Page 1
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
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 4
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?
Page 2
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 6
WHY USE DSP ?
Mathematical abstractions lead to generalization and discovery of new processing techniques
Computer implementations are flexible
Applications provide a physical context
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 7
Fourier Everywhere
TelecommunicationsSound & Music
CDROM, Digital VideoFourier OpticsX-ray Crystallography
Protein Structure & DNAComputerized TomographyNuclear Magnetic Resonance: MRIRadioastronomyRef: Prestini, “The Evolution of Applied Harmonic Analysis”
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 8
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
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 9
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
Page 3
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 10
TUNING FORK A-440 Waveform
ms 3.285.515.8
=−≈T
Hz4353.2/1000
/1
≈== Tf
Time (sec)
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 11
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
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 12
Speech Signal: BAT
Nearly PeriodicPeriodic in Vowel RegionPeriod is (Approximately) T = 0.0065 sec
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 13
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)
Page 4
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 14
STORING DIGITAL SOUND
x[n] is a SAMPLED SINUSOIDA list of numbers stored in memory
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
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 15
Always use the COSINE FORM
Sine is a special case:
))440(2cos( ϕπ +tA
SINES and COSINES
)cos()sin( 2πωω −= tt
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 16
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
Page 5
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 18
PLOT COSINE SIGNAL
Formula defines A, ω, and φ5 0 3 12cos( . . )π πt+
A = 5ω = 0.3πϕ = 1.2π
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 19
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
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 20
PLOT the SINUSOID
Use T=20/3 and the peak location at t=-4
)2.13.0cos(5 ππ +t
→← 320
Page 6
8/22/2003 © 2003, JH McClellan & RW Schafer 1
Signal Processing First
LECTURE #2Phase & Time-ShiftComplex Exponentials
8/22/2003 © 2003, JH McClellan & RW Schafer 3
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
8/22/2003 © 2003, JH McClellan & RW Schafer 4
LECTURE OBJECTIVES
Define Sinusoid Formula from a plotRelate TIME-SHIFT to PHASE
tjXetz ω=)(
Introduce an ABSTRACTION:Complex Numbers represent SinusoidsComplex Exponential Signal
8/22/2003 © 2003, JH McClellan & RW Schafer 5
SINUSOIDAL SIGNAL
FREQUENCYRadians/secor, Hertz (cycles/sec)
PERIOD (in sec)
AMPLITUDEMagnitude
PHASE
)cos( ϕω +tA
f)2( πω =
ωπ21 ==
fT
A
ϕ
ω
Page 7
8/22/2003 © 2003, JH McClellan & RW Schafer 6
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
8/22/2003 © 2003, JH McClellan & RW Schafer 7
ANSWER for the PLOT
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 8
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
8/22/2003 © 2003, JH McClellan & RW Schafer 9
TIME-SHIFTED SINUSOID
))4((3.0cos(5))4(3.0cos(5)4( −−=+=+ tttx ππ
Page 8
8/22/2003 © 2003, JH McClellan & RW Schafer 10
PHASE <--> TIME-SHIFT
Equate the formulas:
and we obtain:
or,
)cos())(cos( ϕωω +=− tAttA m
ϕω =− mt
ωϕ−=mt
8/22/2003 © 2003, JH McClellan & RW Schafer 11
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
8/22/2003 © 2003, JH McClellan & RW Schafer 12
(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)
Page 9
8/22/2003 © 2003, JH McClellan & RW Schafer 14
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
8/22/2003 © 2003, JH McClellan & RW Schafer 15
COMPLEX NUMBERS
To solve: z2 = -1z = jMath and Physics use z = i
Complex number: z = x + j y
x
y zCartesiancoordinatesystem
8/22/2003 © 2003, JH McClellan & RW Schafer 16
PLOT COMPLEX NUMBERS
8/22/2003 © 2003, JH McClellan & RW Schafer 17
COMPLEX ADDITION = VECTORVECTORVECTORVECTOR Addition
26)53()24()52()34(
213
jj
jjzzz
+=+−++=
++−=+=
Page 10
8/22/2003 © 2003, JH McClellan & RW Schafer 18
*** 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
8/22/2003 © 2003, JH McClellan & RW Schafer 19
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
8/22/2003 © 2003, JH McClellan & RW Schafer 20
Euler’s FORMULA
Complex ExponentialReal part is cosineImaginary part is sineMagnitude is one
)sin()cos( θθθ jrrre j +=
)sin()cos( θθθ je j +=
8/22/2003 © 2003, JH McClellan & RW Schafer 21
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 +=
Page 11
8/22/2003 © 2003, JH McClellan & RW Schafer 22
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=+ +
8/22/2003 © 2003, JH McClellan & RW Schafer 23
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
8/22/2003 © 2003, JH McClellan & RW Schafer 24
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 ==)(
Page 12
1/12/2004 © 2003, JH McClellan & RW Schafer 1
Signal Processing First
LECTURE #3Phasor Addition Theorem
1/12/2004 © 2003, JH McClellan & RW Schafer 3
READING ASSIGNMENTS
This Lecture:Chapter 2, Section 2-6
Other Reading:Appendix A: Complex NumbersAppendix B: MATLABNext Lecture: start Chapter 3
1/12/2004 © 2003, JH McClellan & RW Schafer 4
LECTURE OBJECTIVES
Phasors = Complex AmplitudeComplex Numbers represent Sinusoids
Develop the ABSTRACTION:Adding Sinusoids = Complex AdditionPHASOR ADDITION THEOREMPHASOR ADDITION THEOREM
tjjtj eAeXetz ωϕω )()( ==
1/12/2004 © 2003, JH McClellan & RW Schafer 5
Z DRILL (Complex Arith)
Page 13
1/12/2004 © 2003, JH McClellan & RW Schafer 6
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 +=
1/12/2004 © 2003, JH McClellan & RW Schafer 8
Real & Imaginary Part Plots
PHASE DIFFERENCE = ππππ/2
1/12/2004 © 2003, JH McClellan & RW Schafer 9
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 +=
Page 14
1/12/2004 © 2003, JH McClellan & RW Schafer 10
Rotating Phasor
See Demo on CD-ROMChapter 2
1/12/2004 © 2003, JH McClellan & RW Schafer 11
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,
1/12/2004 © 2003, JH McClellan & RW Schafer 12
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
1/12/2004 © 2003, JH McClellan & RW Schafer 13
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= +
Page 15
1/12/2004 © 2003, JH McClellan & RW Schafer 14
WANT to ADD SINUSOIDS
ALL SINUSOIDS have SAME FREQUENCYHOW to GET {Amp,Phase} of RESULT ?
1/12/2004 © 2003, JH McClellan & RW Schafer 15
ADD SINUSOIDS
Sum Sinusoid has SAMESAME Frequency
1/12/2004 © 2003, JH McClellan & RW Schafer 16
PHASOR ADDITION RULE
Get the new complex amplitude by complex addition
1/12/2004 © 2003, JH McClellan & RW Schafer 17
Phasor Addition Proof
Page 16
1/12/2004 © 2003, JH McClellan & RW Schafer 18
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
1/12/2004 © 2003, JH McClellan & RW Schafer 19
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
1/12/2004 © 2003, JH McClellan & RW Schafer 20
ADD SINUSOIDS EXAMPLE
tm1
tm2
tm3
)()()( 213 txtxtx +=
)(1 tx
)(2 tx
1/12/2004 © 2003, JH McClellan & RW Schafer 21
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
Page 17
1/12/2004 © 2003, JH McClellan & RW Schafer 22
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
1/12/2004 © 2003, JH McClellan & RW Schafer 23
ADD SINUSOIDS
VECTOR(PHASOR)ADD
X1
X2
X3
Page 18
8/31/2003 © 2003, JH McClellan & RW Schafer 1
Signal Processing First
Lecture 4Spectrum Representation
8/31/2003 © 2003, JH McClellan & RW Schafer 3
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
8/31/2003 © 2003, JH McClellan & RW Schafer 4
LECTURE OBJECTIVES
Sinusoids with DIFFERENT FrequenciesSYNTHESIZE by Adding Sinusoids
SPECTRUM RepresentationGraphical Form shows DIFFERENTDIFFERENT Freqs
∑=
+=N
kkkk tfAtx
1)2cos()( ϕπ
8/31/2003 © 2003, JH McClellan & RW Schafer 5
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
Page 19
8/31/2003 © 2003, JH McClellan & RW Schafer 6
Another FREQ. DiagramFr
eque
ncy
is th
e ve
rtic
al a
xis
Time is the horizontal axis
A-440
8/31/2003 © 2003, JH McClellan & RW Schafer 7
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?
8/31/2003 © 2003, JH McClellan & RW Schafer 8
Fur Elise WAVEFORM
BeatNotes
8/31/2003 © 2003, JH McClellan & RW Schafer 9
Speech Signal: BAT
Nearly PeriodicPeriodic in Vowel RegionPeriod is (Approximately) T = 0.0065 sec
Page 20
8/31/2003 © 2003, JH McClellan & RW Schafer 10
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 −+=
8/31/2003 © 2003, JH McClellan & RW Schafer 11
INVERSE Euler’s Formula
Solve for cosine (or sine)
)()cos( 21 tjtj eet ωωω −+=
)()sin( 21 tjtjj eet ωωω −−=
8/31/2003 © 2003, JH McClellan & RW Schafer 12
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( −+=
8/31/2003 © 2003, JH McClellan & RW Schafer 13
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
Page 21
8/31/2003 © 2003, JH McClellan & RW Schafer 14
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 ==−
8/31/2003 © 2003, JH McClellan & RW Schafer 15
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 −
8/31/2003 © 2003, JH McClellan & RW Schafer 16
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
8/31/2003 © 2003, JH McClellan & RW Schafer 17
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
Page 22
8/31/2003 © 2003, JH McClellan & RW Schafer 18
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
−−
−−
++
+=
8/31/2003 © 2003, JH McClellan & RW Schafer 19
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
8/31/2003 © 2003, JH McClellan & RW Schafer 20
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(
8/31/2003 © 2003, JH McClellan & RW Schafer 21
FINAL ANSWER
So, we get the general form:
∑=
++=N
kkkk tfAAtx
10 )2cos()( ϕπ
)2/)250(2cos(8)3/)100(2cos(1410)(
ππππ
++−+=tttx
Page 23
8/31/2003 © 2003, JH McClellan & RW Schafer 22
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
8/31/2003 © 2003, JH McClellan & RW Schafer 23
Example: Synthetic Vowel
Sum of 5 Frequency Components
8/31/2003 © 2003, JH McClellan & RW Schafer 24
SPECTRUM of VOWEL
Note: Spectrum has 0.5Xk (except XDC)Conjugates in negative frequency
8/31/2003 © 2003, JH McClellan & RW Schafer 25
SPECTRUM of VOWEL (Polar Format)
φφφφk
0.5Ak
Page 24
8/31/2003 © 2003, JH McClellan & RW Schafer 26
Vowel Waveform(sum of all 5 components)
Page 25
1/28/2005 © 2003, JH McClellan & RW Schafer 1
Signal Processing First
Lecture 5Periodic Signals, Harmonics & Time-Varying Sinusoids
1/28/2005 © 2003, JH McClellan & RW Schafer 3
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
1/28/2005 © 2003, JH McClellan & RW Schafer 4
Problem Solving Skills
Math FormulaSum of CosinesAmp, Freq, Phase
Recorded SignalsSpeechMusicNo simple formula
Plot & SketchesS(t) versus tSpectrum
MATLABNumericalComputationPlotting list of numbers
1/28/2005 © 2003, JH McClellan & RW Schafer 5
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
Page 26
1/28/2005 © 2003, JH McClellan & RW Schafer 6
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
1/28/2005 © 2003, JH McClellan & RW Schafer 7
SPECTRUM for PERIODIC ?
Nearly Periodic in the Vowel RegionPeriod is (Approximately) T = 0.0065 sec
1/28/2005 © 2003, JH McClellan & RW Schafer 8
PERIODIC SIGNALS
Repeat every T secsDefinition
Example:
Speech can be “quasi-periodic”
)()( Ttxtx +=
)3(cos)( 2 ttx =?=T
3π=T3
2π=T
1/28/2005 © 2003, JH McClellan & RW Schafer 9
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 =⎟⎠⎞
⎜⎝⎛==
Page 27
1/28/2005 © 2003, JH McClellan & RW Schafer 10
{ }∑
∑
=
−∗
=
++=
=
++=
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 =
1/28/2005 © 2003, JH McClellan & RW Schafer 11
Define FUNDAMENTAL FREQ
00
1T
f =
(shortest) Periodlfundamenta(largest) Frequencylfundamenta
)2(
)2cos()(
0
0
000
100
==
==
++= ∑=
Tf
ffkf
tkfAAtx
k
N
kkk
πω
ϕπ
1/28/2005 © 2003, JH McClellan & RW Schafer 12
What is the fundamental frequency?
Harmonic Signal (3 Freqs)
3rd5th
10 Hz
1/28/2005 © 2003, JH McClellan & RW Schafer 13
POP QUIZ: FUNDAMENTAL
Here’s another spectrum:
What is the fundamental frequency?
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
Page 28
1/28/2005 © 2003, JH McClellan & RW Schafer 14
SPECIAL RELATIONSHIPto get a PERIODIC SIGNAL
IRRATIONAL SPECTRUM
1/28/2005 © 2003, JH McClellan & RW Schafer 15
Harmonic Signal (3 Freqs)
T=0.1
1/28/2005 © 2003, JH McClellan & RW Schafer 16
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
Page 29
1/28/2005 © 2003, JH McClellan & RW Schafer 18
Time-Varying FREQUENCIES Diagram
Freq
uenc
y is
the
vert
ical
axi
s
Time is the horizontal axis
A-440
1/28/2005 © 2003, JH McClellan & RW Schafer 19
SIMPLE TEST SIGNALC-major SCALE: stepped frequencies
Frequency is constant for each note
IDEAL
1/28/2005 © 2003, JH McClellan & RW Schafer 20
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)
1/28/2005 © 2003, JH McClellan & RW Schafer 21
SPECTROGRAM EXAMPLETwo Constant Frequencies: Beats
))12(2sin())660(2cos( tt ππ
Page 30
1/28/2005 © 2003, JH McClellan & RW Schafer 22
( ) ( )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 −− +−−
1/28/2005 © 2003, JH McClellan & RW Schafer 23
SPECTRUM of AM (Beat)
4 complex exponentials in AM:
What is the fundamental frequency?
648 Hz ? 24 Hz ?
0 648 672 f (in Hz)–672 –648
2/41 πje 2/
41 πje−2/
41 πje−2/
41 πje
1/28/2005 © 2003, JH McClellan & RW Schafer 24
STEPPED FREQUENCIESC-major SCALE: successive sinusoids
Frequency is constant for each note
IDEAL
1/28/2005 © 2003, JH McClellan & RW Schafer 25
SPECTROGRAM of C-Scale
ARTIFACTS at Transitions
Sinusoids ONLY
From SPECGRAMANALYSIS PROGRAM
Page 31
1/28/2005 © 2003, JH McClellan & RW Schafer 26
Spectrogram of LAB SONG
ARTIFACTS at Transitions
Sinusoids ONLY Analysis Frame = 40ms
1/28/2005 © 2003, JH McClellan & RW Schafer 27
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
1/28/2005 © 2003, JH McClellan & RW Schafer 28
)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
1/28/2005 © 2003, JH McClellan & RW Schafer 29
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
==⇒+=
+=
Page 32
1/28/2005 © 2003, JH McClellan & RW Schafer 30
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
1/28/2005 © 2003, JH McClellan & RW Schafer 32
CHIRP WAVEFORM
1/28/2005 © 2003, JH McClellan & RW Schafer 33
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 −==⇒
Page 33
1/28/2005 © 2003, JH McClellan & RW Schafer 34
SINE-WAVE FREQUENCY MODULATION (FM)
Look at CD-ROM Demos in Ch 3
Page 34
9/8/2003 © 2003, JH McClellan & RW Schafer 1
Signal Processing First
Lecture 6Fourier Series Coefficients
9/8/2003 © 2003, JH McClellan & RW Schafer 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: More Fourier Series
9/8/2003 © 2003, JH McClellan & RW Schafer 4
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
9/8/2003 © 2003, JH McClellan & RW Schafer 5
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
Page 35
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
Recall Complex Amplitude vs. Freq
kk aX =21
Xk = Akejϕ k
12 Xk
*
{ }∑=
−∗++=N
k
tfjk
tfjk
kk eXeXXtx1
2212
21
0)( ππka{ *
ka0a
9/8/2003 © 2003, JH McClellan & RW Schafer 8
Harmonic Signal
PERIOD/FREQUENCY of COMPLEX EXPONENTIAL:
tfkj
kkeatx 02)( π∑
∞
−∞==
( )0
00
001or22f
TT
f === πωπ
9/8/2003 © 2003, JH McClellan & RW Schafer 9
kjkk
N
kkk
eAX
tkfAAtx
ϕ
ϕπ
=
++= ∑=1
00 )2cos()(
kjkkk eAXa ϕ
21
21 ==
Fourier Series Synthesis
COMPLEXAMPLITUDE
tfkj
kkeatx 02)( π∑
∞
−∞==
Page 36
9/8/2003 © 2003, JH McClellan & RW Schafer 10
Harmonic Signal (3 Freqs)
T = 0.1
a3 a5
a1
9/8/2003 © 2003, JH McClellan & RW Schafer 11
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
9/8/2003 © 2003, JH McClellan & RW Schafer 12
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
9/8/2003 © 2003, JH McClellan & RW Schafer 13
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 =
Page 37
9/8/2003 © 2003, JH McClellan & RW Schafer 14
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
9/8/2003 © 2003, JH McClellan & RW Schafer 15
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
9/8/2003 © 2003, JH McClellan & RW Schafer 16
SQUARE WAVE EXAMPLE
0–.02 .02 0.04
1
t
x(t)
.01
sec. 04.0for 0
01)(
0
0021
021
=
<≤
<≤=
TTtT
Tttx
9/8/2003 © 2003, JH McClellan & RW Schafer 17
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 )( −−=−
−= −
Page 38
9/8/2003 © 2003, JH McClellan & RW Schafer 18
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
9/8/2003 © 2003, JH McClellan & RW Schafer 19
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
9/8/2003 © 2003, JH McClellan & RW Schafer 20
Spectrum from Fourier Series
=
±±=
±±=−
=
0
,4,20
,3,1
21 k
k
kkj
ak …
…π)25(2)04.0/(20 ππω ==
9/8/2003 © 2003, JH McClellan & RW Schafer 21
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 =−
Page 39
Signal Processing First
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
Recall Complex Amplitude vs. Freq
{ }∑=
−∗++=N
k
tfjk
tfjk
kk eaeaatx1
220)( ππ
kjkk eAa ϕ
21=∗
ka
0a
Page 40
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
Page 41
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
Page 42
9/13/2006 EE-2025 Spring-2005 jMc 16
Spectrum from Fourier Series
⎪⎪
⎩
⎪⎪
⎨
⎧
=
±±=
±±=−
=
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
Page 43
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
Page 44
9/13/2006 EE-2025 Spring-2005 jMc 24
fseriesdemo GUI
Page 45
9/14/2003 © 2003, JH McClellan & RW Schafer 1
Signal Processing First
Lecture 8Sampling & Aliasing
9/14/2003 © 2003, JH McClellan & RW Schafer 3
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
9/14/2003 © 2003, JH McClellan & RW Schafer 4
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)
Page 46
9/14/2003 © 2003, JH McClellan & RW Schafer 6
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
9/14/2003 © 2003, JH McClellan & RW Schafer 7
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]
9/14/2003 © 2003, JH McClellan & RW Schafer 8
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)
9/14/2003 © 2003, JH McClellan & RW Schafer 9
fs = 2 kHz
fs = 500Hz
Hz100=f
Page 47
9/14/2003 © 2003, JH McClellan & RW Schafer 10
SAMPLING THEOREM
HOW OFTEN ?DEPENDS on FREQUENCY of SINUSOIDANSWERED by SHANNON/NYQUIST TheoremALSO DEPENDS on “RECONSTRUCTION”
9/14/2003 © 2003, JH McClellan & RW Schafer 11
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
9/14/2003 © 2003, JH McClellan & RW Schafer 12
STORING DIGITAL SOUND
x[n] is a SAMPLED SINUSOIDA list of numbers stored in memory
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
9/14/2003 © 2003, JH McClellan & RW Schafer 13
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
Page 48
9/14/2003 © 2003, JH McClellan & RW Schafer 14
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
9/14/2003 © 2003, JH McClellan & RW Schafer 15
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
9/14/2003 © 2003, JH McClellan & RW Schafer 16
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
9/14/2003 © 2003, JH McClellan & RW Schafer 17
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 )
Page 49
9/14/2003 © 2003, JH McClellan & RW Schafer 18
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 =−
9/14/2003 © 2003, JH McClellan & RW Schafer 19
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
9/14/2003 © 2003, JH McClellan & RW Schafer 20
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 )
9/14/2003 © 2003, JH McClellan & RW Schafer 21
NORMALIZED FREQUENCY
DIGITAL FREQUENCY
ss f
fT πωω 2ˆ == π2+
Page 50
9/14/2003 © 2003, JH McClellan & RW Schafer 22
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
9/14/2003 © 2003, JH McClellan & RW Schafer 23
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 =
9/14/2003 © 2003, JH McClellan & RW Schafer 24
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 =
9/14/2003 © 2003, JH McClellan & RW Schafer 25
SAMPLING GUI (con2dis)
Page 51
9/14/2003 © 2003, JH McClellan & RW Schafer 26
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
Page 52
8/22/2003 © 2003, JH McClellan & RW Schafer 1
Signal Processing First
Lecture 9D-to-A Conversion
8/22/2003 © 2003, JH McClellan & RW Schafer 3
READING ASSIGNMENTS
This Lecture:Chapter 4: Sections 4-4, 4-5
Other Reading:Recitation: Section 4-3 (Strobe Demo)Next Lecture: Chapter 5 (beginning)
8/22/2003 © 2003, JH McClellan & RW Schafer 4
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
8/22/2003 © 2003, JH McClellan & RW Schafer 5
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
COMPUTER D-to-AA-to-Dx(t) y(t)y[n]x[n]
Page 53
8/22/2003 © 2003, JH McClellan & RW Schafer 6
SAMPLING x(t)
UNIFORM SAMPLING at t = nTsIDEAL: x[n] = x(nTs)
C-to-Dx(t) x[n]
8/22/2003 © 2003, JH McClellan & RW Schafer 7
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
8/22/2003 © 2003, JH McClellan & RW Schafer 8
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
8/22/2003 © 2003, JH McClellan & RW Schafer 9
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 …}
Page 54
8/22/2003 © 2003, JH McClellan & RW Schafer 10
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 =
8/22/2003 © 2003, JH McClellan & RW Schafer 11
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 =
8/22/2003 © 2003, JH McClellan & RW Schafer 12
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
fT πωω 2ˆ == π2+
ππωω 22ˆ +−==s
s ffT FOLDED ALIAS
ω̂
ALIASING
Page 55
8/22/2003 © 2003, JH McClellan & RW Schafer 14
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
8/22/2003 © 2003, JH McClellan & RW Schafer 15
FREQUENCY DOMAINS
D-to-AA-to-Dx(t) y(t)x[n]
ffω
f̂
ω̂
y[n]
sffπ
ω2ˆ
=
π2+sffπω 2ˆ =
8/22/2003 © 2003, JH McClellan & RW Schafer 16
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
8/22/2003 © 2003, JH McClellan & RW Schafer 17
SAMPLING GUI (con2dis)
Page 56
8/22/2003 © 2003, JH McClellan & RW Schafer 19
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)
COMPUTER D-to-AA-to-Dx(t) y(t)y[n]x[n]
8/22/2003 © 2003, JH McClellan & RW Schafer 20
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
8/22/2003 © 2003, JH McClellan & RW Schafer 21
SPECTRUM (ALIASING CASE)
ˆ ω = 2π ffs
fs = 80Hz
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 += nAnx
8/22/2003 © 2003, JH McClellan & RW Schafer 22
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
Page 57
8/22/2003 © 2003, JH McClellan & RW Schafer 23
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
8/22/2003 © 2003, JH McClellan & RW Schafer 24
SQUARE PULSE CASE
8/22/2003 © 2003, JH McClellan & RW Schafer 25
OVER-SAMPLING CASE
EASIER TO RECONSTRUCT
8/22/2003 © 2003, JH McClellan & RW Schafer 26
MATH MODEL for D-to-A
SQUARE PULSE:
Page 58
8/22/2003 © 2003, JH McClellan & RW Schafer 27
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 ) +…
8/22/2003 © 2003, JH McClellan & RW Schafer 28
p(t)
8/22/2003 © 2003, JH McClellan & RW Schafer 29
TRIANGULAR PULSE (2X)
8/22/2003 © 2003, JH McClellan & RW Schafer 30
OPTIMAL PULSE ?
CALLED“BANDLIMITEDINTERPOLATION”
…,2,for 0)(
for sin
)(
ss
TtTt
TTttp
ttps
s
±±==
∞<<∞−= π
π
Page 59
2/18/2005 © 2003, JH McClellan & RW Schafer 1
Signal Processing First
Lecture 10FIR Filtering Intro
2/18/2005 © 2003, JH McClellan & RW Schafer 3
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
2/18/2005 © 2003, JH McClellan & RW Schafer 4
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]
2/18/2005 © 2003, JH McClellan & RW Schafer 5
DIGITAL FILTERING
CONCENTRATE on the COMPUTERPROCESSING ALGORITHMS
SOFTWARE (MATLAB)HARDWARE: DSP chips, VLSI
DSP: DIGITAL SIGNAL PROCESSING
COMPUTER D-to-AA-to-Dx(t) y(t)y[n]x[n]
Page 60
2/18/2005 © 2003, JH McClellan & RW Schafer 6
The TMS32010, 1983
First PC plug-in board from Atlanta Signal Processors Inc.
2/18/2005 © 2003, JH McClellan & RW Schafer 7
Rockland Digital Filter, 1971
For the price of a small house, you could have one of these.
2/18/2005 © 2003, JH McClellan & RW Schafer 8
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]
Page 61
2/18/2005 © 2003, JH McClellan & RW Schafer 10
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]
2/18/2005 © 2003, JH McClellan & RW Schafer 11
DISCRETE-TIME SIGNAL
x[n] is a LIST of NUMBERSINDEXED by “n”
STEM PLOT
2/18/2005 © 2003, JH McClellan & RW Schafer 12
3-PT AVERAGE SYSTEMADD 3 CONSECUTIVE NUMBERS
Do this for each “n”Make a TABLE
n=0
n=1
])2[]1[][(][ 31 ++++= nxnxnxny
2/18/2005 © 2003, JH McClellan & RW Schafer 13
INPUT SIGNAL
OUTPUT SIGNAL
])2[]1[][(][ 31 ++++= nxnxnxny
Page 62
2/18/2005 © 2003, JH McClellan & RW Schafer 14
PAST, PRESENT, FUTURE
“n” is TIME
2/18/2005 © 2003, JH McClellan & RW Schafer 15
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
2/18/2005 © 2003, JH McClellan & RW Schafer 16
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
2/18/2005 © 2003, JH McClellan & RW Schafer 17
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][][
Page 63
2/18/2005 © 2003, JH McClellan & RW Schafer 18
SLIDE a WINDOW across x[n]
x[n]x[n-M]
∑=
−=M
kk knxbny
0][][
GENERAL CAUSAL FIR FILTER
2/18/2005 © 2003, JH McClellan & RW Schafer 19
FILTERED STOCK SIGNAL
OUTPUT
INPUT
50-pt Averager
2/18/2005 © 2003, JH McClellan & RW Schafer 20
⎪⎩
⎪⎨⎧
≠
==
00
01][
n
nnδ
SPECIAL INPUT SIGNALS
x[n] = SINUSOIDx[n] has only one NON-ZERO VALUE
1
n
UNIT-IMPULSE
FREQUENCY RESPONSE (LATER)
2/18/2005 © 2003, JH McClellan & RW Schafer 21
UNIT IMPULSE SIGNAL δ[n]
δ[n] is NON-ZEROWhen its argumentis equal to ZERO
]3[ −nδ3=n
Page 64
2/18/2005 © 2003, JH McClellan & RW Schafer 22
]4[2]3[4]2[6]1[4][2][ −+−+−+−+= nnnnnnx δδδδδ
MATH FORMULA for x[n]
Use SHIFTED IMPULSES to write x[n]
2/18/2005 © 2003, JH McClellan & RW Schafer 23
SUM of SHIFTED IMPULSES
This formula ALWAYS works
2/18/2005 © 2003, JH McClellan & RW Schafer 24
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
2/18/2005 © 2003, JH McClellan & RW Schafer 25
},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
Page 65
2/18/2005 © 2003, JH McClellan & RW Schafer 26
FIR IMPULSE RESPONSE
Convolution = Filter DefinitionFilter Coeffs = Impulse Response
∑=
−=M
kknxkhny
0][][][
CONVOLUTION
∑=
−=M
kk knxbny
0][][
2/18/2005 © 2003, JH McClellan & RW Schafer 27
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
2/18/2005 © 2003, JH McClellan & RW Schafer 28
3-pt AVG EXAMPLE
USE PAST VALUES
400for)4/8/2cos()02.1(][ :Input ≤≤++= nnnx n ππ
2/18/2005 © 2003, JH McClellan & RW Schafer 29
7-pt FIR EXAMPLE (AVG)
CAUSAL: Use Previous
LONGER OUTPUT
400for)4/8/2cos()02.1(][ :Input ≤≤++= nnnx n ππ
Page 66
8/22/2003 © 2003, JH McClellan & RW Schafer 1
Signal Processing First
Lecture 11Linearity & Time-InvarianceConvolution
8/22/2003 © 2003, JH McClellan & RW Schafer 3
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
8/22/2003 © 2003, JH McClellan & RW Schafer 4
LECTURE OBJECTIVES
BLOCK DIAGRAM REPRESENTATIONComponents for HardwareConnect Simple Filters Together to Build More Complicated Systems
LTI SYSTEMS
GENERAL PROPERTIES of FILTERSLLINEARITYTTIME-IINVARIANCE==> CONVOLUTIONCONVOLUTION
8/22/2003 © 2003, JH McClellan & RW Schafer 5
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
Page 67
8/22/2003 © 2003, JH McClellan & RW Schafer 6
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]
8/22/2003 © 2003, JH McClellan & RW Schafer 7
BUILDING BLOCKS
BUILD UP COMPLICATED FILTERSFROM SIMPLE MODULESEx: FILTER MODULE MIGHT BE 3-pt FIR
FILTERy[n]x[n]
FILTER
FILTER
++
INPUT
OUTPUT
8/22/2003 © 2003, JH McClellan & RW Schafer 8
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
8/22/2003 © 2003, JH McClellan & RW Schafer 9
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][][
Page 68
8/22/2003 © 2003, JH McClellan & RW Schafer 10
SPECIAL INPUT SIGNALS
x[n] = SINUSOIDx[n] has only one NON-ZERO VALUE
1
n
UNIT-IMPULSE
FREQUENCY RESPONSE
≠
==
00
01][
n
nnδ
8/22/2003 © 2003, JH McClellan & RW Schafer 11
FIR IMPULSE RESPONSE
Convolution = Filter DefinitionFilter Coeffs = Impulse Response
∑=
−=M
kk knxbny
0][][ ∑
=−=
M
kk knbnh
0][][ δ
8/22/2003 © 2003, JH McClellan & RW Schafer 12
]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 ∗=
Page 69
8/22/2003 © 2003, JH McClellan & RW Schafer 14
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
8/22/2003 © 2003, JH McClellan & RW Schafer 15
GENERAL FIR FILTERSLIDE a Length-L WINDOW over x[n]
x[n]x[n-M]
8/22/2003 © 2003, JH McClellan & RW Schafer 16
DCONVDEMO: MATLAB GUI
8/22/2003 © 2003, JH McClellan & RW Schafer 17
]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=−−=
Page 70
8/22/2003 © 2003, JH McClellan & RW Schafer 18
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][][
8/22/2003 © 2003, JH McClellan & RW Schafer 19
HARDWARE ATOMS
Add, Multiply & Store∑
=−=
M
kk knxbny
0][][
][][ nxny β=
]1[][ −= nxny
][][][ 21 nxnxny +=
8/22/2003 © 2003, JH McClellan & RW Schafer 20
FIR STRUCTURE
Direct Form
SIGNALFLOW GRAPH
∑=
−=M
kk knxbny
0][][
8/22/2003 © 2003, JH McClellan & RW Schafer 21
Moore’s Law for TI DSPs
Double every18 months ?
LOG SCALE
Page 71
8/22/2003 © 2003, JH McClellan & RW Schafer 22
SYSTEM PROPERTIES
MATHEMATICAL DESCRIPTIONMATHEMATICAL DESCRIPTIONTIMETIME--INVARIANCEINVARIANCELINEARITYLINEARITYCAUSALITY
“No output prior to input”
SYSTEMy[n]x[n]
8/22/2003 © 2003, JH McClellan & RW Schafer 23
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
8/22/2003 © 2003, JH McClellan & RW Schafer 24
TESTING Time-Invariance
8/22/2003 © 2003, JH McClellan & RW Schafer 25
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”
Page 72
8/22/2003 © 2003, JH McClellan & RW Schafer 26
TESTING LINEARITY
8/22/2003 © 2003, JH McClellan & RW Schafer 27
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
8/22/2003 © 2003, JH McClellan & RW Schafer 28
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
8/22/2003 © 2003, JH McClellan & RW Schafer 29
CASCADE SYSTEMS
Does the order of S1 & S2 matter?NO, LTI SYSTEMS can be rearranged !!!WHAT ARE THE FILTER COEFFS? {bk}
S1 S2
Page 73
8/22/2003 © 2003, JH McClellan & RW Schafer 30
CASCADE EQUIVALENTFind “overall” h[n] for a cascade ?
S1 S2
S1S2
Page 74
2/25/2005 © 2003, JH McClellan & RW Schafer 1
Signal Processing First
Lecture 12Frequency Response
of FIR Filters
2/25/2005 © 2003, JH McClellan & RW Schafer 3
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
2/25/2005 © 2003, JH McClellan & RW Schafer 4
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
2/25/2005 © 2003, JH McClellan & RW Schafer 5
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
Page 75
2/25/2005 © 2003, JH McClellan & RW Schafer 8
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
2/25/2005 © 2003, JH McClellan & RW Schafer 9
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
2/25/2005 © 2003, JH McClellan & RW Schafer 10
3-pt AVG EXAMPLE
USE PAST VALUES
400for)4/8/2cos()02.1(][ :Input ≤≤++= nnnx n ππ
2/25/2005 © 2003, JH McClellan & RW Schafer 11
7-pt FIR EXAMPLE (AVG)
CAUSAL: Use Previous
LONGER OUTPUT
400for)4/8/2cos()02.1(][ :Input ≤≤++= nnnx n ππ
Page 76
2/25/2005 © 2003, JH McClellan & RW Schafer 12
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
2/25/2005 © 2003, JH McClellan & RW Schafer 13
DCONVDEMO: MATLAB GUI
2/25/2005 © 2003, JH McClellan & RW Schafer 14
COMPLEX EXPONENTIAL
∑∑==
−=−=M
k
M
kk knxkhknxbny
00][][][][
FIR DIFFERENCE EQUATION
∞<<∞−= neAenx njj ωϕ ˆ][x[n] is the input signal—a complex exponential
2/25/2005 © 2003, JH McClellan & RW Schafer 15
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⎟⎟⎠
⎞⎜⎜⎝
⎛= ∑
=
−
Page 77
2/25/2005 © 2003, JH McClellan & RW Schafer 16
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 )( −
=∑=
2/25/2005 © 2003, JH McClellan & RW Schafer 17
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
2/25/2005 © 2003, JH McClellan & RW Schafer 18
PLOT of FREQ RESPONSE
ωω ω ˆˆ )ˆcos22()( jj eeH −+= RESPONSE at π/3
}1,2,1{}{ =kb
(radians)ω̂π− π
ω̂
2/25/2005 © 2003, JH McClellan & RW Schafer 19
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
=
Page 78
2/25/2005 © 2003, JH McClellan & RW Schafer 20
njj eenxny )3/(4/2][ when][ Find ππ=
EXAMPLE 6.2 (answer)
3/ˆat )( 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−=
2/25/2005 © 2003, JH McClellan & RW Schafer 21
EXAMPLE: COSINE INPUT
)cos(2][ andknown is )( when][ Find
43
ˆ
ππ
ω
+= nnxeHny j
ωω ω ˆˆ )ˆcos22()( jj eeH −+=
y[n]x[n] )( ω̂jeHω̂ ω̂
2/25/2005 © 2003, JH McClellan & RW Schafer 22
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
2/25/2005 © 2003, JH McClellan & RW Schafer 23
EX: COSINE INPUT (ans-2)
)cos(2][ when][ Find 43ππ += nnxny
)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
ωω ω ˆˆ )ˆcos22()( jj eeH −+=
Page 79
2/25/2005 © 2003, JH McClellan & RW Schafer 24
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 )( −
=∑=
2/25/2005 © 2003, JH McClellan & RW Schafer 26
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
2/25/2005 © 2003, JH McClellan & RW Schafer 27
BLOCK DIAGRAMS
Equivalent Representations
y[n]x[n]
y[n]x[n])( ω̂jeH
][nh
ω̂ˆ ω
2/25/2005 © 2003, JH McClellan & RW Schafer 28
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
Page 80
2/25/2005 © 2003, JH McClellan & RW Schafer 29
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
2/25/2005 © 2003, JH McClellan & RW Schafer 30
DLTI Demo with Sinusoids
FILTERx[n] y[n]
2/25/2005 © 2003, JH McClellan & RW Schafer 31
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
2/25/2005 © 2003, JH McClellan & RW Schafer 32
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
Page 81
10/6/2003 © 2003, JH McClellan & RW Schafer 1
Signal Processing First
Lecture 13Digital Filtering
of Analog Signals
10/6/2003 © 2003, JH McClellan & RW Schafer 3
READING ASSIGNMENTS
This Lecture:Chapter 6, Sections 6-6, 6-7 & 6-8
Other Reading:Recitation: Chapter 6
FREQUENCY RESPONSE EXAMPLESNext Lecture: Chapter 7
10/6/2003 © 2003, JH McClellan & RW Schafer 4
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
10/6/2003 © 2003, JH McClellan & RW Schafer 5
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
Page 82
10/6/2003 © 2003, JH McClellan & RW Schafer 6
Ex: DELAY by 2 SYSTEM
y[n]x[n] )( ω̂jeH
y[n]x[n] ][nh
]2[][for )( and ][ Find ˆ −= nxnyeHnh jω
]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
]2[][for )( and ][ Find ˆ −= nxnyeHnh jω
∑=
−−=M
k
kjj ekeH0
ˆˆ ]2[)( ωω δ
)( ω̂jeHy[n]x[n] ω̂2je−
ω̂ ω̂
10/6/2003 © 2003, JH McClellan & RW Schafer 8
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
10/6/2003 © 2003, JH McClellan & RW Schafer 9
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
Page 83
10/6/2003 © 2003, JH McClellan & RW Schafer 10
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
10/6/2003 © 2003, JH McClellan & RW Schafer 11
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
10/6/2003 © 2003, JH McClellan & RW Schafer 12
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
10/6/2003 © 2003, JH McClellan & RW Schafer 13
PLOT of FREQ RESPONSE
ωω ω ˆˆ )ˆcos22()( jj eeH −+= RESPONSE at ππππ/3
}1,2,1{}{ =kb
(radians)ω̂π− π
ω̂
Page 84
10/6/2003 © 2003, JH McClellan & RW Schafer 14
EXAMPLE 6.2
ω̂ω̂
y[n]x[n])( ω̂jeH
ωω ω ˆˆ )ˆcos22()( jj eeH −+=
njj
j
eenxeHny
)3/(4/
ˆ
2][ andknown is )( when][ Find
ππ
ω
=
10/6/2003 © 2003, JH McClellan & RW Schafer 15
njj eenxny )3/(4/2][ when][ Find ππ=
EXAMPLE 6.2 (answer)
3/ˆat )( 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−=
10/6/2003 © 2003, JH McClellan & RW Schafer 16
EXAMPLE: COSINE INPUT
)cos(2][ andknown is )( when][ Find
43
ˆ
ππ
ω
+= nnxeHny j
ωω ω ˆˆ )ˆcos22()( jj eeH −+=
y[n]x[n] )( ω̂jeHω̂ ω̂
10/6/2003 © 2003, JH McClellan & RW Schafer 17
EX: COSINE INPUT (ans-1)
)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
nynynyeeHnyeeHny
njj
njj
+=⇒==
+−−
+
πππ
πππ
Page 85
10/6/2003 © 2003, JH McClellan & RW Schafer 18
EX: COSINE INPUT (ans-2)
)cos(2][ when][ Find 43ππ += nnxny
)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
ωω ω ˆˆ )ˆcos22()( jj eeH −+=
10/6/2003 © 2003, JH McClellan & RW Schafer 19
SINUSOID thru FIR
IF Multiply the Magnitudes
Add the Phases
))(ˆcos()(][
)ˆcos(][11 ˆ
1ˆ
1ωω φω
φωjj eHneHAny
nAnx
∠++=⇒
+=
)()( ˆˆ* ωω jj eHeH −=
10/6/2003 © 2003, JH McClellan & RW Schafer 20
LTI Demo with Sinusoids
FILTERx[n]
y[n]
10/6/2003 © 2003, JH McClellan & RW Schafer 21
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
D-to-AA-to-Dx(t) y(t)y[n]x[n] )( ω̂jeH
ω̂ ω̂ω ω
ω̂ω
ω
Page 86
10/6/2003 © 2003, JH McClellan & RW Schafer 22
FREQUENCY SCALING
TIME SAMPLING:IF NO ALIASING:FREQUENCY SCALING
D-to-AA-to-Dx(t) y(t)y[n]x[n] )( ω̂jeH
ˆ ω ˆ ω ω ω
sfsT ωωω ==ˆsnTt =
10/6/2003 © 2003, JH McClellan & RW Schafer 23
11-pt AVERAGER Example
D-to-AA-to-Dx(t) y(t)y[n]x[n] )( ω̂jeH
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
10/6/2003 © 2003, JH McClellan & RW Schafer 24
D-A FREQUENCY SCALING
RECONSTRUCT up to 0.5fsFREQUENCY SCALING
D-to-AA-to-Dx(t) y(t)y[n]x[n] )( ω̂jeH
ˆ ω ˆ ω ω ω
TIME SAMPLING:
sfωω ˆ=
ss ftnnTt ←⇒=
10/6/2003 © 2003, JH McClellan & RW Schafer 25
TRACK the FREQUENCIES
D-to-AA-to-Dx(t) y(t)y[n]x[n] )( ω̂jeH
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
Page 87
10/6/2003 © 2003, JH McClellan & RW Schafer 26
11-pt AVERAGER
NULLS or ZEROS
πω 5.0ˆ =πω 05.0ˆ =
10/6/2003 © 2003, JH McClellan & RW Schafer 27
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 5.0ˆAt =
π
ππ 5.2
)25.0sin(11)75.2sin( je−=
π5.00909.0 je−=
10/6/2003 © 2003, JH McClellan & RW Schafer 28
EVALUATE Freq. Response
fs = 1000MAG SCALE
PHASE CHANGE
)( 1000/)25(2πjeH
)( 1000/)250(2πjeH
10/6/2003 © 2003, JH McClellan & RW Schafer 29
EFFECTIVE RESPONSE
DIGITAL FILTER
LOW-PASS FILTER
)( ω̂jeH
Page 88
10/6/2003 © 2003, JH McClellan & RW Schafer 30
FILTER TYPES
LOW-PASS FILTER (LPF)BLURRINGATTENUATES HIGH FREQUENCIES
HIGH-PASS FILTER (HPF)SHARPENING for IMAGESBOOSTS THE HIGHSREMOVES DC
BAND-PASS FILTER (BPF)
10/6/2003 © 2003, JH McClellan & RW Schafer 31
B & W IMAGE
10/6/2003 © 2003, JH McClellan & RW Schafer 32
B&W IMAGE with COSINEFILTERED: 11-pt AVG
10/6/2003 © 2003, JH McClellan & RW Schafer 33
FILTERED B&W IMAGE
LPF:BLUR
Page 89
10/6/2003 © 2003, JH McClellan & RW Schafer 34
ROW of B&W IMAGE
BLACK = 255
WHITE = 0
10/6/2003 © 2003, JH McClellan & RW Schafer 35
FILTERED ROW of IMAGE
ADJUSTED DELAY by 5 samples
Page 90
8/22/2003 © 2003, JH McClellan & RW Schafer 1
Signal Processing First
Lecture 14Z Transforms: Introduction
8/22/2003 © 2003, JH McClellan & RW Schafer 3
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
8/22/2003 © 2003, JH McClellan & RW Schafer 4
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 →
8/22/2003 © 2003, JH McClellan & RW Schafer 5
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
Page 91
8/22/2003 © 2003, JH McClellan & RW Schafer 6
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 →
8/22/2003 © 2003, JH McClellan & RW Schafer 7
“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
8/22/2003 © 2003, JH McClellan & RW Schafer 8
Z-TRANSFORM IDEA
POLYNOMIAL REPRESENTATION
y[n]x[n]
y[n]x[n] )(zH
][nh∑ −=n
nznhzH ][)(
8/22/2003 © 2003, JH McClellan & RW Schafer 9
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
Page 92
8/22/2003 © 2003, JH McClellan & RW Schafer 10
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
8/22/2003 © 2003, JH McClellan & RW Schafer 12
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][)(
8/22/2003 © 2003, JH McClellan & RW Schafer 13
]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
Page 93
8/22/2003 © 2003, JH McClellan & RW Schafer 14
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 −=
8/22/2003 © 2003, JH McClellan & RW Schafer 16
DELAY PROPERTY
8/22/2003 © 2003, JH McClellan & RW Schafer 17
GENERAL I/O PROBLEMInput is x[n], find y[n] (for FIR, h[n])How to combine X(z) and H(z) ?
Page 94
8/22/2003 © 2003, JH McClellan & RW Schafer 18
FIR Filter = CONVOLUTION
∑∑==
−=−=M
k
M
kk knxkhknxbny
00][][][][
CONVOLUTION 8/22/2003 © 2003, JH McClellan & RW Schafer 19
CONVOLUTION PROPERTYPROOF:
MULTIPLYZ-TRANSFORMS
8/22/2003 © 2003, JH McClellan & RW Schafer 20
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] = ?
Page 95
8/22/2003 © 2003, JH McClellan & RW Schafer 22
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
8/22/2003 © 2003, JH McClellan & RW Schafer 23
CASCADE EQUIVALENT
Multiply the System Functions
x[n] )(1 zH y[n])(2 zH
)()()( 21 zHzHzH =
y[n]x[n] )(zHEQUIVALENTSYSTEM
8/22/2003 © 2003, JH McClellan & RW Schafer 24
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
Page 96
8/22/2003 © 2003, JH McClellan & RW Schafer 1
Signal Processing First
Lecture 15Zeros of H(z) and theFrequency Domain
8/22/2003 © 2003, JH McClellan & RW Schafer 3
READING ASSIGNMENTS
This Lecture:Chapter 7, Section 7-6 to end
Other Reading:Recitation & Lab: Chapter 7
ZEROS (and POLES)Next Lecture:Chapter 8
8/22/2003 © 2003, JH McClellan & RW Schafer 4
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
Page 97
8/22/2003 © 2003, JH McClellan & RW Schafer 6
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
8/22/2003 © 2003, JH McClellan & RW Schafer 7
)()()(][][][ zXzHzYnxnhny =↔∗=
CONVOLUTION PROPERTYConvolution in the n-domain
SAME AS
Multiplication in the z-domain
MULTIPLYz-TRANSFORMS
FIR Filter∑
=−=
∗=M
kknxkh
nhnxny
0][][
][][][
8/22/2003 © 2003, JH McClellan & RW Schafer 8
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
Z-TRANSFORM-DOMAINPOLYNOMIALS: H(z)
FREQ-DOMAIN
kjM
kk
j ebeH ωω ˆ
0
ˆ )( −
=∑=
TIME-DOMAIN
∑=
−=M
kk knxbny
0][][
}{ kb
Page 98
8/22/2003 © 2003, JH McClellan & RW Schafer 10
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
8/22/2003 © 2003, JH McClellan & RW Schafer 11
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.
8/22/2003 © 2003, JH McClellan & RW Schafer 12
ZEROS of H(z)
Find z, where H(z)=01
211)( −−= zzH
21 :at Zero =z
0?01
21
121
=−=− −
zz
8/22/2003 © 2003, JH McClellan & RW Schafer 13
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±
Page 99
8/22/2003 © 2003, JH McClellan & RW Schafer 14
PLOT ZEROS in z-DOMAIN
3 ZEROSH(z) = 0
3 POLES
UNITCIRCLE
8/22/2003 © 2003, JH McClellan & RW Schafer 15
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
8/22/2003 © 2003, JH McClellan & RW Schafer 16
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
8/22/2003 © 2003, JH McClellan & RW Schafer 17
ANGLE is FREQUENCY
ωω
ˆ)()( ˆjez
j zHeH ==
Page 100
8/22/2003 © 2003, JH McClellan & RW Schafer 18
FIR Frequency Response
)( and )( of Zeros ˆ zHeH jω
8/22/2003 © 2003, JH McClellan & RW Schafer 19
3 DOMAINS MOVIE: FIRZEROS MOVE
)( ω̂jeH][nh
)(zH
8/22/2003 © 2003, JH McClellan & RW Schafer 20
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
8/22/2003 © 2003, JH McClellan & RW Schafer 21
PLOT ZEROS in z-DOMAIN
3 ZEROSH(z) = 0
3 POLES
UNITCIRCLE
Page 101
8/22/2003 © 2003, JH McClellan & RW Schafer 22
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
8/22/2003 © 2003, JH McClellan & RW Schafer 23
FIR Frequency Response
)( and )( of Zeros ˆ zHeH jω
8/22/2003 © 2003, JH McClellan & RW Schafer 24
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
8/22/2003 © 2003, JH McClellan & RW Schafer 25
PeZ Demo: Zero Placing
Page 102
8/22/2003 © 2003, JH McClellan & RW Schafer 1
Signal Processing First
Lecture 16IIR Filters: Feedbackand H(z)
8/22/2003 © 2003, JH McClellan & RW Schafer 3
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
8/22/2003 © 2003, JH McClellan & RW Schafer 4
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
8/22/2003 © 2003, JH McClellan & RW Schafer 5
THREE DOMAINS
Z-TRANSFORM-DOMAINPOLYNOMIALS: H(z)
FREQ-DOMAIN
kjM
kk
j ebeH ωω ˆ
0
ˆ )( −
=∑=
TIME-DOMAIN
∑=
−=M
kk knxbny
0][][
}{ kb
Page 103
8/22/2003 © 2003, JH McClellan & RW Schafer 6
Quick Review: Delay by nd
IMPULSE RESPONSE
SYSTEM FUNCTION dnzzH −=)(
][][ dnnxny −=
FREQUENCY RESPONSE
][][ dnnnh −= δ
dnjj eeH ωω ˆˆ )( −=
8/22/2003 © 2003, JH McClellan & RW Schafer 8
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
∞
∑
8/22/2003 © 2003, JH McClellan & RW Schafer 9
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
8/22/2003 © 2003, JH McClellan & RW Schafer 10
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
Page 104
8/22/2003 © 2003, JH McClellan & RW Schafer 11
COMPUTE OUTPUT
8/22/2003 © 2003, JH McClellan & RW Schafer 12
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
8/22/2003 © 2003, JH McClellan & RW Schafer 13
AT REST CONDITION
y[n] = 0, for n<0BECAUSE x[n] = 0, for n<0
8/22/2003 © 2003, JH McClellan & RW Schafer 14
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
∑
Page 105
8/22/2003 © 2003, JH McClellan & RW Schafer 15
COMPUTE MORE y[n]
CONTINUE THE RECURSION:
8/22/2003 © 2003, JH McClellan & RW Schafer 16
PLOT y[n]
8/22/2003 © 2003, JH McClellan & RW Schafer 17
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=
8/22/2003 © 2003, JH McClellan & RW Schafer 18
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
Page 106
8/22/2003 © 2003, JH McClellan & RW Schafer 19
PLOT IMPULSE RESPONSE
h[n] = b0 (a1)n u[n] = 3(0.8)n u[n]
8/22/2003 © 2003, JH McClellan & RW Schafer 20
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
8/22/2003 © 2003, JH McClellan & RW Schafer 21
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
>−
=
==
−
∞
=
−∞
=
− ∑∑
8/22/2003 © 2003, JH McClellan & RW Schafer 22
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=
Page 107
8/22/2003 © 2003, JH McClellan & RW Schafer 23
H(z) = z-Transform{ h[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
8/22/2003 © 2003, JH McClellan & RW Schafer 24
CONVOLUTION PROPERTY
MULTIPLICATION of z-TRANSFORMS
CONVOLUTION in TIME-DOMAIN
Y (z) = H(z)X(z)X(z)
h[n]y[n] = h[n]∗ x[n]x[n]
H(z)
IMPULSE RESPONSE
8/22/2003 © 2003, JH McClellan & RW Schafer 25
STEP RESPONSE: x[n]=u[n]
u[n] =1, for n ≥ 0
8/22/2003 © 2003, JH McClellan & RW Schafer 26
DERIVE STEP RESPONSE
Page 108
8/22/2003 © 2003, JH McClellan & RW Schafer 27
PLOT STEP RESPONSE
y[n] = 15 1 − 0.8n+1( )u[n]y[n] = 0.8y[n −1]+ 3u[n]
Page 109
Signal Processing First
Lecture 17IIR Filters: H(z) and
Frequency Response
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 3
READING ASSIGNMENTS
This Lecture:Chapter 8, Sects. 8-4 8-5 & 8-6
Other Reading:Recitation: Chapter 8, all
POLE-ZERO PLOTS
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 4
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
Z-TRANSFORM-DOMAINPOLYNOMIALS: H(z)
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
Page 110
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 6
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=
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 7
Typical IMPULSE Response
h[n] = b0 (a1)n u[n] = 3(0.8)n u[n]
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 8
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
>−
=
==
−
∞
=
−∞
=
− ∑∑
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 9
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)
Page 111
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 10
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
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 11
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)
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 12
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)
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 13
CONVOLUTION PROPERTY
MULTIPLICATIONMULTIPLICATION of z-TRANSFORMS
CONVOLUTIONCONVOLUTION in TIME-DOMAIN
Y (z) = H(z)X(z)X(z)
h[n]y[n] = h[n]∗ x[n]x[n]
H(z)
IMPULSE RESPONSE
Page 112
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 14
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
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 15
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
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 16
POLE-ZERO PLOT
2 + 2z −1
1 −0.8z−1
ZERO at z = -1
POLE atz = 0.8
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 17
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 )
Page 113
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 18
FREQUENCY RESPONSE
EVALUATE on the UNIT CIRCLE
H(e j ˆ ω ) = H(z ) z = e j ˆ ω
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 19
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
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 20
Frequency Response Plot
ω
ωω
ˆ
ˆˆ
8.0122)( j
jj
eeeH −
−
−+
=
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 21
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
Page 114
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 22
3-D VIEW3-D VIEWPOINT:EVALUATE H(z) EVERYWHERE
UNIT CIRCLE
WHERE isthe POLE ?
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 23
MOVIE for H(z) in 3-D
POLES to H(z) to Frequency ReponseTWO POLES SHOWN
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 24
Frequency Response from H(z)
Walking around the Unit Circle
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 25
3 DOMAINS MOVIE: IIR
POLE MOVES
h[n]
H(z)
)( ω̂jeH
Page 115
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 26
PeZ Demo: Pole-Zero Placing
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 27
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
==
=
=
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 28
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
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 29
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
Page 116
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 30
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
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 31
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
Page 117
4/18/2004 © 2003, JH McClellan & RW Schafer 1
Signal Processing First
Lecture 183-Domains for IIR
4/18/2004 © 2003, JH McClellan & RW Schafer 3
READING ASSIGNMENTS
This Lecture:Chapter 8, all
Other Reading:Recitation: Ch. 8, all
POLES & ZEROSNext Lecture: Chapter 9
4/18/2004 © 2003, JH McClellan & RW Schafer 4
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
∑
4/18/2004 © 2003, JH McClellan & RW Schafer 5
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][][][
Page 118
4/18/2004 © 2003, JH McClellan & RW Schafer 6
Z-TRANSFORM TABLES
4/18/2004 © 2003, JH McClellan & RW Schafer 7
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
4/18/2004 © 2003, JH McClellan & RW Schafer 8
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
4/18/2004 © 2003, JH McClellan & RW Schafer 9
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?
Page 119
4/18/2004 © 2003, JH McClellan & RW Schafer 10
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 −+==
4/18/2004 © 2003, JH McClellan & RW Schafer 11
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
4/18/2004 © 2003, JH McClellan & RW Schafer 12
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
4/18/2004 © 2003, JH McClellan & RW Schafer 13
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]
Page 120
4/18/2004 © 2003, JH McClellan & RW Schafer 14
Complex POLE-ZERO PLOT
1− z −2
1 +0.7225z−2
4/18/2004 © 2003, JH McClellan & RW Schafer 15
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
4/18/2004 © 2003, JH McClellan & RW Schafer 16
FREQUENCY RESPONSEfrom POLE-ZERO PLOT
H(e j ˆ ω ) =1− e− j 2 ˆ ω
1+ 0.7225e− j 2 ˆ ω
4/18/2004 © 2003, JH McClellan & RW Schafer 17
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
Page 121
4/18/2004 © 2003, JH McClellan & RW Schafer 18
Complex POLE-ZERO PLOT
1 −0.45z−1
1 −0.9z−1 +0.81z −2
4/18/2004 © 2003, JH McClellan & RW Schafer 19
h[n]: Decays & Oscillates
1− 0.8227z−1
1 −1.6454z−1 + 0.9025z−2
h[n] = (0.95)n cos(π6 n)u[n]
“PERIOD”=12
4/18/2004 © 2003, JH McClellan & RW Schafer 20
Complex POLE-ZERO PLOT
1− 0.8227z−1
1 −1.6454z−1 + 0.9025z−2
4/18/2004 © 2003, JH McClellan & RW Schafer 21
3 DOMAINS MOVIE: IIRPOLE MOVES
h[n]
H(ωωωω)
H(z)
Page 122
4/18/2004 © 2003, JH McClellan & RW Schafer 22
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]
4/18/2004 © 2003, JH McClellan & RW Schafer 23
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
4/18/2004 © 2003, JH McClellan & RW Schafer 27
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
4/18/2004 © 2003, JH McClellan & RW Schafer 28
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 → ∞
Page 123
4/18/2004 © 2003, JH McClellan & RW Schafer 29
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
4/18/2004 © 2003, JH McClellan & RW Schafer 30
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
4/18/2004 © 2003, JH McClellan & RW Schafer 31
ˆ ω 0 =2π10
Cosine input
(−0.8)n
cos(0.2πn)u[n]
4/18/2004 © 2003, JH McClellan & RW Schafer 32
STABILITY
When Does the TRANSIENT DIE OUT ?
need a1 <1
Page 124
4/18/2004 © 2003, JH McClellan & RW Schafer 33
STABILITY CONDITION
ALL POLES INSIDE the UNIT CIRCLEUNSTABLE EXAMPLE: POLE @ z=1.1
x[n] = cos(0.2πn)u[n]
4/18/2004 © 2003, JH McClellan & RW Schafer 34
BONUS QUESTION
Given:
The input is
Then find y[n]
H(z) =5
1+ 0.8z −1
)5.0cos(4][ ππ −= nnx
?][ =ny
Page 125
Signal Processing First
Lecture 19Continuous-Time Signals and Systems
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 3
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
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 4
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
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 5
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
Page 126
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 6
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
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 8
CT Signals: ONE-SIDED
v(t) = e−tu(t)
Unit step signalu(t) =1 t > 00 t < 0
⎧ ⎨ ⎩
One-SidedSinusoid
“Suddenly applied”Exponential
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 9
CT Signals: FINITE LENGTH
Square Pulse signal
p(t) = u(t − 2) −u(t − 4)
Sinusoid multipliedby a square pulse
Page 127
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 10
What is an Impulse?
A signal that is “concentrated” at one point.
limΔ→0
δΔ (t) = δ (t)δΔ (t)dt = 1
−∞
∞
∫
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 11
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)
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 12
Sampling Property
f (t)δ (t) = f (0)δ (t)
f (t)δ Δ(t) ≈ f (0)δΔ (t)
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 13
General Sampling Property
f (t)δ (t − t0 ) = f (t0 )δ (t − t0 )
Page 128
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 14
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
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 16
CT BUILDING BLOCKS
INTEGRATOR (CIRCUITS)
DIFFERENTIATOR
DELAY by to
MODULATOR (e.g., AM Radio)
MULTIPLIER & ADDER
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 17
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 )
Page 129
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 18
Output of Ideal Delay of 1 secx(t) = e−tu(t)
y(t) = x(t −1) = e−(t−1)u(t − 1)
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 19
Integrator:
Mathematical Definition:
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
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 20
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)
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 21
Graphical Representation
δ(t) =du(t)
dt
u(t) = δ (τ )dτ =1 t > 00 t < 0
⎧ ⎨ ⎩ −∞
t∫
Page 130
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 22
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−=
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 23
Differentiator:
Mathematical Definition:
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
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 24
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
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 25
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)−∞
∞
∫
Page 131
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 26
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)
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 27
Testing Time-Invariance
x(t) x(t − t0 )
y(t)
w(t)
y(t − t0 )
t0
w(t) y(t − t0 )
t0
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 29
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)]
Page 132
11/3/2003 © 2003, JH McClellan & RW Schafer 1
Signal Processing First
Lecture 20Convolution
(Continuous-Time)
11/3/2003 © 2003, JH McClellan & RW Schafer 3
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
11/3/2003 © 2003, JH McClellan & RW Schafer 4
LECTURE OBJECTIVES
Review of C-T LTI systems Evaluating convolutions
ExamplesImpulses
LTI SystemsStability and causalityCascade and parallel connections
11/3/2003 © 2003, JH McClellan & RW Schafer 5
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.
∫∞
∞−
∗=−= )()()()()( thtxdthxty τττ
Page 133
11/3/2003 © 2003, JH McClellan & RW Schafer 6
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)
11/3/2003 © 2003, JH McClellan & RW Schafer 7
Testing Time-Invariance
x(t) x(t − t0 )
y(t)
w(t)
y(t − t0 )
t0
w(t) y(t − t0 )
t0
11/3/2003 © 2003, JH McClellan & RW Schafer 8
Ideal Delay:
Linear
and Time-Invariant
)()( dttxty −=
))(()( 0tttxtw d −−=
))(()( 00 dtttxtty −−=−
)()()()( 2121 tbytayttbxttax dd +=−+−
11/3/2003 © 2003, JH McClellan & RW Schafer 9
00 Let )()( tdtxtwt
−=−= ∫∞−
τσττ
Integrator:Linear
And Time-Invariant
ττ dxtyt
)()( ∫∞−
=
)()(
)()()]()([
21
2121
tbytay
dbxdaxdbxaxttt
+=
+=+ ∫∫∫∞−∞−∞−
τττττττ
)()()( 0
0
ttydxtwtt
−==⇒ ∫−
∞−
σσ
Page 134
11/3/2003 © 2003, JH McClellan & RW Schafer 10
Modulator:
NotNot linear--obvious because
NotNot time-invariant
ttxAty cωcos)]([)( +=
)(cos)]([)( 00 ttytttxAtw c −≠−+= ω
)]([)]([)]()([
21
21tbxAtaxA
tbxtaxA+++
≠++
11/3/2003 © 2003, JH McClellan & RW Schafer 11
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.
∫∞
∞−
∗=−= )()()()()( thtxdthxty τττ
11/3/2003 © 2003, JH McClellan & RW Schafer 12
Convolution of Impulses, etc.
Convolution of two impulses
Convolution of step and shifted impulse
=−∗− )()( 21 tttt δδ )( 21 ttt −−δ
=−∗ )()( 0tttu δ )( 0ttu −
11/3/2003 © 2003, JH McClellan & RW Schafer 13
Evaluating a Convolution
∫∞
∞−
∗=−= )()()()()( txthdtxhty τττ
)()( tueth t−=)1()( −= tutx
Page 135
11/3/2003 © 2003, JH McClellan & RW Schafer 14
“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
11/3/2003 © 2003, JH McClellan & RW Schafer 16
Solution
1 0)( t<ty =
1 1
)(
)1(
1
0
1
0
≥−=
−==
−−
− −−−∫te
edety
t
t tττ τ
11/3/2003 © 2003, JH McClellan & RW Schafer 17
Convolution GUI
Page 136
11/3/2003 © 2003, JH McClellan & RW Schafer 18
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 ≠= − ),()(
11/3/2003 © 2003, JH McClellan & RW Schafer 19
Special Case: u(t)
)()1(1
)()()()()(
tuea
thtxdthxty
at−
∞
∞−
−=
∗=−= ∫ τττ
0),()( ≠= − atuetx at )()( tuth =
)()1(21)(
2 if2 tuety
at−−=
=
11/3/2003 © 2003, JH McClellan & RW Schafer 20
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
Page 137
11/3/2003 © 2003, JH McClellan & RW Schafer 22
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)
11/3/2003 © 2003, JH McClellan & RW Schafer 23
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 < ∞−∞
∞∫
11/3/2003 © 2003, JH McClellan & RW Schafer 24
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
11/3/2003 © 2003, JH McClellan & RW Schafer 25
Substitute x(t)=ax1(t)+bx2(t)
Therefore, convolution is linear.
)()(
)()()()(
)()]()([)(
21
21
21
tbytay
dthxbdthxa
dthbxaxty
+=
−+−=
−+=
∫ ∫
∫∞
∞−
∞
∞−
∞
∞−
ττττττ
ττττ
Convolution is Linear
Page 138
11/3/2003 © 2003, JH McClellan & RW Schafer 26
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 )
Page 139
Signal Processing First
Lecture 21Frequency Response of
Continuous-Time Systems
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 3
READING ASSIGNMENTS
This Lecture:Chapter 10, all
Other Reading:Recitation: Ch. 10 all, start Ch 11Next Lecture: Chapter 11
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 4
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
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 5
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
Page 140
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 6
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
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 7
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
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 8
Complex Exponential Inputtjjtjj eAejHtyeAetx ωϕωϕ ω )()()( ==
∫∞
∞−
−= ττ τωϕ deAehty tjj )()()(
FrequencyResponse∫
∞
∞−
−= ττω ωτ dehjH j)()(
tjjj eAedehty ωϕωτ ττ ⎟⎟⎠
⎞⎜⎜⎝
⎛= ∫
∞
∞−
−)()(
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 9
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) < ∞
Page 141
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 10
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
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 11
Magnitude and Phase Plots
11+ jω
=1
1+ ω 2
∠H( jω ) = −atan(ω)
H( jω ) =1
1 + jω
H(− jω ) = H∗( jω)
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 12
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 )
Page 142
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 14
Ideal Lowpass Filter w/ Delay
HLP( jω ) =e− jωtd ω < ωco
0 ω > ωco
⎧ ⎨ ⎩
fco "cutoff freq."Magnitude
Linear Phase
ω
ω
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 15
Sinusoid in Gives Sinusoid out
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 16
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
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 17
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 ))
Page 143
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 18
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
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 20
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
∫
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 21
Spectrum from Fourier Series
ak =
1− e− jπk
jπk=
2jπk
k = ±1,±3,…
0 k = 0,±2,±4,…
⎧ ⎨ ⎪
⎩ ⎪
ω0 = 2π(25)
Page 144
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 22
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
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 23
Ideal Lowpass Filter (100 Hz)
y(t) =4π
sin 50πt( ) +4
3πsin 150πt( )
H( jω ) =1 ω < ωco0 ω > ωco
⎧ ⎨ ⎩
fco "cutoff freq."
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 24
Ideal Lowpass Filter (200 Hz)
H( jω ) =1 ω < ωco0 ω > ωco
⎧ ⎨ ⎩
fco "cutoff"
y(t) =4π
sin 50πt( ) +4
3πsin 150πt( ) +
45π
sin 250πt( ) +4
7πsin 350πt( )
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 25
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
Page 145
4/3/2006 © 2003-2006, JH McClellan & RW Schafer 26
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 )
Page 146
3/27/2004 © 2003, JH McClellan & RW Schafer 1
Signal Processing First
Lecture 22Introduction to the Fourier Transform
3/27/2004 © 2003, JH McClellan & RW Schafer 3
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
3/27/2004 © 2003, JH McClellan & RW Schafer 4
LECTURE OBJECTIVESReview
Frequency ResponseFourier Series
Definition of Fourier transform
Relation to Fourier SeriesExamples of Fourier transform pairs
∫∞
∞−
−= dtetxjX tjωω )()(
3/27/2004 © 2003, JH McClellan & RW Schafer 5
Everything = Sum of Sinusoids
One Square Pulse = Sum of Sinusoids???????????
Finite LengthNot Periodic
Limit of Square Wave as Period infinityIntuitive Argument
Page 147
3/27/2004 © 2003, JH McClellan & RW Schafer 6
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
3/27/2004 © 2003, JH McClellan & RW Schafer 8
Spectrum from Fourier Series
±±=
±±=≠==
…
…
,4,20
,3,1,00)2/sin(k
k
kkak π
π
3/27/2004 © 2003, JH McClellan & RW Schafer 9
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. !!!!!!!!!
Page 148
3/27/2004 © 2003, JH McClellan & RW Schafer 10
Limiting Behavior of FS
T0=2T
T0=4T
T0=8T
3/27/2004 © 2003, JH McClellan & RW Schafer 11
Limiting Behavior of Spectrum
T0=2T
T0=4T
T0=8T
)(Plot
0 kaT
3/27/2004 © 2003, JH McClellan & RW Schafer 12
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=
∞→
3/27/2004 © 2003, JH McClellan & RW Schafer 13
Fourier Transform Defined
For non-periodic signalsFourier Synthesis
Fourier Analysis
∫∞
∞−
−= dtetxjX tjωω )()(
∫∞
∞−
= ωω ωπ dejXtx tj)()( 21
Page 149
3/27/2004 © 2003, JH McClellan & RW Schafer 14
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
3/27/2004 © 2003, JH McClellan & RW Schafer 15
Frequency Response
Fourier Transform of h(t) is the Frequency Response
ωω
jjHtueth t
+=⇔= −
11)()()(
)()( tueth t−=
3/27/2004 © 2003, JH McClellan & RW Schafer 16
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∫
Page 150
3/27/2004 © 2003, JH McClellan & RW Schafer 18
x(t) =1 t < T / 20 t > T / 2
⇔ X( jω ) = sin(ωT / 2)ω / 2( )
3/27/2004 © 2003, JH McClellan & RW Schafer 19
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)(
3/27/2004 © 2003, JH McClellan & RW Schafer 20
>
<=⇔=
b
bb jXtttx
ωω
ωωω
πω
0
1)()sin()(
3/27/2004 © 2003, JH McClellan & RW Schafer 21
Example 4:
X( jω) = δ (t)e− jωtdt−∞
∞∫ = 1
Shifting Property of the Impulse
)()( 0tttx −= δ
0)()( 0tjtj edtettjX ωωδω −
∞
∞−
− =−= ∫
Page 151
3/27/2004 © 2003, JH McClellan & RW Schafer 22
x(t) = δ (t) ⇔ X( jω ) = 1
3/27/2004 © 2003, JH McClellan & RW Schafer 23
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)
3/27/2004 © 2003, JH McClellan & RW Schafer 24
x(t) = cos(ω0t) ⇔X( jω) = πδ (ω − ω0) + πδ (ω + ω0)
3/27/2004 © 2003, JH McClellan & RW Schafer 25
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 )
Page 152
3/27/2004 © 2003, JH McClellan & RW Schafer 1
Signal Processing First
Lecture 23Fourier TransformProperties
3/27/2004 © 2003, JH McClellan & RW Schafer 3
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)
3/27/2004 © 2003, JH McClellan & RW Schafer 4
LECTURE OBJECTIVESThe Fourier transform
More examples of Fourier transform pairsBasic properties of Fourier transforms
Convolution propertyMultiplication property
∫∞
∞−
−= dtetxjX tjωω )()(
3/27/2004 © 2003, JH McClellan & RW Schafer 5
Fourier Transform
Fourier Analysis(Forward Transform)∫
∞
∞−
−= dtetxjX tjωω )()(
Fourier Synthesis(Inverse Transform)∫
∞
∞−
= ωωπ
ω dejXtx tj)(21)(
)()(Domain-FrequencyDomain-Time
ωjXtx ⇔⇔
Page 153
3/27/2004 © 2003, JH McClellan & RW Schafer 6
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)
3/27/2004 © 2003, JH McClellan & RW Schafer 7
Frequency Response
Fourier Transform of h(t) is the Frequency Response
ωω
jjHtueth t
+=⇔= −
11)()()(
)()( tueth t−=
3/27/2004 © 2003, JH McClellan & RW Schafer 8
2/)2/sin()(
2/0
2/1)(
ωωω TjX
Tt
Tttx =⇔
>
<=
3/27/2004 © 2003, JH McClellan & RW Schafer 9
>
<=⇔=
b
bb jXtttx
ωω
ωωω
πω
0
1)()sin()(
Page 154
3/27/2004 © 2003, JH McClellan & RW Schafer 10
0)()()( 0tjejXtttx ωωδ −=⇔−=
00 =t
3/27/2004 © 2003, JH McClellan & RW Schafer 11
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)()()(
3/27/2004 © 2003, JH McClellan & RW Schafer 12
)()()()cos()( ccc jXttx ωωπδωωπδωω ++−=⇔=
3/27/2004 © 2003, JH McClellan & RW Schafer 13
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
Page 155
3/27/2004 © 2003, JH McClellan & RW Schafer 14
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
∫
3/27/2004 © 2003, JH McClellan & RW Schafer 15
Square Wave Fourier Transform
X( jω ) = 2π akδ(ω − kω0 )k=−∞
∞
∑
x(t) = x(t + T0 )
T0−2T0 −T0 2T00 t
3/27/2004 © 2003, JH McClellan & RW Schafer 16
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
3/27/2004 © 2003, JH McClellan & RW Schafer 17
Scaling Property
expands)(shrinks;)2( 221 ωjXtx
)(
)()(
1
)/(
aa
adajtj
jX
exdteatx
ω
λλωω λ
=
= ∫∫∞
∞−
−∞
∞−
−
)()( 1aa jXatx ω⇔
Page 156
3/27/2004 © 2003, JH McClellan & RW Schafer 18
Scaling Property
)()( 1aa jXatx ω⇔
)2()( 12 txtx =
3/27/2004 © 2003, JH McClellan & RW Schafer 19
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
3/27/2004 © 2003, JH McClellan & RW Schafer 20
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
3/27/2004 © 2003, JH McClellan & RW Schafer 21
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)
Page 157
3/27/2004 © 2003, JH McClellan & RW Schafer 22
Convolution Example
Bandlimited Input Signal“sinc” function
Ideal LPF (Lowpass Filter)h(t) is a “sinc”
Output is BandlimitedConvolve “sincs”
3/27/2004 © 2003, JH McClellan & RW Schafer 23
Ideally Bandlimited Signal
>
<=⇔=
πω
πωω
ππ
1000
1001)()100sin()( jX
tttx
πω 100=b
3/27/2004 © 2003, JH McClellan & RW Schafer 24
Convolution Example
sin(100π t)πt
∗ sin(200πt)π t
=
x(t) ∗h(t) ⇔ H( jω )X( jω )
sin(100π t)πt
3/27/2004 © 2003, JH McClellan & RW Schafer 25
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 ))
Page 158
3/27/2004 © 2003, JH McClellan & RW Schafer 26
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
Ideal Lowpass Filter
y(t) = 4π
sin 50πt( ) + 43π
sin 150πt( )
fco "cutoff freq."
H( jω ) =1 ω < ωco0 ω > ωco
3/27/2004 © 2003, JH McClellan & RW Schafer 28
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
3/27/2004 © 2003, JH McClellan & RW Schafer 29
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
Page 159
3/27/2004 © 2003, JH McClellan & RW Schafer 30
y(t) = x(t)cos(ω0t) ⇔
Y( jω ) = 12X( j(ω − ω0 )) + 1
2X( j(ω + ω0))
x(t)
3/27/2004 © 2003, JH McClellan & RW Schafer 31
Differentiation Property
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
Page 160
4/19/2005 © 2003, JH McClellan & RW Schafer 1
Signal Processing First
Lecture 24Amplitude Modulation (AM)
4/19/2005 © 2003, JH McClellan & RW Schafer 3
LECTURE OBJECTIVES
Review of FT propertiesConvolution <--> multiplicationFrequency shifting
Sinewave Amplitude ModulationAM radio
Frequency-division multiplexingFDM
Reading: Chapter 12, Section 12-2
4/19/2005 © 2003, JH McClellan & RW Schafer 4
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
4/19/2005 © 2003, JH McClellan & RW Schafer 5
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 )
Differentiation Property
Page 161
4/19/2005 © 2003, JH McClellan & RW Schafer 6
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
⎧ ⎨ ⎩
4/19/2005 © 2003, JH McClellan & RW Schafer 7
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)
4/19/2005 © 2003, JH McClellan & RW Schafer 8
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
Ideal Lowpass Filter
Hlp( jω )
ωco−ωco
y(t) = x(t) if ω0 < ωco
y(t) = 0 if ω0 > ωco
Page 162
4/19/2005 © 2003, JH McClellan & RW Schafer 10
Ideal LPF: Fourier Series
y(t) =4π
sin 50πt( ) +4
3πsin 150πt( )
fco "cutoff freq."
H( jω ) =1 ω < ωco0 ω > ωco
⎧ ⎨ ⎩
4/19/2005 © 2003, JH McClellan & RW Schafer 11
The way communication systems work
How do we sharebandwidth ?
4/19/2005 © 2003, JH McClellan & RW Schafer 12
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 )
Differentiation Property
4/19/2005 © 2003, JH McClellan & RW Schafer 13
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
Page 163
4/19/2005 © 2003, JH McClellan & RW Schafer 14
)()()()()()( 21 ωωω π jPjXjYtptxty ∗=⇔=
[ ])()()()()cos()()(
21
cc
c
jXjYttxty
ωωπδωωπδωωω
π ++−∗=⇔=
)()()()cos()(
cc
c
jPttp
ωωπδωωπδωω
++−=⇔=
))(())(()( 21
21
cc jXjXjY ωωωωω ++−=
4/19/2005 © 2003, JH McClellan & RW Schafer 15
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 ))
4/19/2005 © 2003, JH McClellan & RW Schafer 16
))(())(()()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 =⇔
⎪⎩
⎪⎨⎧
>
<=
4/19/2005 © 2003, JH McClellan & RW Schafer 17
x(t)
ωc−ωc
))((21
cjX ωω + ))((21
cjX ωω −
))(())(()()cos()()(
21
21
cc
c
jXjXjYttxty
ωωωωωω
++−=⇔=
Page 164
4/19/2005 © 2003, JH McClellan & RW Schafer 18
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).
y(t) = x(t)cos(ωct)
X( jω)x(t)
cos(ωct)Y( jω ) = 1
2X( j(ω − ωc ))
+ 12
X( j(ω + ωc ))
4/19/2005 © 2003, JH McClellan & RW Schafer 19
DSBAM Waveform
In the time-domain, the “envelope” of sine-wave peaks follows |x(t)|
4/19/2005 © 2003, JH McClellan & RW Schafer 20
Double Sideband AM (DSBAM)“Typical” bandlimitedinput signal
Frequency-shiftedcopies Upper sidebandLower sideband
4/19/2005 © 2003, JH McClellan & RW Schafer 21
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)
y(t) = x(t)cos(ωct)
Page 165
4/19/2005 © 2003, JH McClellan & RW Schafer 22
DSBAM Demodulation
V ( jω ) = H( jω )W( jω ) = X( jω ) if ωb < ωco < 2ωc − ωb
H( jω ) =2 | ω |< ωco0 |ω |> ωco
⎧ ⎨ ⎩
4/19/2005 © 2003, JH McClellan & RW Schafer 23
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)
4/19/2005 © 2003, JH McClellan & RW Schafer 24
FDM Block Diagram (Xmitter)
cos(ωc1t)
cos(ωc 2t)
ωc1 ωc2
Spectrum of inputsmust be bandlimitedNeed ωc2 − ωc1 > 2ωb
4/19/2005 © 2003, JH McClellan & RW Schafer 25
Frequency-Division De-Mux
cos(ωc1t)
cos(ωc 2t)
ωc1 ωc2
Page 166
4/19/2005 © 2003, JH McClellan & RW Schafer 26
Bandpass Filters for De-Mux
4/19/2005 © 2003, JH McClellan & RW Schafer 27
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
Page 167
8/22/2003 © 2003, JH McClellan & RW Schafer 1
Signal Processing First
Lecture 25Sampling and Reconstruction(Fourier View)
8/22/2003 © 2003, JH McClellan & RW Schafer 3
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
8/22/2003 © 2003, JH McClellan & RW Schafer 4
Table of FT Properties
x(t − td ) ⇔ e− jωtd X( jω )
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 )
8/22/2003 © 2003, JH McClellan & RW Schafer 5
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
Page 168
8/22/2003 © 2003, JH McClellan & RW Schafer 6
DSBAM: Frequency-Domain“Typical” bandlimitedinput signal
Frequency-shiftedcopies
))((21
cj jXe ωωϕ +− ))((2
1c
j jXe ωωϕ −
Upper sidebandLower sideband
)( ωjX
8/22/2003 © 2003, JH McClellan & RW Schafer 7
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
8/22/2003 © 2003, JH McClellan & RW Schafer 8
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++++
−−−=
8/22/2003 © 2003, JH McClellan & RW Schafer 9
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
++++
−−−=
Page 169
8/22/2003 © 2003, JH McClellan & RW Schafer 10
Quadrature Modulation: 4 sigs
8700 Hz
3600 Hz
8/22/2003 © 2003, JH McClellan & RW Schafer 11
Ideal C-to-D Converter
• Mathematical Model for A-to-D
x[n] = x(nTs )
FOURIERTRANSFORMof xs(t) ???
8/22/2003 © 2003, JH McClellan & RW Schafer 12
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
8/22/2003 © 2003, JH McClellan & RW Schafer 13
FT of Impulse Train
∑∑∞
−∞=
∞
−∞=−=↔−=
ks
sns k
TjPnTttp )(2)()()( ωωδπωδ
ss T
πω 2=
Page 170
8/22/2003 © 2003, JH McClellan & RW Schafer 14
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=−∞
∞∑
8/22/2003 © 2003, JH McClellan & RW Schafer 15
Illustration of Samplingx(t)
x[n] = x(nTs )
∑∞
−∞=−=
nsss nTtnTxtx )()()( δ
n
t
8/22/2003 © 2003, JH McClellan & RW Schafer 16
Sampling: Freq. Domain
EXPECTFREQUENCYSHIFTING !!!
∑∑∞
−∞=
∞
−∞==−=k
tjkk
ns
seanTttp ωδ )()(
∑∞
−∞==k
tjkk
sea ω
8/22/2003 © 2003, JH McClellan & RW Schafer 17
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
Page 171
8/22/2003 © 2003, JH McClellan & RW Schafer 18
Frequency-Domain Representation of Sampling
Xs ( jω) = 1Ts
X( j(ωk=−∞
∞∑ − kωs ))
“Typical”bandlimited signal
8/22/2003 © 2003, JH McClellan & RW Schafer 19
Aliasing Distortion
If ωs < 2ωb , the copies of X(jω) overlap, and we have aliasing distortion.
“Typical”bandlimited signal
8/22/2003 © 2003, JH McClellan & RW Schafer 20
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 )
Page 172
8/22/2003 © 2003, JH McClellan & RW Schafer 22
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,…
8/22/2003 © 2003, JH McClellan & RW Schafer 23
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=−∞
∞∑
8/22/2003 © 2003, JH McClellan & RW Schafer 24
Shannon Sampling Theorem
“SINC” Interpolation is the idealPERFECT RECONSTRUCTIONof BANDLIMITED SIGNALS
8/22/2003 © 2003, JH McClellan & RW Schafer 25
Reconstruction in Time-Domain
Page 173
8/22/2003 © 2003, JH McClellan & RW Schafer 26
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 ))
Page 174
8/22/2003 © 2003, JH McClellan & RW Schafer 1
Signal Processing First
Lecture 26Review: Digital Filtering
of Analog Signals
8/22/2003 © 2003, JH McClellan & RW Schafer 3
LECTURE OBJECTIVES
Sampling Theorem RevisitedGENERAL: in the FREQUENCY DOMAINFourier transform of sampled signalReconstruction from samples
Effective Frequency Response
Important FT propertiesConvolution multiplicationFrequency shifting
8/22/2003 © 2003, JH McClellan & RW Schafer 4
Sampling: Freq. Domain
EXPECTFREQUENCYSHIFTING !!!
∑∑∞
−∞=
∞
−∞==−=k
tjkk
ns
seanTttp ωδ )()(
∑∞
−∞==k
tjkk
seatp ω)(
8/22/2003 © 2003, JH McClellan & RW Schafer 5
Frequency-Domain Representation of Sampling
Xs ( jω) = 1Ts
X( j(ωk=−∞
∞∑ − kωs ))
“Typical”bandlimited signal
Page 175
8/22/2003 © 2003, JH McClellan & RW Schafer 6
Aliasing Distortion
If ωs < 2ωb , the copies of X(jω) overlap, and we have aliasing distortion.
“Typical”bandlimited signal
8/22/2003 © 2003, JH McClellan & RW Schafer 7
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 8
Reconstruction: Frequency-Domain
)()()(so overlap,not do )(of copies the,2 If
ωωωω
ωω
jXjHjXjX
srr
bs
=
>Hr ( jω )
8/22/2003 © 2003, JH McClellan & RW Schafer 9
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,…
Page 176
8/22/2003 © 2003, JH McClellan & RW Schafer 10
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=−∞
∞∑
8/22/2003 © 2003, JH McClellan & RW Schafer 11
Shannon Sampling Theorem
“SINC” Interpolation is the idealPERFECT RECONSTRUCTIONof BANDLIMITED SIGNALS
8/22/2003 © 2003, JH McClellan & RW Schafer 12
Reconstruction in Time-Domain
8/22/2003 © 2003, JH McClellan & RW Schafer 13
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 ))
Page 177
8/22/2003 © 2003, JH McClellan & RW Schafer 14
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
8/22/2003 © 2003, JH McClellan & RW Schafer 15
DT Filtering of a CT Signal
Spectrum of Discrete-Time Signal
Digital Filter
Reconstruction Filter (Analog)
ω
ω̂
ω̂
ω
Analog Input
Analog Output
8/22/2003 © 2003, JH McClellan & RW Schafer 16
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
8/22/2003 © 2003, JH McClellan & RW Schafer 17
EFFECTIVE RESPONSE
DIGITAL FILTER
H(e j ˆ ω )
Heff ( jω )
fs = 1000 Hz
sin(11 ˆ ω / 2)sin( ˆ ω / 2)
Page 178
8/22/2003 © 2003, JH McClellan & RW Schafer 18
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
8/22/2003 © 2003, JH McClellan & RW Schafer 19
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
8/22/2003 © 2003, JH McClellan & RW Schafer 20
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
8/22/2003 © 2003, JH McClellan & RW Schafer 21
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)
Page 179
8/22/2003 © 2003, JH McClellan & RW Schafer 22
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
8/22/2003 © 2003, JH McClellan & RW Schafer 23
POP QUIZ ANSWER
Given:
When
The output isx(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
y(t) = 3.02 cos(2000π t − 0.452π )
8/22/2003 © 2003, JH McClellan & RW Schafer 24
ANOTHER INPUT FREQ
Given:
Find the output, y(t)When
2 + 2z −1
1 −0.8z−1 D-to-AA-to-Dx(t) y(t)y[n]x[n]
fs = 5000Hz ˆ ω = ?
ˆ ω = ?
))7500(2cos()( ttx π=
8/22/2003 © 2003, JH McClellan & RW Schafer 25
2nd POP QUIZ ANSWER
Given:
When
2 + 2z −1
1 −0.8z−1 D-to-AA-to-Dx(t) y(t)y[n]x[n]
fs = 5000Hz y(t) = ?
ω = ?
))7500(2cos()( ttx π=
)5.1(25000/)7500(2cos(ˆ ππω ==
πω 3ˆ =
Page 180
8/22/2003 © 2003, JH McClellan & RW Schafer 26
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
8/22/2003 © 2003, JH McClellan & RW Schafer 27
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
8/22/2003 © 2003, JH McClellan & RW Schafer 28
THREE DOMAINS
H(z) =
bkz− k∑
1− a z−∑z = e j ˆ ω
8/22/2003 © 2003, JH McClellan & RW Schafer 29
THE FUTURE
Circuits & Laplace Transforms
H(s)
H( jω )h(t)FrequencyResponse
Implementation isRLC-op-amp circuit
Polynomials: Poles & Zeros
Page 181
8/22/2003 © 2003, JH McClellan & RW Schafer 30
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)