Quiz III Review Signals and Systems 6 - MIT · Quiz III Review Signals and Systems 6.003 Massachusetts Institute of Technology April 26th, 2010 (Massachusetts Institute of Technology)Quiz

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


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


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


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


x(t) =1

2π

∫ ∞−∞

X (jω)e jωtdω


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


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


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)


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.


DT Frequency Response

• Similar to the continuous time case, complex geometrics areeigenfunctions of DT LTI systems


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.


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)− · · ·


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)

∗


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


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


x [n] = x [n + N] =∑<N>

ake j 2πNkn

• In the discrete Fourier series there are a finite number ofperiodic harmonics


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


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


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


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.


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


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)


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.


Fourier Relations


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


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.


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.


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.


End of Review

Good luck on Wednesday! :-)


Documents

Quiz III Review Signals and Systems 6 - MIT · Quiz III Review Signals and Systems 6.003 Massachusetts Institute of Technology April 26th, 2010 (Massachusetts Institute of Technology)Quiz