Signal Processing First Lecture Slides

Signal Processing First

LECTURE #1Sinusoids

4/3/2006 © 2003-2006, JH McClellan & RW Schafer 3

READING ASSIGNMENTS

This Lecture:Chapter 2, pp. 9-17

Appendix A: Complex NumbersAppendix B: MATLABChapter 1: Introduction


CONVERGING FIELDS

EECmpE

Math

Applications

Physics

ComputerScience

BIO4/3/2006 © 2003-2006, JH McClellan & RW Schafer 5

COURSE OBJECTIVE

Students will be able to:Understand mathematical descriptions of signal processing algorithms and express those algorithms as computer implementations (MATLAB)

What are your objectives?


WHY USE DSP ?

Mathematical abstractions lead to generalization and discovery of new processing techniques

Computer implementations are flexible

Applications provide a physical context


Fourier Everywhere

TelecommunicationsSound & Music

CDROM, Digital VideoFourier OpticsX-ray Crystallography

Protein Structure & DNAComputerized TomographyNuclear Magnetic Resonance: MRIRadioastronomyRef: Prestini, “The Evolution of Applied Harmonic Analysis”


LECTURE OBJECTIVES

Write general formula for a “sinusoidal”waveform, or signalFrom the formula, plot the sinusoid versus time

What’s a signal?It’s a function of time, x(t)in the mathematical sense


TUNING FORK EXAMPLE

CD-ROM demo“A” is at 440 Hertz (Hz)Waveform is a SINUSOIDAL SIGNALComputer plot looks like a sine waveThis should be the mathematical formula:

))440(2cos( ϕπ +tA


TUNING FORK A-440 Waveform

ms 3.285.515.8

=−≈T

Hz4353.2/1000

/1

≈== Tf

Time (sec)


SPEECH EXAMPLE

More complicated signal (BAT.WAV)Waveform x(t) is NOT a SinusoidTheory will tell us

x(t) is approximately a sum of sinusoidsFOURIER ANALYSIS

Break x(t) into its sinusoidal componentsCalled the FREQUENCY SPECTRUM


Speech Signal: BAT

Nearly PeriodicPeriodic in Vowel RegionPeriod is (Approximately) T = 0.0065 sec


DIGITIZE the WAVEFORM

x[n] is a SAMPLED SINUSOIDA list of numbers stored in memory

Sample at 11,025 samples per secondCalled the SAMPLING RATE of the A/DTime between samples is

1/11025 = 90.7 microsec

Output via D/A hardware (at Fsamp)


STORING DIGITAL SOUND


CD rate is 44,100 samples per second16-bit samplesStereo uses 2 channelsNumber of bytes for 1 minute is

2 X (16/8) X 60 X 44100 = 10.584 Mbytes


Always use the COSINE FORM

Sine is a special case:

))440(2cos( ϕπ +tA

SINES and COSINES

)cos()sin( 2πωω −= tt


SINUSOIDAL SIGNAL

FREQUENCYRadians/secHertz (cycles/sec)

PERIOD (in sec)

AMPLITUDEMagnitude

PHASE

A tcos( )ω ϕ+ω A

ϕω π= ( )2 f

Tf

= =1 2π

ω 4/3/2006 © 2003-2006, JH McClellan & RW Schafer 17

EXAMPLE of SINUSOID

Given the Formula

Make a plot)2.13.0cos(5 ππ +t


PLOT COSINE SIGNAL

Formula defines A, ω, and φ5 0 3 12cos( . . )π πt+

A = 5ω = 0.3πϕ = 1.2π


PLOTTING COSINE SIGNAL from the FORMULA

Determine period:

Determine a peak location by solving

Zero crossing is T/4 before or afterPositive & Negative peaks spaced by T/2

)2.13.0cos(5 ππ +t

3/203.0/2/2 === ππωπT

0)2.13.0(0)( =+⇒=+ ππϕω tt


PLOT the SINUSOID

Use T=20/3 and the peak location at t=-4

)2.13.0cos(5 ππ +t

→← 320

8/22/2003 © 2003, JH McClellan & RW Schafer 1


LECTURE #2Phase & Time-ShiftComplex Exponentials


READING ASSIGNMENTS

This Lecture:Chapter 2, Sects. 2-3 to 2-5

Appendix A: Complex NumbersAppendix B: MATLABNext Lecture: finish Chap. 2,

Section 2-6 to end


LECTURE OBJECTIVES

Define Sinusoid Formula from a plotRelate TIME-SHIFT to PHASE

tjXetz ω=)(

Introduce an ABSTRACTION:Complex Numbers represent SinusoidsComplex Exponential Signal


SINUSOIDAL SIGNAL

FREQUENCYRadians/secor, Hertz (cycles/sec)

PERIOD (in sec)

AMPLITUDEMagnitude

PHASE

)cos( ϕω +tA

f)2( πω =

ωπ21 ==

fT

A

ϕ

ω


PLOTTING COSINE SIGNAL from the FORMULA

Determine period:

Determine a peak location by solving

Peak at t=-4

)2.13.0cos(5 ππ +t

0)( =+ ϕω t

3/203.0/2/2 === ππωπT


ANSWER for the PLOT

Use T=20/3 and the peak location at t=-4

)2.13.0cos(5 ππ +t

→← 320


TIME-SHIFT

In a mathematical formula we can replace t with t-tm

Then the t=0 point moves to t=tm

Peak value of cos(ω(t-tm)) is now at t=tm

))(cos()( mm ttAttx −=− ω


TIME-SHIFTED SINUSOID

))4((3.0cos(5))4(3.0cos(5)4( −−=+=+ tttx ππ


PHASE <--> TIME-SHIFT

Equate the formulas:

and we obtain:

or,

)cos())(cos( ϕωω +=− tAttA m

ϕω =− mt

ωϕ−=mt


SINUSOID from a PLOT

Measure the period, TBetween peaks or zero crossings

Compute frequency: ωωωω = 2π/T

Measure time of a peak: tm

Compute phase: φφφφ = -ω tm

Measure height of positive peak: A

3 steps


(A, ω, φ) from a PLOT

ππωϕ 25.0))(200( =−=−= mm tt

πω ππ 20001.022 === T100

1period1

sec01.0 ==T

sec00125.0−=mt8/22/2003 © 2003, JH McClellan & RW Schafer 13

SINE DRILL (MATLAB GUI)


Ttt

t

mm

m

−=−==

=−+−

−

ωπ

ωϕ

ωπϕ

ωϕ

2)2(2

then, if

PHASE is AMBIGUOUS

The cosine signal is periodic Period is 2π

Thus adding any multiple of 2π leaves x(t) unchanged

)cos()2cos( ϕωπϕω +=++ tAtA


COMPLEX NUMBERS

To solve: z2 = -1z = jMath and Physics use z = i

Complex number: z = x + j y

x

y zCartesiancoordinatesystem


PLOT COMPLEX NUMBERS


COMPLEX ADDITION = VECTORVECTORVECTORVECTOR Addition

26)53()24()52()34(

213

jj

jjzzz

+=+−++=

++−=+=


*** POLAR FORM ***

Vector FormLength =1Angle = θ

Common Valuesj has angle of 0.5π−1 has angle of π−−−− j has angle of 1.5π also, angle of −−−−j could be −0.5π = 1.5π −2πbecause the PHASE is AMBIGUOUS


POLAR <--> RECTANGULAR

Relate (x,y) to (r,θ) rθx

y

Need a notation for POLAR FORM

( )xyyxr

1

222

Tan−=

+=

θ

θθ

sincosryrx

==

Most calculators doPolar-Rectangular


Euler’s FORMULA

Complex ExponentialReal part is cosineImaginary part is sineMagnitude is one

)sin()cos( θθθ jrrre j +=

)sin()cos( θθθ je j +=


COMPLEX EXPONENTIAL

Interpret this as a Rotating Vectorθ = ωθ = ωθ = ωθ = ωtAngle changes vs. timeex: ω=20π rad/sRotates 0.2π in 0.01 secs

)sin()cos( tjte tj ωωω +=

)sin()cos( θθθ je j +=


cos = REAL PART

Real Part of Euler’s}{)cos( tjeet ωω ℜ=

General Sinusoid )cos()( ϕω += tAtx

So,

}{}{)cos( )(

tjj

tj

eAeeAeetA

ωϕ

ϕωϕωℜ=ℜ=+ +


REAL PART EXAMPLE

Answer:

Evaluate:

{ }tjj eAeetA ωϕϕω ℜ=+ )cos(

{ }tjjeetx ω3)( −ℜ=

{ }{ } )5.0cos(33

)3()(5.0 πωωπ

ω

−=ℜ=−ℜ=

− teeeejetxtjj

tj


COMPLEX AMPLITUDE

Then, any Sinusoid = REAL PART of Xejωt

{ } { }tjjtj eAeeXeetx ωϕω ℜ=ℜ=)(

General Sinusoid

{ }tjj eAeetAtx ωϕϕω ℜ=+= )cos()(Complex AMPLITUDE = XComplex AMPLITUDE = X

ϕω jtj AeXXetz ==)(



LECTURE #3Phasor Addition Theorem


READING ASSIGNMENTS

This Lecture:Chapter 2, Section 2-6

Other Reading:Appendix A: Complex NumbersAppendix B: MATLABNext Lecture: start Chapter 3


LECTURE OBJECTIVES

Phasors = Complex AmplitudeComplex Numbers represent Sinusoids

Develop the ABSTRACTION:Adding Sinusoids = Complex AdditionPHASOR ADDITION THEOREMPHASOR ADDITION THEOREM

tjjtj eAeXetz ωϕω )()( ==


Z DRILL (Complex Arith)


AVOID Trigonometry

Algebra, even complex, is EASIER !!!Can you recall cos(θ1+θ2) ?

Use: real part of ej(θ1+θ2) = cos(θ1+θ2)

2121 )( θθθθ jjj eee =+

)sin)(cossin(cos 2211 θθθθ jj ++=

(...))sinsincos(cos 2121 j+−= θθθθ1/12/2004 © 2003, JH McClellan & RW Schafer 7

Euler’s FORMULA

Complex ExponentialReal part is cosineImaginary part is sineMagnitude is one

)sin()cos( tjte tj ωωω +=)sin()cos( θθθ je j +=


Real & Imaginary Part Plots

PHASE DIFFERENCE = ππππ/2


COMPLEX EXPONENTIAL

Interpret this as a Rotating Vectorθ = ωθ = ωθ = ωθ = ωtAngle changes vs. timeex: ω=20π rad/sRotates 0.2π in 0.01 secs

e jjθ θ θ= +cos( ) sin( )

)sin()cos( tjte tj ωωω +=


Rotating Phasor

See Demo on CD-ROMChapter 2


Cos = REAL PART

cos(ωt) = ℜe e jω t{ }Real Part of Euler’s

x(t) = Acos(ωt +ϕ )General Sinusoid

A cos(ω t +ϕ) = ℜe Ae j (ω t+ϕ ){ }= ℜe Ae jϕe jω t{ }

So,


COMPLEX AMPLITUDE

x(t) = Acos(ωt +ϕ ) = ℜe Ae jϕejω t{ }General Sinusoid

z( t) = Xe jωt X = Ae jϕComplex AMPLITUDE = XComplex AMPLITUDE = X

x(t) = ℜe Xe jω t{ }= ℜe z(t){ }Sinusoid = REAL PART of (Aejφ)ejωt


POP QUIZ: Complex AmpFind the COMPLEX AMPLITUDE for:

Use EULER’s FORMULA:

π5.03 jeX =

)5.077cos(3)( ππ += ttx

{ }{ }tjj

tj

eee

eetxππ

ππ

775.0

)5.077(

3

3)(

ℜ=

ℜ= +


WANT to ADD SINUSOIDS

ALL SINUSOIDS have SAME FREQUENCYHOW to GET {Amp,Phase} of RESULT ?


ADD SINUSOIDS

Sum Sinusoid has SAMESAME Frequency


PHASOR ADDITION RULE

Get the new complex amplitude by complex addition


Phasor Addition Proof


POP QUIZ: Add Sinusoids

ADD THESE 2 SINUSOIDS:

COMPLEX ADDITION:

π5.00 31 jj ee +

)5.077cos(3)(

)77cos()(

2

1

ππ

π

+=

=

ttx

ttx


POP QUIZ (answer)

COMPLEX ADDITION:

CONVERT back to cosine form:

j 3 = 3e j0.5π

1

31 j+ 3/231 πjej =+

)77cos(2)( 33ππ += ttx


ADD SINUSOIDS EXAMPLE

tm1

tm2

tm3

)()()( 213 txtxtx +=

)(1 tx

)(2 tx


Convert Time-Shift to Phase

Measure peak times:tm1=-0.0194, tm2=-0.0556, tm3=-0.0394

Convert to phase (T=0.1)φ1=-ωωωωtm1 = -2ππππ(tm1 /T) = 70π/180, φ2= 200π/180

AmplitudesA1=1.7, A2=1.9, A3=1.532


Phasor Add: Numerical

Convert Polar to CartesianX1 = 0.5814 + j1.597X2 = -1.785 - j0.6498sum =

X3 = -1.204 + j0.9476Convert back to Polar

X3 = 1.532 at angle 141.79π/180This is the sum


ADD SINUSOIDS

VECTOR(PHASOR)ADD

X1

X2

X3



Lecture 4Spectrum Representation


READING ASSIGNMENTS

This Lecture:Chapter 3, Section 3-1

Other Reading:Appendix A: Complex Numbers

Next Lecture: Ch 3, Sects 3-2, 3-3, 3-7 & 3-8


LECTURE OBJECTIVES

Sinusoids with DIFFERENT FrequenciesSYNTHESIZE by Adding Sinusoids

SPECTRUM RepresentationGraphical Form shows DIFFERENTDIFFERENT Freqs

∑=

+=N

kkkk tfAtx

1)2cos()( ϕπ


FREQUENCY DIAGRAM

Plot Complex Amplitude vs. Freq

0 100 250–100–250 f (in Hz)

3/7 πje 3/7 πje−2/4 πje− 2/4 πje

10


Another FREQ. DiagramFr

eque

ncy

is th

e ve

rtic

al a

xis

Time is the horizontal axis

A-440


MOTIVATION

Synthesize Complicated SignalsMusical Notes

Piano uses 3 strings for many notesChords: play several notes simultaneously

Human SpeechVowels have dominant frequenciesApplication: computer generated speech

Can all signals be generated this way?Sum of sinusoids?


Fur Elise WAVEFORM

BeatNotes


Speech Signal: BAT

Nearly PeriodicPeriodic in Vowel RegionPeriod is (Approximately) T = 0.0065 sec


Euler’s Formula Reversed

Solve for cosine (or sine))sin()cos( tjte tj ωωω +=

)sin()cos( tjte tj ωωω −+−=−

)sin()cos( tjte tj ωωω −=−

)cos(2 tee tjtj ωωω =+ −

)()cos( 21 tjtj eet ωωω −+=


INVERSE Euler’s Formula

Solve for cosine (or sine)

)()cos( 21 tjtj eet ωωω −+=

)()sin( 21 tjtjj eet ωωω −−=


SPECTRUM Interpretation

Cosine = sum of 2 complex exponentials:

One has a positive frequencyThe other has negative freq.Amplitude of each is half as big

tjAtjA eetA 72

72)7cos( −+=


NEGATIVE FREQUENCY

Is negative frequency real?Doppler Radar provides an example

Police radar measures speed by using the Doppler shift principleLet’s assume 400Hz 60 mph+400Hz means towards the radar-400Hz means away (opposite direction)Think of a train whistle


SPECTRUM of SINE

Sine = sum of 2 complex exponentials:

Positive freq. has phase = -0.5πNegative freq. has phase = +0.5π

tjjtjj

tjjAtj

jA

eAeeAe

eetA75.0

2175.0

21

72

72)7sin(

−−

−

+=

−=ππ

π5.01 jj ej ==−


GRAPHICAL SPECTRUMEXAMPLE of SINE

AMPLITUDE, PHASE & FREQUENCY are shown

ωωωω7-7 0

tjjtjj eAeeAetA 75.02175.0

21)7sin( −− += ππ

π5.021 )( jeA π5.0

21 )( jeA −


SPECTRUM ---> SINUSOID

Add the spectrum components:

What is the formula for the signal x(t)?

0 100 250–100–250 f (in Hz)

3/7 πje 3/7 πje−2/4 πje− 2/4 πje

10


Gather (A,ω,φω,φω,φω,φ) information

Frequencies:-250 Hz-100 Hz0 Hz100 Hz250 Hz

Amplitude & Phase4 -π/27 +π/310 07 -π/34 +π/2

DC is another name for zero-freq componentDC component always has φ=0 φ=0 φ=0 φ=0 or π π π π (for real x(t) )

Note the conjugate phase


Add Spectrum Components-1

Amplitude & Phase4 -ππππ/27 +ππππ/310 07 -ππππ/34 +ππππ/2

Frequencies:-250 Hz-100 Hz0 Hz100 Hz250 Hz

tjjtjj

tjjtjj

eeeeeeee

tx

)250(22/)250(22/

)100(23/)100(23/

4477

10)(

ππππ

ππππ

−−

−−

++

+=


Add Spectrum Components-2

tjjtjj

tjjtjj

eeeeeeee

tx

)250(22/)250(22/

)100(23/)100(23/

4477

10)(

ππππ

ππππ

−−

−−

++

+=

0 100 250–100–250 f (in Hz)

3/7 πje 3/7 πje−2/4 πje− 2/4 πje

10


Use Euler’s Formula to get REAL sinusoids:

Simplify Components

tjjtjj

tjjtjj

eeeeeeee

tx

)250(22/)250(22/

)100(23/)100(23/

4477

10)(

ππππ

ππππ

−−

−−

++

+=

tjjtjj eAeeAetA ωϕωϕϕω −−+=+ 21

21)cos(


FINAL ANSWER

So, we get the general form:

∑=

++=N

kkkk tfAAtx

10 )2cos()( ϕπ

)2/)250(2cos(8)3/)100(2cos(1410)(

ππππ

++−+=tttx


Summary: GENERAL FORM

∗+=ℜ zzze 21

21}{ k

jkk

feAX k

==

Frequency

ϕ{ }∑

=ℜ+=

N

k

tfjk

keXeXtx1

20)( π

{ }∑=

−∗++=N

k

tfjk

tfjk

kk eXeXXtx1

2212

21

0)( ππ

∑=

++=N

kkkk tfAAtx

10 )2cos()( ϕπ


Example: Synthetic Vowel

Sum of 5 Frequency Components


SPECTRUM of VOWEL

Note: Spectrum has 0.5Xk (except XDC)Conjugates in negative frequency


SPECTRUM of VOWEL (Polar Format)

φφφφk

0.5Ak


Vowel Waveform(sum of all 5 components)



Lecture 5Periodic Signals, Harmonics & Time-Varying Sinusoids


READING ASSIGNMENTS

This Lecture:Chapter 3, Sections 3-2 and 3-3Chapter 3, Sections 3-7 and 3-8

Next Lecture:Fourier Series ANALYSISFourier Series ANALYSISSections 3-4, 3-5 and 3-6


Problem Solving Skills

Math FormulaSum of CosinesAmp, Freq, Phase

Recorded SignalsSpeechMusicNo simple formula

Plot & SketchesS(t) versus tSpectrum

MATLABNumericalComputationPlotting list of numbers


LECTURE OBJECTIVESSignals with HARMONICHARMONIC Frequencies

Add Sinusoids with fk = kf0

FREQUENCY can change vs. TIMEChirps:

Introduce Spectrogram Visualization (specgram.m) (plotspec.m)

x(t) = cos(αt2 )

∑=

++=N

kkk tkfAAtx

100 )2cos()( ϕπ


SPECTRUM DIAGRAM

Recall Complex Amplitude vs. Freq

kk aX =21

0 100 250–100–250 f (in Hz)

3/7 πje 3/7 πje−2/4 πje− 2/4 πje

10

)2/)250(2cos(8)3/)100(2cos(1410)(

ππππ

++−+=

tttx

kjkk eAX ϕ=

∗kX2

1


SPECTRUM for PERIODIC ?

Nearly Periodic in the Vowel RegionPeriod is (Approximately) T = 0.0065 sec


PERIODIC SIGNALS

Repeat every T secsDefinition

Example:

Speech can be “quasi-periodic”

)()( Ttxtx +=

)3(cos)( 2 ttx =?=T

3π=T3

2π=T


Period of Complex Exponential

Definition: Period is T

k = integer

tjTtj ee ωω =+ )(

?)()()(

txTtxetx tj

=+= ω

12 =kje π

kTe Tj πωω 21 =⇒=⇒

kkTT

k0

22 ωππω =⎟⎠⎞

⎜⎝⎛==


{ }∑

∑

=

−∗

=

++=

=

++=

N

k

tkfjk

tkfjk

jkk

N

kkk

eXeXXtx

eAX

tkfAAtx

k

1

2212

21

0

100

00)(

)2cos()(

ππ

ϕ

ϕπ

Harmonic Signal Spectrum

0:haveonly can signal Periodic fkfk =

Tf 10 =


Define FUNDAMENTAL FREQ

00

1T

f =

(shortest) Periodlfundamenta(largest) Frequencylfundamenta

)2(

)2cos()(

0

0

000

100

==

==

++= ∑=

Tf

ffkf

tkfAAtx

k

N

kkk

πω

ϕπ


What is the fundamental frequency?

Harmonic Signal (3 Freqs)

3rd5th

10 Hz


POP QUIZ: FUNDAMENTAL

Here’s another spectrum:


100 Hz ? 50 Hz ?

0 100 250–100–250 f (in Hz)

3/7 πje 3/7 πje−2/4 πje− 2/4 πje

10


SPECIAL RELATIONSHIPto get a PERIODIC SIGNAL

IRRATIONAL SPECTRUM



T=0.1


NON-Harmonic Signal

NOTPERIODIC 1/28/2005 © 2003, JH McClellan & RW Schafer 17

FREQUENCY ANALYSIS

Now, a much HARDER problemNow, a much HARDER problemGiven a recording of a song, have the computer write the music

Can a machine extract frequencies?Yes, if we COMPUTE the spectrum for x(t)

During short intervals


Time-Varying FREQUENCIES Diagram

Freq

uenc

y is

the

vert

ical

axi

s

Time is the horizontal axis

A-440


SIMPLE TEST SIGNALC-major SCALE: stepped frequencies

Frequency is constant for each note

IDEAL


R-rated: ADULTS ONLY

SPECTROGRAM ToolMATLAB function is specgram.mSP-First has plotspec.m & spectgr.m

ANALYSIS programTakes x(t) as input & Produces spectrum values Xk

Breaks x(t) into SHORT TIME SEGMENTSThen uses the FFT (Fast Fourier Transform)


SPECTROGRAM EXAMPLETwo Constant Frequencies: Beats

))12(2sin())660(2cos( tt ππ


( ) ( )tjtjj

tjtj eeee )12(2)12(221)660(2)660(2

21 ππππ −− −+

AM Radio SignalSame as BEAT Notes

))12(2sin())660(2cos( tt ππ

))648(2cos())672(2cos( 221

221 ππ ππ ++− tt

( )tjtjtjtjj eeee )648(2)648(2)672(2)672(2

41 ππππ −− +−−


SPECTRUM of AM (Beat)

4 complex exponentials in AM:


648 Hz ? 24 Hz ?

0 648 672 f (in Hz)–672 –648

2/41 πje 2/

41 πje−2/

41 πje−2/

41 πje


STEPPED FREQUENCIESC-major SCALE: successive sinusoids

Frequency is constant for each note

IDEAL


SPECTROGRAM of C-Scale

ARTIFACTS at Transitions

Sinusoids ONLY

From SPECGRAMANALYSIS PROGRAM


Spectrogram of LAB SONG

ARTIFACTS at Transitions

Sinusoids ONLY Analysis Frame = 40ms


Time-Varying Frequency

Frequency can change vs. timeContinuously, not stepped

FREQUENCY MODULATION (FM)FREQUENCY MODULATION (FM)

CHIRP SIGNALSLinear Frequency Modulation (LFM)

))(2cos()( tvtftx c += πVOICE


)2cos()( 02 ϕπα ++= tftAtx

New Signal: Linear FM

Called Chirp Signals (LFM)Quadratic phase

Freq will change LINEARLY vs. timeExample of Frequency Modulation (FM)Define “instantaneous frequency”

QUADRATIC


INSTANTANEOUS FREQDefinition

For Sinusoid:

Derivativeof the “Angle”)()(

))(cos()(tttAtx

dtd

i ψωψ

=⇒=

Makes sense

0

0

0

2)()(2)(

)2cos()(

ftttft

tfAtx

dtd

i πψωϕπψ

ϕπ

==⇒+=

+=


INSTANTANEOUS FREQof the Chirp

Chirp Signals have Quadratic phaseFreq will change LINEARLY vs. time

ϕβαψϕβα

++=⇒

++=

tttttAtx

2

2

)()cos()(

βαψω +==⇒ ttt dtd

i 2)()(1/28/2005 © 2003, JH McClellan & RW Schafer 31

CHIRP SPECTROGRAM


CHIRP WAVEFORM


OTHER CHIRPS

ψ(t) can be anything:

ψ(t) could be speech or music:FM radio broadcast

))cos(cos()( ϕβα += tAtx

)sin()()( ttt dtd

i βαβψω −==⇒


SINE-WAVE FREQUENCY MODULATION (FM)

Look at CD-ROM Demos in Ch 3



Lecture 6Fourier Series Coefficients


READING ASSIGNMENTS

This Lecture:Fourier Series in Ch 3, Sects 3Fourier Series in Ch 3, Sects 3--4, 34, 3--5 & 35 & 3--66

Replaces pp. 62-66 in Ch 3 in DSP FirstNotation: ak for Fourier Series

Other Reading:Next Lecture: More Fourier Series


LECTURE OBJECTIVES

Work with the Fourier Series Integral

ANALYSISANALYSIS via Fourier SeriesFor PERIODIC signals: x(t+T0) = x(t)

Later: spectrum from the Fourier Series

∫ −=0

0

0

0

)/2(1 )(T

dtetxa tTkjTk

π


HISTORY

Jean Baptiste Joseph Fourier1807 thesis (memoir)

On the Propagation of Heat in Solid BodiesHeat !Napoleonic era

http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Fourier.html

9/8/2003 © 2003, JH McClellan & RW Schafer 6 9/8/2003 © 2003, JH McClellan & RW Schafer 7

0 100 250–100–250 f (in Hz)

3/7 πje 3/7 πje−2/4 πje− 2/4 πje

10

SPECTRUM DIAGRAM


kk aX =21

Xk = Akejϕ k

12 Xk

*

{ }∑=

−∗++=N

k

tfjk

tfjk

kk eXeXXtx1

2212

21

0)( ππka{ *

ka0a


Harmonic Signal

PERIOD/FREQUENCY of COMPLEX EXPONENTIAL:

tfkj

kkeatx 02)( π∑

∞

−∞==

( )0

00

001or22f

TT

f === πωπ


kjkk

N

kkk

eAX

tkfAAtx

ϕ

ϕπ

=

++= ∑=1

00 )2cos()(

kjkkk eAXa ϕ

21

21 ==

Fourier Series Synthesis

COMPLEXAMPLITUDE

tfkj

kkeatx 02)( π∑

∞

−∞==



T = 0.1

a3 a5

a1


SYNTHESIS vs. ANALYSIS

SYNTHESISEasyGiven (ωk,Ak,φk) create x(t)

Synthesis can be HARD

Synthesize Speech so that it sounds good

ANALYSISHardGiven x(t), extract (ωωωωk,Ak,φk) How many?Need algorithm for computer


STRATEGY: x(t) ak

ANALYSISGet representation from the signalWorks for PERIODICPERIODIC Signals

Fourier SeriesAnswer is: an INTEGRAL over one period

∫ −=0

0

0

0)(1

T

dtetxa tkjTk

ω


INTEGRAL Property of exp(j)

INTEGRATE over ONE PERIOD

0

)1(2

2

0

0

0

0

0

0

0

)/2(

20

0

)/2(0

0

)/2(

=

−−

=

−=

∫

∫

−

−

−−

TmtTj

mj

TmtTj

TmtTj

dte

emj

T

emj

Tdte

π

π

ππ

π

π

0≠m 00

2Tπω =


ORTHOGONALITY of exp(j)

PRODUCT of exp(+j ) and exp(-j )

=

≠=∫ −

k

kdtee

T

TktTjtTj

1

01 0

00

0

)/2()/2(

0

ππ

∫ −0

0

0

))(/2(

0

1 TtkTj dte

Tπ


Isolate One FS Coefficient

∫

∫∑∫

∫ ∑∫

∑

−

−∞

−∞=

−

−∞

−∞=

−

∞

−∞=

=⇒

=

=

=

=

0

0

0

0

00

0

0

0

0

0

00

0

0

0

0

0

0

)/2(1

0

)/2()/2(1

0

)/2(1

0

)/2()/2(1

0

)/2(1

)/2(

)(

)(

)(

)(

TtkTj

Tk

TtTjtkTj

Tk

k

TtTj

T

TtTjtkTj

kkT

TtTj

T

tkTj

kk

dtetxa

adteeadtetx

dteeadtetx

eatx

π

πππ

πππ

π

=kfor except zero is Integral


SQUARE WAVE EXAMPLE

0–.02 .02 0.04

1

t

x(t)

.01

sec. 04.0for 0

01)(

0

0021

021

=

<≤

<≤=

TTtT

Tttx


FS for a SQUARE WAVE {ak}

)0()(1 0

0

0

)/2(

0≠= ∫ − kdtetx

Ta

TktTj

kπ

02.

0)04./2(

)04./2(04.1

02.

0

)04./2(104.1 ktj

kjktj

k edtea ππ

π −−

− == ∫

kje

kj

kkj

πππ

2)1(1)1(

)2(1 )( −−=−

−= −


DC Coefficient: a0

)0()(1 0

0

0

)/2(

0== ∫ − kdtetx

Ta

TktTj

kπ

)Area(1)(1

0000

0

Tdttx

Ta

T

== ∫

21

02.

00 )002(.

04.11

04.1 =−== ∫ dta


Fourier Coefficients ak

ak is a function of kComplex Amplitude for k-th HarmonicThis one doesn’t depend on the period, T0

=

±±=

±±=

=−−=

0

,4,20

,3,11

2)1(1

21 k

k

kkj

kja

k

k …

…π

π


Spectrum from Fourier Series

=

±±=

±±=−

=

0

,4,20

,3,1

21 k

k

kkj

ak …

…π)25(2)04.0/(20 ππω ==


00 /1 FrequencylFundamenta Tf =

Fourier Series IntegralHOW do you determine ak from x(t) ?

component) (DC)(

)(

0

0

0

0

0

0

10

0

)/2(1

∫

∫

=

= −

T

T

TtkTj

Tk

dttxa

dtetxa π

real is )( when* txaa kk =−


Lecture 7Fourier Series & Spectrum

9/13/2006 EE-2025 Spring-2005 jMc 3

READING ASSIGNMENTS

This Lecture:Fourier Series in Ch 3, Sects 3Fourier Series in Ch 3, Sects 3--4, 34, 3--5 & 35 & 3--66

Replaces pp. 62-66 in Ch 3 in DSP FirstNotation: ak for Fourier Series

Other Reading:Next Lecture: Sampling

9/13/2006 EE-2025 Spring-2005 jMc 4

LECTURE OBJECTIVES

ANALYSISANALYSIS via Fourier SeriesFor PERIODIC signals: x(t+T0) = x(t)

SPECTRUMSPECTRUM from Fourier Seriesak is Complex Amplitude for k-th Harmonic

∫ −=0

0

0

0

)/2(1 )(T

dtetxa tTkjTk

π

9/13/2006 EE-2025 Spring-2005 jMc 5

0 100 250–100–250 f (in Hz)

3/7 πje 3/7 πje−2/4 πje− 2/4 πje

10

SPECTRUM DIAGRAM


{ }∑=

−∗++=N

k

tfjk

tfjk

kk eaeaatx1

220)( ππ

kjkk eAa ϕ

21=∗

ka

0a

9/13/2006 EE-2025 Spring-2005 jMc 6

Harmonic Signal

PERIOD/FREQUENCY of COMPLEX EXPONENTIAL:

tfkj

kkeatx 02)( π∑

∞

−∞=

=

( )0

00

001or22f

TT

f ===πωπ

9/13/2006 EE-2025 Spring-2005 jMc 7

Example )3(sin)( 3 ttx π=

tjtjtjtj ejejejejtx ππππ 9339

883

83

8)( −− ⎟

⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛=

9/13/2006 EE-2025 Spring-2005 jMc 8

Example

In this case, analysisjust requires picking off the coefficients.

)3(sin)( 3 ttx π=

tjtjtjtj ejejejejtx ππππ 9339

883

83

8)( −− ⎟

⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛=

3=k1=k1−=k

3−=k

ka

9/13/2006 EE-2025 Spring-2005 jMc 9

STRATEGY: x(t) ak

ANALYSISGet representation from the signalWorks for PERIODICPERIODIC Signals

Fourier SeriesAnswer is: an INTEGRAL over one period

∫ −=0

0

0

0)(1

T

dtetxa tkjTk

ω

9/13/2006 EE-2025 Spring-2005 jMc 10

FS: Rectified Sine Wave {ak})1()(1 0

0

0

)/2(

0±≠= ∫ − kdtetx

Ta

TktTj

kπ

2/

0))1)(/2((2

)1)(/2(2/

0))1)(/2((2

)1)(/2(

2/

0

)1)(/2(21

2/

0

)1)(/2(21

2/

0

)/2()/2()/2(

1

2/

0

)/2(21

0

00

00

00

0

0

0

0

0

0

0

0

000

0

0

0

00

2

)sin(

T

kTjTj

tkTjT

kTjTj

tkTj

TtkTj

Tj

TtkTj

Tj

TktTj

tTjtTj

T

TktTj

TTk

ee

dtedte

dtejee

dteta

+−

+−

−−

−−

+−−−

−−

−

−=

−=

−=

=

∫∫

∫

∫

π

π

π

π

ππ

πππ

ππ

Half-Wave Rectified Sine

9/13/2006 EE-2025 Spring-2005 jMc 11

( ) ( )( ) ( )

( )( )⎪⎩

⎪⎨

⎧

±==−−−=

−−−=

−−−=

−=

−−

−−−+

+−+

−−−

+−+

−−−

+−

+−

−−

−−

even 1?

odd 01)1(

11

11

)1(1

)1(4)1(1

)1()1(4

1)1()1(4

1

2/)1)(/2()1(4

12/)1)(/2()1(4

1

2/

0))1)(/2((2

)1)(/2(2/

0))1)(/2((2

)1)(/2(

2

2

0000

0

00

00

00

0

kkk

ee

ee

eea

k

kk

kk

kjk

kjk

TkTjk

TkTjk

T

kTjTj

tkTjT

kTjTj

tkTj

k

π

π

ππ

ππ

ππ

ππ

π

π

π

π

FS: Rectified Sine Wave {ak}

41j±

9/13/2006 EE-2025 Spring-2005 jMc 12

SQUARE WAVE EXAMPLE

0–.02 .02 0.04

1

t

x(t)

.01

sec. 04.0for 0

01)(

0

0021

021

=⎪⎩

⎪⎨⎧

<≤

<≤=

TTtT

Tttx

9/13/2006 EE-2025 Spring-2005 jMc 15

Fourier Coefficients ak

ak is a function of kComplex Amplitude for k-th HarmonicThis one doesn’t depend on the period, T0

⎪⎪

⎩

⎪⎪

⎨

⎧

=

±±=

±±=

=−−

=

0

,4,20

,3,11

2)1(1

21 k

k

kkj

kja

k

k K

Kπ

π

9/13/2006 EE-2025 Spring-2005 jMc 16


⎪⎪

⎩

⎪⎪

⎨

⎧

=

±±=

±±=−

=

0

,4,20

,3,1

21 k

k

kkj

ak K

Kπ)25(2)04.0/(20 ππω ==

9/13/2006 EE-2025 Spring-2005 jMc 17

Fourier Series SynthesisHOW do you APPROXIMATE x(t) ?

Use FINITE number of coefficients

∫ −=0

0

00

)/2(1 )(T

tkTjTk dtetxa π

real is )( when* txaa kk =−tfkj

N

Nkkeatx 02)( π∑

−=

=

9/13/2006 EE-2025 Spring-2005 jMc 18

Fourier Series Synthesis

9/13/2006 EE-2025 Spring-2005 jMc 19

Synthesis: 1st & 3rd Harmonics))75(2cos(

32))25(2cos(2

21)( 22

ππ ππ

ππ

−+−+= ttty

9/13/2006 EE-2025 Spring-2005 jMc 20

Synthesis: up to 7th Harmonic)350sin(

72)250sin(

52)150sin(

32)50cos(2

21)( 2 ttttty π

ππ

ππ

ππ

ππ +++−+=

9/13/2006 EE-2025 Spring-2005 jMc 21

Fourier SynthesisK+++= )3sin(

32)sin(2

21)( 00 tttxN ω

πω

π

9/13/2006 EE-2025 Spring-2005 jMc 22

Gibbs’ Phenomenon

Convergence at DISCONTINUITY of x(t)There is always an overshoot9% for the Square Wave case

9/13/2006 EE-2025 Spring-2005 jMc 23

Fourier Series Demos

Fourier Series Java AppletGreg Slabaugh

Interactive

http://users.ece.gatech.edu/mcclella/2025/Fsdemo_Slabaugh/fourier.html

MATLAB GUI: fseriesdemo

http://users.ece.gatech.edu/mcclella/matlabGUIs/index.html

9/13/2006 EE-2025 Spring-2005 jMc 24

fseriesdemo GUI



Lecture 8Sampling & Aliasing


READING ASSIGNMENTS

This Lecture:Chap 4, Sections 4-1 and 4-2

Replaces Ch 4 in DSP First, pp. 83-94

Other Reading:Recitation: Strobe Demo (Sect 4-3)Next Lecture: Chap. 4 Sects. 4-4 and 4-5


LECTURE OBJECTIVES

SAMPLING can cause ALIASINGSampling TheoremSampling Rate > 2(Highest Frequency)

Spectrum for digital signals, x[n]Normalized Frequency

ππωω 22ˆ +==s

s ffT

ALIASING9/14/2003 © 2003, JH McClellan & RW Schafer 5

SYSTEMS Process Signals

PROCESSING GOALS:Change x(t) into y(t)

For example, more BASSImprove x(t), e.g., image deblurringExtract Information from x(t)

SYSTEMx(t) y(t)


System IMPLEMENTATION

DIGITAL/MICROPROCESSORConvert x(t) to numbers stored in memory

ELECTRONICSx(t) y(t)

COMPUTER D-to-AA-to-Dx(t) y(t)y[n]x[n]

ANALOG/ELECTRONIC:Circuits: resistors, capacitors, op-amps


SAMPLING x(t)

SAMPLING PROCESSConvert x(t) to numbers x[n]“n” is an integer; x[n] is a sequence of valuesThink of “n” as the storage address in memory

UNIFORM SAMPLING at t = nTsIDEAL: x[n] = x(nTs)

C-to-Dx(t) x[n]


SAMPLING RATE, fs

SAMPLING RATE (fs)fs =1/Ts

NUMBER of SAMPLES PER SECONDTs = 125 microsec fs = 8000 samples/sec

• UNITS ARE HERTZ: 8000 Hz

UNIFORM SAMPLING at t = nTs = n/fsIDEAL: x[n] = x(nTs)=x(n/fs)

C-to-Dx(t) x[n]=x(nTs)


fs = 2 kHz

fs = 500Hz

Hz100=f


SAMPLING THEOREM

HOW OFTEN ?DEPENDS on FREQUENCY of SINUSOIDANSWERED by SHANNON/NYQUIST TheoremALSO DEPENDS on “RECONSTRUCTION”


Reconstruction? Which One?

)4.0cos(][ nnx π=)4.2cos()4.0cos(

integer an is Whennn

nππ =

Given the samples, draw a sinusoid through the values


STORING DIGITAL SOUND


EXAMPLE: audio CDCD rate is 44,100 samples per second

16-bit samplesStereo uses 2 channels

Number of bytes for 1 minute is2 X (16/8) X 60 X 44100 = 10.584 Mbytes


sfsTnAnx

ωωωϕω

==+=

ˆ)ˆcos(][

)cos()(][)cos()(

ϕωϕω

+==+=

ss nTAnTxnxtAtx

DISCRETE-TIME SINUSOID

Change x(t) into x[n] DERIVATION

))cos((][ ϕω += nTAnx s

DEFINE DIGITAL FREQUENCY


DIGITAL FREQUENCY

VARIES from 0 to 2ππππ, as f varies from 0 to the sampling frequencyUNITS are radians, not rad/sec

DIGITAL FREQUENCY is NORMALIZED

ss f

fT πωω 2ˆ ==

ω̂ω̂


SPECTRUM (DIGITAL)

sffπω 2ˆ =

kHz1=sf ˆ ω

12 X1

2 X*

2π(0.1)π(0.1)π(0.1)π(0.1)–0.2ππππ

))1000/)(100(2cos(][ ϕπ += nAnx


SPECTRUM (DIGITAL) ???

ˆ ω = 2π ffs

fs =100 Hz ˆ ω

12 X1

2 X*

2π(1)π(1)π(1)π(1)–2ππππ

?

x[n] is zero frequency???

))100/)(100(2cos(][ ϕπ += nAnx


The REST of the STORY

Spectrum of x[n] has more than one line for each complex exponential

Called ALIASINGMANY SPECTRAL LINES

SPECTRUM is PERIODIC with period = 2ππππBecause

A cos( ˆ ω n+ ϕ) = A cos(( ˆ ω + 2π )n +ϕ )


ALIASING DERIVATION

Other Frequencies give the same ˆ ω Hz1000at sampled)400cos()(1 == sfttx π

)4.0cos()400cos(][ 10001 nnx n ππ ==

Hz1000at sampled)2400cos()(2 == sfttx π)4.2cos()2400cos(][ 10002 nnx n ππ ==

)4.0cos()24.0cos()4.2cos(][2 nnnnnx ππππ =+==

][][ 12 nxnx =⇒ )1000(24002400 πππ =−


ALIASING DERIVATION–2

Other Frequencies give the same

ss f

fT πωω 2ˆ == π2+

ˆ ω

s

s

ss

s

ff

ff

fff πππω 22)(2ˆ :then +=+=

and we want : x[n] = Acos( ˆ ω n +ϕ ) If x (t) = A cos( 2π( f + f s )t + ϕ ) t ←

nfs


ALIASING CONCLUSIONS

ADDING fs or 2fs or –fs to the FREQ of x(t) gives exactly the same x[n]

The samples, x[n] = x(n/ fs ) are EXACTLY THE SAME VALUES

GIVEN x[n], WE CAN’T DISTINGUISH fo FROM (fo + fs ) or (fo + 2fs )


NORMALIZED FREQUENCY

DIGITAL FREQUENCY

ss f



SPECTRUM for x[n]

PLOT versus NORMALIZED FREQUENCYINCLUDE ALL SPECTRUM LINES

ALIASESADD MULTIPLES of 2ππππSUBTRACT MULTIPLES of 2ππππ

FOLDED ALIASES(to be discussed later)ALIASES of NEGATIVE FREQS


SPECTRUM (MORE LINES)

ˆ ω

12 X1

2 X*

2π(0.1)π(0.1)π(0.1)π(0.1)–0.2ππππ

12 X*

1.8ππππ

12 X

–1.8ππππ

))1000/)(100(2cos(][ ϕπ += nAnx

kHz1=sf

sffπω 2ˆ =


SPECTRUM (ALIASING CASE)

12 X*

–0.5ππππ

12 X

–1.5ππππ

12 X

0.5ππππ 2.5ππππ–2.5ππππˆ ω

12 X1

2 X* 12 X*

1.5ππππ

))80/)(100(2cos(][ ϕπ += nAnxkHz80=sf

sffπω 2ˆ =


SAMPLING GUI (con2dis)


SPECTRUM (FOLDING CASE)

ˆ ω = 2π ffs

fs = 125Hz

12 X*

0.4ππππ

12 X

–0.4ππππ 1.6ππππ–1.6ππππˆ ω

12 X1

2 X*

))125/)(100(2cos(][ ϕπ += nAnx



Lecture 9D-to-A Conversion


READING ASSIGNMENTS

This Lecture:Chapter 4: Sections 4-4, 4-5

Other Reading:Recitation: Section 4-3 (Strobe Demo)Next Lecture: Chapter 5 (beginning)


LECTURE OBJECTIVES

FOLDING: a type of ALIASINGDIGITAL-to-ANALOG CONVERSION is

Reconstruction from samplesSAMPLING THEOREM applies

Smooth InterpolationMathematical Model of D-to-A

SUM of SHIFTED PULSESLinear Interpolation example


A-to-DConvert x(t) to numbers stored in memory

D-to-AConvert y[n] back to a “continuous-time” signal, y(t)y[n] is called a “discrete-time” signal

SIGNAL TYPES



SAMPLING x(t)

UNIFORM SAMPLING at t = nTsIDEAL: x[n] = x(nTs)

C-to-Dx(t) x[n]


NYQUIST RATE

“Nyquist Rate” Samplingfs > TWICE the HIGHEST Frequency in x(t)

“Sampling above the Nyquist rate”BANDLIMITED SIGNALS

DEF: x(t) has a HIGHEST FREQUENCY COMPONENT in its SPECTRUMNON-BANDLIMITED EXAMPLE

TRIANGLE WAVE is NOT BANDLIMITED


SPECTRUM for x[n]

INCLUDE ALL SPECTRUM LINESALIASES

ADD INTEGER MULTIPLES of 2ππππ and -2ππππFOLDED ALIASES

ALIASES of NEGATIVE FREQS

PLOT versus NORMALIZED FREQUENCY

i.e., DIVIDE fo by fs ππω 22ˆ +=sff


EXAMPLE: SPECTRUM

x[n] = Acos(0.2πn+φ)FREQS @ 0.2π and -0.2πALIASES:

{2.2π, 4.2π, 6.2π, …} & {-1.8π,-3.8π,…}EX: x[n] = Acos(4.2πn+φ)

ALIASES of NEGATIVE FREQ: {1.8π,3.8π,5.8π,…} & {-2.2π, -4.2π …}


SPECTRUM (MORE LINES)

ˆ ω

12 X1

2 X*

2π(0.1)π(0.1)π(0.1)π(0.1)–0.2ππππ

12 X*

1.8ππππ

12 X

–1.8ππππ

))1000/)(100(2cos(][ ϕπ += nAnx

kHz1=sf

sffπω 2ˆ =



12 X*

–0.5ππππ

12 X

–1.5ππππ

12 X


12 X1

2 X* 12 X*

1.5ππππ

))80/)(100(2cos(][ ϕπ += nAnxkHz80=sf

sffπω 2ˆ =


FOLDING (a type of ALIASING)

EXAMPLE: 3 different x(t); same x[n]

])1.0(2cos[])1.0(2cos[]2)9.0(2cos[])9.0(2cos[))900(2cos(

])1.0(2cos[])1.1(2cos[))1100(2cos(])1.0(2cos[))100(2cos(

1000

nnnnnt

nntnt

fs

ππππππ

πππππ

=−=−=→

=→→

= )1.0(210001002ˆ ππω ==

900 Hz “folds” to 100 Hz when fs=1kHz8/22/2003 © 2003, JH McClellan & RW Schafer 13

DIGITAL FREQ AGAIN

ss f


ππωω 22ˆ +−==s

s ffT FOLDED ALIAS

ω̂

ALIASING


SPECTRUM (FOLDING CASE)

ˆ ω = 2π ffs

fs = 125Hz

12 X*

0.4ππππ

12 X

–0.4ππππ 1.6ππππ–1.6ππππˆ ω

12 X1

2 X*

))125/)(100(2cos(][ ϕπ += nAnx


FREQUENCY DOMAINS

D-to-AA-to-Dx(t) y(t)x[n]

ffω

f̂

ω̂

y[n]

sffπ

ω2ˆ

=

π2+sffπω 2ˆ =


DEMOS from CHAPTER 4

CD-ROM DEMOSSAMPLING DEMO (con2dis GUI)

Different Sampling RatesAliasing of a Sinusoid

STROBE DEMOSynthetic vs. RealTelevision SAMPLES at 30 fps

Sampling & Reconstruction


SAMPLING GUI (con2dis)


D-to-A Reconstruction

Create continuous y(t) from y[n]IDEALIDEAL

If you have formula for y[n]Replace n in y[n] with fsty[n] = Acos(0.2πn+φ) with fs = 8000 Hzy(t) = Acos(2π(800)t+φ)



D-to-A is AMBIGUOUS !

ALIASINGGiven y[n], which y(t) do we pick ? ? ?INFINITE NUMBER of y(t)

PASSING THRU THE SAMPLES, y[n]D-to-A RECONSTRUCTION MUST CHOOSE ONE OUTPUT

RECONSTRUCT THE SMOOTHESTONE

THE LOWEST FREQ, if y[n] = sinusoid



ˆ ω = 2π ffs

fs = 80Hz

12 X*

–0.5ππππ

12 X

–1.5ππππ

12 X


12 X1

2 X* 12 X*

1.5ππππ

))80/)(100(2cos(][ ϕπ += nAnx


Reconstruction (D-to-A)

CONVERT STREAM of NUMBERS to x(t)“CONNECT THE DOTS”INTERPOLATION

y(t)

y[k]

kTs (k+1)Tst

INTUITIVE,conveys the idea


SAMPLE & HOLD DEVICE

CONVERT y[n] to y(t)y[k] should be the value of y(t) at t = kTs

Make y(t) equal to y[k] forkTs -0.5Ts < t < kTs +0.5Ts

y(t)

y[k]

kTs (k+1)Tst

STAIR-STEPAPPROXIMATION


SQUARE PULSE CASE


OVER-SAMPLING CASE

EASIER TO RECONSTRUCT


MATH MODEL for D-to-A

SQUARE PULSE:


EXPAND the SUMMATION

SUM of SHIFTED PULSES p(t-nTs)“WEIGHTED” by y[n]CENTERED at t=nTs

SPACED by TsRESTORES “REAL TIME”

y[n]p(t − nTs ) =n= −∞

∞

∑…+ y[0]p(t) + y[1]p(t − Ts ) + y[2]p(t − 2Ts ) +…


p(t)


TRIANGULAR PULSE (2X)


OPTIMAL PULSE ?

CALLED“BANDLIMITEDINTERPOLATION”

…,2,for 0)(

for sin

)(

ss

TtTt

TTttp

ttps

s

±±==

∞<<∞−= π

π



Lecture 10FIR Filtering Intro


READING ASSIGNMENTS

This Lecture:Chapter 5, Sects. 5-1, 5-2 and 5-3 (partial)

Other Reading:Recitation: Ch. 5, Sects 5-4, 5-6, 5-7 and 5-8

CONVOLUTIONNext Lecture: Ch 5, Sects. 5-3, 5-5 and 5-6


LECTURE OBJECTIVES

INTRODUCE FILTERING IDEAWeighted AverageRunning Average

FINITE IMPULSE RESPONSE FILTERSFIRFIR FiltersShow how to compute the output y[n] from the input signal, x[n]


DIGITAL FILTERING

CONCENTRATE on the COMPUTERPROCESSING ALGORITHMS

SOFTWARE (MATLAB)HARDWARE: DSP chips, VLSI

DSP: DIGITAL SIGNAL PROCESSING



The TMS32010, 1983

First PC plug-in board from Atlanta Signal Processors Inc.


Rockland Digital Filter, 1971

For the price of a small house, you could have one of these.


Digital Cell Phone (ca. 2000)

Now it plays video2/18/2005 © 2003, JH McClellan & RW Schafer 9

DISCRETE-TIME SYSTEM

OPERATE on x[n] to get y[n]WANT a GENERAL CLASS of SYSTEMS

ANALYZE the SYSTEMTOOLS: TIME-DOMAIN & FREQUENCY-DOMAIN

SYNTHESIZE the SYSTEM

COMPUTERy[n]x[n]


D-T SYSTEM EXAMPLES

EXAMPLES:POINTWISE OPERATORS

SQUARING: y[n] = (x[n])2

RUNNING AVERAGERULE: “the output at time n is the average of three consecutive input values”

SYSTEMy[n]x[n]


DISCRETE-TIME SIGNAL

x[n] is a LIST of NUMBERSINDEXED by “n”

STEM PLOT


3-PT AVERAGE SYSTEMADD 3 CONSECUTIVE NUMBERS

Do this for each “n”Make a TABLE

n=0

n=1

])2[]1[][(][ 31 ++++= nxnxnxny


INPUT SIGNAL

OUTPUT SIGNAL

])2[]1[][(][ 31 ++++= nxnxnxny


PAST, PRESENT, FUTURE

“n” is TIME


ANOTHER 3-pt AVERAGER

Uses “PAST” VALUES of x[n]IMPORTANT IF “n” represents REAL TIME

WHEN x[n] & y[n] ARE STREAMS

])2[]1[][(][ 31 −+−+= nxnxnxny


GENERAL CAUSAL FIR FILTER

FILTER COEFFICIENTS {bk}DEFINE THE FILTER

For example,

∑=

−=M

kk knxbny

0][][

]3[]2[2]1[][3

][][3

0

−+−+−−=

−= ∑=

nxnxnxnx

knxbnyk

k

}1,2,1,3{ −=kb


GENERAL FIR FILTER

FILTER COEFFICIENTS {bk}

FILTER ORDER is MFILTER LENGTH is L = M+1

NUMBER of FILTER COEFFS is L

∑=

−=M

kk knxbny

0][][


SLIDE a WINDOW across x[n]

x[n]x[n-M]

∑=

−=M

kk knxbny

0][][

GENERAL CAUSAL FIR FILTER


FILTERED STOCK SIGNAL

OUTPUT

INPUT

50-pt Averager


⎪⎩

⎪⎨⎧

≠

==

00

01][

n

nnδ

SPECIAL INPUT SIGNALS

x[n] = SINUSOIDx[n] has only one NON-ZERO VALUE

1

n

UNIT-IMPULSE

FREQUENCY RESPONSE (LATER)


UNIT IMPULSE SIGNAL δ[n]

δ[n] is NON-ZEROWhen its argumentis equal to ZERO

]3[ −nδ3=n


]4[2]3[4]2[6]1[4][2][ −+−+−+−+= nnnnnnx δδδδδ

MATH FORMULA for x[n]

Use SHIFTED IMPULSES to write x[n]


SUM of SHIFTED IMPULSES

This formula ALWAYS works


4-pt AVERAGER

CAUSAL SYSTEM: USE PAST VALUES])3[]2[]1[][(][ 4

1 −+−+−+= nxnxnxnxny

INPUT = UNIT IMPULSE SIGNAL = δ[n]

]3[]2[]1[][][][][

41

41

41

41 −+−+−+=

=nnnnny

nnxδδδδ

δ

OUTPUT is called “IMPULSE RESPONSE”},0,0,,,,,0,0,{][ 4

141

41

41 KK=nh


},0,0,,,,,0,0,{][ 41

41

41

41 KK=nh

4-pt Avg Impulse Response

δ[n] “READS OUT” the FILTER COEFFICIENTS

n=0“h” in h[n] denotes Impulse Response

1

n

n=0n=1

n=4n=5

n=–1

NON-ZERO When window overlaps δ[n]

])3[]2[]1[][(][ 41 −+−+−+= nxnxnxnxny


FIR IMPULSE RESPONSE

Convolution = Filter DefinitionFilter Coeffs = Impulse Response

∑=

−=M

kknxkhny

0][][][

CONVOLUTION

∑=

−=M

kk knxbny

0][][


FILTERING EXAMPLE

7-point AVERAGERRemoves cosine

By making its amplitude (A) smaller

3-point AVERAGERChanges A slightly

( )∑=

−=6

071

7 ][][k

knxny

( )∑=

−=2

031

3 ][][k

knxny


3-pt AVG EXAMPLE

USE PAST VALUES

400for)4/8/2cos()02.1(][ :Input ≤≤++= nnnx n ππ


7-pt FIR EXAMPLE (AVG)

CAUSAL: Use Previous

LONGER OUTPUT




Lecture 11Linearity & Time-InvarianceConvolution


READING ASSIGNMENTS

This Lecture:Chapter 5, Sections 5-5 and 5-6

Section 5-4 will be covered, but not “in depth”

Other Reading:Recitation: Ch. 5, Sects 5-6, 5-7 & 5-8

CONVOLUTIONNext Lecture: start Chapter 6


LECTURE OBJECTIVES

BLOCK DIAGRAM REPRESENTATIONComponents for HardwareConnect Simple Filters Together to Build More Complicated Systems

LTI SYSTEMS

GENERAL PROPERTIES of FILTERSLLINEARITYTTIME-IINVARIANCE==> CONVOLUTIONCONVOLUTION


IMPULSE RESPONSE, FIR case: same as

CONVOLUTIONGENERAL:GENERAL CLASS of SYSTEMSLINEAR and TIME-INVARIANT

ALL LTI systems have h[n] & use convolution

OVERVIEW

][][][ nxnhny ∗=

][nh

}{ kb


CONCENTRATE on the FILTER (DSP)DISCRETE-TIME SIGNALS

FUNCTIONS of n, the “time index”INPUT x[n]OUTPUT y[n]

DIGITAL FILTERING

FILTER D-to-AA-to-Dx(t) y(t)y[n]x[n]


BUILDING BLOCKS

BUILD UP COMPLICATED FILTERSFROM SIMPLE MODULESEx: FILTER MODULE MIGHT BE 3-pt FIR

FILTERy[n]x[n]

FILTER

FILTER

++

INPUT

OUTPUT


GENERAL FIR FILTER

FILTER COEFFICIENTS {bk}DEFINE THE FILTER

For example,

∑=

−=M

kk knxbny

0][][

]3[]2[2]1[][3

][][3

0

−+−+−−=

−=∑=

nxnxnxnx

knxbnyk

k

}1,2,1,3{ −=kb


MATLAB for FIR FILTER

yy = conv(bb,xx)

VECTOR bb contains Filter Coefficients

DSP-First: yy = firfilt(bb,xx)

FILTER COEFFICIENTS {bk} conv2()for images

∑=

−=M

kk knxbny

0][][


SPECIAL INPUT SIGNALS

x[n] = SINUSOIDx[n] has only one NON-ZERO VALUE

1

n

UNIT-IMPULSE

FREQUENCY RESPONSE

≠

==

00

01][

n

nnδ


FIR IMPULSE RESPONSE

Convolution = Filter DefinitionFilter Coeffs = Impulse Response

∑=

−=M

kk knxbny

0][][ ∑

=−=

M

kk knbnh

0][][ δ


]4[]3[]2[2]1[][][ −+−−−+−−= nnnnnnh δδδδδ

MATH FORMULA for h[n]

Use SHIFTED IMPULSES to write h[n]

1

n

–1

2

0 4

][nh

}1,1,2,1,1{ −−=kb8/22/2003 © 2003, JH McClellan & RW Schafer 13

∑=

−=M

kknxkhny

0][][][

LTI: Convolution Sum

Output = Convolution of x[n] & h[n]Output = Convolution of x[n] & h[n]NOTATION:Here is the FIR case:

FINITE LIMITS

FINITE LIMITSSame as bk

][][][ nxnhny ∗=


CONVOLUTION Example

n −1 0 1 2 3 4 5 6 7x[n] 0 1 1 1 1 1 1 1 ...h[n] 0 1 −1 2 −1 1 0 0 0

0 1 1 1 1 1 1 1 10 0 −1 −1 −1 −1 −1 −1 −10 0 0 2 2 2 2 2 20 0 0 0 −1 −1 −1 −1 −10 0 0 0 0 1 1 1 1

y[n] 0 1 0 2 1 2 2 2 ...

][][]4[]3[]2[2]1[][][

nunxnnnnnnh

=−+−−−+−−= δδδδδ

]4[]4[]3[]3[]2[]2[]1[]1[

][]0[

−−−−

nxhnxhnxhnxhnxh


GENERAL FIR FILTERSLIDE a Length-L WINDOW over x[n]

x[n]x[n-M]


DCONVDEMO: MATLAB GUI


]1[][][ −−= nununy

POP QUIZ

FIR Filter is “FIRST DIFFERENCE”y[n] = x[n] - x[n-1]

INPUT is “UNIT STEP”

Find y[n]

<

≥=

00

01][

n

nnu

][]1[][][ nnununy δ=−−=


HARDWARE STRUCTURES

INTERNAL STRUCTURE of “FILTER”WHAT COMPONENTS ARE NEEDED?HOW DO WE “HOOK” THEM TOGETHER?

SIGNAL FLOW GRAPH NOTATION

FILTER y[n]x[n]∑

=−=

M

kk knxbny

0][][


HARDWARE ATOMS

Add, Multiply & Store∑

=−=

M

kk knxbny

0][][

][][ nxny β=

]1[][ −= nxny

][][][ 21 nxnxny +=


FIR STRUCTURE

Direct Form

SIGNALFLOW GRAPH

∑=

−=M

kk knxbny

0][][


Moore’s Law for TI DSPs

Double every18 months ?

LOG SCALE


SYSTEM PROPERTIES

MATHEMATICAL DESCRIPTIONMATHEMATICAL DESCRIPTIONTIMETIME--INVARIANCEINVARIANCELINEARITYLINEARITYCAUSALITY

“No output prior to input”

SYSTEMy[n]x[n]


TIME-INVARIANCE

IDEA:“Time-Shifting the input will cause the sametime-shift in the output”

EQUIVALENTLY,We can prove that

The time origin (n=0) is picked arbitrary


TESTING Time-Invariance


LINEAR SYSTEM

LINEARITY = Two PropertiesSCALING

“Doubling x[n] will double y[n]”

SUPERPOSITION:“Adding two inputs gives an output that is the sum of the individual outputs”


TESTING LINEARITY


LTI SYSTEMS

LTI: Linear & Time-Invariant

COMPLETELY CHARACTERIZED by:IMPULSE RESPONSE h[n]CONVOLUTION: y[n] = x[n]*h[n]

The “rule”defining the system can ALWAYS be re-written as convolution

FIR Example: h[n] is same as bk


POP QUIZ

FIR Filter is “FIRST DIFFERENCE”y[n] = x[n] - x[n -1]

Write output as a convolutionNeed impulse response

Then, another way to compute the output:]1[][][ −−= nnnh δδ

( ) ][]1[][][ nxnnny ∗−−= δδ


CASCADE SYSTEMS

Does the order of S1 & S2 matter?NO, LTI SYSTEMS can be rearranged !!!WHAT ARE THE FILTER COEFFS? {bk}

S1 S2


CASCADE EQUIVALENTFind “overall” h[n] for a cascade ?

S1 S2

S1S2



Lecture 12Frequency Response

of FIR Filters


READING ASSIGNMENTS

This Lecture:Chapter 6, Sections 6-1, 6-2, 6-3, 6-4, & 6-5

Other Reading:Recitation: Chapter 6

FREQUENCY RESPONSE EXAMPLESNext Lecture: Chap. 6, Sects. 6-6, 6-7 & 6-8


LECTURE OBJECTIVES

SINUSOIDAL INPUT SIGNALDETERMINE the FIR FILTER OUTPUT

FREQUENCY RESPONSE of FIRPLOTTING vs. FrequencyMAGNITUDE vs. FreqPHASE vs. Freq )(ˆˆ ˆ

)()(ωωω jeHjjj eeHeH ∠=

MAG

PHASE


DOMAINS: Time & Frequency

Time-Domain: “n” = timex[n] discrete-time signalx(t) continuous-time signal

Frequency Domain (sum of sinusoids)Spectrum vs. f (Hz)

• ANALOG vs. DIGITAL

Spectrum vs. omega-hatMove back and forth QUICKLY


DIGITAL “FILTERING”

CONCENTRATE on the SPECTRUMSPECTRUMSINUSOIDAL INPUT

INPUT x[n] = SUM of SINUSOIDSThen, OUTPUT y[n] = SUM of SINUSOIDS

FILTER D-to-AA-to-Dx(t) y(t)y[n]x[n]

ω̂ω̂


FILTERING EXAMPLE

7-point AVERAGERRemoves cosine

By making its amplitude (A) smaller

3-point AVERAGERChanges A slightly

( )∑=

−=6

071

7 ][][k

knxny

( )∑=

−=2

031

3 ][][k

knxny


3-pt AVG EXAMPLE

USE PAST VALUES



7-pt FIR EXAMPLE (AVG)

CAUSAL: Use Previous

LONGER OUTPUT



SINUSOIDAL RESPONSE

INPUT: x[n] = SINUSOIDOUTPUT: y[n] will also be a SINUSOID

Different Amplitude and Phase

SAME Frequency

AMPLITUDE & PHASE CHANGECalled the FREQUENCY RESPONSEFREQUENCY RESPONSE


DCONVDEMO: MATLAB GUI


COMPLEX EXPONENTIAL

∑∑==

−=−=M

k

M

kk knxkhknxbny

00][][][][

FIR DIFFERENCE EQUATION

∞<<∞−= neAenx njj ωϕ ˆ][x[n] is the input signal—a complex exponential


COMPLEX EXP OUTPUT

Use the FIR “Difference Equation”

njj eAeH ωϕω ˆ)ˆ(=

∑∑=

−

=

=−=M

k

knjjk

M

kk eAebknxbny

0

)(ˆ

0][][ ωϕ

njjM

k

kjk eAeeb ωϕω ˆ

0

)(ˆ⎟⎟⎠

⎞⎜⎜⎝

⎛= ∑

=

−


FREQUENCY RESPONSE

Complex-valued formulaHas MAGNITUDE vs. frequencyAnd PHASE vs. frequency

Notation:

FREQUENCYRESPONSE

At each frequency, we can DEFINE

)ˆ(of place in )( ˆ ωω HeH j

kjM

kkebH ωω ˆ

0)ˆ( −

=∑= kjM

kk

j ebeH ωω ˆ

0

ˆ )( −

=∑=


EXAMPLE 6.1

EXPLOITSYMMETRY

}1,2,1{}{ =kb

)ˆcos22()2(

21)(

ˆ

ˆˆˆ

ˆ2ˆˆ

ωω

ωωω

ωωω

+=++=

++=

−

−−

−−

j

jjj

jjj

eeee

eeeH

ω

ωω

ω

ω

ˆ)( is Phaseand

)ˆcos22()( is Magnitude0)ˆcos22( Since

ˆ

ˆ

−=∠

+=

≥+

j

j

eH

eH


PLOT of FREQ RESPONSE

ωω ω ˆˆ )ˆcos22()( jj eeH −+= RESPONSE at π/3

}1,2,1{}{ =kb

(radians)ω̂π− π

ω̂


EXAMPLE 6.2

y[n]x[n])( ω̂jeH

ωω ω ˆˆ )ˆcos22()( jj eeH −+=

njj

j

eenxeHny

)3/(4/

ˆ

2][ andknown is )( when][ Find

ππ

ω

=


njj eenxny )3/(4/2][ when][ Find ππ=

EXAMPLE 6.2 (answer)

3/ât )( evaluate-Step One ˆ πωω =jeHωω ω ˆˆ )ˆcos22()( jj eeH −+=

3/ˆ@3)( 3/ˆ πωπω == − jj eeH

( ) njjj eeeny )3/(4/3/ 23][ πππ ×= − njj ee )3/(12/6 ππ−=


EXAMPLE: COSINE INPUT

)cos(2][ andknown is )( when][ Find

43

ˆ

ππ

ω

+= nnxeHny j


y[n]x[n] )( ω̂jeHω̂ ω̂


EX: COSINE INPUT

)cos(2][ when][ Find 43ππ += nnxny

][][][)cos(2

21

)4/3/()4/3/(43

nxnxnxeen njnj

+=⇒+=+ +−+ ππππππ

][][][)(][

)(][

21

)4/3/(3/2

)4/3/(3/1

nynynyeeHny

eeHnynjj

njj

+=⇒=

=+−−

+

πππ

πππUseLinearity


EX: COSINE INPUT (ans-2)


)4/3/()3/()4/3/(3/2

)4/3/()3/()4/3/(3/1

3)(][3)(][

ππππππ

ππππππ

+−+−−

+−+

==

==njjnjj

njjnjj

eeeeHnyeeeeHny

)cos(6][33][

123

)12/3/()12/3/(

ππ

ππππ

−=⇒+= −−−

nnyeeny njnj



MATLAB:FREQUENCY RESPONSE

HH = freqz(bb,1,ww)VECTOR bb contains Filter Coefficients

SP-First: HH = freekz(bb,1,ww)FILTER COEFFICIENTS {bk}

kjM

kk

j ebeH ωω ˆ

0

ˆ )( −

=∑=


Time & Frequency Relation

Get Frequency Response from h[n]Here is the FIR case:

kjM

k

kjM

kk

j ekhebeH ωωω ˆ

0

ˆ

0

ˆ ][)( −

=

−

=∑∑ ==

IMPULSE RESPONSE


BLOCK DIAGRAMS

Equivalent Representations

y[n]x[n]

y[n]x[n])( ω̂jeH

][nh

ω̂ˆ ω


UNIT-DELAY SYSTEM

y[n]x[n] ]1[ −nδ

y[n]x[n] ω̂je−)( ω̂jeH

ˆ ω ˆ ω

]1[][for )( and ][ Find ˆ −= nxnyeHnh jω

}1,0{}{ =kb


FIRST DIFFERENCE SYSTEM

y[n]x[n] ω̂1 je−−

y[n]x[n] ]1[][ −− nn δδ

)( ω̂jeH

]1[][][ :Equatione Differencfor the )( and ][ Find ˆ

−−= nxnxnyeHnh jω


DLTI Demo with Sinusoids

FILTERx[n] y[n]


CASCADE SYSTEMS

Does the order of S1 & S2 matter?NO, LTI SYSTEMS can be rearranged !!!WHAT ARE THE FILTER COEFFS? {bk}WHAT is the overall FREQUENCY RESPONSE ?

S1 S2][nδ ][1 nh ][][ 21 nhnh ∗][1 nh ][2 nh


CASCADE EQUIVALENT

MULTIPLY the Frequency Responses

y[n]x[n] )( ω̂jeH

x[n] )( ˆ1

ωjeH y[n])( ˆ2

ωjeH

)()()( ˆ2

ˆ1

ˆ ωωω jjj eHeHeH =EQUIVALENTSYSTEM



Lecture 13Digital Filtering

of Analog Signals


READING ASSIGNMENTS

This Lecture:Chapter 6, Sections 6-6, 6-7 & 6-8

Other Reading:Recitation: Chapter 6

FREQUENCY RESPONSE EXAMPLESNext Lecture: Chapter 7


LECTURE OBJECTIVES

Two Domains: Time & FrequencyTrack the spectrum of x[n] thru an FIR Filter: Sinusoid-IN gives Sinusoid-OUTUNIFICATION: How does Frequency Response affect x(t) to produce y(t) ?

D-to-AA-to-Dx(t) y(t)y[n]x[n] )( ω̂jeH

FIRω̂ ω̂


TIME & FREQUENCY

FIR DIFFERENCE EQUATION is the TIME-DOMAIN

∑∑==

−=−=M

k

M

kk knxkhknxbny

00][][][][

∑=

−=M

k

kjj ekheH0

ˆˆ ][)( ωω

++++= −−− ωωωω ˆ3ˆ2ˆˆ ]3[]2[]1[]0[)( jjjj ehehehheH


Ex: DELAY by 2 SYSTEM

y[n]x[n] )( ω̂jeH

y[n]x[n] ][nh


]2[][ −= nnh δ

}1,0,0{=kb

ω̂ ω̂10/6/2003 © 2003, JH McClellan & RW Schafer 7

DELAY by 2 SYSTEM

k = 2 ONLY

y[n]x[n] ]2[ −nδ


∑=

−−=M

k

kjj ekeH0

ˆˆ ]2[)( ωω δ

)( ω̂jeHy[n]x[n] ω̂2je−

ω̂ ω̂


GENERAL DELAY PROPERTY

ONLY ONE non-ZERO TERMfor k at k = nd

dnjM

k

kjd

j eenkeH ωωω δ ˆ

0

ˆˆ ][)( −

=

− =−=∑

][][for )( and ][ Find ˆd

j nnxnyeHnh −=ω

][][ dnnnh −= δ


FREQ DOMAIN --> TIME ??

START with kj bnheH or find and ][)( ω̂

)ˆcos(7)( ˆ2ˆ ωωω jj eeH −=

y[n]x[n] ][nh ?][ =nh

y[n]x[n])( ω̂jeH

ω̂ ω̂


FREQ DOMAIN --> TIME

EULER’s Formula)ˆcos(7)( ˆ2ˆ ωωω jj eeH −=)5.05.0(7 ˆˆˆ2 ωωω jjj eee −− +=

)5.35.3( ˆ3ˆ ωω jj ee −− +=

]3[5.3]1[5.3][ −+−= nnnh δδ}5.3,0,5.3,0{=kb


PREVIOUS LECTURE REVIEW

SINUSOIDAL INPUT SIGNALOUTPUT has SAME FREQUENCYDIFFERENT Amplitude and Phase

FREQUENCY RESPONSE of FIRMAGNITUDE vs. FrequencyPHASE vs. FreqPLOTTING

)(ˆˆ ˆ

)()(ωωω jeHjjj eeHeH ∠=

MAG

PHASE


FREQ. RESPONSE PLOTS

DENSE GRID (ww) from -ππππ to +ππππww = -pi:(pi/100):pi;

HH = freqz(bb,1,ww)VECTOR bb contains Filter CoefficientsDSP-First: HH = freekz(bb,1,ww)

∑=

−=M

k

kjk

j ebeH0

ˆˆ )( ωω


PLOT of FREQ RESPONSE

ωω ω ˆˆ )ˆcos22()( jj eeH −+= RESPONSE at ππππ/3

}1,2,1{}{ =kb

(radians)ω̂π− π

ω̂


EXAMPLE 6.2

ω̂ω̂

y[n]x[n])( ω̂jeH


njj

j

eenxeHny

)3/(4/

ˆ

2][ andknown is )( when][ Find

ππ

ω

=


njj eenxny )3/(4/2][ when][ Find ππ=

EXAMPLE 6.2 (answer)

3/ât )( evaluate -Step One ˆ πωω =jeHωω ω ˆˆ )ˆcos22()( jj eeH −+=

3/ˆ@3)( 3/ˆ πωπω == − jj eeH

( ) njjj eeeny )3/(4/3/ 23][ πππ ×= − njj ee )3/(12/6 ππ−=


EXAMPLE: COSINE INPUT

)cos(2][ andknown is )( when][ Find

43

ˆ

ππ

ω

+= nnxeHny j


y[n]x[n] )( ω̂jeHω̂ ω̂




][][][)cos(2

21

)4/3/()4/3/(43

nxnxnxeen njnj

+=⇒+=+ +−+ ππππππ

][][][)(][

)(][

21

)4/3/(3/2

)4/3/(3/1

nynynyeeHnyeeHny

njj

njj

+=⇒==

+−−

+

πππ

πππ




)4/3/()3/()4/3/(3/2

)4/3/()3/()4/3/(3/1

3)(][3)(][

ππππππ

ππππππ

+−+−−

+−+

====

njjnjj

njjnjj

eeeeHnyeeeeHny

)cos(6][33][

123

)12/3/()12/3/(

ππ

ππππ

−=⇒+= −−−

nnyeeny njnj



SINUSOID thru FIR

IF Multiply the Magnitudes

Add the Phases

))(ˆcos()(][

)ˆcos(][11 ˆ

1ˆ

1ωω φω

φωjj eHneHAny

nAnx

∠++=⇒

+=

)()( ˆˆ* ωω jj eHeH −=


LTI Demo with Sinusoids

FILTERx[n]

y[n]


DIGITAL “FILTERING”

SPECTRUM of x(t) (SUM of SINUSOIDS)SPECTRUM of x[n]

Is ALIASING a PROBLEM ?SPECTRUM y[n] (FIR Gain or Nulls)Then, OUTPUT y(t) = SUM of SINUSOIDS


ω̂ ω̂ω ω

ω̂ω

ω


FREQUENCY SCALING

TIME SAMPLING:IF NO ALIASING:FREQUENCY SCALING


ˆ ω ˆ ω ω ω

sfsT ωωω ==ˆsnTt =


11-pt AVERAGER Example


ˆ ω ˆ ω ω ω

250 Hz

25 Hz ?))250(2cos())25(2cos()( 2

1 πππ −+= tttx

ωω

ωω ˆ5

21

211

ˆ

)ˆsin(11)ˆsin(

)( jj eeH −=

∑=

−=10

0111 ][][

kknxny


D-A FREQUENCY SCALING

RECONSTRUCT up to 0.5fsFREQUENCY SCALING


ˆ ω ˆ ω ω ω

TIME SAMPLING:

sfωω ˆ=

ss ftnnTt ←⇒=


TRACK the FREQUENCIES


ˆ ω ˆ ω ω ω

Fs = 1000 Hz

0.5ππππ

.05ππππ

0.5ππππ

.05ππππ

250 Hz

25 Hz

NO new freqs

250 Hz

25 Hz )(

)(

05.0

5.0

π

π

j

j

eH

eH


11-pt AVERAGER

NULLS or ZEROS

πω 5.0ˆ =πω 05.0ˆ =


EVALUATE Freq. Response

ωω

ωω ˆ5

21

211

ˆ

)ˆsin(11)ˆsin(

)( jj eeH −=

)5.0(5

21

211

ˆ

))5.0(sin(11))5.0(sin(

)( πω

ππ jj eeH −=

πω 5.0Ât =

π

ππ 5.2

)25.0sin(11)75.2sin( je−=

π5.00909.0 je−=


EVALUATE Freq. Response

fs = 1000MAG SCALE

PHASE CHANGE

)( 1000/)25(2πjeH

)( 1000/)250(2πjeH


EFFECTIVE RESPONSE

DIGITAL FILTER

LOW-PASS FILTER

)( ω̂jeH


FILTER TYPES

LOW-PASS FILTER (LPF)BLURRINGATTENUATES HIGH FREQUENCIES

HIGH-PASS FILTER (HPF)SHARPENING for IMAGESBOOSTS THE HIGHSREMOVES DC

BAND-PASS FILTER (BPF)


B & W IMAGE


B&W IMAGE with COSINEFILTERED: 11-pt AVG


FILTERED B&W IMAGE

LPF:BLUR


ROW of B&W IMAGE

BLACK = 255

WHITE = 0


FILTERED ROW of IMAGE

ADJUSTED DELAY by 5 samples



Lecture 14Z Transforms: Introduction


READING ASSIGNMENTS

This Lecture:Chapter 7, Sects 7-1 through 7-5

Other Reading:Recitation: Ch. 7

CASCADING SYSTEMSNext Lecture: Chapter 7, 7-6 to the end


LECTURE OBJECTIVES

INTRODUCE the Z-TRANSFORMGive Mathematical DefinitionShow how the H(z) POLYNOMIAL simplifies analysis

CONVOLUTION is SIMPLIFIED !

Z-Transform can be applied toFIR Filter: h[n] --> H(z)Signals: x[n] --> X(z) ∑ −=

n

nznhzH ][)()(][ zHnh →

)(][ zXnx →


TWO (no, THREE) DOMAINS

Z-TRANSFORM-DOMAINPOLYNOMIALS: H(z)

FREQ-DOMAIN

kjM

kk

j ebeH ωω ˆ

0

ˆ )( −

=∑=

TIME-DOMAIN

∑=

−=M

kk knxbny

0][][

}{ kb


TRANSFORM CONCEPT

Move to a new domain whereOPERATIONS are EASIER & FAMILIARUse POLYNOMIALS

TRANSFORM both waysx[n] ---> X(z) (into the z domain)X(z) ---> x[n] (back to the time domain)

)(][ zXnx →

][)( nxzX →


“TRANSFORM” EXAMPLE

Equivalent Representations

y[n]x[n]

y[n]x[n]

∑ −=n

njj enheH ωω ˆˆ ][)(

ωω ˆˆ 1)( jj eeH −−=

]1[][][ −−= nnnh δδ


Z-TRANSFORM IDEA

POLYNOMIAL REPRESENTATION

y[n]x[n]

y[n]x[n] )(zH

][nh∑ −=n

nznhzH ][)(


Z-Transform DEFINITION

POLYNOMIAL Representation of LTI SYSTEM:

EXAMPLE:

∑ −=n

nznhzH ][)(

APPLIES toAny SIGNAL

POLYNOMIAL in z-1

43210 20302)( −−−−− ++−+= zzzzzzH42 232 −− +−= zz

4121 )(2)(32 −− +−= zz

}2,0,3,0,2{]}[{ −=nh


Z-Transform EXAMPLE

ANY SIGNAL has a z-Transform:

∑ −=n

nznxzX ][)(

4321 24642)( −−−− ++++= zzzzzX?)( =zX8/22/2003 © 2003, JH McClellan & RW Schafer 11

531 321)( −−− −+−= zzzzX

EXPONENT GIVESTIME LOCATION

?][ =nx


Z-Transform of FIR Filter

CALLED the SYSTEM FUNCTIONSYSTEM FUNCTIONh[n] is same as {bk}

FIR DIFFERENCE EQUATION

∑∑==

−=−=M

k

M

kk knxkhknxbny

00][][][][

CONVOLUTION

SYSTEMFUNCTION ∑∑

=

−

=

− ==M

k

kM

k

kk zkhzbzH

00][)(


]2[]1[5][6][ −+−−= nxnxnxny

211 56)( −−− +−==∑ zzzbzH k

Z-Transform of FIR Filter

Get H(z) DIRECTLY from the {bk}Example 7.3 in the book:

}1,5,6{}{ −=kb


Ex. DELAY SYSTEM

UNIT DELAY: find h[n] and H(z)

y[n]x[n]

y[n] = x[n-1]x[n] ]1[ −nδ

∑ −−= nznzH ]1[)( δ 1−= z

1−z8/22/2003 © 2003, JH McClellan & RW Schafer 15

DELAY EXAMPLEUNIT DELAY: find y[n] via polynomials

x[n] = {3,1,4,1,5,9,0,0,0,...}

6543210 95430)( −−−−−− ++++++= zzzzzzzzY

)9543()( 543211 −−−−−− +++++= zzzzzzzY

)()( 1 zXzzY −=


DELAY PROPERTY


GENERAL I/O PROBLEMInput is x[n], find y[n] (for FIR, h[n])How to combine X(z) and H(z) ?


FIR Filter = CONVOLUTION

∑∑==

−=−=M

k

M

kk knxkhknxbny

00][][][][

CONVOLUTION 8/22/2003 © 2003, JH McClellan & RW Schafer 19

CONVOLUTION PROPERTYPROOF:

MULTIPLYZ-TRANSFORMS


CONVOLUTION EXAMPLEMULTIPLY the z-TRANSFORMS:

MULTIPLY H(z)X(z)8/22/2003 © 2003, JH McClellan & RW Schafer 21

CONVOLUTION EXAMPLEFinite-Length input x[n]FIR Filter (L=4) MULTIPLY

Z-TRANSFORMS

y[n] = ?


CASCADE SYSTEMS

Does the order of S1 & S2 matter?NO, LTI SYSTEMS can be rearranged !!!Remember: h1[n] * h2[n]How to combine H1(z) and H2(z) ?

S1 S2


CASCADE EQUIVALENT

Multiply the System Functions

x[n] )(1 zH y[n])(2 zH

)()()( 21 zHzHzH =

y[n]x[n] )(zHEQUIVALENTSYSTEM


CASCADE EXAMPLE

y[n]x[n] )(zH

x[n] )(1 zH y[n])(2 zHw[n]

12 1)( −+= zzH1

1 1)( −−= zzH

211 1)1)(1()( −−− −=+−= zzzzH]2[][][ −−= nxnxny

]1[][][ −−= nxnxnw ]1[][][ −+= nwnwny



Lecture 15Zeros of H(z) and theFrequency Domain


READING ASSIGNMENTS

This Lecture:Chapter 7, Section 7-6 to end

Other Reading:Recitation & Lab: Chapter 7

ZEROS (and POLES)Next Lecture:Chapter 8


LECTURE OBJECTIVES

ZEROS and POLESRelate H(z) to FREQUENCY RESPONSE

THREE DOMAINS:Show Relationship for FIR:

ωω

ˆ)()( ˆjez

j zHeH ==

)()(][ ω̂jeHzHnh ↔↔8/22/2003 © 2003, JH McClellan & RW Schafer 5

DESIGN PROBLEM

Example:Design a Lowpass FIR filter (Find bk)Reject completely 0.7ππππ, 0.8ππππ, and 0.9ππππ

This is NULLINGEstimate the filter length needed to accomplish this task. How many bk ?

Z POLYNOMIALS provide the TOOLS


Z-Transform DEFINITION

POLYNOMIAL Representation of LTI SYSTEM:

EXAMPLE:

∑ −=n

nznhzH ][)(

APPLIES toAny SIGNAL

POLYNOMIAL in z-1

43210 20302)( −−−−− ++−+= zzzzzzH42 232 −− +−= zz

4121 )(2)(32 −− +−= zz

}2,0,3,0,2{]}[{ −=nh


)()()(][][][ zXzHzYnxnhny =↔∗=

CONVOLUTION PROPERTYConvolution in the n-domain

SAME AS

Multiplication in the z-domain

MULTIPLYz-TRANSFORMS

FIR Filter∑

=−=

∗=M

kknxkh

nhnxny

0][][

][][][


CONVOLUTION EXAMPLEy[n]x[n] H(z)

][][][ nhnxny ∗=21 2)( −− += zzzX 11)( −−= zzH

321121 2)1)(2()( −−−−−− −+=−+= zzzzzzzY

]1[][][ −−= nnnh δδ]2[2]1[][ −+−= nnnx δδ

]3[2]2[]1[][ −−−+−= nnnny δδδ8/22/2003 © 2003, JH McClellan & RW Schafer 9

THREE DOMAINS


FREQ-DOMAIN

kjM

kk

j ebeH ωω ˆ

0

ˆ )( −

=∑=

TIME-DOMAIN

∑=

−=M

kk knxbny

0][][

}{ kb


kjM

kk

j

kjM

kk

j

ebeH

ebeH

−

=

−

=

∑

∑

=

=

−

)()(

)(

Domainˆ

ˆ

0

ˆ

ˆ

0

ˆ

ωω

ωω

ω

FREQUENCY RESPONSE ?

Same Form:

SAME COEFFICIENTS

ω̂jez =

kM

kk zbzH

z

−

=∑=

−

0)(

Domain


ANOTHER ANALYSIS TOOL

z-Transform POLYNOMIALS are EASY !ROOTS, FACTORS, etc.

ZEROS and POLES: where is H(z) = 0 ?ZEROS and POLES: where is H(z) = 0 ?

The z-domain is COMPLEXH(z) is a COMPLEX-VALUED function of a COMPLEXVARIABLE z.


ZEROS of H(z)

Find z, where H(z)=01

211)( −−= zzH

21 :at Zero =z

0?01

21

121

=−=− −

zz


ZEROS of H(z)

Find z, where H(z)=0Interesting when z is ON the unit circle.

321 221)( −−− −+−= zzzzH

)1)(1()( 211 −−− +−−= zzzzH

23

21,1 :Roots jz ±= 3/πje±


PLOT ZEROS in z-DOMAIN

3 ZEROSH(z) = 0

3 POLES

UNITCIRCLE


POLES of H(z)

Find z, where Not very interesting for the FIR case

321 221)( −−− −+−= zzzzH

∞→)(zH

3

23 122)(z

zzzzH −+−=

0 :at PolesThree =z


FREQ. RESPONSE from ZEROS

Relate H(z) to FREQUENCY RESPONSEEVALUATE H(z) on the UNIT CIRCLEUNIT CIRCLE

ANGLE is same as FREQUENCY

ωω

ˆ)()( ˆjez

j zHeH ==

1radius CIRCLE,a defines varies)ˆ (asˆ

== ωωjez


ANGLE is FREQUENCY

ωω

ˆ)()( ˆjez

j zHeH ==


FIR Frequency Response

)( and )( of Zeros ˆ zHeH jω


3 DOMAINS MOVIE: FIRZEROS MOVE

)( ω̂jeH][nh

)(zH


njj eeHny )3/()3/( )(][ ππ ⋅=

NULLING PROPERTY of H(z)

When H(z)=0 on the unit circle.Find inputs x[n] that give zero output

y[n]x[n] )(zH

321 221)( −−− −+−= zzzzHωωωω ˆ3ˆ2ˆˆ 221)( jjjj eeeeH −−− −+−=

njenx )3/(][ π=

?)( 3/ =πjeH


PLOT ZEROS in z-DOMAIN

3 ZEROSH(z) = 0

3 POLES

UNITCIRCLE


NULLING PROPERTY of H(z)

Evaluate H(z) at the input “frequency”ωωωω ˆ3ˆ2ˆˆ 221)( jjjj eeeeH −−− −+−=

njj eeHny )3/(3/ )(][ ππ ⋅=njjjj eeeeny )3/(3/33/23/ )221(][ ππππ ⋅−+−= −−−

))1()(2)(21( 23

21

23

21 −−−−+−− jj

0)131311(][ )3/( =⋅+−−+−= njejjny π


FIR Frequency Response

)( and )( of Zeros ˆ zHeH jω


DESIGN PROBLEM

Example:Design a Lowpass FIR filter (Find bk)Reject completely 0.7ππππ, 0.8ππππ, and 0.9ππππEstimate the filter length needed to accomplish this task. How many bk ?

Z POLYNOMIALS provide the TOOLS


PeZ Demo: Zero Placing



Lecture 16IIR Filters: Feedbackand H(z)


READING ASSIGNMENTS

This Lecture:Chapter 8, Sects. 8-1, 8-2 & 8-3

Other Reading:Recitation: Ch. 8, Sects 8-1 thru 8-4

POLES & ZEROSNext Lecture: Chapter 8, Sects. 8-4 8-5 & 8-6


LECTURE OBJECTIVES

Show how to compute the output y[n]FIRST-ORDER CASE (N=1)Z-transform: Impulse Response h[n] H(z)

y[n] = a y[n− ]

=1

N

∑ + bkx[n− k]k=0

M

∑

INFINITE IMPULSE RESPONSE FILTERSDefine IIRIIR DIGITAL Filters Have FEEDBACKFEEDBACK: use PREVIOUS OUTPUTS


THREE DOMAINS


FREQ-DOMAIN

kjM

kk

j ebeH ωω ˆ

0

ˆ )( −

=∑=

TIME-DOMAIN

∑=

−=M

kk knxbny

0][][

}{ kb


Quick Review: Delay by nd

IMPULSE RESPONSE

SYSTEM FUNCTION dnzzH −=)(

][][ dnnxny −=

FREQUENCY RESPONSE

][][ dnnnh −= δ

dnjj eeH ωω ˆˆ )( −=


LOGICAL THREAD

FIND the IMPULSE RESPONSE, h[n]INFINITELY LONGIIRIIR Filters

EXPLOIT THREE DOMAINS:Show Relationship for IIR:

h[n] ↔ H (z) ↔ H(e j ˆ ω )

H(z) = h[n]z− n

n= 0

∞

∑


ONE FEEDBACK TERM

CAUSALITYNOT USING FUTURE OUTPUTS or INPUTS

y[n] = a1y[n −1]+ b0x[n] +b1x[n−1]FIR PART of the FILTER

FEED-FORWARDPREVIOUSFEEDBACK

ADD PREVIOUS OUTPUTS


FILTER COEFFICIENTS

ADD PREVIOUS OUTPUTS

MATLAByy = filter([3,-2],[1,-0.8],xx)

y[n] = 0.8y[n −1]+ 3x[n] −2x[n −1]

SIGN CHANGEFEEDBACK COEFFICIENT


COMPUTE OUTPUT


COMPUTE y[n]

FEEDBACK DIFFERENCE EQUATION:

y[n] = 0.8y[n −1]+ 5x[n]

y[0] = 0.8y[−1] + 5x[0]

NEED y[-1] to get started


AT REST CONDITION

y[n] = 0, for n<0BECAUSE x[n] = 0, for n<0


COMPUTE y[0]

THIS STARTS THE RECURSION:

SAME with MORE FEEDBACK TERMS

y[n] = a1y[n −1]+ a2y[n− 2] + bk x[n− k]k=0

2

∑


COMPUTE MORE y[n]

CONTINUE THE RECURSION:


PLOT y[n]


y[n] = a1y[n −1]+ b0x[n]

IMPULSE RESPONSE

u[n] =1, for n ≥ 0

h[n] = a1h[n −1]+ b0δ[n]

][)(][ 10 nuabnh n=


IMPULSE RESPONSE

DIFFERENCE EQUATION:

Find h[n]

CONVOLUTION in TIME-DOMAIN

h[n]y[n] = h[n]∗ x[n]x[n]

IMPULSE RESPONSE

y[n] = 0.8y[n −1]+ 3x[n]

h[n] = 3(0.8)n u[n]

LTI SYSTEM


PLOT IMPULSE RESPONSE

h[n] = b0 (a1)n u[n] = 3(0.8)n u[n]


Infinite-Length Signal: h[n]POLYNOMIAL Representation

SIMPLIFY the SUMMATION

H(z) = h[n]z −n

n=−∞

∞

∑ APPLIES toAny SIGNAL

∑∑∞

=

−−∞

−∞=

==0

1010 ][)()(n

nnn

n

n zabznuabzH


Derivation of H(z)Recall Sum of Geometric Sequence:

Yields a COMPACT FORMr n

n=0

∞

∑ =1

1− r

111

0

0

110

010

if1

)()(

azza

b

zabzabzHn

n

n

nn

>−

=

==

−

∞

=

−∞

=

− ∑∑


H(z) = z-Transform{ h[n] }

FIRST-ORDER IIR FILTER:y[n] = a1y[n −1]+ b0x[n]

H(z) =b0

1− a1z−1

][)(][ 10 nuabnh n=



ANOTHER FIRST-ORDER IIR FILTER:y[n] = a1y[n −1]+ b0x[n] +b1x[n−1]

H(z) =b0

1− a1z−1 +

b1z−1

1− a1z−1 =

b0 + b1z−1

1 − a1z−1

]1[)(][)(][ 11110 −+= − nuabnuabnh nn

shifta is1−z


CONVOLUTION PROPERTY

MULTIPLICATION of z-TRANSFORMS

CONVOLUTION in TIME-DOMAIN

Y (z) = H(z)X(z)X(z)


H(z)

IMPULSE RESPONSE


STEP RESPONSE: x[n]=u[n]

u[n] =1, for n ≥ 0


DERIVE STEP RESPONSE


PLOT STEP RESPONSE

y[n] = 15 1 − 0.8n+1( )u[n]y[n] = 0.8y[n −1]+ 3u[n]


Lecture 17IIR Filters: H(z) and

Frequency Response


READING ASSIGNMENTS

This Lecture:Chapter 8, Sects. 8-4 8-5 & 8-6

Other Reading:Recitation: Chapter 8, all

POLE-ZERO PLOTS


LECTURE OBJECTIVES

SYSTEM FUNCTION: H(z)H(z) has POLES and ZEROSFREQUENCY RESPONSE of IIR

Get H(z) first

THREE-DOMAIN APPROACHω

ωˆ)()( ˆ

jezj zHeH

==

)()(][ ω̂jeHzHnh ↔↔4/3/2006 © 2003-2006, JH McClellan & RW Schafer 5

THREE DOMAINS


FREQ-DOMAIN

ω

ω

ω

ˆ

1

ˆ

0ˆ

1)(

jN

kjM

kk

j

ea

ebeH

−

=

−

=

∑

∑

−=

TIME-DOMAIN

∑∑==

−+−=M

kk

Nknxbnyany

01][][][

},{ kba

Use H(z) to getFreq. Response

z = e j ˆ ω



FIRST-ORDER IIR FILTER:y[n] = a1y[n −1]+ b0x[n]

H(z) =b0

1− a1z−1

][)(][ 10 nuabnh n=


Typical IMPULSE Response

h[n] = b0 (a1)n u[n] = 3(0.8)n u[n]


First-Order Transform Pair

GEOMETRIC SEQUENCE:

h[n] = banu[n] ↔ H(z) =b

1− a z−1

111

0

0

110

010

if1

)()(

azza

b

zabzabzHn

n

n

nn

>−

=

==

−

∞

=

−∞

=

− ∑∑


DELAY PROPERTY of X(z)

DELAY in TIME<-->Multiply X(z) by z-1

x[n]↔ X(z)

x[n −1] ↔ z −1 X(z)

Proof: x[n −1]z −n

n= −∞

∞

∑ = x[ ]z− ( +1)

=−∞

∞

∑

= z−1 x[ ]z −

= −∞

∞

∑ = z−1X(z)


y[n] = a1y[n −1]+ b0x[n] +b1x[n −1]

Z-Transform of IIR FilterDERIVE the SYSTEM FUNCTION H(z)

Use DELAYDELAY PROPERTY

Y (z) = a1z−1Y(z) + b0X(z) + b1z

−1X(z)EASIER with DELAY PROPERTY


H(z) =Y (z)X(z)

=b0 +b1z

−1

1− a1z−1 =

B(z)A(z)

SYSTEM FUNCTION of IIR

NOTE the FILTER COEFFICIENTS

Y (z) − a1z−1Y(z) = b0 X(z) + b1z

−1X(z)

(1 − a1z−1)Y (z) = (b0 + b1z

−1 )X(z)


SYSTEM FUNCTION

DIFFERENCE EQUATION:

READREAD the FILTER COEFFS:

y[n] = 0.8y[n −1]+ 3x[n] −2x[n −1]

)(8.01

23)( 1

1zX

zzzY ⎟⎟

⎠

⎞⎜⎜⎝

⎛

−−

= −

−

H(z)


CONVOLUTION PROPERTY

MULTIPLICATIONMULTIPLICATION of z-TRANSFORMS

CONVOLUTIONCONVOLUTION in TIME-DOMAIN

Y (z) = H(z)X(z)X(z)


H(z)

IMPULSE RESPONSE


POLES & ZEROS

ROOTS of Numerator & Denominator

POLE: H(z) inf

ZERO: H(z)=0

H(z) =b0 + b1z

−1

1 − a1z−1 → H (z) =

b0z + b1

z − a1

b0z + b1 = 0 ⇒ z = −b1

b0

z − a1 = 0 ⇒ z = a1


EXAMPLE: Poles & Zeros

VALUE of H(z) at POLES is INFINITEINFINITE

POLE at z=0.8

ZERO at z= -1

∞→=−+

=

=−−

−+=

−+

=

−

−

−

−

0)(8.01)(22

)(

0)1(8.01

)1(22)(

8.0122)(

29

154

154

1

1

zH

zH

zzzH


POLE-ZERO PLOT

2 + 2z −1

1 −0.8z−1

ZERO at z = -1

POLE atz = 0.8


FREQUENCY RESPONSE

SYSTEM FUNCTION: H(z)H(z) has DENOMINATORFREQUENCY RESPONSE of IIR

We have H(z)

THREE-DOMAIN APPROACHH(e j ˆ ω ) = H(z ) z = e j ˆ ω

h[n] ↔ H (z) ↔ H(e j ˆ ω )


FREQUENCY RESPONSE

EVALUATE on the UNIT CIRCLE

H(e j ˆ ω ) = H(z ) z = e j ˆ ω


FREQ. RESPONSE FORMULA

ω

ωω

ˆ

ˆˆ

1

1

8.0122)(

8.0122)( j

jj

eeeH

zzzH −

−

−

−

−+

=→−+

=

=2ˆ )( ωjeH ω

ω

ω

ω

ω

ω

ˆ

ˆ

ˆ

ˆ2

ˆ

ˆ

8.0122

8.0122

8.0122

j

j

j

j

j

j

ee

ee

ee

−+

⋅−+

=−+

−

−

−

−

=−−+

+++−

−

ωω

ωω

ˆˆ

ˆˆ

8.08.064.014444

jj

jj

eeee

ωω

ˆcos6.164.1ˆcos88

−+

?ˆ@,40004.0

88)(,0ˆ@2ˆ πωω ω ==

+== jeH


Frequency Response Plot

ω

ωω

ˆ

ˆˆ

8.0122)( j

jj

eeeH −

−

−+

=


UNIT CIRCLE

MAPPING BETWEEN

z = e j ˆ ω

z = 1 ↔ ˆ ω = 0z = −1 ↔ ˆ ω = ±πz = ± j ↔ ˆ ω = ± 1

2 π

z and ˆ ω


3-D VIEW3-D VIEWPOINT:EVALUATE H(z) EVERYWHERE

UNIT CIRCLE

WHERE isthe POLE ?


MOVIE for H(z) in 3-D

POLES to H(z) to Frequency ReponseTWO POLES SHOWN


Frequency Response from H(z)

Walking around the Unit Circle


3 DOMAINS MOVIE: IIR

POLE MOVES

h[n]

H(z)

)( ω̂jeH


PeZ Demo: Pole-Zero Placing


SINUSOIDAL RESPONSE

x[n] = SINUSOID => y[n] is SINUSOIDGet MAGNITUDE & PHASE from H(z)

ωω

ωω

ω

ˆ)()( where)(][ then

][ if

ˆ

ˆˆ

ˆ

jezj

njj

nj

zHeHeeHny

enx

==

=

=


POP QUIZ

Given:

Find the Impulse Response, h[n]

Find the output, y[n]When x[n] = cos(0.25πn)

H(z) =2 + 2z−1

1− 0.8z −1


Evaluate FREQ. RESPONSE

2 + 2z−1

1 − 0.8z −1 at ˆ ω = 0.25π

zero at ω=πˆ ω = 0.25π

0ˆ is 1 == ωz


POP QUIZ: Eval Freq. Resp.

Given:

Find output, y[n], when

Evaluate at

x[n] = cos(0.25πn)

H(z) =2 + 2z−1

1− 0.8z −1

z = e j0.25π

y[n] = 5.182cos(0.25πn − 0.417π )

309.125.0

22

22

182.58.01

)(22)( jj e

ejzH −

− =−

−+= π


CASCADE EQUIVALENT

Multiply the System Functions

y[n]x[n] H(z)

x[n] H1(z) y[n]H2 (z)

H(z) = H1(z)H2 (z)EQUIVALENTSYSTEM



Lecture 183-Domains for IIR


READING ASSIGNMENTS

This Lecture:Chapter 8, all

Other Reading:Recitation: Ch. 8, all

POLES & ZEROSNext Lecture: Chapter 9


LECTURE OBJECTIVES

SECOND-ORDER IIR FILTERSTWO FEEDBACK TERMS

H(z) can have COMPLEX POLES & ZEROSTHREE-DOMAIN APPROACH

BPFs have POLES NEAR THE UNIT CIRCLE

y[n] = a1y[n −1]+ a2 y[n − 2] + bk x[n − k]k=0

2

∑


THREE DOMAINS

a ,bk{ }

Z-TRANSFORM-DOMAIN: poles & zerosPOLYNOMIALS: H(z)

Use H(z) to getFreq. Response

z = e j ˆ ω H(z) =

bkz− k∑

1− a z−∑

FREQ-DOMAIN

ω

ω

ω

ˆ

1

ˆ

0ˆ

1)(

jN

kjM

kk

j

ea

ebeH

−

=

−

=

∑

∑

−=

TIME-DOMAIN

∑∑==

−+−=M

kk

Nknxbnyany

01][][][


Z-TRANSFORM TABLES


SECOND-ORDER FILTERS

Two FEEDBACK TERMS

y[n] = a1y[n −1] + a2y[n − 2]+ b0x[n] + b1x[n −1] + b2x[n − 2]

H(z) = b0 + b1z−1 + b2z−2

1− a1z−1 − a2z−2


MORE POLES

Denominator is QUADRATIC2 Poles: REALor COMPLEX CONJUGATES

H(z) =b0 + b1z

−1 + b2z−2

1 − a1z−1 − a2z

−2 =b0z

2 + b1z + b2

z2 − a1z − a2


TWO COMPLEX POLES

Find Impulse Response ?Can OSCILLATE vs. n“RESONANCE”

Find FREQUENCY RESPONSEFREQUENCY RESPONSEDepends on Pole LocationClose to the Unit Circle?

Make BANDPASS FILTERBANDPASS FILTER

pk( )n= rejθ( )n

= rne jnθ

pole = re jθ

r →1?


2nd ORDER EXAMPLE

H(z) =1− 0.45z −1

1− 0.9z −1 + 0.81z−2

H(z) =0.5

1− 0.9e jπ /3z −1 +0.5

1− 0.9e− jπ / 3z −1

H(z) =1− 0.9cos( π

3 )z−1

(1− 0.9e jπ /3z −1)(1− 0.9e− jπ /3z −1)

][)()9.0(][)cos()9.0(][ 3/3/21

3 nueenunnh njnjnn πππ −+==


h[n]: Decays & Oscillates

1 −0.45z−1

1 −0.9z−1 +0.81z −2

h[n] = (0.9)n cos(π3 n)u[n]

“PERIOD”=6


2nd ORDER Z-transform PAIR

h[n] = rn cos(θn)u[n]

H(z) =1 −r cosθ z −1

1− 2r cosθ z−1 + r 2z −2

h[n] = Arn cos(θn +ϕ )u[n]

H(z) = Acosϕ − rcos(θ −ϕ)z−1

1 −2rcosθz −1 + r2 z−2

GENERAL ENTRY forz-Transform TABLE


2nd ORDER EX: n-Domain

1 −0.45z−1

1 −0.9z−1 +0.81z −2

aa = [ 1, -0.9, 0.81 ];bb = [ 1, -0.45 ];nn = -2:19;hh = filter( bb, aa, (nn==0) );HH = freqz( bb, aa, [-pi,pi/100:pi] );

y[n] = 0.9y[n −1]− 0.81y[n − 2]+ x[n]− 0.45x[n −1]


Complex POLE-ZERO PLOT

1− z −2

1 +0.7225z−2


UNIT CIRCLE

MAPPING BETWEEN

z = e j ˆ ω

z = 1 ↔ ˆ ω = 0z = −1 ↔ ˆ ω = ±πz = ± j ↔ ˆ ω = ± 1

2 π

z and ˆ ω


FREQUENCY RESPONSEfrom POLE-ZERO PLOT

H(e j ˆ ω ) =1− e− j 2 ˆ ω

1+ 0.7225e− j 2 ˆ ω



1 −0.45z−1

1 −0.9z−1 +0.81z −2

h[n] = (0.9)n cos(π3 n)u[n]

“PERIOD”=6



1 −0.45z−1

1 −0.9z−1 +0.81z −2



1− 0.8227z−1

1 −1.6454z−1 + 0.9025z−2

h[n] = (0.95)n cos(π6 n)u[n]

“PERIOD”=12



1− 0.8227z−1

1 −1.6454z−1 + 0.9025z−2


3 DOMAINS MOVIE: IIRPOLE MOVES

h[n]

H(ωωωω)

H(z)


THREE INPUTS

Given:

Find the output, y[n]When

H(z) =5

1+ 0.8z −1

x[n] = cos(0.2πn)x[n] = u[n]x[n] = cos(0.2πn)u[n]


ππ

π 089.02.0

2.0 919.28.015)( j

jj e

eeH =

+= −

SINUSOID ANSWER

Given:

The input:

Then y[n]

x[n] = cos(0.2πn)

y[n] = M cos(0.2πn +ψ )

H(z) =5

1+ 0.8z −1


Step Response

Y(z) = H(z)X(z) = 51+ .8z−1

11− z−1

Y(z) = A1 + .8z−1 + B

1− z−1 = (A + B) + (.8B − A)z−1

1 + .8z−1( )1 − z−1( )⇒ (A + B) = 5 and (.8B − A) = 0

Y(z) = A1 + .8z−1 + B

1− z−1

Partial Fraction Expansion


Step Response

Y(z) =

209

1 + .8z−1 +

259

1− z−1

y[n] = 209

(−.8)nu[n] + 259

u[n]

y[n] → 259

as n → ∞


Stability

Nec. & suff. condition: h[n]n=−∞

∞∑ < ∞

h[n] = b(a)nu[n] ⇔ H(z) = b1− az−1

b a n

n=0

∞∑ < ∞ if a < 1⇒ Pole must be

Inside unit circle


SINUSOID starting at n=0

We’ll look at an example in MATLABcos(0.2ππππn)Pole at –0.8, so an is (–0.8) n

There are two components:TRANSIENT

Start-up region just after n=0; (–0.8) n

STEADY-STATEEventually, y[n] looks sinusoidal.Magnitude & Phase from Frequency Response


ˆ ω 0 =2π10

Cosine input

(−0.8)n

cos(0.2πn)u[n]


STABILITY

When Does the TRANSIENT DIE OUT ?

need a1 <1


STABILITY CONDITION

ALL POLES INSIDE the UNIT CIRCLEUNSTABLE EXAMPLE: POLE @ z=1.1

x[n] = cos(0.2πn)u[n]


BONUS QUESTION

Given:

The input is

Then find y[n]

H(z) =5

1+ 0.8z −1

)5.0cos(4][ ππ −= nnx

?][ =ny


Lecture 19Continuous-Time Signals and Systems


READING ASSIGNMENTS

This Lecture:Chapter 9, Sects 9-1 to 9-5

Other Reading:Recitation: Ch. 9, allNext Lecture: Chapter 9, Sects 9-6 to 9-8


LECTURE OBJECTIVES

Bye bye to D-T Systems for a whileThe UNIT IMPULSE signal

DefinitionProperties

Continuous-time signals and systemsExample systemsReview: LLinearity and TTime-IInvarianceConvolution integral: impulse response


D-T Filtering of C-T Signals

C-to-D D-to-CLTI SystemH(z)

x(t) y(t)

LTI ANALOGSystem

x(t) y(t)

ˆ ω =ω Ts or ω = ˆ ω fs


ANALOG SIGNALS x(t)INFINITE LENGTH

SINUSOIDS: (t = time in secs)PERIODIC SIGNALS

ONE-SIDED, e.g., for t>0UNIT STEP: u(t)

FINITE LENGTHSQUARE PULSE

IMPULSE SIGNAL: δ(t)

DISCRETE-TIME: x[n] is list of numbers4/3/2006 © 2003-2006, JH McClellan & RW Schafer 7

CT Signals: PERIODICx(t) = 10cos(200πt)

Sinusoidal signal

Square Wave INFINITE DURATION


CT Signals: ONE-SIDED

v(t) = e−tu(t)

Unit step signalu(t) =1 t > 00 t < 0

⎧ ⎨ ⎩

One-SidedSinusoid

“Suddenly applied”Exponential


CT Signals: FINITE LENGTH

Square Pulse signal

p(t) = u(t − 2) −u(t − 4)

Sinusoid multipliedby a square pulse


What is an Impulse?

A signal that is “concentrated” at one point.

limΔ→0

δΔ (t) = δ (t)δΔ (t)dt = 1

−∞

∞

∫


Assume the properties apply to the limit:

One “INTUITIVE” definition is:

Defining the Impulse

Unit areaδ(τ )dτ−∞

∞

∫ =1

Concentrated at t=0δ(t) = 0, t ≠ 0

limΔ→0

δΔ (t) = δ (t)


Sampling Property

f (t)δ (t) = f (0)δ (t)

f (t)δ Δ(t) ≈ f (0)δΔ (t)


General Sampling Property

f (t)δ (t − t0 ) = f (t0 )δ (t − t0 )


Properties of the ImpulseConcentrated at one time

Sampling Property

Unit area

Extract one value of f(t)

Derivative of unit step

f (t)δ( t − t0 ) = f (t0 )δ(t − t0)

δ( t − t0 )dt−∞

∞

∫ = 1

δ(t − t0 ) = 0, t ≠ t0

f (t)δ(t − t0 )dt−∞

∞

∫ = f (t0 )

du( t)dt

= δ(t)4/3/2006 © 2003-2006, JH McClellan & RW Schafer 15

Continuous-Time Systems

Examples:Delay

Modulator

Integrator

x(t) y(t)

y(t) = x(t − td )

y(t) = [A + x(t)]cosωct

y(t) = x(τ−∞

t∫ )dτ

Input

Output


CT BUILDING BLOCKS

INTEGRATOR (CIRCUITS)

DIFFERENTIATOR

DELAY by to

MODULATOR (e.g., AM Radio)

MULTIPLIER & ADDER


Ideal Delay:

Mathematical Definition:

To find the IMPULSE RESPONSE, h(t),let x(t) be an impulse, so

h(t) = δ (t − td )

y(t) = x(t − td )


Output of Ideal Delay of 1 secx(t) = e−tu(t)

y(t) = x(t −1) = e−(t−1)u(t − 1)


Integrator:


To find the IMPULSE RESPONSE, h(t),let x(t) be an impulse, so

y(t) = x(τ−∞

t∫ )dτ

h(t) = δ(τ−∞

t

∫ )dτ = u(t)

Running Integral


Integrator:

Integrate the impulse

IF t<0, we get zeroIF t>0, we get one

Thus we have h(t) = u(t) for the integrator

y(t) = x(τ−∞

t∫ )dτ

δ(τ−∞

t

∫ )dτ = u(t)


Graphical Representation

δ(t) =du(t)

dt

u(t) = δ (τ )dτ =1 t > 00 t < 0

⎧ ⎨ ⎩ −∞

t∫


Output of Integrator

)()(

)()(

tutx

dxtyt

∗=

= ∫∞−

ττ

)()1(25.1

0)(

00

)()(

8.0

0

8.0

8.0

tue

tdue

t

duety

t

t

t

−

−

∞−

−

−=

⎪⎩

⎪⎨⎧

≥

<=

=

∫

∫

ττ

ττ

τ

τ

)()( 8.0 tuetx t−=


Differentiator:


To find h(t), let x(t) be an impulse, so

y(t) =dx(t)

dt

h(t) =dδ (t)

dt= δ (1)(t) Doublet


Differentiator Output: y(t) =dx(t)

dt)1()( )1(2 −= −− tuetx t

( )

)1(1)1(2)1()1(2

)1()(

)1(2

)1(2)1(2

)1(2

−+−−=−+−−=

−=

−−

−−−−

−−

ttuetetue

tuedtdty

t

tt

t

δδ


Linear and Time-Invariant (LTI) Systems

If a continuous-time system is both linear and time-invariant, then the output y(t) is related to the input x(t) by a convolution integralconvolution integral

where h(t) is the impulse responseimpulse response of the system.

y(t) = x(τ )h(t − τ )dτ = x(t) ∗ h(t)−∞

∞

∫


Testing for Linearity

x1(t)

x2 (t)

y1(t)

y2 (t)w(t)

y(t)x(t)x2 (t)

x1(t)w(t) y(t)


Testing Time-Invariance

x(t) x(t − t0 )

y(t)

w(t)

y(t − t0 )

t0

w(t) y(t − t0 )

t0


Integrator:

Linear

And Time-Invariant

y(t) = x(τ−∞

t∫ )dτ

[ax1(τ−∞

t∫ ) + bx2 (τ )]dτ = ay1(t) + by2 (t)

w(t) = x(τ − t0−∞

t

∫ )dτ let σ = τ − t0

⇒ w(t) = x(σ )dσ−∞

t−t 0

∫ = y(t - t0 )4/3/2006 © 2003-2006, JH McClellan & RW Schafer 30

Modulator:

NotNot linear--obvious because

NotNot time-invariant

y(t) = [A + x(t)]cosωct

w(t) = [A + x(t − t0 )]cosωct ≠ y(t − t0 )

[A + ax1(t) + bx2 (t)] ≠

[A + ax1(t)]+ [A + bx2 (t)]



Lecture 20Convolution

(Continuous-Time)


READING ASSIGNMENTS

This Lecture:Chapter 9, Sects. 9-6, 9-7, and 9-8

Other Reading:Recitation: Ch. 9, allNext Lecture: Start reading Chapter 10


LECTURE OBJECTIVES

Review of C-T LTI systems Evaluating convolutions

ExamplesImpulses

LTI SystemsStability and causalityCascade and parallel connections





∫∞

∞−

∗=−= )()()()()( thtxdthxty τττ


Testing for Linearity

x1(t)

x2 (t)

y1(t)

y2 (t)w(t)

y(t)x(t)x2 (t)

x1(t)w(t) y(t)


Testing Time-Invariance

x(t) x(t − t0 )

y(t)

w(t)

y(t − t0 )

t0

w(t) y(t − t0 )

t0


Ideal Delay:

Linear

and Time-Invariant

)()( dttxty −=

))(()( 0tttxtw d −−=

))(()( 00 dtttxtty −−=−

)()()()( 2121 tbytayttbxttax dd +=−+−


00 Let )()( tdtxtwt

−=−= ∫∞−

τσττ

Integrator:Linear

And Time-Invariant

ττ dxtyt

)()( ∫∞−

=

)()(

)()()]()([

21

2121

tbytay

dbxdaxdbxaxttt

+=

+=+ ∫∫∫∞−∞−∞−

τττττττ

)()()( 0

0

ttydxtwtt

−==⇒ ∫−

∞−

σσ


Modulator:

NotNot linear--obvious because

NotNot time-invariant

ttxAty cωcos)]([)( +=

)(cos)]([)( 00 ttytttxAtw c −≠−+= ω

)]([)]([)]()([

21

21tbxAtaxA

tbxtaxA+++

≠++





∫∞

∞−

∗=−= )()()()()( thtxdthxty τττ


Convolution of Impulses, etc.

Convolution of two impulses

Convolution of step and shifted impulse

=−∗− )()( 21 tttt δδ )( 21 ttt −−δ

=−∗ )()( 0tttu δ )( 0ttu −


Evaluating a Convolution

∫∞

∞−

∗=−= )()()()()( txthdtxhty τττ

)()( tueth t−=)1()( −= tutx


“Flipping and Shifting”)(τx

g(τ ) = x(−τ ) = u(−τ −1)

g(τ − t) = x(−(τ − t)) = x(t − τ )

“flipping”

“flipping and shifting”

t1−t 11/3/2003 © 2003, JH McClellan & RW Schafer 15

Evaluating the Integral

>−

<−= ∫

−− 01

010)( 1

0

tde

tty t

ττ


Solution

1 0)( t<ty =

1 1

)(

)1(

1

0

1

0

≥−=

−==

−−

− −−−∫te

edety

t

t tττ τ


Convolution GUI


General Convolution Example

)(00

0

00

0)()(

)()()()()(

0)(

tuabee

t

tba

ee

t

tdeeedtueue

thtxdthxty

btatbtat

tbabt

tba

−−=

<

>+−

−=

<

>=−=

∗=−=

−−−−

−−∞

∞−

−−−

∞

∞−

∫∫

∫

ττττ

τττ

ττττ

)()( tuetx at−= abtueth bt ≠= − ),()(


Special Case: u(t)

)()1(1

)()()()()(

tuea

thtxdthxty

at−

∞

∞−

−=

∗=−= ∫ τττ

0),()( ≠= − atuetx at )()( tuth =

)()1(21)(

2 if2 tuety

at−−=

=


Convolve Unit Steps

)(00

000

01)()(

)()()()()(

0

tutt

ttt

tddtuu

thtxdthxty

t

=

<

>=

<

>=−=

∗=−=

∫∫

∫

∞

∞−

∞

∞−

ττττ

τττ

)()( tutx = )()( tuth =

Unit Ramp 11/3/2003 © 2003, JH McClellan & RW Schafer 21

Convolution is Commutative

∫

∫

∫

∞

∞−

∞−

∞

∞

∞−

∗=−=

−−=∗

−=−=

−=∗

)()()()(

)()()()(

and let

)()()()(

thtxdxth

dxthtxth

ddt

dtxhtxth

σσσ

σσσ

τστσ

τττ


Cascade of LTI Systems

δ(t) h1(t) h1(t)∗ h2(t)

δ(t) h2 (t) h2 (t)∗h1(t)

h(t) = h1(t)∗ h2 (t) = h2(t) ∗h1(t)


Stability

A system is stable if every bounded input produces a bounded output.A continuous-time LTI system is stable if and only if

h(t)dt < ∞−∞

∞∫


Causal Systems

A system is causal if and only if y(t0)depends only on x(τ) for τ< t0 .

An LTI system is causal if and only if

0for 0)( <= tth


Substitute x(t)=ax1(t)+bx2(t)

Therefore, convolution is linear.

)()(

)()()()(

)()]()([)(

21

21

21

tbytay

dthxbdthxa

dthbxaxty

+=

−+−=

−+=

∫ ∫

∫∞

∞−

∞

∞−

∞

∞−

ττττττ

ττττ

Convolution is Linear


Convolution is Time-Invariant

Substitute x(t-t0)

w(t) = h(τ )x((t − τ) − to)dτ−∞

∞

∫

= h(τ )x((t − to ) −τ )dτ−∞

∞

∫= y(t − to )


Lecture 21Frequency Response of

Continuous-Time Systems


READING ASSIGNMENTS

This Lecture:Chapter 10, all

Other Reading:Recitation: Ch. 10 all, start Ch 11Next Lecture: Chapter 11


LECTURE OBJECTIVES

Review of convolutionTHETHE operation for LTILTI Systems

Complex exponential input signalsFrequency ResponseCosine signals

Real part of complex exponential

Fourier Series thru H(jω)These are Analog Filters


LTI Systems

Convolution defines an LTI system

Response to a complex exponential gives frequency response H(jω)

y(t) = h( t)∗ x(t) = h(τ )−∞

∞

∫ x(t −τ )dτ


Thought Process #1

SUPERPOSITION (Linearity)Make x(t) a weighted sum of signalsThen y(t) is also a sum—same weights

• But DIFFERENT OUTPUT SIGNALS usually

Use SINUSOIDS• “SINUSOID IN GIVES SINUSOID OUT”

Make x(t) a weighted sum of sinusoidsThen y(t) is also a sum of sinusoids

Different Magnitudes and Phase

LTI SYSTEMS: Sinusoidal Response


Thought Process #2

SUPERPOSITION (Linearity)Make x(t) a weighted sum of signals

Use Use SINUSOIDSSINUSOIDSAnyAny x(t) = weighted sum of sinusoidsx(t) = weighted sum of sinusoidsHOW?HOW? Use FOURIER ANALYSIS INTEGRALUse FOURIER ANALYSIS INTEGRAL

To find the weights from x(t)To find the weights from x(t)

LTI SYSTEMS:Frequency Response changes each sinusoidal component


Complex Exponential Inputtjjtjj eAejHtyeAetx ωϕωϕ ω )()()( ==

∫∞

∞−

−= ττ τωϕ deAehty tjj )()()(

FrequencyResponse∫

∞

∞−

−= ττω ωτ dehjH j)()(

tjjj eAedehty ωϕωτ ττ ⎟⎟⎠

⎞⎜⎜⎝

⎛= ∫

∞

∞−

−)()(


When does H(jω) Exist?

When is ?

Thus the frequency response exists if the LTI system is a stable system.

H( jω) = h(τ )−∞

∞

∫ e− jωτ dτ ≤ h(τ )−∞

∞

∫ e− jωτ dτ

H( jω ) ≤ h(τ )−∞

∞∫ dτ < ∞

H( jω) < ∞


Suppose that h(t) is:h(t) = e− tu(t)

H( jω ) = e− aτu(−∞

∞

∫ τ )e− jωτdτ = e−(a+ jω)τdτ0

∞

∫

H( jω ) =e−(a+ jω)τ

−(a + jω ) 0

∞

=e− aτ e− jωτ

−(a + jω ) 0

∞

=1

a + jω

h(t) = e− atu(t) ⇔ H( jω ) =1

a + jω

a = 1

a > 0


Magnitude and Phase Plots

11+ jω

=1

1+ ω 2

∠H( jω ) = −atan(ω)

H( jω ) =1

1 + jω

H(− jω ) = H∗( jω)


Freq Response of Integrator?

Impulse Responseh(t) = u(t)

NOT a Stable SystemFrequency response H(jω) does NOT exist

h(t) = e− atu(t) ⇔ H( jω ) =1

a + jω→

1jω

?

Need another term

a → 0“Leaky” Integrator (a is small)

Cannot build a perfect Integral

4/3/2006 © 2003-2006, JH McClellan & RW Schafer 13( ) tjtjttj

tj

eeety

etxdd ωωω

ω

−− ==

=)()(

)(

Ideal Delay: y(t) = x(t − td )

H( jω ) = e− jωtd

H( jω ) = δ(τ − td )−∞

∞

∫ e− jωτdτ = e− jωtd

H( jω )


Ideal Lowpass Filter w/ Delay

HLP( jω ) =e− jωtd ω < ωco

0 ω > ωco

⎧ ⎨ ⎩

fco "cutoff freq."Magnitude

Linear Phase

ω

ω


Sinusoid in Gives Sinusoid out


Example: Ideal Low Pass

HLP( jω ) =e− j 3ω ω < 2

0 ω > 2⎧ ⎨ ⎩

== )(10)( 5.13/ tyeetx tjjπ tjj eejH 5.13/10)5.1( π

( ) )3(5.13/5.13/5.4 1010)( −− == tjjtjjj eeeeety ππ


Cosine Input

x(t) = Acos(ω0t +φ) =A2

e jφe jω0 t +A2

e− jφ e− jω0 t

y(t) = H( jω0 )A2

ejφ ejω 0 t + H(− jω 0)A2

e− jφe− jω0 t

Since H(− jω0 ) = H∗ (jω0 )

y(t) = A H( jω0 ) cos(ω0t + φ + ∠H( jω0 ))


Review Fourier SeriesANALYSIS

Get representation from the signalWorks for PERIODIC Signals

Fourier SeriesINTEGRAL over one period

ak =1T0

x(t)e− jω 0ktdt0

T0

∫4/3/2006 © 2003-2006, JH McClellan & RW Schafer 19

General Periodic Signalsx(t) = x(t + T0 )

T0−2T0 −T0 2T00 t

x(t) = akejω 0k t

k =−∞

∞

∑

ak =1T0

x(t)e− jω 0ktdt0

T0

∫

Fundamental Freq.ω0 = 2π / T0 = 2πf0

Fourier Synthesis

Fourier Analysis


Square Wave Signal x(t) = x(t + T0 )

T0−2T0 −T0 2T00 t

ak =e− jω0kt

− jω0kT0 0

T0 / 2

−e− jω 0kt

− jω0kT0 T0 /2

T0

=1− e− jπk

jπk

ak =1T0

(1)e− jω0 ktdt +1T0

(−1)e− jω 0ktdtT0 / 2

T0

∫0

T0 / 2

∫



ak =

1− e− jπk

jπk=

2jπk

k = ±1,±3,…

0 k = 0,±2,±4,…

⎧ ⎨ ⎪

⎩ ⎪

ω0 = 2π(25)


LTI Systems with Periodic Inputs

By superposition,ake

jω0kt H( jω0k )akejω 0kt

y(t) = ak H( jω 0k )e jω 0k t = bkejω0k t

k = −∞

∞

∑k= −∞

∞

∑

bk = akH( jω 0k)

Output has same frequencies


Ideal Lowpass Filter (100 Hz)

y(t) =4π

sin 50πt( ) +4

3πsin 150πt( )

H( jω ) =1 ω < ωco0 ω > ωco

⎧ ⎨ ⎩

fco "cutoff freq."


Ideal Lowpass Filter (200 Hz)


⎧ ⎨ ⎩

fco "cutoff"

y(t) =4π

sin 50πt( ) +4

3πsin 150πt( ) +

45π

sin 250πt( ) +4

7πsin 350πt( )


Ideal Bandpass Filter

y(t) = 2j 7π ej 2π (175) t − 2

j 7π e− j 2π (175)t =4

7πcos 2π(175)t − 1

2 π( )

H( jω ) =1 ω ± ωc < 1

2 ωB

0 elsewhere⎧ ⎨ ⎩

What is the ouput signal ?

Passband Passband


Example

y(t) = ake− jω 0k td e jω0k t = ake

jω0k ( t −td )

k =−∞

∞

∑k= −∞

∞

∑

bk = akH( jω 0k) = ake− jω0k t d

H( jω ) = e− jω td

x(t) = ake

jω 0k t

k =−∞

∞

∑ y(t) = bkejω0k t

k = −∞

∞

∑

∴ y(t) = x(t − td )



Lecture 22Introduction to the Fourier Transform


READING ASSIGNMENTS

This Lecture:Chapter 11, Sects. 11-1 to 11-4

Other Reading:Recitation: Ch. 10

And Chapter 11, Sects. 11-1 to 11-4Next Lecture: Chapter 11, Sects. 11-5, 11-6


LECTURE OBJECTIVESReview

Frequency ResponseFourier Series

Definition of Fourier transform

Relation to Fourier SeriesExamples of Fourier transform pairs

∫∞

∞−

−= dtetxjX tjωω )()(


Everything = Sum of Sinusoids

One Square Pulse = Sum of Sinusoids???????????

Finite LengthNot Periodic

Limit of Square Wave as Period infinityIntuitive Argument


Fourier Series: Periodic x(t)x(t) = x(t + T0 )

T0−2T0 −T0 2T00 t

x(t) = akejω 0k t

k =−∞

∞

∑

ak =1T0

x(t)e− jω0ktdt−T0/ 2

T0 / 2

∫

Fundamental Freq.ω0 = 2π / T0 = 2πf0

Fourier Synthesis

Fourier Analysis3/27/2004 © 2003, JH McClellan & RW Schafer 7

Square Wave Signal

ak =e− jω0kt

− jω0kT0 −T0 /4

T0 / 4

=e− jπk / 2 − ejπk / 2

− j2πk=

sin(πk / 2)πk

x(t) = x(t + T0 )

T0−2T0 −T0 2T00 t

∫−

−=4/

4/0

0

0

0)1(1 T

T

tkjk dteT

a ω



±±=

±±=≠==

…

…

,4,20

,3,1,00)2/sin(k

k

kkak π

π


What if x(t) is not periodic?

Sum of Sinusoids?Non-harmonically related sinusoids Would not be periodic, but would probably be non-zero for all t.

Fourier transformgives a “sum” (actually an integralintegral) that involves ALLALL frequenciescan represent signals that are identically zero for negative t. !!!!!!!!!


Limiting Behavior of FS

T0=2T

T0=4T

T0=8T


Limiting Behavior of Spectrum

T0=2T

T0=4T

T0=8T

)(Plot

0 kaT


FS in the LIMIT (long period)

Fourier Synthesis

Fourier Analysis

( ) ( ) ∫∑∞

∞−

∞

−∞=== ωω ω

ππω

π dejXtxeaTtx tjT

tkj

kkT )()()( 2

1202

10

0

0

∫∫∞

∞−

−

−

− == dtetxjXdtetxaT tjT

T

tkjTk

ωω ω )()()(2/

2/0

0

0

0

0

ωπ dTT

=∞→ 0

2lim0

ωπ =∞→

kTT 0

2lim0

)(lim 00

ωjXaT kT=

∞→


Fourier Transform Defined

For non-periodic signalsFourier Synthesis

Fourier Analysis

∫∞

∞−


∫∞

∞−

= ωω ωπ dejXtx tj)()( 21


Example 1: x(t) = e−atu(t)

X( jω) = 1a + jω

X( jω ) = e−at

0

∞

∫ e− jω tdt =0

∞

∫ e− (a+ jω )tdt

X( jω ) = −e−ate− jω t

a + jω 0

∞

=1

a + jω

a > 0


Frequency Response

Fourier Transform of h(t) is the Frequency Response

ωω

jjHtueth t

+=⇔= −

11)()()(

)()( tueth t−=


Magnitude and Phase Plots

ωω

jajH

+= 1)(

−=∠ −

ajH ωω 1tan)(

22

11

ωω +=

+ aja

)()( ωω jHjH ∗=− 3/27/2004 © 2003, JH McClellan & RW Schafer 17

X( jω) = sin(ωT / 2)ω / 2( )

Example 2: x(t) =1 t < T / 20 t > T / 2

X( jω) = e− jω t

− jω −T / 2

T /2

= e− jωT / 2 − e jωT /2

− jω

X( jω) = (1)e− jωtdt−T / 2

T /2∫ = e− jωtdt

−T / 2

T /2∫


x(t) =1 t < T / 20 t > T / 2

⇔ X( jω ) = sin(ωT / 2)ω / 2( )


Example 3:

>

<=

b

bjX

ωω

ωωω

0

1)(

tttx b

πω )sin()( =

∫∫−

∞

∞−

==b

b

dedejXtx tjtjω

ω

ωω ωπ

ωωπ

121)(

21)(

jtee

jtetx

tjtjtj bbb

b

ωωω

ω

ω

ππ

−

−

−==21

21)(


>

<=⇔=

b

bb jXtttx

ωω

ωωω

πω

0

1)()sin()(


Example 4:

X( jω) = δ (t)e− jωtdt−∞

∞∫ = 1

Shifting Property of the Impulse

)()( 0tttx −= δ

0)()( 0tjtj edtettjX ωωδω −

∞

∞−

− =−= ∫


x(t) = δ (t) ⇔ X( jω ) = 1


Example 5: X( jω) = 2πδ (ω − ω0 )

x(t) = 12π 2πδ (ω − ω0 )e jωtdω

−∞

∞∫ = e jω0t

x(t) = 1 ⇔ X( jω) = 2πδ (ω)

x(t) = e jω0 t ⇔ X( jω) = 2πδ (ω − ω0 )

x(t) = cos(ω0t) ⇔X( jω) = πδ (ω − ω0) + πδ (ω + ω0)


x(t) = cos(ω0t) ⇔X( jω) = πδ (ω − ω0) + πδ (ω + ω0)


Table of Fourier Transforms

x(t) = e−atu(t) ⇔ X( jω ) = 1a + jω

x(t) =1 t < T / 20 t > T / 2

⇔ X( jω ) = sin(ωT / 2)ω / 2( )

x(t) = sin(ω0t)π t( ) ⇔ X( jω ) =

1 ω < ω00 ω > ω0

x(t) = δ (t − t0) ⇔ X( jω ) = e− jωt0

x(t) = e jω0 t ⇔ X( jω) = 2πδ (ω − ω0 )



Lecture 23Fourier TransformProperties


READING ASSIGNMENTS

This Lecture:Chapter 11, Sects. 11-5 to 11-9Tables in Section 11-9

Other Reading:Recitation: Chapter 11, Sects. 11-1 to 11-9Next Lectures: Chapter 12 (Applications)


LECTURE OBJECTIVESThe Fourier transform

More examples of Fourier transform pairsBasic properties of Fourier transforms

Convolution propertyMultiplication property

∫∞

∞−



Fourier Transform

Fourier Analysis(Forward Transform)∫

∞

∞−


Fourier Synthesis(Inverse Transform)∫

∞

∞−

= ωωπ

ω dejXtx tj)(21)(

)()(Domain-FrequencyDomain-Time

ωjXtx ⇔⇔


WHY use the Fourier transform?

Manipulate the “Frequency Spectrum”Analog Communication Systems

AM: Amplitude Modulation; FMWhat are the “Building Blocks” ?

Abstract Layer, not implementation

Ideal Filters: mostly BPFsFrequency Shifters

aka Modulators, Mixers or Multipliers: x(t)p(t)


Frequency Response

Fourier Transform of h(t) is the Frequency Response

ωω

jjHtueth t

+=⇔= −

11)()()(

)()( tueth t−=


2/)2/sin()(

2/0

2/1)(

ωωω TjX

Tt

Tttx =⇔

>

<=


>

<=⇔=

b

bb jXtttx

ωω

ωωω

πω

0

1)()sin()(


0)()()( 0tjejXtttx ωωδ −=⇔−=

00 =t


Table of Fourier Transforms

)(2)()( ctj jXetx c ωωπδωω −=⇔=

0)()()( 0tjejXtttx ωωδ −=⇔−=

>

<=⇔=

b

bb jXtttx

ωω

ωωω

πω

0

1)()sin()(

2/)2/sin()(

2/0

2/1)(

ωωω TjX

Tt

Tttx =⇔

>

<=

ωω

jjXtuetx t

+=⇔= −

11)()()(


)()()()cos()( ccc jXttx ωωπδωωπδωω ++−=⇔=


Fourier Transform of a General Periodic Signal

If x(t) is periodic with period T0 ,

∫∑ −∞

−∞===

0

00

00)(1)(

Ttjk

kk

tjkk dtetx

Taeatx ωω

)(2 since Therefore, 00 ωωπδω ke tjk −⇔

∑∞

−∞=−=

kk kajX )(2)( 0ωωδπω


Square Wave Signal x(t) = x(t + T0 )

T0−2T0 −T0 2T00 t

ak =e− jω0kt

− jω0kT0 0

T0 / 2

−e− jω 0kt

− jω0kT0 T0 /2

T0

=1− e− jπk

jπk

ak =1T0

(1)e− jω0 ktdt +1T0

(−1)e− jω 0ktdtT0 / 2

T0

∫0

T0 / 2

∫


Square Wave Fourier Transform

X( jω ) = 2π akδ(ω − kω0 )k=−∞

∞

∑

x(t) = x(t + T0 )

T0−2T0 −T0 2T00 t


Table of Easy FT Properties

ax1(t) + bx2(t) ⇔ aX1( jω) + bX2( jω )

x(t − td ) ⇔ e− jωtd X( jω )

x(t)e jω0t ⇔ X( j(ω − ω0))

Delay Property

Frequency Shifting

Linearity Property

x(at) ⇔ 1|a | X( j(ω

a ))Scaling


Scaling Property

expands)(shrinks;)2( 221 ωjXtx

)(

)()(

1

)/(

aa

adajtj

jX

exdteatx

ω

λλωω λ

=

= ∫∫∞

∞−

−∞

∞−

−

)()( 1aa jXatx ω⇔


Scaling Property

)()( 1aa jXatx ω⇔

)2()( 12 txtx =


Uncertainty Principle

Try to make x(t) shorterThen X(jωωωω) will get widerNarrow pulses have wide bandwidth

Try to make X(jωωωω) narrowerThen x(t) will have longer duration

Cannot simultaneously reduce time Cannot simultaneously reduce time duration and bandwidthduration and bandwidth


Significant FT Properties

x(t) ∗h(t) ⇔ H( jω )X( jω )

x(t)e jω0t ⇔ X( j(ω − ω0 ))

x(t)p(t) ⇔ 12π X( jω )∗P( jω )

dx(t)dt

⇔ ( jω)X( jω)

Differentiation Property


Convolution Property

Convolution in the time-domain

corresponds to MULTIPLICATIONMULTIPLICATION in the frequency-domain

y(t) = h(t) ∗ x(t) = h(τ )−∞

∞∫ x(t − τ )dτ

Y( jω ) = H( jω )X( jω )

y(t) = h(t) ∗ x(t)x(t)

Y( jω ) = H( jω )X( jω )X( jω)


Convolution Example

Bandlimited Input Signal“sinc” function

Ideal LPF (Lowpass Filter)h(t) is a “sinc”

Output is BandlimitedConvolve “sincs”


Ideally Bandlimited Signal

>

<=⇔=

πω

πωω

ππ

1000

1001)()100sin()( jX

tttx

πω 100=b


Convolution Example

sin(100π t)πt

∗ sin(200πt)π t

=

x(t) ∗h(t) ⇔ H( jω )X( jω )

sin(100π t)πt


Cosine Input to LTI SystemY (jω) = H( jω )X( jω)

= H( jω )[πδ(ω − ω0 ) +πδ(ω +ω 0)]

= H( jω0 )πδ (ω −ω0 ) + H(− jω0 )πδ (ω +ω0 )

y(t) = H ( jω0 ) 12 e

jω0t +H(− jω0 ) 12 e

− jω 0t

= H( jω0 ) 12 e

jω0t +H *( jω 0)12 e

− jω0t

= H( jω0 ) cos(ω 0t +∠H( jω0 ))


Ideal Lowpass Filter

Hlp( jω)

ωco−ωco

y(t) = x(t) if ω0 < ωco

y(t) = 0 if ω0 > ωco3/27/2004 © 2003, JH McClellan & RW Schafer 27


y(t) = 4π

sin 50πt( ) + 43π

sin 150πt( )

fco "cutoff freq."



Signal Multiplier (Modulator)

Multiplication in the time-domain corresponds to convolution in the frequency-domain.

Y( jω ) = 12π X( jω) ∗P( jω )

y(t) = p(t)x(t)

X( jω )x(t)

p(t)

Y( jω ) = 12π

X( jθ )−∞

∞∫ P( j(ω −θ ))dθ


Frequency Shifting Property

x(t)e jω0t ⇔ X( j(ω − ω0))

y(t) =sin 7t

πtejω 0 t ⇔ Y ( jω ) =

1 ω 0−7 < ω <ω0 +70 elsewhere

))((

)()(

0

)( 00

ωω

ωωωω

−=

= ∫∫∞

∞−

−−∞

∞−

−

jX

dtetxdtetxe tjtjtj


y(t) = x(t)cos(ω0t) ⇔

Y( jω ) = 12X( j(ω − ω0 )) + 1

2X( j(ω + ω0))

x(t)



dx (t )dt

=ddt

12π

X ( jω )e jω t dω−∞

∞∫

=1

2π ( jω )X( jω )e jω tdω−∞

∞∫

ddte−atu(t)( )= −ae−atu(t) + e−atδ (t)

= δ (t) − ae−atu(t)⇔ jωa + jω

Multiply by jωωωω



Lecture 24Amplitude Modulation (AM)


LECTURE OBJECTIVES

Review of FT propertiesConvolution <--> multiplicationFrequency shifting

Sinewave Amplitude ModulationAM radio

Frequency-division multiplexingFDM

Reading: Chapter 12, Section 12-2


Table of Easy FT Properties

ax1(t) + bx2(t) ⇔ aX1( jω) + bX2( jω )


x(t)e jω0t ⇔ X( j(ω − ω0))

Delay Property

Frequency Shifting

Linearity Property

x(at) ⇔ 1|a | X( j(ω

a ))Scaling


Table of FT Properties

x(t) ∗h(t) ⇔ H( jω )X( jω )

x(t)e jω0t ⇔ X( j(ω − ω0))

x(t)p(t) ⇔ 12π

X( jω )∗ P( jω )

dx(t)dt

⇔ ( jω )X( jω )



Frequency Shifting Property

e− jω0 t x(t )e− jω tdt−∞

∞

∫ = x(t)e− j (ω −ω 0 )t dt−∞

∞

∫= X( j(ω −ω 0))

x(t)e jω0t ⇔ X( j(ω − ω0))

y(t) =sin 7t

πtejω 0 t ⇔ Y ( jω ) =

1 ω 0−7 < ω <ω0 +70 elsewhere

⎧ ⎨ ⎩


Convolution Property

Convolution in the time-domain

corresponds to MULTIPLICATIONMULTIPLICATION in the frequency-domain

y(t) = h(t) ∗ x(t) = h(τ )−∞

∞∫ x(t − τ )dτ

Y( jω ) = H( jω )X( jω )

y(t) = h(t) ∗ x(t)x(t)

Y( jω ) = H( jω )X( jω )X( jω)


Cosine Input to LTI SystemY (jω) = H( jω )X(jω)

= H( jω )[πδ(ω − ω0 ) +πδ(ω +ω 0)]

= H( jω0 )πδ (ω −ω0 ) + H(− jω0 )πδ (ω +ω0 )

y(t) = H ( jω0 ) 12 e jω0t + H(− jω0 ) 1

2 e− jω 0t

= H( jω0 ) 12 e jω0t + H *( jω 0)

12 e− jω0t

= H( jω0 ) cos(ω 0t +∠H( jω0 )) 4/19/2005 © 2003, JH McClellan & RW Schafer 9


Hlp( jω )

ωco−ωco

y(t) = x(t) if ω0 < ωco

y(t) = 0 if ω0 > ωco


Ideal LPF: Fourier Series

y(t) =4π

sin 50πt( ) +4

3πsin 150πt( )

fco "cutoff freq."


⎧ ⎨ ⎩


The way communication systems work

How do we sharebandwidth ?



x(t) ∗h(t) ⇔ H( jω )X( jω )

x(t)e jω0t ⇔ X( j(ω − ω0))

x(t)p(t) ⇔ 12π

X( jω )∗ P( jω )

dx(t)dt

⇔ ( jω )X( jω )



Signal Multiplier (Modulator)

Multiplication in the time-domain corresponds to convolution in the frequency-domain.

Y( jω ) = 12π

X( jω ) ∗ P( jω )

y(t) = p(t)x(t)

X( jω)x(t)

p(t)

Y( jω ) =1

2πX( jθ )

−∞

∞∫ P( j(ω −θ ))dθ


)()()()()()( 21 ωωω π jPjXjYtptxty ∗=⇔=

[ ])()()()()cos()()(

21

cc

c

jXjYttxty

ωωπδωωπδωωω

π ++−∗=⇔=

)()()()cos()(

cc

c

jPttp

ωωπδωωπδωω

++−=⇔=

))(())(()( 21

21

cc jXjXjY ωωωωω ++−=


Amplitude Modulator

x(t) modulates the amplitude of the cosine wave. The result in the frequency-domain is two shifted copies of X(jω).

y(t) = x(t)cos(ωct)

X( jω)x(t)

cos(ωct)Y( jω ) = 1

2X( j(ω − ωc ))

+ 12

X( j(ω + ωc ))


))(())(()()cos()()(

21

21

cc

c

jXjXjYttxty

ωωωωωω

++−=⇔=

)())sin((

)())sin(()(

)cos()()(

c

c

c

c

c

TTjY

ttxty

ωωωω

ωωωωω

ω

++

+−−

=

⇔=

)()sin(2)(

0

1)(

ωωω TjX

Tt

Tttx =⇔

⎪⎩

⎪⎨⎧

>

<=


x(t)

ωc−ωc

))((21

cjX ωω + ))((21

cjX ωω −

))(())(()()cos()()(

21

21

cc

c

jXjXjYttxty

ωωωωωω

++−=⇔=


DSBAM Modulator

If X(jω)=0 for |ω|>ωb and ωc >ωb,the result in the frequency-domain is two shifted and scaled exact copies of X(jω).


X( jω)x(t)

cos(ωct)Y( jω ) = 1

2X( j(ω − ωc ))

+ 12

X( j(ω + ωc ))


DSBAM Waveform

In the time-domain, the “envelope” of sine-wave peaks follows |x(t)|


Double Sideband AM (DSBAM)“Typical” bandlimitedinput signal

Frequency-shiftedcopies Upper sidebandLower sideband


DSBAM DEmodulator

w(t) = x(t)[cos(ωct)]2 = 1

2x(t) + 1

2x(t)cos(2ωct)

W( jω ) = 12

X( jω ) + 14

X( j(ω − 2ωc )) + 14

X( j(ω + 2ωc ))

V ( jω ) = H( jω )W( jω )

w(t) v(t)x(t)

cos(ωct) cos(ωct)



DSBAM Demodulation

V ( jω ) = H( jω )W( jω ) = X( jω ) if ωb < ωco < 2ωc − ωb

H( jω ) =2 | ω |< ωco0 |ω |> ωco

⎧ ⎨ ⎩


Frequency-Division Multiplexing (FDM)

Shifting spectrum of signal to higher frequency:

Permits transmission of low-frequency signals with high-frequency EM wavesBy allocating a frequency band to each signal multiple bandlimited signals can share the same channelAM radio: 530-1620 kHz (10 kHz bands)FM radio: 88.1-107.9 MHz (200 kHz bands)


FDM Block Diagram (Xmitter)

cos(ωc1t)

cos(ωc 2t)

ωc1 ωc2

Spectrum of inputsmust be bandlimitedNeed ωc2 − ωc1 > 2ωb


Frequency-Division De-Mux

cos(ωc1t)

cos(ωc 2t)

ωc1 ωc2


Bandpass Filters for De-Mux


Pop Quiz: FT thru LPF

∑∞

−∞=

−=↔k

kjXtx )30(4)()(Input πωδπω

cofor a value find then,2)( isoutput theIf ω=ty

1

coωcoω−

)(LP ωjH

ω



Lecture 25Sampling and Reconstruction(Fourier View)


LECTURE OBJECTIVES

Sampling Theorem RevisitedGENERAL: in the FREQUENCY DOMAINFourier transform of sampled signalReconstruction from samples

Reading: Chap 12, Section 12-3

Review of FT propertiesConvolution multiplicationFrequency shiftingReview of AM




x(t)e jω0t ⇔ X( j(ω − ω0))

Delay Property

Frequency Shifting

x(at) ⇔ 1|a | X( j(ω

a ))Scaling

x(t) ∗h(t) ⇔ H( jω )X( jω )


Amplitude Modulator

x(t) modulates the amplitude of the cosine wave. The result in the frequency-domain is two SHIFTED copies of X(jω).

y(t) = x(t)cos(ωct +ϕ )

X( jω )x(t)

cos(ωct + ϕ)Y ( jω) = 1

2 e jϕX( j(ω −ωc))

+ 12 e− jϕX( j(ω + ωc))

Phase


DSBAM: Frequency-Domain“Typical” bandlimitedinput signal

Frequency-shiftedcopies

))((21

cj jXe ωωϕ +− ))((2

1c

j jXe ωωϕ −

Upper sidebandLower sideband

)( ωjX


DSBAM Demod Phase Synch

w(t) v(t)x(t)

cos(ωct) )cos( ϕω +tc

)cos()()( ttxty cω=

))2(())2(()()()(

41

41

41

41

cj

cj

jj

jXejXejXejXejW

ωωωωωωω

ϕϕ

ϕϕ

++−++=

−

−

? ifwhat )()cos()( 21

21 πϕωϕω == jXjV


Quadrature Modulator

TWO signals on ONE channel: “out of phase” Can you “separate” them in the demodulator ?

))(())(())(())(()(

22121

22121

cj

c

cj

c

jXjXjXjXjY

ωωωωωωωωω++++

−−−=


Demod: Quadrature System

)cos( ϕω +tc

)()()()()(

22/

41

141

22/

41

141

ωωωωω

ϕπϕ

ϕπϕ

jXeejXejXeejXejV

jjj

jjj

+++= −−−

0 if )()( 1 == ϕtxtv

2/ if )()( 2 πϕ −== txtv

))(())((

))(())(()(

22121

22121

cj

c

cj

c

jXjXjXjXjY

ωωωωωωωωω

++++

−−−=


Quadrature Modulation: 4 sigs

8700 Hz

3600 Hz


Ideal C-to-D Converter

• Mathematical Model for A-to-D

x[n] = x(nTs )

FOURIERTRANSFORMof xs(t) ???


Periodic Impulse Train

ωs = 2πTs

∑∑∞

−∞=

∞

−∞==−=k

tjkk

ns

seanTttp ωδ )()(

s

T

T

tjk

sk T

dtetT

as

s

s1)(1 2/

2/

== ∫−

− ωδFourier Series


FT of Impulse Train

∑∑∞

−∞=

∞

−∞=−=↔−=

ks

sns k

TjPnTttp )(2)()()( ωωδπωδ

ss T

πω 2=


Impulse Train Sampling

xs (t) = x(t) δ (t − nTs )n=−∞

∞∑ = x(t)δ (t − nTs )

n=−∞

∞∑

xs(t) = x(nTs)δ(t−nTs)n=−∞

∞∑


Illustration of Samplingx(t)

x[n] = x(nTs )

∑∞

−∞=−=

nsss nTtnTxtx )()()( δ

n

t


Sampling: Freq. Domain

EXPECTFREQUENCYSHIFTING !!!

∑∑∞

−∞=

∞

−∞==−=k

tjkk

ns

seanTttp ωδ )()(

∑∞

−∞==k

tjkk

sea ω


Frequency-Domain Analysis

xs (t) = x(t) δ (t − nTs )n=−∞

∞∑ = x(nTs )δ (t − nTs )

n=−∞

∞∑

xs (t) = x(t) 1Tsk=−∞

∞∑ e jkωst = 1

Tsx(t)

k=−∞

∞∑ e jkωst

Xs ( jω) = 1Ts

X( j(ωk=−∞

∞∑ − kωs ))

ωs = 2πTs


Frequency-Domain Representation of Sampling

Xs ( jω) = 1Ts

X( j(ωk=−∞

∞∑ − kωs ))

“Typical”bandlimited signal


Aliasing Distortion

If ωs < 2ωb , the copies of X(jω) overlap, and we have aliasing distortion.



Reconstruction of x(t)

xs (t) = x(nTs )δ (t − nTs )n=−∞

∞∑

Xs ( jω) = 1Ts

X( j(ωk=−∞

∞∑ − kωs ))

Xr ( jω) = Hr ( jω)Xs ( jω )8/22/2003 © 2003, JH McClellan & RW Schafer 21

Reconstruction: Frequency-Domain

)()()(so overlap,not do )(of copies the,2 If

ωωωω

ωω

jXjHjXjX

srr

bs

=

>Hr ( jω )


Ideal Reconstruction Filter

hr (t) =sin π

Tst

πTst

Hr ( jω) =Ts ω < π

Ts0 ω > π

Ts

hr (0) = 1

hr (nTs ) = 0, n = ±1,±2,…


Signal Reconstruction

xr (t) = hr (t) ∗ xs (t) = hr (t)∗ x(nTs )δ (t − nTs )n=−∞

∞∑

xr (t) = x(nTs )sin π

Ts(t − nTs )

πTs

(t − nTs )n=−∞

∞∑

Ideal bandlimited interpolation formula

xr (t) = x(nTs )hr (t − nTs )n=−∞

∞∑


Shannon Sampling Theorem

“SINC” Interpolation is the idealPERFECT RECONSTRUCTIONof BANDLIMITED SIGNALS


Reconstruction in Time-Domain


Ideal C-to-D and D-to-C

x[n] = x(nTs )xr (t) = x[n]

sin πTs

(t − nTs )πTs

(t − nTs )n=−∞

∞∑

Ideal Sampler Ideal bandlimited interpolator

Xr ( jω) = Hr ( jω)Xs ( jω )Xs ( jω) = 1Ts

X( j(ωk=−∞

∞∑ − kωs ))



Lecture 26Review: Digital Filtering

of Analog Signals


LECTURE OBJECTIVES

Sampling Theorem RevisitedGENERAL: in the FREQUENCY DOMAINFourier transform of sampled signalReconstruction from samples

Effective Frequency Response

Important FT propertiesConvolution multiplicationFrequency shifting


Sampling: Freq. Domain

EXPECTFREQUENCYSHIFTING !!!

∑∑∞

−∞=

∞

−∞==−=k

tjkk

ns

seanTttp ωδ )()(

∑∞

−∞==k

tjkk

seatp ω)(


Frequency-Domain Representation of Sampling

Xs ( jω) = 1Ts

X( j(ωk=−∞

∞∑ − kωs ))



Aliasing Distortion

If ωs < 2ωb , the copies of X(jω) overlap, and we have aliasing distortion.



Reconstruction of x(t)

xs (t) = x(nTs )δ (t − nTs )n=−∞

∞∑

Xs ( jω) = 1Ts

X( j(ωk=−∞

∞∑ − kωs ))

Xr ( jω) = Hr ( jω)Xs ( jω )


Reconstruction: Frequency-Domain

)()()(so overlap,not do )(of copies the,2 If

ωωωω

ωω

jXjHjXjX

srr

bs

=

>Hr ( jω )


Ideal Reconstruction Filter

hr (t) =sin π

Tst

πTst

Hr ( jω) =Ts ω < π

Ts0 ω > π

Ts

hr (0) = 1

hr (nTs ) = 0, n = ±1,±2,…


Signal Reconstruction

xr (t) = hr (t) ∗ xs (t) = hr (t)∗ x(nTs )δ (t − nTs )n=−∞

∞∑

xr (t) = x(nTs )sin π

Ts(t − nTs )

πTs

(t − nTs )n=−∞

∞∑

Ideal bandlimited interpolation formula

xr (t) = x(nTs )hr (t − nTs )n=−∞

∞∑


Shannon Sampling Theorem

“SINC” Interpolation is the idealPERFECT RECONSTRUCTIONof BANDLIMITED SIGNALS


Reconstruction in Time-Domain


Ideal C-to-D and D-to-C

x[n] = x(nTs )xr (t) = x[n]

sin πTs

(t − nTs )πTs

(t − nTs )n=−∞

∞∑

Ideal Sampler Ideal bandlimited interpolator

Xr ( jω) = Hr ( jω)Xs ( jω )Xs ( jω) = 1Ts

X( j(ωk=−∞

∞∑ − kωs ))


DT Filtering of CT Signals

D-to-CC-to-Dx(t) y(t)y[n]x[n] H(e j ˆ ω )X( jω ) Y( jω )

X(e j ˆ ω ) Y(e j ˆ ω )

If no aliasing occurs in sampling x(t), then it follows that

Y( jω ) = Heff ( jω)X( jω)

Heff ( jω ) =H(ejωTs ) ω < 1

2 ω s

0 ω > 12 ω s

UNDEFINEDNOT LTI


DT Filtering of a CT Signal

Spectrum of Discrete-Time Signal

Digital Filter

Reconstruction Filter (Analog)

ω

ω̂

ω̂

ω

Analog Input

Analog Output


EFFECTIVE Freq. Response

Assume NO Aliasing, thenANALOG FREQ <--> DIGITAL FREQ

So, we can plot:Scaled Freq. Axis

ˆ ω =ωTs = ωfs

H(e jωTs ) vs. ωANALOG FREQUENCY

DIGITAL FILTER


EFFECTIVE RESPONSE

DIGITAL FILTER

H(e j ˆ ω )

Heff ( jω )

fs = 1000 Hz

sin(11 ˆ ω / 2)sin( ˆ ω / 2)


Frequency Response for Discrete-time

Analog Frequency Response

Heff for 11-pt Averager

1000ˆ ωωωω ===sfsT

)2/ˆsin()2/ˆ11sin()( ˆ

ωωω =jeH

)2000/sin()2000/11sin()(

ωωω =jH


POP QUIZ

Given:

Find the output, y(t)When x(t) = cos(2000π t)

2 + 2z −1

1 −0.8z−1 D-to-AA-to-Dx(t) y(t)y[n]x[n]

fs = 5000Hz


POP QUIZ BECOMES

Given:

Find the output, y[n]When

Because

x[n] = cos(0.4πn)

H(z) =2 + 2z−1

1− 0.8z −1

ωTs = 2000π / 5000 = 0.4πNO Aliasing


ωω

ωω

ω

ˆ)()( where)(][ then][ if

ˆ

ˆˆ

ˆ

jezj

njj

nj

zHeHeeHny

enx

==

==

SINUSOIDAL RESPONSE

x[n] = SINUSOID => y[n] is SINUSOIDGet MAGNITUDE & PHASE from H(z)


POP QUIZ INSIDE ANSWER

Given:

The input:

Then y[n]

x[n] = cos(0.4πn)

H(z) =2 + 2z−1

1− 0.8z −1

y[n] = M cos(0.4πn +ψ )

H(e j 0.4π ) =2 + 2e− j0.4π

1− 0.8e− j 0.4π = 3.02e− j 0.452π


POP QUIZ ANSWER

Given:

When

The output isx(t) = cos(2000π t)

2 + 2z −1


fs = 5000Hz

y(t) = 3.02 cos(2000π t − 0.452π )


ANOTHER INPUT FREQ

Given:

Find the output, y(t)When

2 + 2z −1


fs = 5000Hz ˆ ω = ?

ˆ ω = ?

))7500(2cos()( ttx π=


2nd POP QUIZ ANSWER

Given:

When

2 + 2z −1


fs = 5000Hz y(t) = ?

ω = ?

))7500(2cos()( ttx π=

)5.1(25000/)7500(2cos(ˆ ππω ==

πω 3ˆ =


IMPORTANT CONCEPTS

ALL Signals have Frequency ContentSum of SinusoidsComplex ExponentialsImpulses, Square Pulses

FILTERS alter the Frequency ContentImage Processing Example: BlurLinear Time-Invariant Processing

3 Domains for Analysis


Superficial Knowledge

It depends how carefully you think about it. If you don’t think very carefully it’s obvious; but if you think about it in depth, you’ll get confused and it won’t be obvious.

…anon


THREE DOMAINS

H(z) =

bkz− k∑

1− a z−∑z = e j ˆ ω


THE FUTURE

Circuits & Laplace Transforms

H(s)

H( jω )h(t)FrequencyResponse

Implementation isRLC-op-amp circuit

Polynomials: Poles & Zeros


Mathematical Elegance

x(t ) =1

2πX( jω )e jω t dω

−∞

∞∫ Fourier Synthesis

(Inverse Transform)

Time - domain ⇔ Frequency - domain x(t) ⇔ X( jω )

X( jω) = x(t)−∞

∞∫ e− jωtdt Fourier Analysis

(Forward Transform)

Documents

Signal Processing First Lecture Slides