Quiz III Review
Signals and Systems
6.003
Massachusetts Institute of Technology
April 26th, 2010
(Massachusetts Institute of Technology) Quiz III Review April 26th, 2010 1 / 1
Quiz 3 Details
• Date: Wednesday April 28th, 2010
• Time: 7.30pm–9.30pm
• Where: 34-101
• Content: (boundaries inclusive)• Lectures 1–20• Recitations 1–20• Homeworks 1–11
(Massachusetts Institute of Technology) Quiz III Review April 26th, 2010 2 / 1
Review Outline
• CT Fourier Series
• CT Fourier Transforms
• DT Frequency Response
• DT Fourier Series
• DT Fourier Transforms
• Fourier Relations
• The Impulse Train and Periodic Extension
• Filters
(Massachusetts Institute of Technology) Quiz III Review April 26th, 2010 3 / 1
CT Fourier Series
• Periodic signals can be represented by a sum of harmonics
• The integral over one period of a harmonic is equal to zero,except for k=0. ∫
T
e jkω0tdt = T δ[k]
• The ”analysis” equation gives us the Fourier coefficients
ak =1
T
∫T
x(t)e−j2πTktdt
• The ”synthesis” equation reconstructs the periodic signal
x(t) = x(t + T ) =∞∑
k=−∞
ake j 2πTkt
(Massachusetts Institute of Technology) Quiz III Review April 26th, 2010 4 / 1
CT Fourier Transform
• The aperiodic extension: The Fourier series can be generalizedto an aperiodic signal by viewing the signal as a periodic signalwith an infinite period.
• The ”analysis” equation gives us the Fourier Transform
X (jω) =
∫ ∞−∞
x(t)e−jωtdt
• The ”synthesis” equation reconstructs the periodic signal
x(t) =1
2π
∫ ∞−∞
X (jω)e jωtdω
(Massachusetts Institute of Technology) Quiz III Review April 26th, 2010 5 / 1
Fourier and Laplace Transform Properties
X (jω) = X (s)|s=jω
6.003: Signals and Systems Lecture 16 April 6, 2010
2
Fourier Transform
As T →∞, discrete harmonic amplitudes → a continuum E(ω).
xT (t)
t−S ST
ak = 1T
T/2
−T/2xT (t)e−j
2πT ktdt = 1
T
S
−Se−j 2πT ktdt =
sin 2πkSTπk
= 2T
sinωSω
2 sinωSω
ω0 = 2π/T
ω = kω0 = k2πT
Tak
kω
limT→∞
Tak = limT→∞
T/2
−T/2x(t)e−jωtdt = 2
ωsinωS = E(ω)
Fourier Transform
As T →∞, synthesis sum → integral.
xT (t)
t−S ST
2 sinωSω
ω0 = 2π/T
ω = kω0 = k2πT
Tak
kω
limT→∞
Tak = limT→∞
T/2
−T/2x(t)e−jωtdt = 2
ωsinωS = E(ω)
x(t) =∞
k=−∞
1TE(ω)
ak
ej 2πT kt =
∞
k=−∞
ω02πE(ω)ejωt → 1
2π
∞
−∞E(ω)ejωtdω
Fourier Transform
Replacing E(ω) by X(jω) yields the Fourier transform relations.
E(ω) = X(s)|s=jω ≡ X(jω)
Fourier transform
X(jω)= ∞
−∞x(t)e−jωtdt (“analysis” equation)
x(t)= 12π
∞
−∞X(jω)ejωtdω (“synthesis” equation)
Form is similar to that of Fourier series
→ provides alternate view of signal.
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:
X(s) = ∞
−∞x(t)e−stdt
Fourier transform:
X(jω) = ∞
−∞x(t)e−jωtdt = H(s)|s=jω
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−st0X(s) e
−jωt0X(jω)
Time scale x(at) 1|a|X
s
a
1|a|X
jω
a
Differentiationdx(t)dt
sX(s) jωX(jω)
Multiply by t tx(t) − ddsX(s) −1
j
d
dωX(jω)
Convolution x1(t) ∗ x2(t) X1(s)×X2(s) X1(jω)×X2(jω)
Relation between Fourier and Laplace Transforms
There are also important differences.
Compare Fourier and Laplace transforms of x(t) = e−tu(t).
x(t)
t
Laplace transform
X(s) = ∞
−∞e−tu(t)e−stdt =
∞
0e−(s+1)t
dt = 11 + s ; Re(s) > −1
a complex-valued function of complex domain.
Fourier transform
X(jω) = ∞
−∞e−tu(t)e−jωtdt =
∞
0e−(jω+1)t
dt = 11 + jω
a complex-valued function of real domain.* The form of the Fourier transform and its inverse are very similar.We can therefore use duality to find new transform pairs.
(Massachusetts Institute of Technology) Quiz III Review April 26th, 2010 6 / 1
DT Frequency Response
• Similar to the continuous time case, complex geometrics areeigenfunctions of DT LTI systems
6.003: Signals and Systems Lecture 18 April 13, 2010
2
Signal Processing
Modern audio systems process sounds digitally.
A/D DT filter D/Ax(t) y(t)x[n] y[n]
Signal Processing
Modern audio systems process sounds digitally.
Texas Instruments TAS3004
• 2 channels
• 24 bit ADC, 24 bit DAC
• 48 kHz sampling rate
• 100 MIPS
• $7.70 ($4.81 in bulk)
Con
trolPWR_DN
TEST
AINRP
AINRM
AINLM
AINLP
24-BitStereoADC
RINA
RINB
AINRM
AINRP
AINLM
AINLP
LINA
LINB
Con
trolCS1
SDA
SCL
Con
trol
ler
GPI0
GPI1
GPI2
GPI3
GPI4
GPI5
24-BitStereo DAC
CA
P_P
LL
MC
LKX
TALO
MC
LKO
CLK
SE
L
SD
IN2
SD
IN1
SDATAControl
LRC
LK/O
SC
LK/O
SDOUT1
L
L+RSDOUT2
32-Bit Audio SignalProcessor
AOUTL
VCOM
AOUTR
L+R
R32-Bit Audio Signal
Processor
OSC/CLKSelect PLL
ReferenceVoltage
SuppliesAnalog
SuppliesDigital
IFM
/S
RESET
INPA
ALLPASS
XTA
LI/
Figure 1- 1. TAS3004 Block Diagram
AV
SS
(RE
F)
VR
FILT
AV
DD
AV
SS
VR
EFM
DV
DD
DV
SS
VR
EFP
I2C
ControlAnalog
FormatOutput
SDOUT0
LogicControl
Register
DT Fourier Series and Frequency Response
Today: frequency representations for DT signals and systems.
Complex Geometric Sequences
Complex geometric sequences are eigenfunctions of DT LTI systems.
Find response of DT LTI system (h[n]) to input x[n] = zn.
y[n] = (h ∗ x)[n] =∞
k=−∞h[k]zn−k = zn
∞
k=−∞h[k]z−k = H(z) zn .
Complex geometrics (DT): analogous to complex exponentials (CT)
h[n]zn H(z) zn
h(t)est H(s) est
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) = 2z−2 + 7z−1 + 8z−2 + 3z−1 + 4 = 2 + 7z + 8z2
1 + 3z + 4z2 ≡N(z)D(z)
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) · · ·
q0q0
z0 − q0z0
z-planez0
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.
6.003: Signals and Systems Lecture 18 April 13, 2010
3
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)− · · ·
DT Frequency Response
Response to eternal sinusoids.
Let x[n] = cos Ω0n (for all time):
x[n] = 12
ejΩ0n + e−jΩ0n
= 1
2
zn0 + zn1
where z0 = e jΩ0 and z1 = e−jΩ0.
The response to a sum is the sum of the responses:
y[n] = 12
H(z0) zn0 +H(z1) zn1
= 12
H(e jΩ0) e jΩ0n +H(e−jΩ0) e−jΩ0n
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:
H(z) =∞
n=−∞h[n]z−n
where h[n] is a real-valued function of n for physical systems.
H(e jΩ) =∞
n=−∞h[n]e−jΩn
H(e−jΩ) =∞
n=−∞h[n]e jΩn ≡
H(e jΩ)
∗
DT Frequency Response
Response to eternal sinusoids.
Let x[n] = cos Ω0n (for all time), which can be written as
x[n] = 12
ejΩ0n + e−jΩ0n
.
Then
y[n] = 12
H(e jΩ0)e jΩ0n +H(e−jΩ0)e−jΩ0n
= ReH(e jΩ0)e jΩ0n
= Re|H(e jΩ0)|e j∠H(e jΩ0)
ejΩ0n
= |H(e jΩ0)|ReejΩ0n+j∠H(e jΩ0)
y[n] =H(e jΩ0)
cos
Ω0n+ ∠H(e jΩ0)
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.
H(z)cos(Ωn) |H(e jΩ)| cos
Ωn+ ∠H(e jΩ)
H(e jΩ) = H(z)|z=e jΩ
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.
s-plane
σ
ω
−5 0 5
|H(jω)|
z-plane
−π 0 π
1
H(e jΩ)
• The DT frequency response is equivalent to the z-transformevaluated along the unit circle. It is periodic with period 2π.
H(e j(Ω+2πk)) = H(e jΩ)
• The ”highest” DT frequency is Ω = π
(Massachusetts Institute of Technology) Quiz III Review April 26th, 2010 7 / 1
DT Fourier Series
• Discrete time periodic signals can also be represented by a sumof harmonics
• The ”analysis” equation gives us the Fourier coefficients
ak =1
N
∑<N>
x [n]e−j2πNkn
• The ”synthesis” equation reconstructs the periodic signal
x [n] = x [n + N] =∑<N>
ake j 2πNkn
• In the discrete Fourier series there are a finite number ofperiodic harmonics
(Massachusetts Institute of Technology) Quiz III Review April 26th, 2010 8 / 1
DT Fourier TransformThe aperiodic extension of the discrete Fourier series.
X (e jΩ) =∞∑
n=−∞
x [n]e−jΩn
x [n] =1
2π
∫2π
X (e jΩ)e jΩndΩ
6.003: Signals and Systems Lecture 16 April 6, 2010
2
Fourier Transform
As T →∞, discrete harmonic amplitudes → a continuum E(ω).
xT (t)
t−S ST
ak = 1T
T/2
−T/2xT (t)e−j
2πT ktdt = 1
T
S
−Se−j 2πT ktdt =
sin 2πkSTπk
= 2T
sinωSω
2 sinωSω
ω0 = 2π/T
ω = kω0 = k2πT
Tak
kω
limT→∞
Tak = limT→∞
T/2
−T/2x(t)e−jωtdt = 2
ωsinωS = E(ω)
Fourier Transform
As T →∞, synthesis sum → integral.
xT (t)
t−S ST
2 sinωSω
ω0 = 2π/T
ω = kω0 = k2πT
Tak
kω
limT→∞
Tak = limT→∞
T/2
−T/2x(t)e−jωtdt = 2
ωsinωS = E(ω)
x(t) =∞
k=−∞
1TE(ω)
ak
ej 2πT kt =
∞
k=−∞
ω02πE(ω)ejωt → 1
2π
∞
−∞E(ω)ejωtdω
Fourier Transform
Replacing E(ω) by X(jω) yields the Fourier transform relations.
E(ω) = X(s)|s=jω ≡ X(jω)
Fourier transform
X(jω)= ∞
−∞x(t)e−jωtdt (“analysis” equation)
x(t)= 12π
∞
−∞X(jω)ejωtdω (“synthesis” equation)
Form is similar to that of Fourier series
→ provides alternate view of signal.
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:
X(s) = ∞
−∞x(t)e−stdt
Fourier transform:
X(jω) = ∞
−∞x(t)e−jωtdt = H(s)|s=jω
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−st0X(s) e
−jωt0X(jω)
Time scale x(at) 1|a|X
s
a
1|a|X
jω
a
Differentiationdx(t)dt
sX(s) jωX(jω)
Multiply by t tx(t) − ddsX(s) −1
j
d
dωX(jω)
Convolution x1(t) ∗ x2(t) X1(s)×X2(s) X1(jω)×X2(jω)
Relation between Fourier and Laplace Transforms
There are also important differences.
Compare Fourier and Laplace transforms of x(t) = e−tu(t).
x(t)
t
Laplace transform
X(s) = ∞
−∞e−tu(t)e−stdt =
∞
0e−(s+1)t
dt = 11 + s ; Re(s) > −1
a complex-valued function of complex domain.
Fourier transform
X(jω) = ∞
−∞e−tu(t)e−jωtdt =
∞
0e−(jω+1)t
dt = 11 + jω
a complex-valued function of real domain.
(Massachusetts Institute of Technology) Quiz III Review April 26th, 2010 9 / 1
Fourier Relations
6.003: Signals and Systems Lecture 20 April 22, 2010
6
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.
Ω
Xp(ejΩ)
−π4π4−2π 2π
π2
Ω
ak
−1 1−8 8
14
Relation between Fourier Transforms and Series
The effect of periodic extension was to sample the frequency repre-
sentation.
Ω
X(ejΩ)
2π−2π
2
Ω
ak
−1 1−8 8
14
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
aperiodic DT
DTFT
periodic CT
CTFS
aperiodic CT
CTFT
N →∞
(sampling in frequency)periodic extension
T →∞
periodic extension
interpolate sample interpolate sample
Relations among Fourier Representations
Different Fourier representations are related because they apply to
signals that are related.
DTFS (discrete-time Fourier series): periodic DT
DTFT (discrete-time Fourier transform): aperiodic DT
CTFS (continuous-time Fourier series): periodic CT
CTFT (continuous-time Fourier transform): aperiodic CT
periodic DT
DTFS
aperiodic DT
DTFT
periodic CT
CTFS
aperiodic CT
CTFT
N →∞
periodic extension
T →∞
periodic extension
interpolate sample interpolate sample
(Massachusetts Institute of Technology) Quiz III Review April 26th, 2010 10 / 1
Impulse Trains and Periodic Extension
• Periodic extension can be accomplished by convolving a signalwith an impulse train. This is equivalent to multiplying by animpulse train in frequency.
6.003: Signals and Systems Lecture 17 April 8, 2010
5
Fourier Transforms in Physics: Crystallography
The phase of light scattered from different parts of the target un-
dergo different amounts of phase delay.
θx sin θ
x
Phase at a point x is delayed (i.e., negative) relative to that at 0:
φ = −2πx sin θλ
Fourier Transforms in Physics: Crystallography
Total light F (θ) at angle θ is the integral of amount scattered from
each part of the target (f(x)) appropriately shifted in phase.
F (θ) =f(x)e−j2π
x sin θλ dx
Assume small angles so sin θ ≈ θ.
Let ω = 2π θλ .
Then the pattern of light at the detector is
F (ω) =f(x)e−jωxdx
which is the Fourier transform of f(x) !
Fourier Transforms in Physics: Diffraction
There is a Fourier transform relation between this structure and the
far-field intensity pattern.
· · ·· · ·
grating ≈ impulse train with pitch D
t0 D
· · ·· · ·
far-field intensity ≈ impulse train with reciprocal pitch ∝ λD
ω0 2πD
Impulse Train
The Fourier transform of an impulse train is an impulse train.
· · ·· · ·
x(t) =∞
k=−∞δ(t− kT )
t0 T
1
· · ·· · ·
ak = 1T ∀ k
k
1T
· · ·· · ·
X(jω) =∞
k=−∞
2πTδ(ω − k2π
T)
ω0 2πT
2πT
Two Dimensions
Demonstration: 2D grating.
An Historic Fourier Transform
Taken by Rosalind Franklin, this image sparked Watson and Crick’s
insight into the double helix.
(Massachusetts Institute of Technology) Quiz III Review April 26th, 2010 11 / 1
Filters
• Fourier representations allow us to think of systems as filters
• LTI systems cannot create new frequencies
• LTI systems scale the magnitude and shift the phase of existingfrequency components.
(Massachusetts Institute of Technology) Quiz III Review April 26th, 2010 12 / 1
End of Review
Good luck on Wednesday! :-)
(Massachusetts Institute of Technology) Quiz III Review April 26th, 2010 13 / 1