 102 0 − π2 1
1. 40 s
40 s 2. 2π
1 s 3. 40
4. 2π 40 s
0. none of the above 9
10
102 103 ω [log scale]
104
1 5. 2π×40 s
105
Check Yourself
Determine the step response of an LM741. System function: αK0 K(s) = s+α Impulse response: h(t) = αK0 e−αt u(t) Step response: t
s(t) =
t
h(τ )dτ = −∞
t
αK0 e−ατ dτ =
0
Parameters: A = K0 = 2 × 105 1 1 s τ= = α 40 10
αK0 e−ατ −αt )u(t)
= K0 (1 − e −α 0
Check Yourself
s(t) A
τ
t [seconds]
∠K(jω)| |K(jω)| [log scale]
Determine τ for the unit-step response s(t) of an LM741. 105 104 103 102 0 − π2 1
1. 40 s
40 s 2. 2π
1 s 3. 40
4. 2π 40 s
0. none of the above 11
10
102 103 ω [log scale]
104
1 5. 2π×40 s
105
Op-Amp
Performance parameters for real op-amps fall short of the ideal.
|K(jω)| [log scale]
Frequency Response: high gain but only at low frequencies.
105
audio frequencies
104 103 102
ω [log scale] 1
10
102
103
104
105
106
Step Response: slow by electronic standards.
2 × 105 t [seconds] 1/40 12
Op-Amp
We can use feedback to improve performance of op-amps. circuit
Vi
6.003 model
K(s)
Vo
Vi
+
−
K(s)
R1 βV0
β
R2 K(s) Vo = Vi 1 + βK(s) V− = βVo =
R2 R1 + R2
Vo
13
Vo
Dominant Pole
Op-amps are designed to have a dominant pole at low frequencies: → simplifies the application of feedback.
s-plane
−40
α = 40 rad/s =
40 rad/s ≈ 6.4 Hz 2π rad/cycle
14
Improving Performance
Using feedback to improve performance parameters. circuit
Vi
6.003 model
K(s)
Vo
Vi
R1
+
K(s)
−
βV0
β
K(s) Vo = Vi 1 + βK(s)
R2
V− = βVo =
R2 R1 + R2
=
αK0 s+α 0 1 + β αK s+α
=
αK0 s + α + αβK0
Vo
15
Vo
Check Yourself
What is the most negative value of the closed-loop pole that can be achieved with feedback?
s-plane
?
−α
1. −α(1 + β) 2. −α(1 + βK0 ) 3. −α(1 + K0 ) 4. −∞ 5. none of the above 16
Improving Performance
Using feedback to improve performance parameters. circuit
Vi
6.003 model
K(s)
Vo
Vi
R1
+
K(s)
−
βV0
β
K(s) Vo = Vi 1 + βK(s)
R2
V− = βVo =
R2 R1 + R2
=
αK0 s+α 0 1 + β αK s+α
=
αK0 s + α + αβK0
Vo
17
Vo
Check Yourself
What is the most negative value of the closed-loop pole that can be achieved with feedback? Open loop system function:
αK0 s+α
→ pole: s = −α. Closed-loop system function:
αK0 s + α + αβK0
→ pole: s = −α(1 + βK0 ). The feedback constant is 0 ≤ β ≤ 1. → most negative value of the closed-loop pole is s = −α(1 + K0 ).
18
Check Yourself
What is the most negative value of the closed-loop pole that can be achieved with feedback? 3
s-plane
−α(1 + K0 )
−α
1. −α(1 + β) 2. −α(1 + βK0 ) 3. −α(1 + K0 ) 4. −∞ 5. none of the above 19
Improving Performance
∠H(jω)|
|H(jω)| [log scale] |H(j0)|
Feedback extends frequency response by a factor of 1 + βK0 (K0 = 2 × 105 ).
1 0.1
1 + βK0
0.01 0.001
ω [log scale] 1
10
102
103
0
104
105
106
1 + βK0
− π2
ω [log scale] 1
10
102
103
104
20
105
106
Improving Performance
Feedback produces higher bandwidths by reducing the gain at low
frequencies. It trades gain for bandwidth.
β = 0 (open loop)
|H(jω)| [log scale]
105 104 103 102
β = 10−4
β = 10−2
10 1
β=1
0.1 ω [log scale] 1
102
104 21
106
108
Improving Performance
Feedback makes the time response faster by a factor of 1 + βK0 (K0 = 2 × 105 ). Step response s(t) =
K0 (1 − e−α(1+βK0 )t )u(t) 1 + βK0
K0 1 + βK0
s(t) β=1 β=0 t [seconds] 1/40
22
Improving Performance
Feedback produces faster responses by reducing the final value of the step response. It trades gain for speed. Step response s(t) =
K0 (1 − e−α(1+βK0 )t )u(t) 1 + βK0
s(t)
β 0
2 × 105
0.5 × 10−5 1.5 × 10−5 t [seconds] 1/40 ds(t) The maximum rate of voltage change is not increased. dt t=0+
23
Improving Performance
Feedback improves performance parameters of op-amp circuits. • •
can extend frequency response can increase speed
Performance enhancements are achieved through a reduction of gain.
24
Motor Controller
We wish to build a robot arm (actually its elbow). The input should
be voltage v(t), and the output should be the elbow angle θ(t).
v(t)
robotic arm
θ(t) ∝ v(t)
We wish to build the robot arm with a DC motor.
v(t)
DC motor
θ(t)
This problem is similar to the head-turning servo in 6.01 !
25
Check Yourself
What is the relation between v(t) and θ(t) for a DC motor?
v(t)
1. 2. 3. 4. 5.
DC motor
θ(t) ∝ v(t) cos θ(t) ∝ v(t) θ(t) ∝ v(t) ˙ cos θ(t) ∝ v(t) ˙ none of the above
26
θ(t)
Check Yourself
What is the relation between v(t) and θ(t) for a DC motor?
To first order, the rotational speed θ˙(t) of a DC motor is proportional
to the input voltage v(t).
v(t)
θ(t)
DC motor
v(t)
θ(t)
t
t
First-order model: integrator
V
γA 27
Θ
Check Yourself
What is the relation between v(t) and θ(t) for a DC motor?
v(t)
1. 2. 3. 4. 5.
DC motor
θ(t) ∝ v(t) cos θ(t) ∝ v(t) θ(t) ∝ v(t) ˙ cos θ(t) ∝ v(t) ˙ none of the above
28
θ(t)
Motor Controller
Use proportional feedback to control the angle of the motor’s shaft.
v(t)
+
amplifier
DC motor
α
γA
−
β feedback (potentiometer) αγ 1s αγ Θ αγA = = = 1 V 1 + αβγA s + αβγ 1 + αβγ s
29
θ(t)
Motor Controller
The closed loop system has a single pole at s = −αβγ. Θ αγ = V s + αβγ
ω
s-plane
σ
As α increases, the closed-loop pole becomes increasingly negative.
30
Motor Controller
Find the impulse and step response. The system function is Θ αγ = . V s + αβγ The impulse response is h(t) = αγe−αβγt u(t) and the step response is therefore 1p s(t) = 1 − e−αβγt u(t) . β
θ(t) 1 β α↑
t
The response is faster for larger values of α. Try it: Demo.
31
Motor Controller
The speed of a DC motor does not change instantly if the voltage
is stepped. There is lag due to rotational inertia. Second-order model integrator with lag pA γA Θ2 V 1 + pA
First-order model integrator
V
γA
Θ1
Step response of the models:
v(t)
θ(t) θ1 (t) = γtu(t) θ2 (t) = γt − γp (1 − e−pt ) u(t) t
1 t
32
Motor Controller
Analyze second-order model. amplifier
v(t)
+
−
DC motor γpA2 1 + pA
α
θ(t)
β feedback (potentiometer) αγpA2
αγp Θ αγpA2 1+pA = = 2 2 = s2 + ps + αβγp αβγpA V 1 + pA + αβγpA 1 + 1+pA p p 2 p s=− ± − αβγp 2 2
33
Motor Controller
For second-order model, increasing α causes the poles at 0 and −p to approach each other, collide at s = −p/2, then split into two poles with imaginary parts.
ω
s-plane
σ
−p
Increasing the gain α does not increase speed of convergence.
34
Motor Controller
Step response.
s(t)
1 β
t
35
Motor Controller
Step response.
s(t)
1 β
t
36
Motor Controller
Step response.
s(t)
1 β
t
37
Motor Controller
Step response.
s(t)
1 β
t
38
Motor Controller
Step response.
s(t)
1 β
e−pt/2 cos(ωd t + φ)
39
t
Feedback and Control: Summary
CT feedback is useful for many reasons. Today we saw two: • •
increasing speed and bandwidth controlling position instead of speed
Next time we will look at several others: • • •
reduce sensitivity to parameter variation reduce distortion stabilize unstable systems − magnetic levitation − inverted pendulum
40
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6.003 Signals and Systems Fall 2011
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
6.003: Signals and Systems CT Feedback and Control
October 25, 2011
1
Mid-term Examination #2
Tomorrow, October 26, 7:30-9:30pm, No recitations on the day of the exam. Coverage:
Lectures 1–12 Recitations 1–12 Homeworks 1–7
Homework 7 will not be collected or graded. Solutions are posted. Closed book: 2 pages of notes (8 12 × 11 inches; front and back). No calculators, computers, cell phones, music players, or other aids. Designed as 1-hour exam; two hours to complete. Old exams and solutions are posted on the 6.003 website.
2
Feedback and Control
Using feedback to enhance performance. Examples: • • • • •
improve performance of an op amp circuit. control position of a motor. reduce sensitivity to unwanted parameter variation. reduce distortions. stabilize unstable systems − magnetic levitation − inverted pendulum
3
Feedback and Control
Reducing sensitivity to unwanted parameter variation.
Example: power amplifier power amplifier
F0
MP3 player
speaker
8 < F0 < 12 Changes in F0 (due to changes in temperature, for example) lead to undesired changes in sound level.
4
Feedback and Control
Feedback can be used to compensate for parameter variation. power amplifier MP3 player
X
+
−
F0
K
8 < F0 < 12 β H(s) =
KF0 1 + βKF0
If K is made large, so that βKF0 » 1, then 1 H(s) ≈ β independent of K or F0 ! 5
Y speaker
Feedback and Control
Feedback reduces the change in gain due to change in F0 . MP3 player
X
+
−
F0
100
Y
8 < F0 < 12 1 10
Gain to Speaker
20 F0 (no feedback) 100F0
(feedback) 0 1 + 100F 10
10 8 < F0 < 12 0
F0 0
10
6
20
Check Yourself power amplifier MP3 player
X
+
−
F0
K
8 < F0 < 12
Y speaker
β Feedback greatly reduces sensitivity to variations in K or F0 . lim H(s) =
K→∞
KF0 1 → 1 + βKF0 β
What about variations in β? Aren’t those important?
7
Check Yourself
What about variations in β? Aren’t those important? The value of β is typically determined with resistors, whose values are quite stable (compared to semiconductor devices).
8
Crossover Distortion
Feedback can compensate for parameter variation even when the variation occurs rapidly. Example: using transistors to amplify power.
+50V
MP3 player speaker
−50V
9
Crossover Distortion
This circuit introduces “crossover distortion.”
For the upper transistor to conduct, Vi − Vo > VT . For the lower transistor to conduct, Vi − Vo < −VT .
Vo
+50V
Vi
−VT
Vo
VT −50V
10
Vi
Crossover Distortion
Crossover distortion changes the shapes of signals. Example: crossover distortion when the input is Vi (t) = B sin(ω0 t).
Vo (t)
+50V
Vi
Vo
t
−50V
11
Crossover Distortion
Feedback can reduce the effects of crossover distortion.
+50V
MP3 player
+
−
K speaker
−50V
12
Crossover Distortion
When K is small, feedback has little effect on crossover distortion.
+50V
Vi
+
−
Vo (t)
Vo
K
−50V
13
K=1
t
Crossover Distortion Feedback reduces crossover distortion.
+50V
Vi
+
−
Vo (t)
Vo
K
−50V
14
K=2
t
Crossover Distortion Feedback reduces crossover distortion.
+50V
Vi
+
−
Vo (t)
Vo
K
−50V
15
K=4
t
Crossover Distortion Feedback reduces crossover distortion.
+50V
Vi
+
−
Vo (t)
Vo
K
−50V
16
K = 10
t
Crossover Distortion
+50V
Demo
• • • • • • •
original no feedback K=2 K=4 K=8 K = 16 original
Vi
+
−
Vo
K
−50V Vo (t)
t
J.S. Bach, Sonata No. 1 in G minor Mvmt. IV. Presto Nathan Milstein, violin 17
Feedback and Control
Using feedback to enhance performance. Examples: • • • • •
improve performance of an op amp circuit. control position of a motor. reduce sensitivity to unwanted parameter variation. reduce distortions. stabilize unstable systems − magnetic levitation − inverted pendulum
18
Control of Unstable Systems
Feedback is useful for controlling unstable systems.
Example: Magnetic levitation.
i(t) = io
y(t)
19
Control of Unstable Systems
Magnetic levitation is unstable.
i(t) = io
fm (t) y(t)
Mg Equilibrium (y = 0): magnetic force fm (t) is equal to the weight M g.
Increase y → increased force → further increases y.
Decrease y → decreased force → further decreases y.
Positive feedback!
20
Modeling Magnetic Levitation
The magnet generates a force that depends on the distance y(t).
i(t) = io
fm (t) y(t)
Mg fm (t) i(t) = i0 Mg y(t) 21
Modeling Magnetic Levitation
The net force f (t) = fm (t) − M g accelerates the mass.
i(t) = io
fm (t) y(t)
Mg f (t) = fm (t) − M g = M a = M y¨(t) i(t) = i0 y(t)
22
Modeling Magnetic Levitation
Represent the magnet as a system: input y(t) and output f (t).
i(t) = io
fm (t) y(t)
Mg f (t) = fm (t) − M g = M a = M y¨(t) i(t) = i0 y(t)
23
y(t)
magnet
f (t)
Modeling Magnetic Levitation
The magnet system is part of a feedback system.
f (t) = fm (t) − M g = M a = M y¨(t) i(t) = i0 y(t)
y(t)
magnet
f (t)
y(t)
magnet
y¨(t)
1 M
A
24
A
y(t)
f (t)
Modeling Magnetic Levitation
For small distances, force grows approximately linearly with distance.
f (t) = fm (t) − M g = M a = M y¨(t) i(t) = i0 K
y(t)
f (t) K
1 M
y(t)
y(t)
K
A
y(t)
y¨(t) A
25
f (t)
“Levitation” with a Spring
Relation between force and distance for a spring is opposite in sign. F = K x(t) − y(t) = M y¨(t)
x(t)
y(t) f (t)
Mg −K y(t) 26
Block Diagrams
Block diagrams for magnetic levitation and spring/mass are similar. Spring and mass F = K x(t) − y(t) = M y¨(t)
x(t)
+
K M
−
y¨(t)
˙ y(t)
A
y(t)
A
Magnetic levitation F = Ky(t) = M y¨(t)
x(t) = 0
+ +
K M
y¨(t)
27
A
˙ y(t)
A
y(t)
Check Yourself
How do the poles of these two systems differ? Spring and mass F = K x(t) − y(t) = M y¨(t)
x(t)
+
K M
−
y¨(t)
˙ y(t)
A
y(t)
A
Magnetic levitation F = Ky(t) = M y¨(t)
x(t) = 0
+ +
K M
y¨(t)
28
A
˙ y(t)
A
y(t)
Check Yourself
How do the poles of the two systems differ?
Spring and mass
F = K x(t) − y(t) = M y¨(t) K Y M = K X s2 + M
r → s = ±j
s-plane
K M
Magnetic levitation
s-plane
F = Ky(t) = M y¨(t) r K K → s=± s = M M 2
29
Magnetic Levitation is Unstable
i(t) = io
fm (t) y(t)
Mg y(t)
magnet
f (t)
y¨(t)
1 M
A
30
A
y(t)
Magnetic Levitation
We can stabilize this system by adding an additional feedback loop to control i(t).
f (t) i(t) = 1.1i0 i(t) = i0 i(t) = 0.9i0 Mg y(t)
31
Stabilizing Magnetic Levitation
Stabilize magnetic levitation by controlling the magnet current.
i(t) = io
fm (t) y(t)
Mg i(t)
y(t)
magnet
α f (t)
y¨(t)
1 M
A
32
A
y(t)
Stabilizing Magnetic Levitation
Stabilize magnetic levitation by controlling the magnet current.
i(t) = io
fm (t) y(t)
Mg fi (t)
+
fo (t)
−K2 1 M
A K 33
A
y(t)
Magnetic Levitation
Increasing K2 moves poles toward the origin and then onto jω axis.
x(t)
+
K−K2 M
y¨(t)
A
˙ y(t)
A
y(t)
s-plane
But the poles are still marginally stable. 34
Magnetic Levitation
Adding a zero makes the poles stable for sufficiently large K2 .
x(t)
+
K−K2 M
(s + z0 )
y¨(t)
A
˙ y(t)
A
s-plane
Try it: Demo [designed by Prof. James Roberge]. 35
y(t)
Inverted Pendulum
As a final example of stabilizing an unstable system, consider an
inverted pendulum.
m
θ(t)
d2 x(t) dt2
mg
θ(t)
l
mg l
x(t)
lab frame (inertial)
cart frame (non-inertial)
2 d2 x(t) 2 d θ(t) = mg l sin θ(t) − m l cos θ(t) ml m-l2 dt2 m-l2 m -l 2 dt2 2 m -l 2 m -l I
force distance
force
36
distance
Check Yourself: Inverted Pendulum
Where are the poles of this system?
θ(t)
m mg
d2 x(t) dt2
θ(t)
l
mg
x(t)
ml2
l
d2 x(t) d2 θ(t) = mgl sin θ(t) − m l cos θ(t) dt2 dt2
37
Check Yourself: Inverted Pendulum
Where are the poles of this system?
θ(t)
m mg
d2 x(t) dt2
θ(t)
l
mg
x(t)
l
ml2
d2 x(t) d2 θ(t) = mgl sin θ(t) − m l cos θ(t) dt2 dt2
ml2
d2 θ(t) d2 x(t) − mglθ(t) = −ml dt2 dt2
H(s) =
−s2 /l Θ −mls2 = = X s2 − g/l ml2 s2 − mgl 38
poles at s = ±
g l
Inverted Pendulum
This unstable system can be stablized with feedback.
θ(t)
m mg
d2 x(t) dt2
θ(t)
l
mg
x(t)
l
Try it. Demo. [originally designed by Marcel Gaudreau]
39
Feedback and Control
Using feedback to enhance performance. Examples: • • • • •
improve performance of an op amp circuit. control position of a motor. reduce sensitivity to unwanted parameter variation. reduce distortions. stabilize unstable systems − magnetic levitation − inverted pendulum
40
MIT OpenCourseWare http://ocw.mit.edu
6.003 Signals and Systems Fall 2011
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
6.003: Signals and Systems Fourier Representations
October 27, 2011 1
Fourier Representations Fourier series represent signals in terms of sinusoids. → leads to a new representation for systems as filters.
2
Fourier Series Representing signals by their harmonic components.
1
2
3
4
5
fundamental →
second harmonic →
third harmonic →
fourth harmonic →
fifth harmonic →
6 sixth harmonic →
0 DC →
ω0 2ω0 3ω0 4ω0 5ω0 6ω0
3
ω ← harmonic #
Musical Instruments Harmonic content is natural way to describe some kinds of signals.
Ex: musical instruments (http://theremin.music.uiowa.edu/MIS.html)
piano
cello t
t oboe
bassoon
horn t
t altosax
t violin
bassoon t
4
t
t 1 seconds 252
Musical Instruments Harmonic content is natural way to describe some kinds of signals. Ex: musical instruments (http://theremin.music.uiowa.edu/MIS.html)
piano
cello
k oboe
bassoon
k horn
k
altosax
k violin
k 5
k
k
Musical Instruments Harmonic content is natural way to describe some kinds of signals. Ex: musical instruments (http://theremin.music.uiowa.edu/MIS.html )
piano piano t k violin violin t k bassoon bassoon t k 6
Harmonics Harmonic structure determines consonance and dissonance. octave (D+D’)
fifth (D+A)
D+E
0 1 2 3 4 5 6 7 8 9 101112
0 1 2 3 4 5 6 7 8 9 101112
0 1 2 3 4 5 6 7 8 9 101112
1
0
–1
ti me( per iods of "D")
D'
A
E
0 1 2 3 4 5 6 7 8 9
0 1 2 3 4 5 6 7 8 9
0123456789
0123456789
0123456789
D
D
D
0
1
2
3
4
5
6
7
har monics 7
Harmonic Representations What signals can be represented by sums of harmonic components?
ω0
ω
2ω0 3ω0 4ω0 5ω0 6ω0
t 2π T= ω 0
t T=
2π ω0
Only periodic signals: all harmonics of ω0 are periodic in T = 2π/ω0 . 8
Harmonic Representations Is it possible to represent ALL periodic signals with harmonics? What about discontinuous signals?
t 2π ω0
t 2π ω0
Fourier claimed YES — even though all harmonics are continuous! Lagrange ridiculed the idea that a discontinuous signal could be written as a sum of continuous signals. We will assume the answer is YES and see if the answer makes sense. 9
Separating harmonic components Underlying properties. 1. Multiplying two harmonics produces a new harmonic with the same fundamental frequency: e jkω0 t × e jlω0 t = e j(k+l)ω0 t . 2. The integral of a harmonic over any time interval with length equal to a period T is zero unless the harmonic is at DC: t0 +T t0
e jkω0 t dt ≡
e jkω0 t dt = T
0, k = 0 T, k = 0
= T δ[k]
10
Separating harmonic components Assume that x(t) is periodic in T and is composed of a weighted sum of harmonics of ω0 = 2π/T . ∞ f x(t) = x(t + T ) = ak e jω0 kt k=−∞
Then Z x(t)e
−jlω0 t
Z dt =
T
=
=
∞ f
ak e jω0 kt e−jω0 lt dt
T k=−∞ Z ∞ f
ak
k=−∞ ∞ f
e jω0 (k−l)t dt
T
ak T δ[k − l] = T al
k=−∞
Therefore Z 1 ak = x(t)e−jω0 kt dt T T
=
Z 2π 1 x(t)e−j T kt dt T T 11
Fourier Series Determining harmonic components of a periodic signal.
Z 2π 1 ak = x(t)e−j T kt dt T T x(t)= x(t + T ) =
∞ f
ak e j
(“analysis” equation) 2π kt T
(“synthesis” equation)
k=−∞
12
Check Yourself Let ak represent the Fourier series coefficients of the following square wave. 1 2
0
1
t
− 12 How many of the following statements are true? 1. 2. 3. 4. 5.
ak = 0 if k is even ak is real-valued |ak | decreases with k 2 there are an infinite number of non-zero ak all of the above 13
Check Yourself Let ak represent the Fourier series coefficients of the following square wave. 1 2
0
1
− 12 Z 1 Z 1 0 −j2πkt 1 2 −j2πkt ak = x(t)e dt = − e dt + e dt 2 −1 2 0 T 2 � � 1 = 2 − e jπk − e−jπk j4πk ⎧ ⎨ 1 ; if k is odd = jπk ⎩ 0 ; otherwise Z
−j 2π T kt
14
t
Check Yourself Let ak represent the Fourier series coefficients of the following square wave. 1 ; if k is odd ak = jπk 0 ; otherwise How many of the following statements √ are true? 1. ak = 0 if k is even 2. ak is real-valued X 3. |ak | decreases with k 2 X 4. there are an infinite number of non-zero ak 5. all of the above X
15
√
Check Yourself Let ak represent the Fourier series coefficients of the following square wave. 1 2
0
t
1
− 12 How many of the following statements are true? 1. 2. 3. 4. 5.
2
√ ak = 0 if k is even ak is real-valued X |ak | decreases with k 2 X there are an infinite number of non-zero ak all of the above X 16
√
Fourier Series Properties If a signal is differentiated in time, its Fourier coefficients are multi plied by j 2π T k. Proof: Let x(t) = x(t + T ) =
∞ f
ak e j
2π kt T
k=−∞
then x˙ (t) = x˙ (t + T ) =
∞ f k=−∞
j
2π 2π k ak e j T kt T
17
Check Yourself Let bk represent the Fourier series coefficients of the following triangle wave. 1 8
0
1
t
− 18 How many of the following statements are true? 1. 2. 3. 4. 5.
bk = 0 if k is even bk is real-valued |bk | decreases with k 2 there are an infinite number of non-zero bk all of the above 18
Check Yourself The triangle waveform is the integral of the square wave. 1 2
0
1
t
− 12 1 8
0
1
t
− 18 Therefore the Fourier coefficients of the triangle waveform are times those of the square wave. 1 1 −1 bk = × = 2 2 ; k odd jkπ j2πk 2k π 19
1 j2πk
Check Yourself Let bk represent the Fourier series coefficients of the following tri angle wave. −1 bk = 2 2 ; k odd 2k π How many of the following statements √ are true? 1. bk = 0 if k is even √ 2. bk is real-valued √ 3. |bk | decreases with k 2 4. there are an infinite√number of non-zero bk 5. all of the above
20
√
Check Yourself Let bk represent the Fourier series coefficients of the following triangle wave. 1 8
0
t
1
− 18 How many of the following statements are true? 1. 2. 3. 4. 5.
5
√ bk = 0 if k is even √ bk is real-valued √ |bk | decreases with k 2 there are an infinite√number of non-zero bk all of the above 21
√
Fourier Series One can visualize convergence of the Fourier Series by incrementally adding terms. Example: triangle waveform 0 X k = −0 k odd
1 8
0
−1 j2πkt e 2k 2 π 2
1
− 18
22
t
Fourier Series One can visualize convergence of the Fourier Series by incrementally adding terms. Example: triangle waveform 1 X k = −1 k odd
1 8
0
−1 j2πkt e 2k 2 π 2
1
− 18
23
t
Fourier Series One can visualize convergence of the Fourier Series by incrementally adding terms. Example: triangle waveform 3 X k = −3 k odd
1 8
0
−1 j2πkt e 2k 2 π 2
1
− 18
24
t
Fourier Series One can visualize convergence of the Fourier Series by incrementally adding terms. Example: triangle waveform 5 X k = −5 k odd
1 8
0
−1 j2πkt e 2k 2 π 2
1
− 18
25
t
Fourier Series One can visualize convergence of the Fourier Series by incrementally adding terms. Example: triangle waveform 7 X k = −7 k odd
1 8
0
−1 j2πkt e 2k 2 π 2
1
− 18
26
t
Fourier Series One can visualize convergence of the Fourier Series by incrementally adding terms. Example: triangle waveform 9 X k = −9 k odd
1 8
0
−1 j2πkt e 2k 2 π 2
1
− 18
27
t
Fourier Series One can visualize convergence of the Fourier Series by incrementally adding terms. Example: triangle waveform 19 X k = −19 k odd
1 8
0
−1 j2πkt e 2k 2 π 2
1
− 18
28
t
Fourier Series One can visualize convergence of the Fourier Series by incrementally adding terms. Example: triangle waveform 29 X k = −29 k odd
1 8
0
−1 j2πkt e 2k 2 π 2
1
− 18
29
t
Fourier Series One can visualize convergence of the Fourier Series by incrementally adding terms. Example: triangle waveform 39 X k = −39 k odd
1 8
0
−1 j2πkt e 2k 2 π 2
1
t
− 18 Fourier series representations of functions with discontinuous slopes converge toward functions with discontinuous slopes.
30
Fourier Series One can visualize convergence of the Fourier Series by incrementally adding terms. Example: square wave 0 X k = −0 k odd
1 2
0
1 j2πkt e jkπ
1
− 12
31
t
Fourier Series One can visualize convergence of the Fourier Series by incrementally adding terms. Example: square wave 1 X k = −1 k odd
1 2
0
1 j2πkt e jkπ
1
− 12
32
t
Fourier Series One can visualize convergence of the Fourier Series by incrementally adding terms. Example: square wave 3 X k = −3 k odd
1 2
0
1 j2πkt e jkπ
1
− 12
33
t
Fourier Series One can visualize convergence of the Fourier Series by incrementally adding terms. Example: square wave 5 X k = −5 k odd
1 2
0
1 j2πkt e jkπ
1
− 12
34
t
Fourier Series One can visualize convergence of the Fourier Series by incrementally adding terms. Example: square wave 7 X k = −7 k odd
1 2
0
1 j2πkt e jkπ
1
− 12
35
t
Fourier Series One can visualize convergence of the Fourier Series by incrementally adding terms. Example: square wave 9 X k = −9 k odd
1 2
0
1 j2πkt e jkπ
1
− 12
36
t
Fourier Series One can visualize convergence of the Fourier Series by incrementally adding terms. Example: square wave 19 X k = −19 k odd
1 2
0
1 j2πkt e jkπ
1
− 12
37
t
Fourier Series One can visualize convergence of the Fourier Series by incrementally adding terms. Example: square wave 29 X k = −29 k odd
1 2
0
1 j2πkt e jkπ
1
− 12
38
t
Fourier Series One can visualize convergence of the Fourier Series by incrementally adding terms. Example: square wave 39 X k = −39 k odd
1 2
0
1 j2πkt e jkπ
1
− 12
39
t
Fourier Series Partial sums of Fourier series of discontinuous functions “ring” near discontinuities: Gibb’s phenomenon.
9% 1 2
0
t
1
− 12
This ringing results because the magnitude of the Fourier coefficients is only decreasing as k1 (while they decreased as 12 for the triangle). k
You can decrease (and even eliminate the ringing) by decreasing the magnitudes of the Fourier coefficients at higher frequencies.
40
Fourier Series: Summary Fourier series represent periodic signals as sums of sinusoids. • valid for an extremely large class of periodic signals • valid even for discontinuous signals such as square wave However, convergence as # harmonics increases can be complicated.
41
Filtering The output of an LTI system is a “filtered” version of the input. Input: Fourier series → sum of complex exponentials. ∞ f 2π x(t) = x(t + T ) = ak e j T kt k=−∞
Complex exponentials: eigenfunctions of LTI systems. 2π 2π 2π e j T kt → H(j k)e j T kt T Output: same eigenfunctions, amplitudes/phases set by system. ∞ ∞ f f 2π 2π j 2Tπ kt x(t) = ak e → y(t) = ak H(j k)e j T kt T k=−∞
k=−∞
42
Filtering Notion of a filter. LTI systems • cannot create new frequencies. • can scale magnitudes and shift phases of existing components.
Example: Low-Pass Filtering with an RC circuit
R + vi
+ −
C
vo −
43
Lowpass Filter Calculate the frequency response of an RC circuit.
R +
vi (t)
= Ri(t) + vo (t)
C:
i(t)
= Cv˙ o (t)
Solving: vi (t)
vo
C
= RC v˙ o (t) + vo (t)
Vi (s) = (1 + sRC)Vo (s) 1 Vo (s) H(s) = = 1 + sRC Vi (s)
− 1 |H(jω)|
+ −
0.1 0.01
∠H(jω)|
vi
KVL:
0.01
0.1
1
10
ω 100 1/RC
10
ω 100 1/RC
0 − π2 0.01
0.1
1
44
Lowpass Filtering Let the input be a square wave. 1 2
0 − 12 1 jω0 kt e ; jπk k odd 1
|X(jω)|
f
ω0 =
2π T
0.1 0.01
∠X(jω)|
x(t) =
t
T
0.01
0.1
1
10
ω 100 1/RC
10
ω 1/RC 100
0 − π2 0.01
0.1
1 45
Lowpass Filtering Low frequency square wave: ω0 1/RC. 1 2
0 − 12 1 jω0 kt e ; jπk k odd 1
|H(jω)|
f
ω0 =
2π T
0.1 0.01
∠H(jω)|
x(t) =
t
T
0.01
0.1
1
10
ω 100 1/RC
10
ω 1/RC 100
0 − π2 0.01
0.1
1 49
Fourier Series: Summary Fourier series represent signals by their frequency content. Representing a signal by its frequency content is useful for many signals, e.g., music. Fourier series motivate a new representation of a system as a filter.
50
MIT OpenCourseWare http://ocw.mit.edu
6.003 Signals and Systems Fall 2011
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
6.003: Signals and Systems Fourier Series
November 1, 2011
1
Last Time: Describing Signals by Frequency Content
Harmonic content is natural way to describe some kinds of signals.
Ex: musical instruments (http://theremin.music.uiowa.edu/MIS .html )
piano piano t k violin violin t k bassoon bassoon t k 2
Last Time: Fourier Series
Determining harmonic components of a periodic signal.
ak =
� 2π 1 x(t)e −j T kt dt
T T
x(t)= x(t + T ) =
∞ 0
ak e j
(“analysis” equation)
2π kt T
(“synthesis” equation)
k=−∞
We can think of Fourier series as an orthogonal decomposition.
3
Orthogonal Decompositions
Vector representation of 3-space: let r¯ represent a vector with components {x, y, and z} in the {x ˆ, yˆ, and zˆ} directions, respectively. x = r¯ · x ˆ y = r¯ · yˆ z = r¯ · zˆ
(“analysis” equations)
r¯ = xx ˆ + yyˆ + zzˆ
(“synthesis” equation)
Fourier series: let x(t) represent a signal with harmonic components {a0 , a1 , . . ., ak } for harmonics {e j0t , e j
Z� 2π 1 ak = x(t)e −j T kt dt
T T x(t)= x(t + T ) =
∞ 0
ak e j
2π t T ,
. . ., e j
2π kt T }
respectively.
(“analysis” equation)
2π kt T
(“synthesis” equation)
k=−∞ 4
Orthogonal Decompositions
Vector representation of 3-space: let r¯ represent a vector with components {x, y, and z} in the {x ˆ, yˆ, and zˆ} directions, respectively. x = r¯ · x ˆ y = r¯ · yˆ z = r¯ · zˆ
(“analysis” equations)
r¯ = xˆ x + y yˆ + z zˆ
(“synthesis” equation)
Fourier series: let x(t) represent a signal with harmonic components {a0 , a1 , . . ., ak } for harmonics {e j0t , e j
Z� 2π 1 ak = x(t)e −j T kt dt
T T x(t)= x(t + T ) =
∞ 0
ak e j
2π t T ,
. . ., e j
2π kt T }
respectively.
(“analysis” equation)
2π kt T
(“synthesis” equation)
k=−∞ 5
Orthogonal Decompositions
Vector representation of 3-space: let r¯ represent a vector with components {x, y, and z} in the {x ˆ, yˆ, and zˆ} directions, respectively. x = r¯ · x ˆ y = r¯ · yˆ z = r¯ · zˆ
(“analysis” equations)
r¯ = xx ˆ + yyˆ + zzˆ
(“synthesis” equation)
Fourier series: let x(t) represent a signal with harmonic components {a0 , a1 , . . ., ak } for harmonics {e j0t , e j
Z � 2π 1 x(t)e−j T kt dt ak = T T x(t)= x(t + T ) =
∞ 0
ak e j
2π t T ,
. . ., e j
2π kt T }
respectively.
(“analysis” equation)
2π kt T
(“synthesis” equation)
k=−∞ 6
Orthogonal Decompositions
Integrating over a period sifts out the k th component of the series. Sifting as a dot product: x = r¯ · x ˆ ≡ |r¯||x ˆ| cos θ Sifting as an inner product: Z� 2π 1 j 2Tπ kt ak = e · x(t) ≡ x(t)e−j T kt dt
T T Z where � 1 a(t) · b(t) = a∗ (t)b(t)dt . T T The complex conjugate (∗ ) makes the inner product of the k th and mth components equal to 1 iff k Z = m: Z � � 2π 2π 1 1 1 if k = m
j 2Tπ kt ∗ j 2Tπ mt e
e
dt =
e −j T kt e j T mt dt =
T T T T 0 otherwise
7
Check Yourself
How many of the following pairs of functions are orthogonal (⊥) in T = 3? 1. cos 2πt ⊥ sin 2πt ? 2. cos 2πt ⊥ cos 4πt ? 3. cos 2πt ⊥ sin πt ? 4. cos 2πt ⊥ e j2πt ?
8
Check Yourself
How many of the following are orthogonal (⊥) in T = 3? cos 2πt ⊥ sin 2πt ?
cos 2πt t 1
2
3
1
2
3
sin 2πt t
cos 2πt sin 2πt = 12 sin 4πt t 2
1 Z � 3 dt = 0 therefore YES 0
9
3
Check Yourself
How many of the following are orthogonal (⊥) in T = 3? cos 2πt ⊥ cos 4πt ?
cos 2πt t 1
2
3
1
2
3
cos 4πt t
cos 2πt cos 4πt = 12 cos 6πt + 21 cos 2πt t 2
1 Z 3 � dt = 0 therefore YES 0
10
3
Check Yourself
How many of the following are orthogonal (⊥) in T = 3? cos 2πt ⊥ sin πt ?
cos 2πt t 1
2
3
1
2
3
sin πt t
cos 2πt sin πt = 12 sin 3πt − 12 sin πt t 2
1 Z� 3 dt = 0 therefore NO 0
11
3
Check Yourself
How many of the following are orthogonal (⊥) in T = 3? cos 2πt ⊥ e2πt ? e2πt = cos 2πt + j sin 2πt cos 2πt ⊥ sin 2πt but not cos 2πt Therefore NO
12
Check Yourself
How many of the following pairs of functions are orthogonal (⊥) in T = 3? 2 √ 1. cos 2πt ⊥ sin 2πt ? √ 2. cos 2πt ⊥ cos 4πt ? 3. cos 2πt ⊥ sin πt ?
X
4. cos 2πt ⊥ e j2πt ?
X
13
Speech
Vowel sounds are quasi-periodic.
bat
bait t
bit
bet t
t bought
bite t
beet
boat t
t but
boot t
t
14
t
t
Speech
Harmonic content is natural way to describe vowel sounds.
bat
bait
k bit
bet
k
k bought
bite
k
beet
but
boat
k
k boot
k
k
15
k
k
Speech
Harmonic content is natural way to describe vowel sounds.
bat bat t k beet beet t k boot boot t k 16
Speech Production
Speech is generated by the passage of air from the lungs, through the vocal cords, mouth, and nasal cavity.
Nasal cavity Hard palate Soft palate (velum)
Tongue
Lips
Pharynx Epiglottis Larynx
Vocal cords (glottis)
Esophogus
Trachea Stomach Lungs
A dapt ed f rom T. F. W eis s 17
Speech Production
Controlled by complicated muscles, vocal cords are set in vibration by the passage of air from the lungs. Loo ki ng d ow n th e th ro at :
Vocal cords open Glottis
Vocal cords closed Vocal cords
Gr a y 's A na t o m y
Ada p t e d f ro m T. F. We i s s 18
Speech Production
Vibrations of the vocal cords are “filtered” by the mouth and nasal cavities to generate speech.
19
Filtering
Notion of a filter.
LTI systems • cannot create new frequencies. • can only scale magnitudes & shift phases of existing components.
Example: Low-Pass Filtering with an RC circuit
R + vi
+ −
C
vo −
20
Lowpass Filter
Calculate the frequency response of an RC circuit.
R +
vi (t)
= Ri(t) + vo (t)
C:
i(t)
= Cv˙ o (t)
Solving: vi (t)
vo
C
= RC v˙ o (t) + vo (t)
Vi (s) = (1 + sRC)Vo (s) 1 Vo (s) H(s) = = 1 + sRC Vi (s)
− 1 |H(jω)|
+ −
0.1 0.01
∠H(jω)|
vi
KVL:
0.01
0.1
1
10
ω 100 1/RC
10
ω 100 1/RC
0 − π2 0.01
0.1
1
21
Lowpass Filtering Let the input be a square wave. 1 2
0 − 12 1 jω0 kt e ; jπk k odd 1
|X(jω)|
0
ω0 =
2π T
0.1 0.01
∠X(jω)|
x(t) =
t
T
0.01
0.1
1
10
ω 100 1/RC
10
ω 1/RC 100
0 − π2 0.01
0.1
1 22
Lowpass Filtering Low frequency square wave: ω0 1/RC. 1 2
0 − 12 1 jω0 kt e ; jπk k odd 1
|H(jω)|
0
ω0 =
2π T
0.1 0.01
∠H(jω)|
x(t) =
t
T
0.01
0.1
1
10
ω 100 1/RC
10
ω 1/RC 100
0 − π2 0.01
0.1
1 26
Source-Filter Model of Speech Production
Vibrations of the vocal cords are “filtered” by the mouth and nasal
cavities to generate speech.
buzz from vocal cords
throat and nasal cavities 27
speech
Speech Production
X-ray movie showing speech in production.
Courtesy of Kenneth N. Stevens. Used with permission.
28
Demonstration
Artificial speech.
buzz from vocal cords
throat and nasal cavities 29
speech
Formants
Resonant frequencies of the vocal tract. amplitude
F1 F2 F3 frequency
Men
Women
Children
Formant F1 F2 F3 F1 F2 F3 F1 F2 F3
heed 270 2290 3010 310 2790 3310 370 3200 3730
head 530 1840 2480 610 2330 2990 690 2610 3570
had 660 1720 2410 860 2050 2850 1010 2320 3320
hod 730 1090 2440 850 1220 2810 1030 1370 3170
http://www.sfu.ca/sonic-studio/handbook/Formant.html 30
haw’d 570 840 2410 590 920 2710 680 1060 3180
who’d 300 870 2240 370 950 2670 430 1170 3260
Speech Production
Same glottis signal + different formants → different vowels. glottis signal
vocal tract filter
vowel sound
ak
bk
ak
bk
We detect changes in the filter function to recognize vowels.
31
Singing
We detect changes in the filter function to recognize vowels
... at least sometimes.
Demonstration.
“la” scale.
“lore” scale.
“loo” scale.
“ler” scale.
“lee” scale.
Low Frequency: “la” “lore” “loo” “ler” “lee”.
High Frequency: “la” “lore” “loo” “ler” “lee”.
http://www.phys.unsw.edu.au/jw/soprane.html 32
Speech Production
We detect changes in the filter function to recognize vowels.
lo w i n t er me dia t e hig h
33
Speech Production
We detect changes in the filter function to recognize vowels.
lo w i n t er me dia t e hig h
34
Continuous-Time Fourier Series: Summary
Fourier series represent signals by their frequency content.
Representing a signal by its frequency content is useful for many
signals, e.g., music.
Fourier series motivate a new representation of a system as a filter.
Representing a system as a filter is useful for many systems, e.g.,
speech synthesis.
35
MIT OpenCourseWare http://ocw.mit.edu
6.003 Signals and Systems Fall 2011
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
6.003: Signals and Systems Fourier Transform
November 3, 2011
1
Last Time: Fourier Series
Representing periodic signals as sums of sinusoids. → new representations for systems as filters.
Today: generalize for aperiodic signals.
2
Fourier Transform
An aperiodic signal can be thought of as periodic with infinite period. Let x(t) represent an aperiodic signal.
x(t)
−S “Periodic extension”: xT (t) =
t
S
∞ 0
x(t + kT )
k=−∞
xT (t)
−S
S
Then x(t) = lim xT (t). T →∞
3
t T
Fourier Transform
Represent xT (t) by its Fourier series.
xT (t)
−S
ak =
t
S
T
2π sin 2πkS 2 sin ωS 1 T /2 1 S −j 2π kt T = xT (t)e−j T kt dt = e T dt = T ω T −T /2 πk T −S
T ak
2 sin ωS ω
ω = kω0 = k
2π T k ω
ω0 = 2π/T
4
Fourier Transform
Doubling period doubles # of harmonics in given frequency interval.
xT (t)
−S
t
S
T
Z Z sin 2πkS 2 sin ωS 1 T /2 1 S −j 2π kt −j 2π kt T = ak = xT (t)e T dt = e T dt = T ω T −T /2 πk T −S T ak
2 sin ωS ω
ω = kω0 = k
2π T k ω
ω0 = 2π/T
5
Fourier Transform
As T → ∞, discrete harmonic amplitudes → a continuum E(ω).
xT (t)
−S
t
S
T
Z Z sin 2πkS 2 sin ωS 1 T /2 1 S −j 2π kt −j 2π kt T = ak = xT (t)e T dt = e T dt = T ω T −T /2 πk T −S T ak
2 sin ωS ω
ω = kω0 = k
2π T k ω
ω0 = 2π/T 2 lim T ak = lim x(t)e−jωt dt = sin ωS = E(ω) ω T →∞ T →∞ −T /2 Z T /2
6
Fourier Transform
As T → ∞, synthesis sum → integral.
xT (t)
−S
t
S
T ak
T 2 sin ωS ω
ω = kω0 = k
2π T k ω
ω0 = 2π/T 2 lim T ak = lim x(t)e−jωt dt = sin ωS = E(ω) ω T →∞ T →∞ −T /2 Z ∞ ∞ 0 1 0 ω0 1 ∞ j 2π kt jωt T x(t) = E(ω) e
=
E(ω)e → E(ω)e jωt dω T 2π
2π
−∞ k=−∞ � �� �
k=−∞ Z T /2
ak
7
Fourier Transform
Replacing E(ω) by X(jω) yields the Fourier transform relations. E(ω) = X(jω) Fourier transform Z ∞ X(jω)= x(t)e−jωt dt
(“analysis” equation)
−∞
Z ∞ 1 x(t)= X(jω)ejωt dω 2π −∞
(“synthesis” equation)
Form is similar to that of Fourier series → provides alternate view of signal. 8
Relation between Fourier and Laplace Transforms
If the Laplace transform of a signal exists and if the ROC includes the jω axis, then the Fourier transform is equal to the Laplace transform evaluated on the jω axis. Laplace transform: Z ∞ X(s) = x(t)e−st dt −∞
Fourier transform: Z ∞ X(jω) = x(t)e−jωt dt = X(s)|s=jω −∞
9
Relation between Fourier and Laplace Transforms
Fourier transform “inherits” properties of Laplace transform.
Property
x(t)
X(s)
X(jω)
Linearity
ax1 (t) + bx2 (t)
aX1 (s) + bX2 (s)
aX1 (jω) + bX2 (jω)
Time shift
x(t − t0 )
e−st0 X(s)
e−jωt0 X(jω)
Time scale
x(at)
1 s X |a| a
Differentiation
dx(t) dt
sX(s)
Multiply by t
tx(t)
Convolution
x1 (t) ∗ x2 (t)
−
d X(s) ds
X1 (s) × X2 (s)
10
1 X |a|
jω a
jωX(jω) −
1 d X(jω) j dω
X1 (jω) × X2 (jω)
Relation between Fourier and Laplace Transforms
There are also important differences. Compare Fourier and Laplace transforms of x(t) = e−t u(t).
x(t)
t Laplace transform
Z ∞ Z ∞ −t −st X(s) = e u(t)e dt = e−(s+1)t dt = −∞
0
1 ; Re(s) > −1 1+s
a complex-valued function of complex domain. Fourier transform Z ∞ Z ∞ −t −jωt X(jω) = e u(t)e dt = e−(jω+1)t dt = −∞
0
a complex-valued function of real domain.
11
1 1 + jω
Laplace Transform
The Laplace transform maps a function of time t to a complex-valued
function of complex-valued domain s.
x(t)
t
Magnitude
1 |X(s)| = 1 + s
10
0 Ima 1 0 gin -1 ary (s)
1 -1 0eal(s) R 12
Fourier Transform
The Fourier transform maps a function of time t to a complex-valued function of real-valued domain ω.
x(t)
t X(j ω) = 1 1 + jω
ω 0
1
Frequency plots provide intuition that is difficult to otherwise obtain. 13
Check Yourself
Find the Fourier transform of the following square pulse.
x1 (t) 1 −1
1
t
1. X1 (jω) =
1 ω e − e−ω ω
2. X1 (jω) =
1 sin ω ω
3. X1 (jω) =
2 ω e − e−ω ω
4. X1 (jω) =
2 sin ω ω
5. none of the above
14
Fourier Transform
Compare the Laplace and Fourier transforms of a square pulse.
x1 (t) 1 −1
1
Laplace transform:
Z 1 1 1
e−st = e s − e−s e −st dt = X1 (s) =
s −s −1 −1 Fourier transform
Z 1 X1 (jω) =
e −jωt dt =
−1
t
[function of s = σ + jω]
1
2 sin ω
e−jωt =
−jω −1 ω
15
[function of ω]
Check Yourself
Find the Fourier transform of the following square pulse. 4
x1 (t) 1 −1
1
t
1. X1 (jω) =
1 ω e − e−ω ω
2. X1 (jω) =
1 sin ω ω
3. X1 (jω) =
2 ω e − e−ω ω
4. X1 (jω) =
2 sin ω ω
5. none of the above
16
Laplace Transform
Laplace transform: complex-valued function of complex domain.
x1 (t) 1 −1
t
1
1 s −s |X(s)| = (e − e ) s 30 20 10 0 5 5 0
0 -5 -5
17
Fourier Transform
The Fourier transform is a function of real domain: frequency ω. Time representation:
x1 (t) 1 −1
1
t
Frequency representation:
X1 (jω) =
2 sin ω ω 2 ω π
18
Check Yourself
Signal x2 (t) and its Fourier transform X2 (jω) are shown below.
x2 (t)
X2 (jω) 1
−2
2
b ω
t ω0
Which is true?
1. 2. 3. 4. 5.
b = 2 and ω0 b = 2 and ω0 b = 4 and ω0 b = 4 and ω0 none of the
= π/2 = 2π = π/2 = 2π above 19
Check Yourself
Find the Fourier transform.
Z 2 X2 (jω) =
e −2
−jωt
2 e−jωt 2 sin 2ω 4 sin 2ω dt = = = −jω −2 ω 2ω 4
ω π/2
20
Check Yourself
Signal x2 (t) and its Fourier transform X2 (jω) are shown below.
x2 (t)
X2 (jω) 1
−2
2
Which is true?
1. 2. 3. 4. 5.
b ω
t ω0
3
b = 2 and ω0 b = 2 and ω0 b = 4 and ω0 b = 4 and ω0 none of the
= π/2 = 2π = π/2 = 2π above 21
Fourier Transforms
Stretching time compresses frequency.
X1 (jω) =
x1 (t)
2
1 −1
2 sin ω ω
ω
t
1
π X2 (jω) =
4 sin 2ω 2ω 4
x2 (t) 1 −2
2
ω
t π/2 22
Check Yourself
Stretching time compresses frequency. Find a general scaling rule. Let x2 (t) = x1 (at).
If time is stretched in going from x1 to x2 , is a > 1 or a < 1?
23
Check Yourself
Stretching time compresses frequency.
Find a general scaling rule.
Let x2 (t) = x1 (at).
If time is stretched in going from x1 to x2 , is a > 1 or a < 1?
x2 (2) = x1 (1)
x2 (t) = x1 (at)
Therefore a = 1/2, or more generally, a < 1.
24
Check Yourself
Stretching time compresses frequency. Find a general scaling rule. Let x2 (t) = x1 (at). If time is stretched in going from x1 to x2 , is a > 1 or a < 1? a 0). ∞ 1 jω −jωτ /a 1 x1 (τ )e dτ = X1 X2 (jω) = a a a −∞ If a < 0 the sign of dτ would change along with the limits of integra tion. In general, 1 jω x1 (at) ↔ X1 . |a| a If time is stretched (a < 1) then frequency is compressed and ampli tude increases (preserving area). 26
Moments
The value of X(jω) at ω = 0 is the integral of x(t) over time t.
Z ∞ X(jω)|ω=0 =
x(t)e
−jωt
Z ∞ dt =
−∞
x(t)e
Z ∞ dt =
−∞
area = 2
1 1
x(t) dt −∞
X1 (jω) =
x1 (t)
−1
j0t
2 sin ω ω 2 ω
t π
27
Moments
The value of x(0) is the integral of X(jω) divided by 2π.
Z ∞ Z ∞ 1 1 jωt x(0) = X(jω) e dω = X(jω) dω 2π −∞ 2π −∞
X1 (jω) =
x1 (t)
area =1 2π
1
2 +
−1
1
t
+
−
28
−
2 sin ω ω
+
π
−
+
−
ω
Moments
The value of x(0) is the integral of X(jω) divided by 2π. Z ∞ Z ∞ 1 1 x(0) = X(jω) e jωt dω = X(jω) dω 2π −∞ 2π −∞
X1 (jω) =
x1 (t)
area =1 2π
1
2 +
−1
1
t
+
−
2 sin ω ω
+
−
π
−
+
−
ω
equal areas !
2 ω π 29
Stretching to the Limit
Stretching time compresses frequency and increases amplitude (preserving area). 2 sin ω X1 (jω) = x1 (t) ω
2
1 −1
ω
t
1
4
π
1 −2
2
ω
t π 1
2π ω
t New way to think about an impulse! 30
Fourier Transform
One of the most useful features of the Fourier transform (and Fourier series) is the simple “inverse” Fourier transform.
Z ∞ X(jω)=
x(t)e−jωt dt
(Fourier transform)
−∞
Z ∞ 1 x(t)= X(jω)ejωt dω 2π −∞
(“inverse” Fourier transform)
31
Inverse Fourier Transform
Find the impulse reponse of an “ideal” low pass filter.
H(jω) 1 ω ω0 −ω0 ω Z ∞ Z ω0 1 1 ejωt 0 sin ω0 t 1 jωt jωt = h(t) = H(jω)e dω = e dω = πt 2π −∞ 2π −ω0 2π jt −ω0 h(t) ω0 /π t π ω0 This result is not so easily obtained without inverse relation. 32
Fourier Transform
The Fourier transform and its inverse have very similar forms.
Z ∞ X(jω)= x(t)e−jωt dt (Fourier transform) −∞
Z ∞ 1 x(t)= X(jω)ejωt dω 2π −∞
(“inverse” Fourier transform)
Convert one to the other by • t→ω • ω → −t • scale by 2π
33
Duality
The Fourier transform and its inverse have very similar forms.
Z ∞ X(jω) =
x(t)e−jωt dt
Z−∞
∞
x(t) =
1 X(jω)ejωt dω 2π −∞
Two differences: • •
minus sign: flips time axis (or equivalently, frequency axis) divide by 2π (or multiply in the other direction)
x1 (t) = f (t) ↔ X1 (jω) = g(ω) t → ω ; flip ; ×2π
ω→t
x2 (t) = g(t) ↔ X2 (jω) = 2πf (−ω)
34
Duality
Using duality to find new transform pairs.
x1 (t) = f (t) ↔ X1 (jω) = g(ω) t → ω ; flip ; ×2π
ω→t
x2 (t) = g(t) ↔ X2 (jω) = 2πf (−ω) f (t) = δ(t)
g(ω) = 1
1
↔ ω
t
↓
g(t) = 1
↔
2πf (−ω) = 2πδ(ω) 2π ω
t The function g(t) = 1 does not have a Laplace transform! 35
More Impulses
Fourier transform of delayed impulse: δ(t − T ) ↔ e−jωT .
x(t) = δ(t − T ) 1 T
t
R∞ X(jω) = −∞ δ(t − T )e−jωt dt = e−jωT X(jω) = 1 1 ω ∠X(jω) = −ωT ω −T 36
Eternal Sinusoids
Using duality to find the Fourier transform of an eternal sinusoid.
↔
δ(t − T )
e−jωT t → ω ; flip ; ×2π
ω→t e−jtT
↔
2πδ(ω + T )
e−jω0 t
↔
2πδ(ω + ω0 )
T → ω0 :
x(t) = x(t + T ) =
∞ 0
2π kt T
CTFS
←→
j 2Tπ kt
CTFT
←→
a k ej
k=−∞
x(t) = x(t + T ) =
∞ 0
ak e
k=−∞
37
{ak } ∞ 0 k=−∞
2πak δ ω −
2π
k
T
Relation between Fourier Transform and Fourier Series Each term in the Fourier series is replaced by an impulse. ∞ X x(t) = xp (t − kT ) k=−∞
···
··· t
0 ak
T
···
··· k X(jω) =
∞ X
2π ak δ(ω − k
k=−∞
···
2π ) T ··· ω
0 38
2π T
Summary
Fourier transform generalizes ideas from Fourier series to aperiodic signals. Fourier transform is strikingly similar to Laplace transform • •
similar properties (linearity, differentiation, ...) but has a simple inverse (great for computation!)
Next time – applications (demos) of Fourier transforms
39
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6.003 Signals and Systems Fall 2011
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
6.003: Signals and Systems Discrete-Time Frequency Representations
November 8, 2011
1
Mid-term Examination #3
Wednesday, November 16, 7:30-9:30pm, No recitations on the day of the exam. Coverage:
Lectures 1–18 Recitations 1–16 Homeworks 1–10
Homework 10 will not be collected or graded. Solutions will be posted. Closed book: 3 pages of notes (8 12 × 11 inches; front and back). No calculators, computers, cell phones, music players, or other aids. Designed as 1-hour exam; two hours to complete. Review session Monday at 3pm and at open office hours. Prior term midterm exams have been posted on the 6.003 website. Conflict? Contact before Friday, Nov. 11, 5pm. 2
Signal Processing: From CT to DT
Signal-processing problems first conceived & addressed in CT: •
•
audio − radio (noise/static reduction, automatic gain control, etc.) − telephone (equalizers, echo-suppression, etc.) − hi-fi (bass, treble, loudness, etc.) imaging − television (brightness, tint, etc.) − photography (image enhancement, gamma) − x-rays (noise reduction, contrast enhancement) − radar and sonar (noise reduction, object detection)
Such problems are increasingly solved with DT signal processing: • • •
MP3 JPEG MPEG 3
Signal Processing: Acoustical
Mechano-acoustic components to optimize frequency response of loudspeakers: e.g., “bass-reflex” system.
driver
reflex port
4
Signal Processing: Acoustico-Mechanical
Passive radiator for improved low-frequency preformance.
driver
passive radiator
5
Signal Processing: Electronic
Low-cost electronics → new ways to overcome frequency limitations.
Magnitude (dB)
110
100
90
80
70 1
10
2
10
3
10
4
10
5
10
Frequency (Hz)
Small speakers (4 inch): eight facing wall, one facing listener. Electronic “equalizer” compensated for limited frequency response. 6
Signal Processing
Modern audio systems process sounds digitally.
x(t)
A/D
x[n]
DT filter
7
y[n]
D/A
y(t)
Signal Processing
DVSS
DVDD
AVSS
AVDD
VRFILT
VREFM
VREFP
AVSS(REF)
Modern audio systems process sounds digitally. AINRP AINRM RINA RINB
Voltage
Analog
Digital
Reference
Supplies
Supplies
Analog
AINRP
Control Register
AINRM Output
Texas Instruments TAS3004
Format
24Bit
AINLP
Control
Stereo
AINLM LINA
AINLP
LINB
• 2 channels
AINLM
VCOM
ALLPASS INPA GPI5 GPI4 GPI3
• 48 kHz sampling rate
GPI2
AOUTL Controller
• 24 bit ADC, 24 bit DAC
SDOUT0
Logic
ADC
AOUTR
24Bit Stereo DAC
GPI1 GPI0
SDA SCL
I2C Control
SDOUT2 CS1
32Bit Audio Signal Processor
SDOUT1
Processor
PLL
CAP_PLL
XTALO
MCLKO
XTALI/ MCLK
Select
CLKSEL
OSC/CLK
Control
IFM/S
SDATA
SCLK/O
SDIN1
TEST
LRCLK/O
RESET
32Bit Audio Signal
R
L
SDIN2
PWR_DN
Control
• $9.63 ($5.20 in bulk)
L+R
L+R
• 100 MIPS
Figure 1 1. TAS3004 Block Diagram
8
Courtesy of Texas Instruments. Used with permission.
DT Fourier Series and Frequency Response
Today: frequency representations for DT signals and systems.
9
Review: Complex Geometric Sequences
Complex geometric sequences are eigenfunctions of DT LTI systems. Find response of DT LTI system (h[n]) to input x[n] = z n . y[n] = (h ∗ x)[n] =
∞ 0
h[k]z n−k = z n
k=−∞
∞ 0
h[k]z −k = H(z) z n .
k=−∞
Complex geometrics (DT): analogous to complex exponentials (CT)
zn
h[n]
H(z) z n
est
h(t)
H(s) est
10
Review: Rational System Functions
A system described by a linear difference equation with constant coefficients → system function that is a ratio of polynomials in z. Example: y[n − 2] + 3y[n − 1] + 4y[n] = 2x[n − 2] + 7x[n − 1] + 8x[n]
H(z) =
2 + 7z + 8z 2 N (z)
2z −2 + 7z −1 + 8 ≡ = −2 −1 2 D(z) 1 + 3z + 4z z + 3z + 4
11
DT Vector Diagrams
Factor the numerator and denominator of the system function to make poles and zeros explicit. H(z0 ) = K
(z0 − q0 )(z0 − q1 )(z0 − q2 ) · · · (z0 − p0 )(z0 − p1 )(z0 − p2 ) · · ·
z0 z0 − q0
z-plane z0
q q0 0
Each factor in the numerator/denominator corresponds to a vector from a zero/pole (here q0 ) to z0 , the point of interest in the z-plane. Vector diagrams for DT are similar to those for CT. 12
DT Vector Diagrams
Value of H(z) at z = z0 can be determined by combining the contri butions of the vectors associated with each of the poles and zeros. H(z0 ) = K
(z0 − q0 )(z0 − q1 )(z0 − q2 ) · · · (z0 − p0 )(z0 − p1 )(z0 − p2 ) · · ·
The magnitude is determined by the product of the magnitudes. |H(z0 )| = |K|
|(z0 − q0 )||(z0 − q1 )||(z0 − q2 )| · · · |(z0 − p0 )||(z0 − p1 )||(z0 − p2 )| · · ·
The angle is determined by the sum of the angles. ∠H(z0 ) = ∠K + ∠(z0 − q0 ) + ∠(z0 − q1 ) + · · · − ∠(z0 − p0 ) − ∠(z0 − p1 ) − · · ·
13
DT Frequency Response
Response to eternal sinusoids. Let x[n] = cos Ω0 n (for all time): x[n] =
1 jΩ0 n 1 n e + e−jΩ0 n = z + z1n 2 2 0
where z0 = e jΩ0 and z1 = e−jΩ0 .
The response to a sum is the sum of the responses:
1 y[n] = H(z0 ) z0n + H(z1 ) z1n 2 1 = H(e jΩ0 ) e jΩ0 n + H(e−jΩ0 ) e−jΩ0 n 2
14
Conjugate Symmetry
For physical systems, the complex conjugate of H(e jΩ ) is H(e−jΩ ).
The system function is the Z transform of the unit-sample response:
∞ 0 H(z) = h[n]z −n n=−∞
where h[n] is a real-valued function of n for physical systems. ∞ 0
H(e jΩ ) =
h[n]e−jΩn
n=−∞ ∞ 0
H(e −jΩ ) =
h[n]e jΩn ≡ H(e jΩ )
n=−∞
15
∗
DT Frequency Response
Response to eternal sinusoids. Let x[n] = cos Ω0 n (for all time), which can be written as 1 jΩ0 n + e−jΩ0 n . x[n] = e 2 Then
1 H(e jΩ0 )e jΩ0 n + H(e−jΩ0 )e−jΩ0 n 2 = Re H(e jΩ0 )e jΩ0 n
y[n] =
= Re |H(e jΩ0 )|e j∠H(e
jΩ0 ) jΩ n 0
e
jΩ0
) = |H(e jΩ0 )|Re e jΩ0 n+j∠H(e y[n] = H(e jΩ0 ) cos Ω0 n + ∠H(e jΩ0 )
16
DT Frequency Response
The magnitude and phase of the response of a system to an eternal
cosine signal is the magnitude and phase of the system function evaluated on the unit circle.
cos(Ωn)
H(z)
|H(e jΩ )| cos Ωn + ∠H(e jΩ )
H(e jΩ ) = H(z)|z=e jΩ
17
Finding Frequency Response with Vector Diagrams H(z) =
H(e jΩ )
z − q1 z − p1
1 z-plane −π
0
π
∠H(e jΩ ) π/2
π
−π −π/2
18
Finding Frequency Response with Vector Diagrams H(z) =
H(e jΩ )
z − q1 z − p1
1 z-plane −π
0
π
∠H(e jΩ ) π/2
π
−π −π/2
19
Finding Frequency Response with Vector Diagrams H(z) =
H(e jΩ )
z − q1 z − p1
1 z-plane −π
0
π
∠H(e jΩ ) π/2
π
−π −π/2
20
Finding Frequency Response with Vector Diagrams H(z) =
H(e jΩ )
z − q1 z − p1
1 z-plane −π
0
π
∠H(e jΩ ) π/2
π
−π −π/2
21
Finding Frequency Response with Vector Diagrams H(z) =
H(e jΩ )
z − q1 z − p1
1 z-plane −π
0
π
∠H(e jΩ ) π/2
π
−π −π/2
22
Finding Frequency Response with Vector Diagrams H(z) =
H(e jΩ )
z − q1 z − p1
1 z-plane −π
0
π
∠H(e jΩ ) π/2
π
−π −π/2
23
Finding Frequency Response with Vector Diagrams H(z) =
H(e jΩ )
z − q1 z − p1
1 z-plane −π
0
π
∠H(e jΩ ) π/2
π
−π −π/2
24
Finding Frequency Response with Vector Diagrams H(z) =
H(e jΩ )
z − q1 z − p1
1 z-plane −π
0
π
∠H(e jΩ ) π/2
π
−π −π/2
25
Finding Frequency Response with Vector Diagrams H(z) =
H(e jΩ )
z − q1 z − p1
1 z-plane −π
0
π
∠H(e jΩ ) π/2
π
−π −π/2
26
Finding Frequency Response with Vector Diagrams H(z) =
H(e jΩ )
z − q1 z − p1
1 z-plane −π
0
π
∠H(e jΩ ) π/2
π
−π −π/2
27
Comparision of CT and DT Frequency Responses
CT frequency response: H(s) on the imaginary axis, i.e., s = jω. DT frequency response: H(z) on the unit circle, i.e., z = e jΩ . ω
s-plane
z-plane
σ
H(e jΩ )
|H(jω)|
1
−5
0
−π
5 28
0
π
DT Periodicity
DT frequency responses are periodic functions of Ω, with period 2π. If Ω2 = Ω1 + 2πk where k is an integer then H(e jΩ2 ) = H(e j(Ω1 +2πk) ) = H(e jΩ1 e j2πk ) = H(e jΩ1 ) The periodicity of H(e jΩ ) results because H(e jΩ ) is a function of e jΩ , which is itself periodic in Ω. Thus DT complex exponentials have many “aliases.” e jΩ2 = e j(Ω1 +2πk) = e jΩ1 e j2πk = e jΩ1 Because of this aliasing, there is a “highest” DT frequency: Ω = π.
29
Comparision of CT and DT Frequency Responses
CT frequency response: H(s) on the imaginary axis, i.e., s = jω. DT frequency response: H(z) on the unit circle, i.e., z = e jΩ . ω
s-plane
z-plane
σ
H(e jΩ )
|H(jω)|
1
−5
0
−π
5 30
0
π
Check Yourself
Consider 3 CT signals: x1 (t) = cos(3000t)
;
x2 (t) = cos(4000t)
;
x3 (t) = cos(5000t)
;
x3 [n] = x3 (nT )
Each of these is sampled so that x1 [n] = x1 (nT )
;
x2 [n] = x2 (nT )
where T = 0.001. Which list goes from lowest to highest DT frequency? 0. x1 [n] x2 [n] x3 [n]
1. x1 [n] x3 [n] x2 [n]
2. x2 [n] x1 [n] x3 [n]
3. x2 [n] x3 [n] x1 [n]
4. x3 [n] x1 [n] x2 [n]
5. x3 [n] x2 [n] x1 [n]
31
Check Yourself
The discrete signals are x1 [n] = cos[3n]
x2 [n] = cos[4n]
x3 [n] = cos[5n]
and the corresponding discrete frequencies are Ω = 3, 4 and 5, repre sented below with × marking e jΩ and o marking e−jΩ ).
5 4 3 3 4 5 32
Check Yourself
Ω = 0.25
x[n] = cos(0.25n)
n
33
Check Yourself
Ω = 0.5
x[n] = cos(0.5n)
n
34
Check Yourself
Ω=1
x[n] = cos(n)
n
35
Check Yourself
Ω=2
x[n] = cos(2n)
n
36
Check Yourself
Ω=3
x[n] = cos(3n)
n
37
Check Yourself
Ω=4
x[n] = cos(4n) = cos(2π − 4n) ≈ cos(2.283n)
n
38
Check Yourself
Ω=5
x[n] = cos(5n) = cos(2π − 5n) ≈ cos(1.283n)
n
39
Check Yourself
Ω=6
x[n] = cos(6n) = cos(2π − 6n) ≈ cos(0.283n)
n
40
Check Yourself
The discrete signals are x1 [n] = cos[3n]
x2 [n] = cos[4n]
x3 [n] = cos[5n]
and the corresponding discrete frequencies are Ω = 3, 4 and 5, repre sented below with × marking e jΩ and o marking e−jΩ ).
5 4 3 3 4 5 41
Check Yourself
Consider 3 CT signals: x1 (t) = cos(3000t)
;
x2 (t) = cos(4000t)
;
x3 (t) = cos(5000t)
;
x3 [n] = x3 (nT )
Each of these is sampled so that x1 [n] = x1 (nT )
;
x2 [n] = x2 (nT )
where T = 0.001. Which list goes from lowest to highest DT frequency? 0. x1 [n] x2 [n] x3 [n]
1. x1 [n] x3 [n] x2 [n]
2. x2 [n] x1 [n] x3 [n]
3. x2 [n] x3 [n] x1 [n]
4. x3 [n] x1 [n] x2 [n]
5. x3 [n] x2 [n] x1 [n]
42
5
Check Yourself
What kind of filtering corresponds to the following?
z-plane
1. high pass 3. band pass 5. none of above
2. low pass 4. band stop (notch)
43
Check Yourself
What kind of filtering corresponds to the following?
z-plane
1. high pass 3. band pass 5. none of above
2. low pass 4. band stop (notch)
44
1
DT Fourier Series
DT Fourier series represent DT signals in terms of the amplitudes and phases of harmonic components. 0 x[n] = ak e jkΩ0 n The period N of all harmonic components is the same (as in CT).
45
DT Fourier Series
There are (only) N distinct complex exponentials with period N . (There were an infinite number in CT!) If y[n] = e jΩn is periodic in N then y[n] = e jΩn = y[n + N ] = e jΩ(n+N ) = e jΩn e jΩN and e jΩN must be 1, and ejΩ must be one of the N th roots of 1. Example: N = 8
z-plane
46
DT Fourier Series
There are N distinct complex exponentials with period N . These can be combined via Fourier series to produce periodic time signals with N independent samples.
Example: periodic in N=3
n 3 samples repeated in time
3 complex exponentials
Example: periodic in N=4
n 4 samples repeated in time
4 complex exponentials 47
DT Fourier Series
DT Fourier series represent DT signals in terms of the amplitudes and phases of harmonic components.
x[n] = x[n + N ] =
N −1 0
ak e jkΩ0 n
; Ω0 =
k=0
2π N
N equations (one for each point in time n) in N unknowns (ak ). Example: N = 4 ⎡
⎤ ⎡ j 2π 0·0 x[0]
e N ⎢ x[1] ⎥ ⎢ e j 2Nπ 0·1 ⎢ ⎥ ⎢ ⎢ ⎥ = ⎢ 2π ⎣ x[2] ⎦ ⎣ e j N 0·2 2π x[3] e j N 0·3
2π
e j N 1·0 2π e j N 1·1 2π e j N 1·2 2π e j N 1·3
2π
e j N 2·0 2π e j N 2·1 2π e j N 2·2 2π e j N 2·3
48
⎤⎡ ⎤ 2π a0 e j N 3·0 j 2Nπ 3·1 ⎥ ⎢ a ⎥ e
⎥⎢ 1⎥ ⎢ ⎥ j 2Nπ 3·2 ⎥ ⎦ ⎣ a2 ⎦
e
2π a3 e j N 3·3
DT Fourier Series
DT Fourier series represent DT signals in terms of the amplitudes and phases of harmonic components.
x[n] = x[n + N ] =
N −1 0
ak e jkΩ0 n
; Ω0 =
k=0
2π N
N equations (one for each point in time n) in N unknowns (ak ). Example: N = 4 ⎡
⎤ ⎡
x[0]
1 ⎢ x[1] ⎥ ⎢ 1 ⎢ ⎥ ⎢ ⎢ ⎥ = ⎢ ⎣ x[2] ⎦ ⎣ 1 x[3] 1
1 j −1 −j
1 −1 1 −1
⎤⎡ ⎤ 1 a0 ⎥ ⎢ −j ⎥ ⎢ a1 ⎥ ⎥ ⎥⎢ ⎥ −1 ⎦ ⎣ a2 ⎦
j a3
49
Orthogonality
DT harmonics are orthogonal to each other (as were CT harmonics).
N −1 0
e
jΩ0 kn −jΩ0 ln
e
=
n=0
−1 N 0
e jΩ0 (k−l)n
n=0
=
⎧ ⎨N ⎩
; k=l
1−e jΩ0 (k−l)N 1−e jΩ0 (k−l)
= N δ[k − l]
50
=
j 2π (k−l)N 1−e N j 2π (k−l) 1−e N
=0
; k= � l
Sifting
Use orthogonality property of harmonics to sift out FS coefficients. Assume x[n] =
N −1
0
ak e jkΩ0 n
k=0
Multiply both sides by the complex conjugate of the lth harmonic, and sum over time. N −1 0
−jlΩ0 n
x[n]e
=
n=0
=
−1 N −1 N 0 0
ak e
jkΩ0 n −jlΩ0 n
e
n=0 k=0 N −1 0
k=0
ak N δ[k − l] = N al
k=0
ak =
=
N −1 0
N −1 1 0 x[n]e−jkΩ0 n N n=0
51
ak
N −1 0 n=0
e jkΩ0 n e−jlΩ0 n
DT Fourier Series
Since both x[n] and ak are periodic in N , the sums can be taken over any N successive indices. Notation. If f [n] is periodic in N , then N −1 N N +1 0 0 0 0 f [n] = f [n] = f [n] = · · · = n=0
n=1
n=2
f [n]
n=
DT Fourier Series
ak = ak+N =
1 N
x[n]= x[n + N ] =
0
x[n]e−jkΩ0 n ; Ω0 =
n=
0
ak e jkΩ0 n
2π N
(“analysis” equation)
(“synthesis” equation)
k=
52
DT Fourier Series
DT Fourier series have simple matrix interpretations.
0
x[n] = x[n + 4] =
ak e jkΩ0 n =
k=
⎡
⎤ ⎡
x[0]
1 ⎢ x[1] ⎥ ⎢ 1 ⎢ ⎥ ⎢ ⎢ ⎥ = ⎢ ⎣ x[2] ⎦ ⎣ 1 x[3] 1 ak = ak+4 =
1 j −1 −j
1 −1 1 −1
⎤
⎡
ak e jk
k=
2π n 4
=
0
ak j kn
k=
⎤⎡ ⎤ 1 a0 ⎥ ⎢ −j ⎥ ⎢ a1 ⎥ ⎥ ⎥⎢ ⎥ −1 ⎦ ⎣ a2 ⎦
j a3
1 0
1 0 −jk 2π n 1 0
x[n]e −jkΩ0 n =
e N =
x[n]j −kn 4
4
4
n=
⎡
0
a0 1 ⎢ a ⎥ 1 ⎢ 1 ⎢ 1⎥ ⎢ ⎢ ⎥= ⎢ ⎣ a2 ⎦ 4 ⎣ 1 a3 1
1 −j −1 j
n=
1 −1
1 −1
⎤⎡
⎤
1
x[0]
⎢ ⎥ j ⎥ ⎥ ⎢ x[1] ⎥ ⎥⎢ ⎥ −1 ⎦ ⎣ x[2] ⎦
−j x[3]
These matrices are inverses of each other. 53
n=
Discrete-Time Frequency Representations
Similarities and differences between CT and DT. DT frequency response • vector diagrams (similar to CT) • frequency response on unit circle in z-plane (jω axis in CT) DT Fourier series • represent signal as sum of harmonics (similar to CT) • finite number of periodic harmonics (unlike CT) • finite sum (unlike CT) The finite length of DT Fourier series make them especially useful for signal processing! (more on this next time)
54
MIT OpenCourseWare http://ocw.mit.edu
6.003 Signals and Systems Fall 2011
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6.003: Signals and Systems DT Fourier Representations
November 10, 2011
1
Mid-term Examination #3
Wednesday, November 16, 7:30-9:30pm, No recitations on the day of the exam. Coverage:
Lectures 1–18 Recitations 1–16 Homeworks 1–10
Homework 10 will not be collected or graded. Solutions will be posted. Closed book: 3 pages of notes (8 12
Conflict? Contact before Friday, Nov. 11, 5pm. 2
Review: DT Frequency Response
The frequency response of a DT LTI system is the value of the system function evaluated on the unit circle.
cos(Ωn)
H(z)
|H(ejΩ )| cos Ωn + ∠H(ejΩ )
H(e jΩ ) = H(z)|z=e jΩ
3
Comparision of CT and DT Frequency Responses
CT frequency response: H(s) on the imaginary axis, i.e., s = jω. DT frequency response: H(z) on the unit circle, i.e., z = e jΩ . ω
s-plane
z -plane
σ
H(ejΩ )
|H(jω)|
1
0
ω
−π 4
0
πΩ
Check Yourself
A system H(z) =
1 − az has the following pole-zero diagram. z−a
z -plane
Classify this system as one of the following filter types. 1. high pass 3. band pass 5. band stop
2. low pass 4. all pass 0. none of the above 5
Check Yourself
Classify the system ... 1 − az H(z) = z−a Find the frequency response: 1 − ae jΩ e−jΩ − a = e jΩ jΩ H(e jΩ ) = jΩ e −a e −a
← complex ← conjugates
jΩ ) = 1.
Because complex conjugates have equal magnitudes, H(e
→ all-pass filter
6
Check Yourself
A system H(z) =
1 − az has the following pole-zero diagram. z−a
z -plane
Classify this system as one of the following filter types. 1. high pass 3. band pass 5. band stop
2. low pass 4. all pass 0. none of the above 7
4
Effects of Phase
8
Effects of Phase
9
Effects of Phase
http://public.research.att.com/~ttsweb/tts/demo.php
10
Effects of Phase
artificial speech synthesized by Robert Donovan
11
Effects of Phase
x[n]
y[n] = x[−n]
???
artificial speech synthesized by Robert Donovan
12
Effects of Phase
x[n]
y[n] = x[−n]
???
How are the phases of X and Y related?
13
Effects of Phase
How are the phases of X and Y related?
ak = x[n]e −jkΩ0 n n
x[−n]e−jkΩ0 n =
bk = n
x[m]e jkΩ0 m = a−k
m
Flipping x[n] about n = 0 flips ak about k = 0. Because x[n] is real-valued, ak is conjugate symmetric: a−k = a∗k . bk = a−k = a∗k = |ak |e−j∠ak The angles are negated at all frequencies. 14
Review: Periodicity
DT frequency responses are periodic functions of Ω, with period 2π. If Ω2 = Ω1 + 2πk where k is an integer then H(e jΩ2 ) = H(e j(Ω1 +2πk) ) = H(e jΩ1 e j2πk ) = H(e jΩ1 ) The periodicity of H(e jΩ ) results because H(e jΩ ) is a function of e jΩ , which is itself periodic in Ω. Thus DT complex exponentials have many “aliases.” e jΩ2 = e j(Ω1 +2πk) = e jΩ1 e j2πk = e jΩ1 Because of this aliasing, there is a “highest” DT frequency: Ω = π.
15
Review: Periodic Sinusoids
There are (only) N distinct complex exponentials with period N . (There were an infinite number in CT!) If y[n] = e jΩn is periodic in N then y[n] = e jΩn = y[n + N ] = e jΩ(n+N ) = e jΩn e jΩN and e jΩN must be 1, and ejΩ must be one of the N th roots of 1. Example: N = 8
z -plane
16
Review: DT Fourier Series
DT Fourier series represent DT signals in terms of the amplitudes and phases of harmonic components.
DT Fourier Series ak = ak+N =
1 N
x[n]= x[n + N ] =
X
x[n]e−jkΩ0 n ; Ω0 =
n=
X
ak ejkΩ0 n
2π N
(“analysis” equation)
(“synthesis” equation)
k=
17
DT Fourier Series
DT Fourier series have simple matrix interpretations.
x[n] = x[n + 4] =
X
ak ejkΩ0 n =
k=
⎡
⎤ ⎡
x[0]
1 ⎢ x[1] ⎥ ⎢ 1 ⎢ ⎥ ⎢ ⎢ ⎥ = ⎢ ⎣ x[2] ⎦ ⎣ 1 x[3] 1 1 ak = ak+4 = 4
⎡
⎤ ⎡ a0 1 ⎢ a ⎥ 1 ⎢ 1 ⎢ ⎢ 1⎥ ⎢ ⎥= ⎢ ⎣ a2 ⎦ 4 ⎣ 1 1 a3
X
ak ejk
k=
⎤⎡
2π n 4
=
X
ak j kn
k=
⎤
1 1 1 a0 ⎥ ⎢ j −1 −j ⎥ ⎢ a1 ⎥ ⎥ ⎥⎢ ⎥ −1 1 −1 ⎦ ⎣ a2 ⎦
a3 −j −1 j X 1 X −jk 2π n 1 X x[n]e− e N =
x[n]j −kn
jkΩ0 n =
4
4
n=
1 −j −1 j
n=
1 −1
1 −1
⎤⎡ ⎤
1
x[0]
⎢ ⎥ j ⎥ ⎥ ⎢ x[1] ⎥ ⎥⎢ ⎥ −1 ⎦ ⎣ x[2] ⎦
−j x[3]
These matrices are inverses of each other. 18
n=
Scaling
DT Fourier series are important computational tools.
However, the DT Fourier series do not scale well with the length N.
ak = ak+2 =
1 X 1 X −jk 2π n 1 X 2 x[n]e−jkΩ0 n = e = x[n](−1)−kn 2 2 2 n=
a0 1 1 = 2 1 a1
1 −1
n=
x[0] x[1]
n=
1 X 1 X −jk 2π n 1 X 4 ak = ak+4 = x[n]e−jkΩ0 n = e = x[n]j −kn 4 4 4 n=
⎡
⎤ ⎡ 1 a0 ⎢ a ⎥ 1 ⎢ 1 ⎢ 1⎥ ⎢ ⎢ ⎥= ⎢ ⎣ a2 ⎦ 4 ⎣ 1 a3 1
1 −j −1 j
n=
1 −1
1 −1
⎤⎡ ⎤
1
x[0]
⎢ ⎥ j ⎥ ⎥ ⎢ x[1] ⎥ ⎥⎢ ⎥ −1 ⎦ ⎣ x[2] ⎦
−j x[3]
Number of multiples increases as N 2 . 19
n=
Fast Fourier “Transform”
Exploit structure of Fourier series to simplify its calculation.
Divide FS of length 2N into two of length N (divide and conquer).
Matrix formulation of 8-point FS:
⎡ ⎤ ⎡ 0 c0 W 8 ⎢ c1 ⎥ ⎢ W 0 ⎢ ⎥ ⎢ 8 ⎢ c ⎥ ⎢ W 0 ⎢ 2⎥ ⎢ 8 ⎢ ⎥ ⎢ 0 ⎢ c3 ⎥ ⎢ W 8 ⎢ ⎥ = ⎢ ⎢ c ⎥ ⎢ W 0 ⎢ 4⎥ ⎢ 8 ⎢ ⎥ ⎢ 0 ⎢ c5 ⎥ ⎢ W 8 ⎢ ⎥ ⎢ ⎣ c6 ⎦ ⎣ W 0 8 c7 W 80
W 80 W 8 1 W 8 2 W 8 3 W 8 4 W 8 5 W 8 6 W 8 7
W 80 W 8 2 W 8 4 W 8 6 W 80 W 8 2 W 8 4 W 8 6
W 80 W 8 3 W 8 6 W 8 1 W 8 4 W 8 7 W 8 2 W 8 5
W 80 W 8 4 W 80 W 8 4 W 80 W 8 4 W 80 W 8 4
2π
where WN = e−j N
8 × 8 = 64 multiplications 20
W 80 W 8 5 W 8 2 W 8 7 W 8 4 W 8 1 W 8 6 W 8 3
W 80 W 8 6 W 8 4 W 8 2 W 80 W 8 6 W 8 4 W 8 2
⎤⎡ ⎤
W 80 x[0]
⎢ ⎥ W 8 7 ⎥ ⎥ ⎢ x[1] ⎥ ⎥ ⎢ 6 W 8 ⎥ ⎢ x[2] ⎥ ⎥ ⎥⎢ ⎥ ⎢ x[3] ⎥ W 8 5 ⎥ ⎥⎢ ⎥ ⎢ ⎥ W 8 4 ⎥ ⎥ ⎢ x[4] ⎥ ⎥⎢ ⎥ W 8 3 ⎥ ⎢ x[5] ⎥ ⎥⎢ ⎥ W 2 ⎦ ⎣ x[6] ⎦
8
W 8 1
x[7]
FFT
Divide into two 4-point series (divide and conquer).
Even-numbered entries ⎡ ⎤ ⎡ 0 W4 W 40 a0 ⎢ a ⎥ ⎢ W 0 W 1 ⎢ 1⎥ ⎢ 4 4 ⎢ ⎥ = ⎢ 0 ⎣ a2 ⎦ ⎣ W 4 W 42 a3 W 40 W 43
in x[n]:
Odd-numbered entries ⎡ ⎤ ⎡ 0 b0 W4 W 40 ⎢ b ⎥ ⎢ W 0 W 1 ⎢ 1⎥ ⎢ 4 4 ⎢ ⎥ = ⎢ 0 ⎣ b2 ⎦ ⎣ W 4 W 42 b3 W40 W 43
in x[n]:
W 40 W 42 W 40 W 42
W 40 W 42 W 40 W 42
⎤⎡ ⎤
W 40 x[0]
⎢ ⎥ W 43 ⎥ ⎥ ⎢ x[2] ⎥ ⎥ ⎢ ⎥ W 42 ⎦ ⎣ x[4] ⎦
W 41 x[6]
⎤⎡ ⎤
W 40 x[1]
⎢ ⎥ W 43 ⎥ ⎥ ⎢ x[3] ⎥ ⎥⎢ ⎥ W 42 ⎦ ⎣ x[5] ⎦
W 41 x[7]
Sum of multiplications = 2 × (4 × 4) = 32: fewer than the previous 64.
21
FFT
Break the original 8-point DTFS coefficients ck into two parts: ck = dk + ek where dk comes from the even-numbered x[n] (e.g., ak ) and ek comes from the odd-numbered x[n] (e.g., bk )
22
FFT
The 4-point DTFS coefficients ak of the even-numbered ⎡ ⎤ ⎡ 0 ⎤⎡ ⎤ ⎡ 0 a0 W 4 W 40 W 40 W 40 W 8 W 80 W 80 x[0]
⎢ a ⎥ ⎢ W 0 W 1 W 2 W 3 ⎥⎢ x[2] ⎥ ⎢ W 0 W 2 W 4 ⎥ ⎢ 8 ⎢ 1⎥ ⎢ 4 4
4
4 ⎥⎢ 8
8
⎥=⎢ ⎢ ⎥=⎢ 0 ⎥⎢ ⎣ a2 ⎦ ⎣ W 4 W 4 2 W 40 W 4 2 ⎦⎣ x[4] ⎦ ⎣ W 80 W 8 4 W 80 a3 W 40 W 4 3 W 4 2 W 4 1 x[6]
W 80 W 8 6 W 8 4
x[n]
⎤⎡ ⎤
W 80 x[0]
⎢ ⎥ W 8 6 ⎥ ⎥⎢ x[2] ⎥ ⎥⎢ ⎥ W 8 4 ⎦⎣ x[4] ⎦
W 8 2 x[6]
contribute to the 8-point DTFS coefficients dk : ⎡ ⎤ ⎡ 0 d0 W 8 ⎢ d1 ⎥ ⎢ W 0 ⎢ ⎥ ⎢ 8 ⎢ d ⎥ ⎢ W 0 ⎢ 2⎥ ⎢ 8 ⎢ ⎥ ⎢ 0 ⎢ d3 ⎥ ⎢ W 8 ⎢ ⎥ = ⎢ ⎢ d ⎥ ⎢ W 0 ⎢ 4⎥ ⎢ 8 ⎢ ⎥ ⎢ 0 ⎢ d5 ⎥ ⎢ W 8 ⎢ ⎥ ⎢ ⎣ d6 ⎦ ⎣ W 0 8 d7 W80
W 80 W 8 1 W 8 2 W 8 3 W 8 4 W 8 5 W 8 6 W 8 7
W 80 W 8 2 W 8 4 W 8 6 W 80 W 8 2 W 8 4 W 8 6
W 80 W 8 3 W 8 6 W 8 1 W 8 4 W 8 7 W 8 2 W 8 5
W 80 W 8 4 W 80 W 8 4 W 80 W 8 4 W 80 W 8 4 23
W 80 W 8 5 W 8 2 W 8 7 W 8 4 W 8 1 W 8 6 W 8 3
W 80 W 8 6 W 8 4 W 8 2 W 80 W 8 6 W 8 4 W 8 2
⎤
⎤ ⎡
x[0]
W 80 ⎢ ⎥ W 8 7 ⎥ ⎥ ⎢ x[1] ⎥ ⎥ ⎢ 6 W 8 ⎥ ⎢ x[2] ⎥ ⎥ ⎥⎢ ⎥ ⎢ x[3] ⎥ W 8 5 ⎥ ⎥⎢ ⎥ ⎢ x[4] ⎥ W 8 4 ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ W 8 3 ⎥ ⎢ x[5] ⎥ ⎥⎢ ⎥ W 8 2 ⎦ ⎣ x[6] ⎦
x[7]
W 81
FFT
The 4-point DTFS coefficients ak of the even-numbered ⎡ ⎤ ⎡ 0 ⎤⎡ ⎤ ⎡ 0 a0 W 4 W 40 W 40 W 40 W 8 W 80 W 80 x[0]
⎢ a ⎥ ⎢ W 0 W 1 W 2 W 3 ⎥⎢ x[2] ⎥ ⎢ W 0 W 2 W 4 ⎥ ⎢ 8 ⎢ 1⎥ ⎢ 4 4 4 4 ⎥⎢ 8 8 ⎥=⎢ ⎢ ⎥=⎢ 0 ⎥⎢ ⎣ a2 ⎦ ⎣ W 4 W 42 W 40 W 42 ⎦⎣ x[4] ⎦ ⎣ W 80 W 84 W 80 a3 W 40 W 43 W 42 W 41 x[6]
W 80 W 86 W 84
x[n]
⎤⎡ ⎤
W 80 x[0]
⎢ ⎥ W 86 ⎥ ⎥⎢ x[2] ⎥ ⎥⎢ ⎥ W 84 ⎦⎣ x[4] ⎦
W 82 x[6]
contribute to the 8-point DTFS coefficients dk : ⎡ ⎤ ⎡ 0 d0 W 8 ⎢ d1 ⎥ ⎢ W 0 ⎢ ⎥ ⎢ 8 ⎢ d ⎥ ⎢ W 0 ⎢ 2⎥ ⎢ 8 ⎢ ⎥ ⎢ 0 ⎢ d3 ⎥ ⎢ W 8 ⎢ ⎥ = ⎢ ⎢ d ⎥ ⎢ W 0 ⎢ 4⎥ ⎢ 8 ⎢ ⎥ ⎢ 0 ⎢ d5 ⎥ ⎢ W 8 ⎢ ⎥ ⎢ ⎣ d6 ⎦ ⎣ W 0 8 W80 d7
W 80 W 82 W 84 W 86 W 80 W 82 W 84 W 86
W 80 W 84 W 80 W 84 W 80 W 84 W 80 W 84 24
W 80 W 86 W 84 W 82 W 80 W 86 W 84 W 82
⎤
⎤ ⎡
x[0]
⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ⎢ x[2] ⎥ ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ⎢ x[4] ⎥ ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ⎥⎢ ⎦ ⎣ x[6] ⎦
FFT
The 4-point DTFS coefficients ak of the even-numbered ⎡ ⎤ ⎡ 0 ⎤⎡ ⎤ ⎡ 0 a0 W 4 W 40 W 40 W 40 W 8 W 80 W 80 x[0]
⎢ a ⎥ ⎢ W 0 W 1 W 2 W 3 ⎥⎢ x[2] ⎥ ⎢ W 0 W 2 W 4 ⎥ ⎢ 8 ⎢ 1⎥ ⎢ 4 4 4 4 ⎥⎢ 8 8 ⎥=⎢ ⎢ ⎥=⎢ 0 ⎥⎢ ⎣ a2 ⎦ ⎣ W 4 W 42 W 40 W 42 ⎦⎣ x[4] ⎦ ⎣ W 80 W 84 W 80 a3 W 40 W 43 W 42 W 41 x[6]
W 80 W 86 W 84
x[n]
⎤⎡ ⎤
W 80 x[0]
⎢ ⎥ W 86 ⎥ ⎥⎢ x[2] ⎥ ⎥⎢ ⎥ W 84 ⎦⎣ x[4] ⎦
W 82 x[6]
contribute to the 8-point DTFS coefficients dk : ⎡ ⎤ ⎡ ⎤ ⎡ 0 d0 a0 W 8 ⎢ d1 ⎥ ⎢ a1 ⎥ ⎢ W 0 ⎢ ⎥ ⎢ ⎥ ⎢ 8 ⎢ d ⎥ ⎢ a ⎥ ⎢ W 0 ⎢ 2⎥ ⎢ 2⎥ ⎢ 8 ⎢ ⎥ ⎢ ⎥ ⎢ 0 ⎢ d3 ⎥ ⎢ a3 ⎥ ⎢ W 8 ⎢ ⎥ = ⎢ ⎥ = ⎢ ⎢ d ⎥ ⎢ a ⎥ ⎢ W 0 ⎢ 4⎥ ⎢ 0⎥ ⎢ 8 ⎢ ⎥ ⎢ ⎥ ⎢ 0 ⎢ d5 ⎥ ⎢ a1 ⎥ ⎢ W 8 ⎢ ⎥ ⎢ ⎥ ⎢ ⎣ d6 ⎦ ⎣ a2 ⎦ ⎣ W 0 8 W80 a3 d7
W 80 W 82 W 84 W 86 W 80 W 82 W 84 W 86
W 80 W 84 W 80 W 84 W 80 W 84 W 80 W 84 25
W 80 W 86 W 84 W 82 W 80 W 86 W 84 W 82
⎤
⎤ ⎡
x[0]
⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ⎢ x[2] ⎥ ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ⎢ x[4] ⎥ ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ⎥⎢ ⎦ ⎣ x[6] ⎦
FFT The 4-point DTFS coefficients ak of the even-numbered ⎡ ⎤ ⎡ 0 ⎤⎡ ⎤ ⎡ 0 a0 W 4 W 40 W 40 W 40 W 8 W 80 W 80 x[0]
⎢ a ⎥ ⎢ W 0 W 1 W 2 W 3 ⎥⎢ x[2] ⎥ ⎢ W 0 W 2 W 4 ⎥ ⎢ 8 ⎢ 1⎥ ⎢ 4 4 4 4 ⎥⎢ 8 8 ⎥=⎢ ⎢ ⎥=⎢ 0 ⎥⎢ ⎣ a2 ⎦ ⎣ W 4 W 42 W 40 W 42 ⎦⎣ x[4] ⎦ ⎣ W 80 W 84 W 80 a3 W 40 W 43 W 42 W 41 x[6]
W 80 W 86 W 84
x[n]
⎤⎡ ⎤
W 80 x[0]
⎢ ⎥ W 86 ⎥ ⎥⎢ x[2] ⎥ ⎥⎢ ⎥ W 84 ⎦⎣ x[4] ⎦
W 82 x[6]
contribute to the 8-point DTFS coefficients dk : ⎤ ⎡ ⎤ ⎡ 0 W 8 a0 d0 ⎢ d1 ⎥ ⎢ a1 ⎥ ⎢ W 0 ⎢ ⎥ ⎢ ⎥ ⎢ 8 ⎢ d ⎥ ⎢ a ⎥ ⎢ W 0 ⎢ 2⎥ ⎢ 2⎥ ⎢ 8 ⎢ ⎥ ⎢ ⎥ ⎢ 0 ⎢ d3 ⎥ ⎢ a3 ⎥ ⎢ W 8 ⎢ ⎥ = ⎢ ⎥ = ⎢ ⎢ d ⎥ ⎢ a ⎥ ⎢ W 0 ⎢ 4⎥ ⎢ 0⎥ ⎢ 8 ⎢ ⎥ ⎢ ⎥ ⎢ 0 ⎢ d5 ⎥ ⎢ a1 ⎥ ⎢ W 8 ⎢ ⎥ ⎢ ⎥ ⎢ ⎣ d6 ⎦ ⎣ a2 ⎦ ⎣ W 0 8 W 80 a3 d7 ⎡
W 80 W 82 W 84 W 86 W 80 W 82 W 84 W 86
W 80 W 84 W 80 W 84 W 80 W 84 W 80 W 84 26
W 80 W 86 W 84 W 82 W 80 W 86 W 84 W 82
⎤
⎤ ⎡
x[0]
⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ⎢ x[2] ⎥ ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ⎢ x[4] ⎥ ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ ⎥⎢ ⎦ ⎣ x[6] ⎦
FFT
The ek components result ⎡ ⎤ ⎡ 0 b0 W4 W 40 W 40 ⎢ b ⎥ ⎢ W 0 W 1 W 2 ⎢ 1⎥ ⎢ 4 4
4
⎢ ⎥=⎢ 0 ⎣ b2 ⎦ ⎣ W 4 W 4 2 W 40 b3 W40 W 4 3 W 4 2 ⎡ ⎤ ⎡ 0 e0 W 8 ⎢ e1 ⎥ ⎢ W 0 ⎢ ⎥ ⎢ 8 ⎢ e ⎥ ⎢ W 0 ⎢ 2⎥ ⎢ 8 ⎢ ⎥ ⎢ 0 ⎢ e3 ⎥ ⎢ W 8 ⎢ ⎥ = ⎢ ⎢ e ⎥ ⎢ W 0 ⎢ 4⎥ ⎢ 8 ⎢ ⎥ ⎢ 0 ⎢ e5 ⎥ ⎢ W 8 ⎢ ⎥ ⎢ ⎣ e6 ⎦ ⎣ W 0 8 W 80 e7
W 80 W 8 1 W 8 2 W 8 3 W 8 4 W 8 5 W 8 6 W 8 7
from the odd-number entries in x[n].
W 80 W 8 2 W 8 4 W 8 6 W 80 W 8 2 W 8 4 W 8 6
⎤⎡ ⎤ ⎡ 0 x[1]
W 40 W 8 ⎢ ⎥ ⎥ ⎢ 3 W 4 ⎥⎢ x[3] ⎥ ⎢ W 80 ⎥=⎢ ⎥⎢ W 4 2 ⎦⎣ x[5] ⎦ ⎣ W 80 W 4 1 x[7]
W 80 W 80 W 8 3 W 8 6 W 8 1 W 8 4 W 8 7 W 8 2 W 8 5
W 80 W 8 4 W 80 W 8 4 W 80 W 8 4 W 80 W 8 4
27
W 80 W 8 5 W 8 2 W 8 7 W 8 4 W 8 1 W 8 6 W 8 3
W 80 W 8 2 W 8 4 W 8 6 W 80 W 8 6 W 8 4 W 8 2 W 80 W 8 6 W 8 4 W 8 2
W 80 W 8 4 W 80 W 8 4
⎤⎡ ⎤
x[1]
W 80 ⎢ ⎥ W 8 6 ⎥ ⎥⎢ x[3] ⎥ ⎥⎢ ⎥ W 8 4 ⎦⎣ x[5] ⎦
W 8 2 x[7]
⎤ ⎡
⎤
W 80 x[0]
⎢ ⎥ W 8 7 ⎥ ⎥ ⎢ x[1] ⎥ ⎢ ⎥ W 8 6 ⎥ ⎥ ⎢ x[2] ⎥ ⎥ ⎢ ⎥ ⎢ ⎥ W 8 5 ⎥ ⎥ ⎢ x[3] ⎥ ⎥ ⎢ 4 W 8 ⎥ ⎢ x[4] ⎥ ⎥ ⎥⎢ ⎥ W 8 3 ⎥ ⎢ x[5] ⎥ ⎥⎢ ⎥ W 8 2 ⎦ ⎣ x[6] ⎦
W 81 x[7]
FFT
The ek components result ⎡ ⎤ ⎡ 0 b0 W4 W 40 W 40 ⎢ b ⎥ ⎢ W 0 W 1 W 2 ⎢ 1⎥ ⎢ 4 4
4
⎢ ⎥=⎢ 0 ⎣ b2 ⎦ ⎣ W 4 W 4 2 W 40 b3 W40 W 4 3 W 4 2 ⎡ ⎤ ⎡
e0 ⎢ e1 ⎥ ⎢ ⎢ ⎥ ⎢ ⎢e ⎥ ⎢ ⎢ 2⎥ ⎢ ⎢ ⎥ ⎢ ⎢ e3 ⎥ ⎢ ⎢ ⎥ = ⎢ ⎢e ⎥ ⎢ ⎢ 4⎥ ⎢ ⎢ ⎥ ⎢ ⎢ e5 ⎥ ⎢ ⎢ ⎥ ⎢ ⎣ e6 ⎦ ⎣
e7
W 80 W 8 1 W 8 2 W 8 3 W 8 4 W 8 5 W 8 6 W 8 7
from the odd-number entries in x[n]. ⎤⎡ ⎤ ⎡ 0 x[1]
W 40 W 8 ⎢ ⎥ ⎥ ⎢ 3 W 4 ⎥⎢ x[3] ⎥ ⎢ W 80 ⎥=⎢ ⎥⎢ W 4 2 ⎦⎣ x[5] ⎦ ⎣ W 80 W 4 1 x[7]
W 80 W 80 W 8 3 W 8 6 W 8 1 W 8 4 W 8 7 W 8 2 W 8 5
W 80 W 8 5 W 8 2 W 8 7 W 8 4 W 8 1 W 8 6 W 8 3
28
W 80 W 8 2 W 8 4 W 8 6
W 80 W 8 4 W 80 W 8 4
⎤⎡ ⎤
x[1]
W 80 ⎢ ⎥ W 8 6 ⎥ ⎥⎢ x[3] ⎥ ⎥⎢ ⎥ W 8 4 ⎦⎣ x[5] ⎦
W 8 2 x[7]
⎤⎡ ⎤ W 80 ⎢ ⎥ W 8 7 ⎥ ⎥ ⎢ x[1] ⎥ ⎢ ⎥ W 8 6 ⎥ ⎥ ⎥⎢ ⎥ ⎥ ⎢ 5 ⎢ ⎥ W 8 ⎥ ⎥ ⎢ x[3] ⎥ ⎥ ⎥ ⎢ 4 W 8 ⎥ ⎢ ⎥ ⎥ ⎢ ⎥ W 8 3 ⎥ ⎢ x[5] ⎥ ⎥ ⎥⎢ ⎦
W 8 2 ⎦ ⎣ W 81 x[7]
FFT
The ek components result ⎡ ⎤ ⎡ 0 b0 W4 W 40 W 40 ⎢ b ⎥ ⎢ W 0 W 1 W 2 ⎢ 1⎥ ⎢ 4 4 4 ⎢ ⎥=⎢ 0 ⎣ b2 ⎦ ⎣ W 4 W 42 W 40 b3 W40 W 43 W 42 ⎡ ⎤ ⎡ 0 ⎤ ⎡
W 8 b0 e0 ⎢ e1 ⎥ ⎢ W 1 b1 ⎥ ⎢ ⎢ ⎥ ⎢ 8 ⎥ ⎢ ⎢ e ⎥ ⎢ W 2 b ⎥ ⎢ ⎢ 2⎥ ⎢ 8 2⎥ ⎢ ⎢ ⎥ ⎢ 3 ⎥ ⎢ ⎢ e3 ⎥ ⎢ W 8 b3 ⎥ ⎢ ⎢ ⎥ = ⎢ ⎥ ⎢ ⎢ e ⎥ ⎢ W 4 b ⎥ = ⎢ 4 0 ⎢ ⎥ ⎢ 8 ⎥ ⎢ ⎢ ⎥ ⎢ 5 ⎥ ⎢ ⎢ e5 ⎥ ⎢ W 8 b1 ⎥ ⎢ ⎢ ⎥ ⎢ ⎥ ⎢ ⎣ e6 ⎦ ⎣ W 6 b2 ⎦ ⎣
8 e7 W 87 b3
from the odd-number entries in x[n]. ⎤⎡ ⎤ ⎡ 0 x[1]
W 40 W 8 ⎢ ⎥ ⎥ ⎢ 3 W 4 ⎥⎢ x[3] ⎥ ⎢ W 80 ⎥=⎢ ⎥⎢ W 42 ⎦⎣ x[5] ⎦ ⎣ W 80 W 41 x[7]
W 80 W 80 W 81 W 82 W 83 W 84 W 85 W 86 W87
W 80 W 83 W 86 W 81 W 84 W 87 W 82 W 85
29
W 80 W 82 W 84 W 86
W 80 W 85 W 82 W 87 W 84 W 81 W 86 W 83
W 80 W 84 W 80 W 84
⎤⎡ ⎤
x[1]
W 80 ⎢ ⎥ W 86 ⎥ ⎥⎢ x[3] ⎥ ⎥⎢ ⎥ W 84 ⎦⎣ x[5] ⎦
W 82 x[7]
⎤⎡ ⎤ W 80 ⎥ ⎢ W 87 ⎥ ⎥ ⎢ x[1] ⎥ ⎥ ⎢ W 86 ⎥ ⎥ ⎥⎢ ⎥ ⎥ ⎢ 5 ⎥ ⎢ W 8 ⎥ ⎥ ⎢ x[3] ⎥ ⎥ ⎥ ⎢ 4 W 8 ⎥ ⎢ ⎥ ⎥ ⎥ ⎢ W 83 ⎥ ⎢ x[5] ⎥ ⎥ ⎥⎢ ⎦
W 82 ⎦ ⎣ W 81 x[7]
FFT The ek components result ⎡ ⎤ ⎡ 0 b0 W4 W 40 W 40 ⎢ b ⎥ ⎢ W 0 W 1 W 2 ⎢ 1⎥ ⎢ 4 4 4 ⎢ ⎥=⎢ 0 ⎣ b2 ⎦ ⎣ W 4 W 42 W 40 b3 W40 W 43 W 42 ⎤ ⎡ 0 ⎤ ⎡
W8 b0 e0 ⎢ e1 ⎥ ⎢ W 1 b1 ⎥ ⎢ ⎢ ⎥ ⎢ 8 ⎥ ⎢ ⎢e ⎥ ⎢W2 b ⎥ ⎢ ⎢ 2⎥ ⎢ 8 2⎥ ⎢ ⎢ ⎥ ⎢ 3 ⎥ ⎢ ⎢ e3 ⎥ ⎢ W8 b3 ⎥ ⎢ ⎢ ⎥=⎢ ⎥ ⎢ ⎢ e ⎥ ⎢ W 4 b ⎥ = ⎢ 4 0 ⎢ ⎥ ⎢ 8 ⎥ ⎢ ⎢ ⎥ ⎢ 5 ⎥ ⎢ ⎢ e5 ⎥ ⎢ W8 b1 ⎥ ⎢ ⎢ ⎥ ⎢ ⎥ ⎢ ⎣ e6 ⎦ ⎣ W 6 b2 ⎦ ⎣
8 e7 W87 b3 ⎡
from the odd-number entries in x[n]. ⎤⎡ ⎤ ⎡ 0 x[1]
W 40 W 8 ⎢ ⎥ ⎥ ⎢ 3 W 4 ⎥⎢ x[3] ⎥ ⎢ W 80 ⎥=⎢ ⎥⎢ W 42 ⎦⎣ x[5] ⎦ ⎣ W 80 W 41 x[7]
W 80 W 80 W 81 W 82 W 83 W 84 W 85 W 86 W 87
W 80 W 83 W 86 W 81 W 84 W 87 W 82 W 85
30
W 80 W 82 W 84 W 86
W 80 W 85 W 82 W 87 W 84 W 81 W 86 W 83
W 80 W 84 W 80 W 84
⎤⎡ ⎤
x[1]
W 80 ⎢ ⎥ W 86 ⎥ ⎥⎢ x[3] ⎥ ⎥⎢ ⎥ W 84 ⎦⎣ x[5] ⎦
W 82 x[7]
⎤⎡ ⎤ W 80 ⎥ ⎢ W 87 ⎥ ⎥ ⎢ x[1] ⎥ ⎥ ⎢ W 86 ⎥ ⎥ ⎥⎢ ⎥ ⎥ ⎢ 5 ⎥ ⎢ W 8 ⎥ ⎥ ⎢ x[3] ⎥ ⎥ ⎥ ⎢ 4 W 8 ⎥ ⎢ ⎥ ⎥ ⎥ ⎢ W 83 ⎥ ⎢ x[5] ⎥ ⎥ ⎥⎢ ⎦
W 82 ⎦ ⎣ W 81 x[7]
FFT
Combine ak and bk to get ck . ⎡ ⎤ ⎡
⎤ ⎡ ⎤ ⎡ 0 ⎤
a0 d0 + e0 c0 W8 b0 ⎢ c1 ⎥ ⎢ d1 + e1 ⎥ ⎢ a1 ⎥ ⎢ W 1 b1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 8 ⎥ ⎢c ⎥ ⎢d + e ⎥ ⎢a ⎥ ⎢W2 b ⎥ ⎢ 2⎥ ⎢ 2 ⎢ 2⎥ ⎢ 2⎥ 2⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 8 ⎥ ⎢ c3 ⎥ ⎢ d3 + e3 ⎥ ⎢ a3 ⎥ ⎢ W83 b3 ⎥ ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ c ⎥ ⎢ d + e ⎥ = ⎢ a ⎥ + ⎢ W 4 b ⎥ 4⎥ ⎢ 4⎥ ⎢ 4 ⎢ 0⎥ ⎢ 8 0⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ c5 ⎥ ⎢ d5 + e5 ⎥ ⎢ a1 ⎥ ⎢ W85 b1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ c6 ⎦ ⎣ d6 + e6 ⎦ ⎣ a2 ⎦ ⎣ W 6 b2 ⎦
8
c7
d7 + e7
a3
W87 b3
FFT procedure: • compute ak and bk : 2 × (4 × 4) = 32 multiplies • combine ck = ak + W8k bk : 8 multiples • total 40 multiplies: fewer than the orginal 8 × 8 = 64 multiplies
31
Scaling of FFT algorithm
How does the new algorithm scale?
Let M (N ) = number of multiplies to perform an N point FFT.
M (1) = 0
M (2) = 2M (1) + 2 = 2
M (4) = 2M (2) + 4 = 2 × 4
M (8) = 2M (4) + 8 = 3 × 8
M (16) = 2M (8) + 16 = 4 × 16
M (32) = 2M (16) + 32 = 5 × 32
M (64) = 2M (32) + 64 = 6 × 64
M (128) = 2M (64) + 128 = 7 × 128
. . .
M (N ) = (log2 N ) × N
Significantly smaller than N 2 for N large.
32
Fourier Transform: Generalize to Aperiodic Signals
An aperiodic signal can be thought of as periodic with infinite period. Let x[n] represent an aperiodic signal DT signal.
x[n] 1 n 0 “Periodic extension”: xN [n] =
∞ X
x[n + kN ]
k=−∞
xN [n] 1 n N Then x[n] = lim xN [n]. N →∞
33
Fourier Transform
Represent xN [n] by its Fourier series.
xN [n] 1 n −N1 N1
ak =
N 2π 1 X 1 X1 xN [n]e−j N kn = N N N
N 2π
e−j N kn =
n=−N1
1 sin N1 + 12 Ω N sin 12 Ω
sin 23 Ω
N ak
sin 21 Ω k Ω Ω0 =
34
2π N
Ω = kΩ0 = k
2π N
Fourier Transform
Doubling period doubles # of harmonics in given frequency interval.
xN [n] 1 n −N1 N1 2π 1 X 1 ak = xN [n]e−j N kn = N N
N
N
N1 X
e
n=−N1
−j 2π N kn
1 sin N1 + 12 Ω = N sin 12 Ω
sin 23 Ω
N ak
sin 21 Ω k Ω Ω0 =
35
2π N
Ω = kΩ0 = k
2π N
Fourier Transform
As N → ∞, discrete harmonic amplitudes → a continuum E(Ω).
xN [n] 1 n −N1 N1 2π 1 X 1 ak = xN [n]e−j N kn = N N
N
N
N1 X
e
n=−N1
−j 2π N kn
1 sin N1 + 12 Ω = N sin 12 Ω
sin 23 Ω
N ak
sin 21 Ω k Ω Ω0 = N ak =
X n=
2π
x[n]e−j N kn =
2π N
X n= 36
Ω = kΩ0 = k
x[n]e−jΩn = E(Ω)
2π N
Fourier Transform
As N → ∞, synthesis sum → integral.
xN [n] 1 n −N1 N1
N sin 23 Ω
N ak
sin 21 Ω k Ω Ω0 = N ak =
2π
x[n]e−j N kn =
X
n=
x[n] =
X
k=
2π N
X n=
2π N
x[n]e−jΩn = E(Ω)
1 Ω0 jΩn E(Ω)e
→ E(Ω)e jΩn dΩ 2π
2π 2π k=
X 2π 1 E(Ω) e j N kn =
�N
ak
Ω = kΩ0 = k
37
Fourier Transform
Replacing E(Ω) by X(e jΩ ) yields the DT Fourier transform relations.
X(e jΩ )=
∞ X
x[n]e−jΩn
(“analysis” equation)
n=−∞
x[n]=
Z 1 X(e jΩ )e jΩn dΩ 2π 2π
(“synthesis” equation)
38
Relation between Fourier and Z Transforms
If the Z transform of a signal exists and if the ROC includes the unit circle, then the Fourier transform is equal to the Z transform evaluated on the unit circle. Z transform: ∞ X X(z) = x[n]z −n n=−∞
DT Fourier transform: ∞ X X(e jΩ ) = x[n]e−jΩn = H(z) z=e jΩ n=−∞
39
Relation between Fourier and Z Transforms
Fourier transform “inherits” properties of Z transform.
Property
x[n]
X(z)
X(e jΩ )
Linearity
ax1 [n] + bx2 [n]
aX1 (s) + bX2 (s)
aX1 (e jΩ ) + bX2 (e jΩ )
Time shift
x[n − n0 ]
z −n0 X(z)
e−jΩn0 X(e jΩ )
Multiply by n
nx[n]
Convolution
(x1 ∗ x2 )[n]
−z
d X(z) dz
X1 (z) × X2 (z)
40
j
d X(e jΩ ) dΩ
X1 (e jΩ ) × X2 (e jΩ )
DT Fourier Series of Images Magnitude
Angle
41
DT Fourier Series of Images Magnitude
Uniform Angle
42
DT Fourier Series of Images Uniform Magnitude
Angle
43
DT Fourier Series of Images Different Magnitude
Angle
44
DT Fourier Series of Images Magnitude
Angle
45
DT Fourier Series of Images Magnitude
Angle
46
DT Fourier Series of Images Different Magnitude
Angle
47
Fourier Representations: Summary
Thinking about signals by their frequency content and systems as filters has a large number of practical applications.
48
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6.003: Signals and Systems Relations among Fourier Representations
November 15, 2011
1
Mid-term Examination #3
Wednesday, November 16, 7:30-9:30pm, No recitations on the day of the exam. Coverage:
Lectures 1–18 Recitations 1–16 Homeworks 1–10
Homework 10 will not be collected or graded. Solutions are posted. Closed book: 3 pages of notes (8 12 × 11 inches; front and back). No calculators, computers, cell phones, music players, or other aids. Designed as 1-hour exam; two hours to complete. Prior term midterm exams have been posted on the 6.003 website.
2
Fourier Representations
We’ve seen a variety of Fourier representations: • • • •
CT CT DT DT
Fourier Fourier Fourier Fourier
series transform series transform
Today: relations among the four Fourier representations.
3
Four Fourier Representations
We have discussed four closely related Fourier representations. DT Fourier Series
1 ak = ak+N = N
X
DT Fourier transform 2π
x[n]e−j N kn
n=
x[n] = x[n + N ] =
X
∞ X
X(e jΩ ) =
x[n]e−jΩn
n=−∞
ak e
j 2π N kn
Z
x[n] =
k=
CT Fourier Series
1 X(e jΩ )e jΩn dΩ 2π
CT Fourier transform
Z ∞
Z 2π 1 x(t)e−j T kt dt ak = T T ∞ X 2π x(t) = x(t + T ) = ak e j T kt
X(jω) = −∞ Z ∞
x(t) =
k=−∞ 4
x(t)e−jωt dt
1 X(jω)e jωt dω 2π −∞
Four Types of “Time”
discrete vs. continuous (t) and periodic vs aperiodic (↔)
DT Fourier Series
DT Fourier transform
n
n
CT Fourier Series
CT Fourier transform
t
t 5
Four Types of “Frequency”
discrete vs. continuous (↔) and periodic vs aperiodic (t)
DT Fourier Series
DT Fourier transform
2π k N
Ω
CT Fourier Series
CT Fourier transform
ω
2π k T
6
Relation between Fourier Series and Transform
A periodic signal can be represented by a Fourier series or by an equivalent Fourier transform. Series: represent periodic signal as weighted sum of harmonics ∞ 0 2π x(t) = x(t + T ) = ak ejω0 kt ; ω0 = T k=−∞
The Fourier transform of a sum is the sum of the Fourier transforms: ∞ 0 X(jω) = 2πak δ(ω − kω0 ) k=−∞
Therefore periodic signals can be equivalently represented as Fourier transforms (with impulses!).
7
Relation between Fourier Series and Transform
A periodic signal can be represented by a Fourier series or by an equivalent Fourier transform. Fourier Series
a a−2 −1a0 a1 a2 a3 a4 a−4a−3 0 1 ak ejω0 kt
k
↔ Fourier Transform
k=−∞
2πa−4 2πa−3 2πa−2 2πa−1 2πa0 2πa1 2πa2 2πa3 2πa4
x(t) = x(t + T ) =
∞ X
0 ω0
8
ω
Relations among Fourier Representations
Explore other relations among Fourier representations.
Start with an aperiodic CT signal. Determine its Fourier transform.
Convert the signal so that it can be represented by alternate Fourier
representations and compare.
periodic DT DTFS interpolate
N →∞ periodic extension
sample
periodic CT CTFS
aperiodic DT DTFT
interpolate
T →∞ periodic extension
9
sample
aperiodic CT CTFT
Start with the CT Fourier Transform
Determine the Fourier transform of the following signal.
x(t) 1 −1 0 1
t
Could calculate Fourier transform from the definition. ∞
X(jω) =
x(t)ejωt dt
−∞
Easier to calculate x(t) by convolution of two square pulses:
y(t)
y(t) 1
1 ∗
− 12 12
t 10
− 12 12
t
Start with the CT Fourier Transform
The transform of y(t) is Y (jω) =
sin(ω/2)
ω/2
y(t) 1 ↔
Y (jω) 1
t
− 12 12
ω −2π
2π
so the transform of x(t) = (y ∗ y)(t) is X(jω) = Y (jω) × Y (jω).
x(t) 1 −1
1
↔
X(jω) 1 ω
t
−2π
11
2π
Relation between Fourier Transform and Series
What is the effect of making a signal periodic in time?
Find Fourier transform of periodic extension of x(t) to period T = 4.
z(t) =
∞ X
x(t + 4k)
k=−∞
1 t −4
−1
1
4
Could calculate Z(jω) for the definition ... ugly.
12
Relation between Fourier Transform and Series
Easier to calculate z(t) by convolving x(t) with an impulse train.
z(t) =
∞ X
x(t + 4k)
k=−∞
1 t z(t) =
∞ 0
−4
−1
1
x(t + 4k) = (x ∗ p)(t)
k=−∞
where p(t) =
∞ 0
δ(t + 4k)
k=−∞
Then Z(jω) = X(jω) × P (jω) 13
4
Check Yourself
What’s the Fourier transform of an impulse train?
x(t) =
∞ X
δ(t − kT )
k=−∞
1
··· 0
14
··· t T
Check Yourself
What’s the Fourier transform of an impulse train?
x(t) =
∞ X
δ(t − kT )
k=−∞
1
···
··· t
0 ak = T1 ···
T ∀ k 1 T
··· k
∞ X 2π 2π X(jω) = δ(ω − k ) T T k=−∞
2π T
··· 0 15
2π T
··· ω
Relation between Fourier Transform and Series
Easier to calculate z(t) by convolving x(t) with an impulse train.
z(t) =
∞ X
x(t + 4k)
k=−∞
1 t z(t) =
∞ 0
−4
−1
1
x(t + 4k) = (x ∗ p)(t)
k=−∞
where p(t) =
∞ 0
δ(t + 4k)
k=−∞
Then Z(jω) = X(jω) × P (jω) 16
4
Relation between Fourier Transform and Series
Convolving in time corresponds to multiplying in frequency.
X(jω) 1 ω −2π
2π P (jω) π 2
− π2 π2
ω
Z(jω) π/2
− π2 π2 17
ω
Relation between Fourier Transform and Series
The Fourier transform of a periodically extended function is a dis crete function of frequency ω.
z(t) =
∞ X
x(t + 4k)
k=−∞
1 t −4
−1
1
4
Z(jω) π/2
− π2 π2 18
ω
Relation between Fourier Transform and Series
The weight (area) of each impulse in the Fourier transform of a
periodically extended function is 2π times the corresponding Fourier
series coefficient.
Z(jω) π/2
− π2 π2
ω
ak 1/4
−1 1
19
k
Relation between Fourier Transform and Series
The effect of periodic extension of x(t) to z(t) is to sample the
frequency representation.
X(jω) 1 ω −2π
2π Z(jω) π/2
− π2 π2
ω
ak 1/4
−1 1 20
k
Relation between Fourier Transform and Series
Periodic extension of CT signal → discrete function of frequency. Periodic extension = convolving with impulse train in time = multiplying by impulse train in frequency → sampling in frequency
N →∞
periodic DT DTFS interpolate
periodic extension
sample
periodic CT CTFS
aperiodic DT DTFT
interpolate
T →∞ periodic extension (sampling in frequency) 21
sample
aperiodic CT CTFT
Four Types of “Time”
discrete vs. continuous (t) and periodic vs aperiodic (↔)
DT Fourier Series
DT Fourier transform
n
n
CT Fourier Series
CT Fourier transform
t
t 22
Four Types of “Frequency”
discrete vs. continuous (↔) and periodic vs aperiodic (t)
DT Fourier Series
DT Fourier transform
2π k N
Ω
CT Fourier Series
CT Fourier transform
ω
2π k T
23
Relation between Fourier Transform and Series
Periodic extension of CT signal → discrete function of frequency. Periodic extension = convolving with impulse train in time = multiplying by impulse train in frequency → sampling in frequency
N →∞
periodic DT DTFS interpolate
periodic extension
sample
periodic CT CTFS
aperiodic DT DTFT
interpolate
T →∞ periodic extension (sampling in frequency) 24
sample
aperiodic CT CTFT
Relations among Fourier Representations
Compare to sampling in time.
periodic DT DTFS interpolate
N →∞ periodic extension
sample
periodic CT CTFS
aperiodic DT DTFT
interpolate
T →∞ periodic extension
25
sample
aperiodic CT CTFT
Relations between CT and DT transforms
Sampling a CT signal generates a DT signal.
x[n] = x(nT )
x(t) 1 −1 0 1
t
Take T = 1
2 .
x[n]
n −1 1 What is the effect on the frequency representation? 26
Relations between CT and DT transforms
We can generate a signal with the same shape by multiplying x(t) by an impulse train with T = 1
2 .
∞ 0
xp (t) = x(t) × p(t) where p(t) =
δ(t + kT )
k=−∞
x(t) 1 −1 0 1
t xp (t)
x[n]
n
t
−1 1
−1 27
1
Relations between CT and DT transforms
We can generate a signal with the same shape by multiplying x(t) by an impulse train with T = 12 . ∞ 0
xp (t) = x(t) × p(t) where p(t) =
δ(t + kT )
k=−∞
x(t) 1 −1 0 1
t
xp (t)
t −1
1 28
Relations between CT and DT transforms
Multiplying x(t) by an impulse train in time is equivalent to convolving
X(jω) by an impulse train in frequency (then ÷2π).
X(jω) 1 ω −2π
2π P (jω) 4π ω
−4π
4π Xp (jω) 2 ω
−4π
4π 29
Relations between CT and DT transforms
Fourier transform of sampled signal xp (t) is periodic in ω, period 4π.
xp (t)
t −1
1
Xp (jω) 2 ω −4π
4π
30
Relations between CT and DT transforms
Fourier transform of sampled signal xp (t) has same shape as DT Fourier transform of x[n].
x[n]
n −1 1 X(ejΩ ) 2
−2π
2π
31
Ω
DT Fourier transform
CT Fourier transform of sampled signal xp (t) = DT Fourier transform of samples x[n] where Ω = ωT , i.e., X(jω) = X(ejΩ )
xp (t)
Ω=ωT
.
Xp (jω) 2 ↔ ω
t −1
−4π
1
4π X(ejΩ ) 2
x[n] ↔ n −1 1
−2π
Ω = ωT = 12 ω 32
2π
Ω
Relation between CT and DT Fourier Transforms
Compare the definitions: X(ejΩ ) =
0
x[n]e−jΩn
n
Z Xp (jω) = =
xp (t)e−jωt dt
Z 0
x[n]δ(t − nT )e−jωt dt
n
=
0
=
0
Z x[n]
δ(t − nT )e−jωt dt
n
x[n]e−jωnT
n
Ω = ωT
33
Relation Between DT Fourier Transform and Series
Periodic extension of a DT signal is equivalent to convolution of the
signal with an impulse train.
x[n]
n −1 1 p[n]
n −8
8 xp [n] = (x ∗ p)[n]
n −8
−1 1 34
8
Relation Between DT Fourier Transform and Series
Convolution by an impulse train in time is equivalent to multiplication
by an impulse train in frequency.
X(ejΩ ) 2
−2π
2π
Ω
P (ejΩ ) π 4
Ω Xp (ejΩ ) π 2
−2π
− π4 π4 35
Ω 2π
Relation Between DT Fourier Transform and Series
Periodic extension of a discrete signal (x[n]) results in a signal (xp [n]) that is both periodic and discrete. Its transform (Xp (ejΩ )) is also periodic and discrete.
xp [n] = (x ∗ p)[n]
n −8
−1 1
8
Xp (ejΩ ) π 2
−2π
− π4 π4
36
Ω 2π
Relation Between DT Fourier Transform and Series
The weight of each impulse in the Fourier transform of a periodi cally extended function is 2π times the corresponding Fourier series
coefficient.
Xp (ejΩ ) π 2
−2π
− π4 π4
Ω 2π
ak 1 4
−8
−1 1
37
Ω 8
Relation between Fourier Transforms and Series
The effect of periodic extension was to sample the frequency repre sentation.
X(ejΩ ) 2
−2π
2π
Ω
ak 1 4
−8
−1 1
38
Ω 8
Four Types of “Time”
discrete vs. continuous (t) and periodic vs aperiodic (↔)
DT Fourier Series
DT Fourier transform
n
n
CT Fourier Series
CT Fourier transform
t
t 39
Four Types of “Frequency”
discrete vs. continuous (↔) and periodic vs aperiodic (t)
DT Fourier Series
DT Fourier transform
2π k N
Ω
CT Fourier Series
CT Fourier transform
ω
2π k T
40
Relation between Fourier Transforms and Series
Periodic extension of a DT signal produces a discrete function of frequency. Periodic extension = convolving with impulse train in time = multiplying by impulse train in frequency → sampling in frequency
periodic DT DTFS interpolate
N →∞
aperiodic DT DTFT
periodic extension (sampling in frequency) sample interpolate
periodic CT CTFS
T →∞ periodic extension 41
sample
aperiodic CT CTFT
Four Fourier Representations
Underlying structure → view as one transform, not four. DT Fourier Series
1 ak = ak+N = N
X
DT Fourier transform 2π
x[n]e−j N kn
n=
x[n] = x[n + N ] =
X
∞ X
X(e jΩ ) =
x[n]e−jΩn
n=−∞
ak e
j 2π N kn
Z
x[n] =
k=
CT Fourier Series
1 X(e jΩ )e jΩn dΩ 2π
CT Fourier transform
Z ∞
Z 2π 1 x(t)e−j T kt dt ak = T T ∞ X 2π x(t) = x(t + T ) = ak e j T kt
X(jω) = −∞ Z ∞
x(t) =
k=−∞ 42
x(t)e−jωt dt
1 X(jω)e jωt dω 2π −∞
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6.003 Signals and Systems Fall 2011
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6.003: Signals and Systems Applications of Fourier Transforms
November 17, 2011
1
Filtering
Notion of a filter.
LTI systems • cannot create new frequencies. • can only scale magnitudes and shift phases of existing components.
Example: Low-Pass Filtering with an RC circuit
R + vi
+ −
C
vo −
2
Lowpass Filter
Calculate the frequency response of an RC circuit.
R +
vi (t)
= Ri(t) + vo (t)
C:
i(t)
= Cv˙ o (t)
Solving: vi (t)
vo
C
= RC v˙ o (t) + vo (t)
Vi (s) = (1 + sRC)Vo (s) 1 Vo (s) H(s) = = 1 + sRC Vi (s)
− 1 |H(jω)|
+ −
0.1 0.01
∠H(jω)|
vi
KVL:
0.01
0.1
1
10
ω 100 1/RC
10
ω 100 1/RC
0 − π2 0.01
0.1
1 3
Lowpass Filtering Let the input be a square wave. 1 2
0 − 12
|X(jω)|
1 jω0 kt e ; jπk k odd 1
ω0 =
2π T
0.1 0.01
∠X(jω)|
x(t) =
t
T
0.01
0.1
1
10
ω 100 1/RC
10
ω 1/RC 100
0 − π2 0.01
0.1
1 4
Lowpass Filtering Low frequency square wave: ω0 1/RC. 1 2
0 − 12 1 jω0 kt e ; jπk k odd 1
|H(jω)|
X
ω0 =
2π T
0.1 0.01
∠H(jω)|
x(t) =
t
T
0.01
0.1
1
10
ω 100 1/RC
10
ω 1/RC 100
0 − π2 0.01
0.1
1 8
Source-Filter Model of Speech Production
Vibrations of the vocal cords are “filtered” by the mouth and nasal
cavities to generate speech.
buzz from vocal cords
throat and nasal cavities 9
speech
Filtering
LTI systems “filter” signals based on their frequency content. Fourier transforms represent signals as sums of complex exponen tials. ∞ 1 x(t) = X(jω)e jωt dω 2π −∞
Complex exponentials are eigenfunctions of LTI systems. e jωt → H(jω)e jωt LTI systems “filter” signals by adjusting the amplitudes and phases of each frequency component. ∞ ∞ 1 1 x(t) = X(jω)e jωt dω → y(t) = H(jω)X(jω)e jωt dω 2π −∞ 2π −∞
10
Filtering
Systems can be designed to selectively pass certain frequency bands.
Examples: low-pass filter (LPF) and high-pass filter (HPF).
LPF
0
HPF
ω
LPF
t
HPF
t
t
11
Filtering Example: Electrocardiogram
An electrocardiogram is a record of electrical potentials that are generated by the heart and measured on the surface of the chest.
x(t) [mV] 2 1 0 −1
t [s] 0
ECG and analysis by T. F. Weiss
12
10
20
30
40
50
60
Filtering Example: Electrocardiogram
In addition to electrical responses of heart, electrodes on the skin also pick up other electrical signals that we regard as “noise.” We wish to design a filter to eliminate the noise.
x(t)
filter
13
y(t)
Filtering Example: Electrocardiogram
We can identify “noise” using the Fourier transform. x(t) [mV] 2 1 0 −1
t [s] 0
10
20
30
40
50
60
1000
60 Hz
|X(jω)| [µV]
100 10 1 0.1 0.01
low-freq. noise
0.001
cardiac signal
0.0001 0.01
0.1 f= 14
10 1 ω [Hz] 2π
high-freq. noise 100
Filtering Example: Electrocardiogram
Filter design: low-pass flter + high-pass filter + notch.
|H(jω)|
1 0.1 0.01 0.001 0.01
0.1
1 10 ω [Hz] f= 2π
15
100
Electrocardiogram: Check Yourself
Which poles and zeros are associated with • the high-pass filter? • the low-pass filter? • the notch filter? s-plane
2
( )
2 2
( )( )
16
Electrocardiogram: Check Yourself
Which poles and zeros are associated with • the high-pass filter? • the low-pass filter? • the notch filter?
s-plane notch
low-pass
2
2 2
( )
( )( )
notch
17
high-pass
Filtering Example: Electrocardiogram
Filtering is a simple way to reduce unwanted noise.
x(t) [mV ]
Unfiltered ECG
2 1 t [s]
0 0
10
20
30
40
50
60
y(t) [mV ]
Filtered ECG
1 t [s]
0 0
10
20
30
40
50 18
60
Fourier Transforms in Physics: Diffraction
A diffraction grating breaks a laser beam input into multiple beams.
Demonstration.
19
Fourier Transforms in Physics: Diffraction
grating
Multiple beams result from periodic structure of grating (period D).
λ sin θ = D
λ D
θ
Viewed at a distance from angle θ, scatterers are separated by D sin θ. Constructive interference if D sin θ = nλ, i.e., if sin θ = nλ D → periodic array of dots in the far field 20
Fourier Transforms in Physics: Diffraction
CD demonstration.
21
Check Yourself
CD demonstration.
3 feet
1 feet laser pointer λ = 500 nm CD screen What is the spacing of the tracks on the CD? 1. 160 nm
2. 1600 nm
3. 16µm
22
4. 160µm
Check Yourself
What is the spacing of the tracks on the CD?
grating
tan θ
θ
sin θ
CD
1 3
0.32
0.31
D=
500 nm sin θ
1613 nm
23
manufacturing spec. 1600 nm
Check Yourself
Demonstration.
3 feet
1 feet laser pointer λ = 500 nm CD screen What is the spacing of the tracks on the CD? 1. 160 nm
2. 1600 nm
3. 16µm
24
2.
4. 160µm
Fourier Transforms in Physics: Diffraction
DVD demonstration.
25
Check Yourself
DVD demonstration.
1 feet
1 feet laser pointer λ = 500 nm DVD screen What is track spacing on DVD divided by that for CD? 1. 4×
3. 12 ×
2. 2×
26
4. 14 ×
Check Yourself
What is spacing of tracks on DVD divided by that for CD?
D=
500 nm sin θ
grating
tan θ
θ
sin θ
CD
1 3
0.32
0.31
1613 nm
1600 nm
DVD
1
0.78
0.71
704 nm
740 nm
27
manufacturing spec.
Check Yourself
DVD demonstration.
1 feet
1 feet laser pointer λ = 500 nm DVD screen What is track spacing on DVD divided by that for CD? 3 1. 4×
3. 12 ×
2. 2×
28
4. 14 ×
Fourier Transforms in Physics: Diffraction
Macroscopic information in the far field provides microscopic (invis ible) information about the grating.
λ sin θ = D
θ
29
λ D
Fourier Transforms in Physics: Crystallography
What if the target is more complicated than a grating?
target
image?
30
Fourier Transforms in Physics: Crystallography
Part of image at angle θ has contributions for all parts of the target.
θ target
image?
31
Fourier Transforms in Physics: Crystallography
The phase of light scattered from different parts of the target un dergo different amounts of phase delay.
x sin
θ θ
x
Phase at a point x is delayed (i.e., negative) relative to that at 0: φ = −2π
x sin θ λ
32
Fourier Transforms in Physics: Crystallography
Total light F (θ) at angle θ is integral of light scattered from each part of target f (x), appropriately shifted in phase. Z x sin θ
F (θ) = f (x) e−j2π λ dx
Assume small angles so sin θ ≈ θ.
θ , then the pattern of light at the detector is
Let ω = 2π λ
Z F (ω) = f (x) e−jωx dx
which is the Fourier transform of f (x) !
33
Fourier Transforms in Physics: Diffraction
Fourier transform relation between structure of object and far-field intensity pattern.
grating ≈ impulse train with pitch D ···
··· t
0
D
λ far-field intensity ≈ impulse train with reciprocal pitch ∝ D
···
··· ω 0 34
2π D
Impulse Train
The Fourier transform of an impulse train is an impulse train.
x(t) =
∞ X
δ(t − kT )
k=−∞
1
···
··· t
0 ak = T1 ···
X(jω) =
T ∀ k 1 T
··· k
∞ X 2π 2π δ(ω − k ) T T
k=−∞
2π T
··· 0 35
2π T
··· ω
Two Dimensions
Demonstration: 2D grating.
36
An Historic Fourier Transform
Taken by Rosalind Franklin, this image sparked Watson and Crick’s
insight into the double helix.
Reprinted by permission from Macmillan Publishers Ltd: Nature. Source: Franklin, R., and R. G. Gosling. "Molecular Configuration in Sodium Thymonucleate." Nature 171 (1953): 740-741. (c) 1953. 37
An Historic Fourier Transform
This is an x-ray crystallographic image of DNA, and it shows the
Fourier transform of the structure of DNA.
Reprinted by permission from Macmillan Publishers Ltd: Nature. Source: Franklin, R., and R. G. Gosling. "Molecular Configuration in Sodium Thymonucleate." Nature 171 (1953): 740-741. (c) 1953. 38
An Historic Fourier Transform
High-frequency bands indicate repeating structure of base pairs.
b 1/b
Reprinted by permission from Macmillan Publishers Ltd: Nature. Source: Franklin, R., and R. G. Gosling. "Molecular Configuration in Sodium Thymonucleate." Nature 171 (1953): 740-741. (c) 1953.
39
An Historic Fourier Transform
Low-frequency bands indicate a lower frequency repeating structure.
1/h
h
Reprinted by permission from Macmillan Publishers Ltd: Nature. Source: Franklin, R., and R. G. Gosling. "Molecular Configuration in Sodium Thymonucleate." Nature 171 (1953): 740-741. (c) 1953.
40
An Historic Fourier Transform
Tilt of low-frequency bands indicates tilt of low-frequency repeating
structure: the double helix!
θ θ
Reprinted by permission from Macmillan Publishers Ltd: Nature. Source: Franklin, R., and R. G. Gosling. "Molecular Configuration in Sodium Thymonucleate." Nature 171 (1953): 740-741. (c) 1953. 41
Simulation
Easy to calculate relation between structure and Fourier transform.
42
Fourier Transform Summary
Represent signals by their frequency content.
Key to “filtering,” and to signal-processing in general.
Important in many physical phenomenon: x-ray crystallography.
43
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6.003: Signals and Systems Sampling
November 22, 2011
1
Sampling
Conversion of a continuous-time signal to discrete time.
x[n]
x(t)
n
t 0
2
4
6
8
10
0
2
4
6
We have used sampling a number of times before.
Today: new insights from Fourier representations.
2
8
10
Sampling
Sampling allows the use of modern digital electronics to process, record, transmit, store, and retrieve CT signals. • • • •
audio: MP3, CD, cell phone pictures: digital camera, printer video: DVD everything on the web
3
Sampling
Sampling is pervasive.
Example: digital cameras record sampled images.
y
I(x, y)
n
x
I[m, n]
m
4
Sampling
Photographs in newsprint are “half-tone” images. black or white and the average conveys brightness.
5
Each point is
Sampling
Zoom in to see the binary pattern.
6
Sampling
Even high-quality photographic paper records discrete images. When AgBr crystals (0.04 − 1.5µm) are exposed to light, some of the Ag is reduced to metal. During “development” the exposed grains are completely reduced to metal and unexposed grains are removed.
Images of discrete grains in photographic paper removed due to copyright restrictions.
7
Sampling
Every image that we see is sampled by the retina, which contains ≈ 100 million rods and 6 million cones (average spacing ≈ 3µm) which act as discrete sensors.
Courtesy of Helga Kolb, Eduardo Fernandez, and Ralph Nelson. Used with permission.
http://webvision.med.utah.edu/imageswv/sagschem.jpeg 8
Check Yourself
Your retina is sampling this slide, which is composed of 1024×768 pixels. Is the spatial sampling done by your rods and cones ade quate to resolve individual pixels in this slide?
9
Check Yourself
The spacing of rods and cones limits the angular resolution of your retina to approximately θeye =
rod/cone spacing 3 × 10−6 m ≈ ≈ 10−4 radians diameter of eye 3 cm
The angle between pixels viewed from the center of the classroom is approximately screen size / 1024 3 m/1024 θpixels = ≈ ≈ 3 × 10−4 radians distance to screen 10 m Light from a single pixel falls upon multiple rods and cones.
10
Sampling
How does sampling affect the information contained in a signal?
11
Sampling
We would like to sample in a way that preserves information, which may not seem possible.
x(t)
t
Information between samples is lost. Therefore, the same samples can represent multiple signals.
cos 7π 3 n?
cos π3 n?
t 12
Sampling and Reconstruction
To determine the effect of sampling, compare the original signal x(t) to the signal xp (t) that is reconstructed from the samples x[n]. Uniform sampling (sampling interval T ).
x[n] = x(nT ) t n
Impulse reconstruction.
xp (t) =
X
x[n]δ(t − nT )
n
t n 13
Reconstruction
Impulse reconstruction maps samples x[n] (DT) to xp (t) (CT). xp (t) = =
=
∞ 0
x[n]δ(t − nT )
n=−∞ ∞ 0 n=−∞ ∞ 0
x(nT )δ(t − nT ) x(t)δ(t − nT )
n=−∞ ∞ 0
= x(t)
δ(t − nT )
n=−∞
-
,,
≡ p(t)
"
Resulting reconstruction xp (t) is equivalent to multiplying x(t) by impulse train.
14
Sampling
Multiplication by an impulse train in time is equivalent to convolution by an impulse train in frequency. → generates multiple copies of original frequency content.
X(jω) 1 ω
−W W P (jω) 2π T
ω
ωs
−ωs
1 (X(j · ) ∗ P (j · ))(ω) Xp (jω) = 2π 1 T
ω −ωs
ωs = 15
2π T
Check Yourself
What is the relation between the DTFT of x[n] = x(nT ) and the CTFT of xp (t) = x[n]δ(t − nT ) for X(jω) below.
X(jω) 1 ω
−W W 1. Xp (jω) = X(e jΩ )|Ω=ω 2. Xp (jω) = X(e jΩ )|Ω= ω T
3. Xp (jω) = X(e
jΩ
)|Ω=ωT
4. none of the above
16
Check Yourself
DTFT
X(e jΩ ) =
∞ 0
x[n]e−jΩn
n=−∞
CTFT of xp (t) ∞ 0
∞
Xp (jω) =
=
∞ 0
∞
x[n]
∞ 0
δ(t − nT )e−jωt dt
−∞
n=−∞
=
x[n]δ(t − nT )e−jωt dt
−∞ n=−∞
x[n]e−jωnT
n=−∞ jΩ
= X(e
� � )�
Ω=ωT
17
Check Yourself
� � Xp (jω) = X(e jΩ )�
Ω=ωT
X(jω) 1 ω
−W W
1 (X(j · ) ∗ P (j · ))(ω) Xp (jω) = 2π 1 T
ω −ωs
ωs =
2π T
X(ejΩ ) = Xp (jω) ω= Ω T
1 T
Ω
−2π
2π 18
Check Yourself
What is the relation between the DTFT of x[n] = x(nT ) and the CTFT of xp (t) = x[n]δ(t − nT ) for X(jω) below.
X(jω) 1 ω
−W W 1. Xp (jω) = X(e jΩ )|Ω=ω 2. Xp (jω) = X(e jΩ )|Ω= ω T
3. Xp (jω) = X(e
jΩ
)|Ω=ωT
4. none of the above
19
Sampling
The high frequency copies can be removed with a low-pass filter (also multiply by T to undo the amplitude scaling). 1 (X(j · ) ∗ P (j · ))(ω) Xp (jω) = 2π 1 T T
ωs − 2
ωs 2
ω
Impulse reconstruction followed by ideal low-pass filtering is called bandlimited reconstruction.
20
The Sampling Theorem
If signal is bandlimited → sample without loosing information. If x(t) is bandlimited so that X(jω) = 0 for |ω| > ωm then x(t) is uniquely determined by its samples x(nT ) if 2π > 2ωm . ωs = T The minimum sampling frequency, 2ωm , is called the “Nyquist rate.”
21
Summary
Three important ideas. Sampling x(t) → x[n] = x(nT ) Bandlimited Reconstruction
x[n]
Impulse Reconstruction
ωs =
2π T
LPF T
xp (t) = P
x[n]δ(t − nT )
Sampling Theorem: If X(jω) = 0 ∀ |ω| >
22
− ω2s
ωs 2
ω
ωs then xr (t) = x(t). 2
xr (t)
Check Yourself
We can hear sounds with frequency components between 20 Hz and 20 kHz. What is the maximum sampling interval T that can be used to sample a signal without loss of audible information? 1. 100 µs 3. 25 µs 5. 50π µs
2. 50 µs 4. 100π µs 6. 25π µs
23
Check Yourself
ωs 2π = 2 2T 1 1 T < = = 25 µs 2 × 20 kHz 2fm 2πfm = ωm
X(jω)
π ? T
ω P (jω) 2π T
ωs
−ωs
ω
1 (X(j · ) ∗ P (j · ))(ω) Xp (jω) = 2π 1 T
ω − ω2s
ωs 2
27
Aliasing
What happens if X contains frequencies |ω| >
X(jω)
π ? T
ω P (jω) 2π T
ωs
−ωs
ω
1 (X(j · ) ∗ P (j · ))(ω) Xp (jω) = 2π 1 T
ω − ω2s
ωs 2
28
Aliasing
What happens if X contains frequencies |ω| >
X(jω)
π ? T
ω P (jω) 2π T
ωs
−ωs
ω
1 (X(j · ) ∗ P (j · ))(ω) Xp (jω) = 2π 1 T
ω − ω2s
ωs 2
29
Aliasing
What happens if X contains frequencies |ω| >
X(jω)
π ? T
ω P (jω) 2π T
ωs
−ωs
ω
1 (X(j · ) ∗ P (j · ))(ω) Xp (jω) = 2π 1 T
ω − ω2s
ωs 2
30
Aliasing
The effect of aliasing is to wrap frequencies.
Output frequency ωs 2
Input frequency ωs 2
X(jω)
ω
ω 1 T
ω − ω2s
ωs 2 31
Aliasing
The effect of aliasing is to wrap frequencies.
Output frequency ωs 2
Input frequency ωs 2
X(jω)
ω
ω 1 T
ω − ω2s
ωs 2 32
Aliasing
The effect of aliasing is to wrap frequencies.
Output frequency ωs 2
Input frequency ωs 2
X(jω)
ω
ω 1 T
ω − ω2s
ωs 2 33
Aliasing
The effect of aliasing is to wrap frequencies.
Output frequency ωs 2
Input frequency ωs 2
X(jω)
ω
ω 1 T
ω − ω2s
ωs 2 34
Check Yourself
A periodic signal, period of 0.1 ms, is sampled at 44 kHz. To what frequency does the third harmonic alias?
1. 2. 3. 4. 5. 0.
18 kHz 16 kHz 14 kHz 8 kHz 6 kHz none of the above
35
Check Yourself Output frequency (kHz) 88 44 22
Input frequency (kHz) 22 44 66 88
36
Check Yourself Output frequency (kHz) 88 44 22
Input frequency (kHz) 22 44 66 88
Harmonic 10 kHz 20 kHz 30 kHz
Alias 10 kHz 20 kHz 44 kHz-30 kHz =14 kHz
37
Check Yourself
A periodic signal, period of 0.1 ms, is sampled at 44 kHz. To what frequency does the third harmonic alias? 3
1. 2. 3. 4. 5. 0.
18 kHz 16 kHz 14 kHz 8 kHz 6 kHz none of the above
38
Check Yourself Output frequency (kHz) 88 44 22
Input frequency (kHz) 22 44 66 88
Harmonic 10 kHz 20 kHz 30 kHz 40 kHz 50 kHz 60 kHz 70 kHz 80 kHz
Alias 10 kHz 20 kHz 44 kHz-30 44 kHz-40 50 kHz-44 60 kHz-44 88 kHz-70 88 kHz-80
kHz kHz kHz kHz kHz kHz
=14 = 4 = 6 =16 =18 = 8 39
kHz kHz kHz kHz kHz kHz
Check Yourself
Scrambled harmonics.
ω
ω
40
Aliasing
High frequency components of complex signals also wrap.
X(jω) 1 ω P (jω) 2π T
ω
ωs
−ωs
1 (X(j · ) ∗ P (j · ))(ω) Xp (jω) = 2π 1 T
ω − ω2s
ωs 2
41
Aliasing
High frequency components of complex signals also wrap.
X(jω) 1 ω P (jω) 2π T
ω
ωs
−ωs
1 (X(j · ) ∗ P (j · ))(ω) Xp (jω) = 2π 1 T
ω − ω2s
ωs 2
42
Aliasing
High frequency components of complex signals also wrap.
X(jω) 1 ω P (jω) 2π T
ω
ωs
−ωs
1 (X(j · ) ∗ P (j · ))(ω) Xp (jω) = 2π 1 T
ω − ω2s
ωs 2
43
Aliasing
High frequency components of complex signals also wrap.
X(jω) 1 ω P (jω) 2π T
ω
ωs
−ωs
1 (X(j · ) ∗ P (j · ))(ω) Xp (jω) = 2π 1 T
ω − ω2s
ωs 2
44
Aliasing
Aliasing increases as the sampling rate decreases.
X(jω) 1 ω P (jω) 2π T
ω
ωs
−ωs
1 (X(j · ) ∗ P (j · ))(ω) Xp (jω) = 2π 1 T
ω − ω2s
ωs 2
45
Aliasing
Aliasing increases as the sampling rate decreases.
X(jω) 1 ω P (jω) 2π T
ω
ωs
−ωs
1 (X(j · ) ∗ P (j · ))(ω) Xp (jω) = 2π 1 T
ω − ω2s
ωs 2
46
Aliasing
Aliasing increases as the sampling rate decreases.
X(jω) 1 ω P (jω) 2π T
ω
ωs
−ωs
1 (X(j · ) ∗ P (j · ))(ω) Xp (jω) = 2π 1 T
ω − ω2s
ωs 2
47
Aliasing
Aliasing increases as the sampling rate decreases.
X(jω) 1 ω P (jω) 2π T
ω
ωs
−ωs
1 (X(j · ) ∗ P (j · ))(ω) Xp (jω) = 2π 1 T
ω − ω2s ω2s
48
Aliasing Demonstration
Sampling Music
ωs =
2π = 2πfs T
• fs = 44.1 kHz • fs = 22 kHz • fs = 11 kHz • fs = 5.5 kHz • fs = 2.8 kHz J.S. Bach, Sonata No. 1 in G minor Mvmt. IV. Presto Nathan Milstein, violin
49
Aliasing
Aliasing increases as the sampling rate decreases.
X(jω) 1 ω P (jω) 2π T
ω
ωs
−ωs
1 (X(j · ) ∗ P (j · ))(ω) Xp (jω) = 2π 1 T
ω − ω2s
ωs 2
50
Aliasing
Aliasing increases as the sampling rate decreases.
X(jω) 1 ω P (jω) 2π T
ω
ωs
−ωs
1 (X(j · ) ∗ P (j · ))(ω) Xp (jω) = 2π 1 T
ω − ω2s
ωs 2
51
Aliasing
Aliasing increases as the sampling rate decreases.
X(jω) 1 ω P (jω) 2π T
ω
ωs
−ωs
1 (X(j · ) ∗ P (j · ))(ω) Xp (jω) = 2π 1 T
ω − ω2s
ωs 2
52
Aliasing
Aliasing increases as the sampling rate decreases.
X(jω) 1 ω P (jω) 2π T
ω
ωs
−ωs
1 (X(j · ) ∗ P (j · ))(ω) Xp (jω) = 2π 1 T
ω − ω2s ω2s
53
Anti-Aliasing Filter
To avoid aliasing, remove frequency components that alias before sampling.
x(t)
Anti-aliasing Filter 1 ω ωs − ω2s 2
× p(t)
54
Reconstruction Filter T xp (t) ω ωs − ω2s 2
xr (t)
Aliasing
Aliasing increases as the sampling rate decreases.
X(jω) 1 ω P (jω) 2π T
ω
ωs
−ωs
1 (X(j · ) ∗ P (j · ))(ω) Xp (jω) = 2π 1 T
ω − ω2s
ωs 2
55
Aliasing
Aliasing increases as the sampling rate decreases.
Anti-aliased X(jω)
ω P (jω) 2π T
ω
ωs
−ωs
1 (X(j · ) ∗ P (j · ))(ω) Xp (jω) = 2π 1 T
ω − ω2s
ωs 2
56
Aliasing
Aliasing increases as the sampling rate decreases.
Anti-aliased X(jω)
ω P (jω) 2π T
ω
ωs
−ωs
1 (X(j · ) ∗ P (j · ))(ω) Xp (jω) = 2π 1 T
ω − ω2s
ωs 2
57
Aliasing
Aliasing increases as the sampling rate decreases.
Anti-aliased X(jω)
ω P (jω) 2π T
ω
ωs
−ωs
1 (X(j · ) ∗ P (j · ))(ω) Xp (jω) = 2π 1 T
ω − ω2s ω2s
58
Anti-Aliasing Demonstration
Sampling Music
ωs =
2π = 2πfs T
• fs = 11 kHz without anti-aliasing • fs = 11 kHz with anti-aliasing • fs = 5.5 kHz without anti-aliasing • fs = 5.5 kHz with anti-aliasing • fs = 2.8 kHz without anti-aliasing • fs = 2.8 kHz with anti-aliasing J.S. Bach, Sonata No. 1 in G minor Mvmt. IV. Presto Nathan Milstein, violin 59
Sampling: Summary
Effects of sampling are easy to visualize with Fourier representations. Signals that are bandlimited in frequency (e.g., −W < ω < W ) can be sampled without loss of information. The minimum sampling frequency for sampling without loss of in formation is called the Nyquist rate. The Nyquist rate is twice the highest frequency contained in a bandlimited signal. Sampling at frequencies below the Nyquist rate causes aliasing. Aliasing can be eliminated by pre-filtering to remove frequency com ponents that would otherwise alias.
60
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6.003 Signals and Systems Fall 2011
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6.003: Signals and Systems Sampling and Quantization
November 29, 2011
1
Last Time: Sampling
Sampling allows the use of modern digital electronics to process, record, transmit, store, and retrieve CT signals. • • • •
audio: MP3, CD, cell phone pictures: digital camera, printer video: DVD everything on the web
2
Last Time: Sampling Theory
Sampling x(t) → x[n] = x(nT ) Impulse Reconstruction LPF
x[n]
Impulse Reconstruction
xp (t) = P
Bandlimited Reconstruction
x[n]
Impulse Reconstruction
xr (t)
x[n]δ(t − nT )
ωs =
2π T
LPF
T
xp (t) = P
x[n]δ(t − nT )
Sampling Theorem: If X(jω) = 0 ∀ |ω| > 3
− ω2s
ωs 2
ω
ωs then xr (t) = x(t). 2
xr (t)
Aliasing
Frequencies outside the range
−ωs ωs 100 km
6
5
Check Yourself
What frequency E/M wave is well matched to an antenna with a length of 10 cm (about 4 inches)? 1. 2. 3. 4. 5.
< 100 kHz 1 MHz 10 MHz 100 MHz > 1 GHz
7
Check Yourself
A wavelength of 10 cm corresponds to a frequency of
f=
c 3 × 108 m/s ∼ ≈ 3 GHz . λ 10 cm
Modern cell phones use frequencies near 2 GHz.
8
Check Yourself
What frequency E/M wave is well matched to an antenna with a length of 10 cm (about 4 inches)? 5 1. 2. 3. 4. 5.
< 100 kHz 1 MHz 10 MHz 100 MHz > 1 GHz
9
Wireless Communication
Speech is not well matched to the wireless medium. Many applications require the use of signals that are not well matched to the required media. signal
applications
audio
telephone, radio, phonograph, CD, cell phone, MP3
video
television, cinema, HDTV, DVD
internet
coax, twisted pair, cable TV, DSL, optical fiber, E/M
We can often modify the signals to obtain a better match. Today we will introduce simple matching strategies based on modulation.
10
Check Yourself
Construct a signal Y that codes the audio frequency information in X using frequency components near 2 GHz. |X(jω)|
ω |Y (jω)|
ωc
ω
Determine an expression for Y in terms of X. 1. y(t) = x(t) e jωc t 2. y(t) = x(t) ∗ e jωc t 3. y(t) = x(t) cos(ωc t) 4. y(t) = x(t) ∗ cos(ωc t) 5. none of the above 11
Check Yourself
Construct a signal Y that codes the audio frequency information in X using frequency components near 2 GHz. |X(jω)|
ω |Y (jω)|
ωc Determine an expression for Y in terms of X.
ω
1
1. y(t) = x(t) e jωc t 2. y(t) = x(t) ∗ e jωc t 3. y(t) = x(t) cos(ωc t) 4. y(t) = x(t) ∗ cos(ωc t) 5. none of the above 12
Amplitude Modulation
Multiplying a signal by a sinusoidal carrier signal is called amplitude
modulation (AM). AM shifts the frequency components of X by ±ωc .
x(t)
×
y(t)
cos ωc t |X(jω)|
ω
ωc
−ωc
ω
|Y (jω)|
−ωc
ωc 13
ω
Amplitude Modulation
Multiplying a signal by a sinusoidal carrier signal is called amplitude
modulation. The signal “modulates” the amplitude of the carrier.
x(t)
×
y(t)
cos ωc t
x(t)
t
cos ωc t
t
x(t) cos ωc t
t 14
Amplitude Modulation
How could you recover x(t) from y(t)?
x(t)
× cos ωc t
15
y(t)
Synchronous Demodulation
X can be recovered by multiplying by the carrier and then low-pass filtering. This process is called synchronous demodulation. y(t) = x(t) cos ωc t � z(t) = y(t) cos ωc t = x(t) × cos ωc t × cos ωc t = x(t)
16
� 1 1 + cos(2ωc t) 2 2
Synchronous Demodulation
Synchronous demodulation: convolution in frequency.
|Y (jω)|
−ωc
ωc
−ωc
ωc
ω
ω
|Z(jω)|
ω −2ωc
2ωc
17
Synchronous Demodulation
We can recover X by low-pass filtering.
|Y (jω)|
−ωc
ωc
−ωc
ωc
ω
ω
|Z(jω)| 2 ω −2ωc
2ωc
18
Frequency-Division Multiplexing
Multiple transmitters can co-exist, as long as the frequencies that they transmit do not overlap.
z1(t)
x1(t)
cos w1t z2(t)
x2(t)
z(t)
cos wct
cos w2t z3(t)
x3(t)
cos w3t
19
LPF
y(t)
Frequency-Division Multiplexing
Multiple transmitters simply sum (to first order). z1(t)
x1(t)
cos w1t z2(t)
x2(t)
z(t)
cos wct
cos w2t z3(t)
x3(t)
LPF
cos w3t
20
y(t)
Frequency-Division Multiplexing
The receiver can select the transmitter of interest by choosing the corresponding demodulation frequency.
w)
Z(j
w w w)
w
X1(j
w1 w2 w3 21
w
Frequency-Division Multiplexing
The receiver can select the transmitter of interest by choosing the corresponding demodulation frequency.
w)
Z(j
w w w)
w
X2(j
w1 w2 w3 22
w
Frequency-Division Multiplexing
The receiver can select the transmitter of interest by choosing the corresponding demodulation frequency.
w)
Z(j
w w w)
w
X3(j
w1 w2 w3 23
w
Broadcast Radio
“Broadcast” radio was championed by David Sarnoff, who previously worked at Marconi Wireless Telegraphy Company (point-to-point). • envisioned “radio music boxes” • analogous to newspaper, but at speed of light • receiver must be cheap (as with newsprint) • transmitter can be expensive (as with printing press)
24
Inexpensive Radio Receiver
The problem with making an inexpensive radio receiver is that you must know the carrier signal exactly!
x(t)
LPF
z(t)
wc t)
wc t+f)
cos(
cos(
25
y(t)
Check Yourself
The problem with making an inexpensive radio receiver is that you must know the carrier signal exactly!
x(t)
LPF
z(t)
wc t)
y(t)
wc t+f)
cos(
cos(
What happens if there is a phase shift φ between the signal used to modulate and that used to demodulate?
26
Check Yourself
y(t) = x(t) × cos(ωc t) × cos(ωc t + φ) � � 1 1 = x(t) × cos φ + cos(2ωc t + φ) 2 2 Passing y(t) through a low pass filter yields 12 x(t) cos φ. If φ = π/2, the output is zero! If φ changes with time, then the signal “fades.”
27
AM with Carrier
One way to synchronize the sender and receiver is to send the carrier
along with the message.
x(t)
×
z(t)
+ C
cos ωc t z(t) = x(t) cos ωc t + C cos ωc t = (x(t) + C) cos ωc t
x(t) + C t z(t) Adding carrier is equivalent to shifting the DC value of x(t). If we shift the DC value sufficiently, the message is easy to decode: it is just the envelope (minus the DC shift). 28
Inexpensive Radio Receiver
If the carrier frequency is much greater than the highest frequency in the message, AM with carrier can be demodulated with a peak detector. y(t)
z(t)
R
C
t
y(t) z(t)
In AM radio, the highest frequency in the message is 5 kHz and the
carrier frequency is between 500 kHz and 1500 kHz.
This circuit is simple and inexpensive.
But there is a problem.
29
Inexpensive Radio Receiver
AM with carrier requires more power to transmit the carrier than to transmit the message! x(t) xp
xp > 35xrms
xrms
t
Speech sounds have high crest factors (peak value divided by rms value). The DC offset C must be larger than xp for simple envelope detection to work.
The power needed to transmit the carrier can be 352 ≈ 1000× that
needed to transmit the message.
Okay for broadcast radio (WBZ: 50 kwatts).
Not for point-to-point (cell phone batteries wouldn’t last long!).
30
Inexpensive Radio Receiver
Envelope detection also cannot separate multiple senders.
y(t)
z(t)
R
C
t
y(t) z(t)
31
Superheterodyne Receiver
Edwin Howard Armstrong invented the superheterodyne receiver, which made broadcast AM practical.
Edwin Howard Armstrong also invented and patented the “regenerative” (positive feedback) circuit for amplifying radio signals (while he was a junior at Columbia University). He also in vented wide-band FM. 32
Digital Radio
Could we implement a radio with digital electronics?
Commercial AM radio • • •
106 channels each channel is allocated 10 kHz bandwidth center frequencies from 540 to 1600 kHz
xa (t) Uniform xd [n] Digital Signal yd [n] ya (t) Bandlimited Processor Sampler Reconstruction
T
T
33
Check Yourself
Determine T to decode commercial AM radio. Commercial AM radio • • •
106 channels each channel is allocated 10 kHz bandwidth center frequencies from 540 to 1600 kHz xa (t) Uniform xd [n] Digital Signal yd [n] ya (t) Bandlimited Processor Sampler Reconstruction
T
T
The maximum value of T is approximately 1. 0.3 fs 2. 0.3 ns 3. 0.3µs 5. none of these 34
4. 0.3 ms
Check Yourself
Determine T to decode commercial AM radio.
3.
Commercial AM radio • • •
106 channels each channel is allocated 10 kHz bandwidth center frequencies from 540 to 1600 kHz xa (t) Uniform xd [n] Digital Signal yd [n] ya (t) Bandlimited Processor Sampler Reconstruction
T
T
The maximum value of T is approximately 1. 0.3 fs 2. 0.3 ns 3. 0.3µs 5. none of these 35
4. 0.3 ms
Digital Radio
The digital electronics must implement a bandpass filter, multipli cation by cos ωc t, and a lowpass filter.
xd [n]
BPF
× cos ωc t
36
LPF
yd [n]
Check Yourself
Which of following systems implement a bandpass filter?
System 1
×
LPF
×
cos Ωc n
cos Ωc n
cos Ωc n
cos Ωc n
×
LPF
× +
System 2
×
LPF
sin Ωc n System 3
× sin Ωc n
h[n] = hLP F [n] cos Ωc n 37
Check Yourself
h[n] = hLP F [n] cos Ωc n y[n] = x[n] ∗ hLP F [n] cos Ωc n x[k]hLP F [n − k] cos Ωc (n − k)
= k
x[k]hLP F [n − k] (cos Ωc n cos Ωc k + sin Ωc n sin Ωc k)
= k
x[k] cos Ωc k hLP F [n − k] cos Ωc n
= k
x[k] sin Ωc k hLP F [n − k] sin Ωc n
+ k
= (x[n] cos Ωc n) ∗ hLP F [n] cos Ωc n + (x[n] sin Ωc n) ∗ hLP F [n] sin Ωc n 38
Check Yourself
Which of following systems implement a bandpass filter?
System 1
×
LPF
×
cos Ωc n
cos Ωc n
cos Ωc n
cos Ωc n
×
LPF
× +
System 2
×
LPF
sin Ωc n System 3
× sin Ωc n
h[n] = hLP F [n] cos Ωc n 39
MIT OpenCourseWare http://ocw.mit.edu
6.003 Signals and Systems Fall 2011
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
6.003: Signals and Systems Modulation
December 6, 2011
1
Communications Systems
Signals are not always well matched to the media through which we wish to transmit them. signal
applications
audio
telephone, radio, phonograph, CD, cell phone, MP3
video
television, cinema, HDTV, DVD
internet
coax, twisted pair, cable TV, DSL, optical fiber, E/M
Modulation can improve match based on frequency.
2
Amplitude Modulation
Amplitude modulation can be used to match audio frequencies to
radio frequencies. It allows parallel transmission of multiple channels.
z1(t)
x1(t)
cos w1t z2(t)
x2(t)
z(t)
cos wct
cos w2t z3(t)
x3(t)
cos w3t
3
LPF
y(t)
Superheterodyne Receiver
Edwin Howard Armstrong invented the superheterodyne receiver, which made broadcast AM practical.
Edwin Howard Armstrong also invented and patented the “regenerative” (positive feedback) circuit for amplifying radio signals (while he was a junior at Columbia University). He also in vented wide-band FM. 4
Amplitude, Phase, and Frequency Modulation
There are many ways to embed a “message” in a carrier.
Amplitude Modulation (AM) + carrier: y1 (t) = x(t) + C cos(ωc t) Phase Modulation (PM):
y2 (t) = cos(ωc t + kx(t))
Frequency Modulation (FM):
y3 (t) = cos ωc t + k −∞ x(τ )dτ
t
PM: signal modulates instantaneous phase of the carrier. y2 (t) = cos(ωc t + kx(t)) FM: signal modulates instantaneous frequency of carrier. t
y3 (t) = cos ωc t + k x(τ )dτ −∞ ' v " φ(t)
d ωi (t) = ωc + φ(t) = ωc + kx(t) dt
5
Frequency Modulation
Compare AM to FM for x(t) = cos(ωm t). AM: y1 (t) = x(t) + C cos(ωc t) = (cos(ωm t) + 1.1) cos(ωc t)
t
Rt FM: y3 (t) = cos ωc t + k −∞ x(τ )dτ = cos(ωc t + ωkm sin(ωm t)) t
Advantages of FM: • constant power • no need to transmit carrier (unless DC important) • bandwidth? 6
Frequency Modulation
Early investigators thought that narrowband FM could have arbitrar ily narrow bandwidth, allowing more channels than AM.
Z t y3 (t) = cos ωc t + k x(τ )dτ ' −∞v " φ(t)
d ωi (t) = ωc + φ(t) = ωc + kx(t) dt Small k → small bandwidth. Right?
7
Frequency Modulation
Early investigators thought that narrowband FM could have arbitrar ily narrow bandwidth, allowing more channels than AM. Wrong! 0 Z t y3 (t) = cos ωc t + k x(τ )dτ −∞ 0 Z t 0 Z t = cos(ωc t) × cos k x(τ )dτ − sin(ωc t) × sin k x(τ )dτ −∞
If k → 00 then Z t cos k x(τ )dτ −∞ 0 Z t sin k x(τ )dτ
−∞
→1
Z t →k
x(τ )dτ 0 Z t y3 (t) ≈ cos(ωc t) − sin(ωc t) × k x(τ )dτ −∞
−∞
−∞
Bandwidth of narrowband FM is the same as that of AM! (integration does not change the highest frequency in the signal) 8
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) , therefore cos(m sin(ωm t)) is periodic in T . x(t) is periodic in T = ω2π m
1 sin(ωm t) 1 t
0 −1 cos(1 sin(ωm t)) 1
t
0 −1 9
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) , therefore cos(m sin(ωm t)) is periodic in T . x(t) is periodic in T = ω2π m
2 sin(ωm t) 2 t
0 −2 cos(2 sin(ωm t)) 1
t
0 −1 10
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) , therefore cos(m sin(ωm t)) is periodic in T . x(t) is periodic in T = ω2π m
3 sin(ωm t) 3 t
0 −3 cos(3 sin(ωm t)) 1
t
0 −1 11
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) , therefore cos(m sin(ωm t)) is periodic in T . x(t) is periodic in T = ω2π m
4 sin(ωm t) 4 t
0 −4 cos(4 sin(ωm t)) 1
t
0 −1 12
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) , therefore cos(m sin(ωm t)) is periodic in T . x(t) is periodic in T = ω2π m
5 sin(ωm t) 5 t
0 −5 cos(5 sin(ωm t)) 1
t
0 −1 13
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) , therefore cos(m sin(ωm t)) is periodic in T . x(t) is periodic in T = ω2π m
6 sin(ωm t) 6 t
0 −6 cos(6 sin(ωm t)) 1
t
0 −1 14
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) , therefore cos(m sin(ωm t)) is periodic in T . x(t) is periodic in T = ω2π m
7 sin(ωm t) 7 t
0 −7 cos(7 sin(ωm t)) 1
t
0 −1 15
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) , therefore cos(m sin(ωm t)) is periodic in T . x(t) is periodic in T = ω2π m
8 sin(ωm t) 8 t
0 −8 cos(8 sin(ωm t)) 1
t
0 −1 16
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) , therefore cos(m sin(ωm t)) is periodic in T . x(t) is periodic in T = ω2π m
9 sin(ωm t) 9 t
0 −9 cos(9 sin(ωm t)) 1
t
0 −1 17
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) , therefore cos(m sin(ωm t)) is periodic in T . x(t) is periodic in T = ω2π m
10 sin(ωm t) 10 t
0 −10 cos(10 sin(ωm t)) 1
t
0 −1 18
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) , therefore cos(m sin(ωm t)) is periodic in T . x(t) is periodic in T = ω2π m
20 sin(ωm t) 20 t
0 −20 cos(20 sin(ωm t)) 1
t
0 −1 19
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) , therefore cos(m sin(ωm t)) is periodic in T . x(t) is periodic in T = ω2π m
50 sin(ωm t) 50 t
0 −50 cos(50 sin(ωm t)) 1
t
0 −1 20
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) , therefore cos(m sin(ωm t)) is periodic in T . x(t) is periodic in T = ω2π m
m sin(ωm t) m t
0 −m cos(m sin(ωm t)) 1
t increasing m
0 −1 21
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) x(t) is periodic in T = ω2π , therefore cos(m sin(ωm t)) is periodic in T . m
cos(m sin(ωm t)) 1 t
0 −1 m=0 |ak | 0
10
20
30 22
40
50
60
k
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) x(t) is periodic in T = ω2π , therefore cos(m sin(ωm t)) is periodic in T . m
cos(m sin(ωm t)) 1 t
0 −1 m=1 |ak | 0
10
20
30 23
40
50
60
k
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) x(t) is periodic in T = ω2π , therefore cos(m sin(ωm t)) is periodic in T . m
cos(m sin(ωm t)) 1 t
0 −1 m=2 |ak | 0
10
20
30 24
40
50
60
k
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) x(t) is periodic in T = ω2π , therefore cos(m sin(ωm t)) is periodic in T . m
cos(m sin(ωm t)) 1 t
0 −1 m=5 |ak | 0
10
20
30 25
40
50
60
k
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) x(t) is periodic in T = ω2π , therefore cos(m sin(ωm t)) is periodic in T . m
cos(m sin(ωm t)) 1 t
0 −1 m = 10 |ak | 0
10
20
30 26
40
50
60
k
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) x(t) is periodic in T = ω2π , therefore cos(m sin(ωm t)) is periodic in T . m
cos(m sin(ωm t)) 1 t
0 −1 m = 20 |ak | 0
10
20
30 27
40
50
60
k
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) x(t) is periodic in T = ω2π , therefore cos(m sin(ωm t)) is periodic in T . m
cos(m sin(ωm t)) 1 t
0 −1 m = 30 |ak | 0
10
20
30 28
40
50
60
k
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) x(t) is periodic in T = ω2π , therefore cos(m sin(ωm t)) is periodic in T . m
cos(m sin(ωm t)) 1 t
0 −1 m = 40 |ak | 0
10
20
30 29
40
50
60
k
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) x(t) is periodic in T = ω2π , therefore cos(m sin(ωm t)) is periodic in T . m
cos(m sin(ωm t)) 1 t
0 −1 m = 50 |ak | 0
10
20
30 30
40
50
60
k
Phase/Frequency Modulation
Fourier transform of first part. x(t) = sin(ωm t) y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) ' v " ya (t)
|Ya (jω)|
m = 50
ω ωc
ωc 100ωm
31
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) , therefore sin(m sin(ωm t)) is periodic in T . x(t) is periodic in T = ω2π m
m sin(ωm t) m t
0 −m sin(m sin(ωm t)) 1
increasing m t increasing m
0 −1 32
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) x(t) is periodic in T = ω2π , therefore sin(m sin(ωm t)) is periodic in T . m
sin(m sin(ωm t)) 1 t
0 −1 m=0 |bk | 0
10
20
30 33
40
50
60
k
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) x(t) is periodic in T = ω2π , therefore sin(m sin(ωm t)) is periodic in T . m
sin(m sin(ωm t)) 1 t
0 −1 m=1 |bk | 0
10
20
30 34
40
50
60
k
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) x(t) is periodic in T = ω2π , therefore sin(m sin(ωm t)) is periodic in T . m
sin(m sin(ωm t)) 1 t
0 −1 m=2 |bk | 0
10
20
30 35
40
50
60
k
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) x(t) is periodic in T = ω2π , therefore sin(m sin(ωm t)) is periodic in T . m
sin(m sin(ωm t)) 1 t
0 −1 m=5 |bk | 0
10
20
30 36
40
50
60
k
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) x(t) is periodic in T = ω2π , therefore sin(m sin(ωm t)) is periodic in T . m
sin(m sin(ωm t)) 1 t
0 −1 m = 10 |bk | 0
10
20
30 37
40
50
60
k
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) x(t) is periodic in T = ω2π , therefore sin(m sin(ωm t)) is periodic in T . m
sin(m sin(ωm t)) 1 t
0 −1 m = 20 |bk | 0
10
20
30 38
40
50
60
k
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) x(t) is periodic in T = ω2π , therefore sin(m sin(ωm t)) is periodic in T . m
sin(m sin(ωm t)) 1 t
0 −1 m = 30 |bk | 0
10
20
30 39
40
50
60
k
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) x(t) is periodic in T = ω2π , therefore sin(m sin(ωm t)) is periodic in T . m
sin(m sin(ωm t)) 1 t
0 −1 m = 40 |bk | 0
10
20
30 40
40
50
60
k
Phase/Frequency Modulation
Find the Fourier transform of a PM/FM signal. y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) x(t) is periodic in T = ω2π , therefore sin(m sin(ωm t)) is periodic in T . m
sin(m sin(ωm t)) 1 t
0 −1 m = 50 |bk | 0
10
20
30 41
40
50
60
k
Phase/Frequency Modulation
Fourier transform of second part. x(t) = sin(ωm t) y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) ' v " ' v " ya (t)
yb (t)
|Yb (jω)|
m = 50
ω ωc
ωc 100ωm
42
Phase/Frequency Modulation
Fourier transform. x(t) = sin(ωm t) y(t) = cos(ωc t + mx(t)) = cos(ωc t + m sin(ωm t)) = cos(ωc t) cos(m sin(ωm t))) − sin(ωc t) sin(m sin(ωm t))) ' v " ' v " ya (t)
yb (t)
|Y (jω)|
m = 50
ω ωc
ωc 100ωm
43
Frequency Modulation
Wideband FM is useful because it is robust to noise. AM: y1 (t) = (cos(ωm t) + 1.1) cos(ωc t)
t
FM: y3 (t) = cos(ωc t + m sin(ωm t))
t
FM generates a redundant signal that is resilient to additive noise.
44
Summary
Modulation is useful for matching signals to media. Examples: commercial radio (AM and FM)
Close with unconventional application of modulation – in microscopy.
45
6.003 Microscopy
Dennis M. Freeman Stanley S. Hong Jekwan Ryu Michael S. Mermelstein Berthold K. P. Horn Courtesy of Stanley Hong, Jekwan Ryu, Michael Mermelstein, and Berthold K. P. Horn. Used with permission. 46
6.003 Model of a Microscope
microscope
Microscope = low-pass filter Courtesy of Stanley Hong, Jekwan Ryu, Michael Mermelstein, and Berthold K. P. Horn. Used with permission. 47
Phase-Modulated Microscopy
microscope
Courtesy of Stanley Hong, Jekwan Ryu, Michael Mermelstein, and Berthold K. P. Horn. Used with permission.
48
Courtesy of Stanley Hong, Jekwan Ryu, Michael Mermelstein, and Berthold K. P. Horn. Used with permission. 49
Courtesy of Stanley Hong, Jekwan Ryu, Michael Mermelstein, and Berthold K. P. Horn. Used with permission. 50
Courtesy of Stanley Hong, Jekwan Ryu, Michael Mermelstein, and Berthold K. P. Horn. Used with permission. 51
Courtesy of Stanley Hong, Jekwan Ryu, Michael Mermelstein, and Berthold K. P. Horn. Used with permission. 52
Courtesy of Stanley Hong, Jekwan Ryu, Michael Mermelstein, and Berthold K. P. Horn. Used with permission. 53
Courtesy of Stanley Hong, Jekwan Ryu, Michael Mermelstein, and Berthold K. P. Horn. Used with permission.
54
Courtesy of Stanley Hong, Jekwan Ryu, Michael Mermelstein, and Berthold K. P. Horn. Used with permission.
55
Courtesy of Stanley Hong, Jekwan Ryu, Michael Mermelstein, and Berthold K. P. Horn. Used with permission. 56
Courtesy of Stanley Hong, Jekwan Ryu, Michael Mermelstein, and Berthold K. P. Horn. Used with permission. 57
Courtesy of Stanley Hong, Jekwan Ryu, Michael Mermelstein, and Berthold K. P. Horn. Used with permission. 58
Courtesy of Stanley Hong, Jekwan Ryu, Michael Mermelstein, and Berthold K. P. Horn. Used with permission.
59
Courtesy of Stanley Hong, Jekwan Ryu, Michael Mermelstein, and Berthold K. P. Horn. Used with permission. 60
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6.003 Signals and Systems Fall 2011
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6.003: Signals and Systems From LPs to CDs – and how 6.003 helps get you there
December 8, 2011
1
Edison’s Phonograph
":;'
•
2
Edison’s Phonograph
3
Edison’s Phonograph
4
Edison’s Phonograph
5
Edison’s Phonograph
Image by Infrogmation on Wikimedia Commons. 6
Edison’s Phonograph
Photo of Pioneer record player removed due to copyright restrictions.
7
Edison’s Phonograph
8
Edison’s Phonograph
Photo of Grado phono cartridge removed due to copyright restrictions.
9
Edison’s Phonograph LPs: 100 years of optimization → good fidelity, but • fragile: easily scratched • lots of distortions: e.g., wow and flutter • expensive CDs: much higher fidelity • nearly indestructible • very low distortion • inexpensive → many of these advantages made possible by concepts from Signals and Systems!
10
Edison’s Phonograph
Image by Dante Alighieri on Wikimedia Commons.
11
What’s on a CD?
0.83-3.56 Jlm
Image by Dante Alighieri on Wikimedia Commons.
k"'" protective layer (plus label) ~~------~ +-- reflective layer (typically aluminum) ~~------~ +-- polycarbonate (injection molded) 12
Edison’s Phonograph LPs: 100 years of optimization → good fidelity, but • fragile: easily scratched • lots of distortions: e.g., wow and flutter • expensive CDs: much higher fidelity √ • nearly indestructible • very low distortion • inexpensive → many of these advantages made possible by concepts from Signals and Systems!
13
What’s on a CD?
0.83-3.56 Jlm Image by Dante Alighieri on Wikimedia Commons.
k"'" protective layer (plus label) ~~------~ +-- reflective layer (typically aluminum) ~~------~ +-- polycarbonate (injection molded) 14
What’s on a CD? Continuous signal (audio) Discrete storage (pits and lands) → sampling!
15
What’s on a CD? 120
•...•
Threshold (dB SPL)
100 80
Dadson & King, 1952 Yeowart, Brian, & Tempest, 1967 Green, Kidd, & Stevens, 1987
60 40 20 0
10
100
1000 10000 Frequency (Hz) 16
100000
What’s on a CD?
is = 44.1kHz Xm Ct)
Anti-aliasing CT filter
x Ct)
!
Sample and hold
17
x Ct)
Analog-todigital converter
x [n)
What’s on a CD?
Xm(f) Ideal anti-al iasi ng filter
-20
20 18
f (kHz)
What’s on a CD?
XU)
JMM IM M( J(\ 11 (\ 111 (\ (JI(\ l1i( o
44.1
f
(kHz)
Without anti-aliasing filter
X(!)
o
20 t 44.1 24.1
19
f
(kHz)
With ideal anti-aliasing filter
What’s on a CD?
---co
---
"0
"0
:::J
c 0> co
~
---0>
0 -20 -40 -60 -80 0
---
-500
«
-1000
"0
0>
1
C
10°
1
10 Frequency (kHz) 20
10
2
What’s on a CD?
jm
21
What’s on a CD?
---co
---
"0
"0
:::J
c 0> co
~
---0>
0 -20 -40 -60 -80 0
---
-500
«
-1000
"0
0>
1
C
10°
1
10 Frequency (kHz) 22
10
2
What’s on a CD?
Is =
176.4kHz
23
What’s on a CD?
0 ,
~
,
~
XU)
~ 0
24
176.4
f (kHz)
,
,
~
176.4
A f
(kHz)
What’s on a CD?
Q)
-
-g
-40
c
0)
~
~
-0)
-80 ~~~~~--~~~~~~~~~
0
r=~~~~-:~~-:~~
Q)
::s.-200 Q)
0)
~ -400 ~~~~~~~ ____~;::::::~~~ 2 1 3 10°
10 10 Frequency (kHz) 25
10
What’s on a CD?
Is =
176.4kHz
26
What’s on a CD?
20. r 24.1 kHz
o Q)
"0 ::::l
.~
0)
88.2 kHz.
-40 -80
~ -120 L-~~-!---+.L.....UJL...1...-~LJ.J..:..~'..LL..----1L~..L.l.l.':"'-
!-40 N "'v..---------O ~~~~~~~--~~~--~-
0)
c
«
-80
: -120 L-~~-'-'--~~~_ _~~~_ _~_
o
O.
0.2 0.3 0.4 DT frequency, Q /21t 27
0.5
What’s on a CD?
24 .1 176.4
..cfIII/J-......... / , /
1
t:: co
= 0.1366 20
176.4
= 0.1134
0..
>.
~o c
o
en co E
-1
-1
o
1 Real part 28
2
3
What’s on a CD?
0.2