Statistical and Adaptive Signal Processing - Solution Manual

467 Pages·2009·4.69 MB·English

by  Dimitris G. Manolakis

Solutions Manual
for
Statistical and Adaptive Signal Processing
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Preface
This solutions manual provides solutions to the problems contained in the first
edition of our book STATISTICAL AND ADAPTIVE SIGNAL PROCESSING.
Thesolutions havebeenpreparedandwrittenbyDavidMardenandourselves. We
have attempted to provide very detailed solutions to the problems with notation
consistent with that used in the book. Where applicable, we have also given a
printout ofMatlabcodeandfiguresfortheproblemsolution.
Despite our efforts, however, you may find that some of the solutions may be less
detailed and refined than others. Inevitably through the use of this solutions man-
ual, omissions and errors will be discovered. Our goal is to continually improve
upon thismanual usingyour comments. Periodically, wewillissue changes tothe
solutionsmanualandnewproblemstoyouuponrequest. Inthisrespect,wewould
appreciate any feedback including improved solutions, suggestions for new prob-
lems,corrections whichcanbesenttoProf. [email protected]
orat
Prof. VinayK.Ingle
DepartmentofElectricalandComputerEngineering
Northeastern University
360HuntingtonAvenue
Boston,MA02115
Thankyou.
DimitrisG.Manolakis
VinayK.Ingle
Stephen K.Kogon
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Chapter2
Discrete-Time Signals and Systems
2.1 Samplingfrequency F = 100sam/sec
s
(a) Continuous-time signal x (t) = 2cos(40πt +π/3)hasfrequency of20Hz. Hence
c
(cid:1) (cid:2)
40πn
x(n) = x (t)| = 2cos +π/3
c t=n/Fs 100
whichimpliesthatω = 40π = 2π.
0 100 5
(b) Steady-state response yc,ss(t): Giventhath(n) = 0.8nu(n),thefrequency response functionis
1
H(ejω) =
1−0.8e−jω
Sinceω = 2π/5,thesystemresponse atω is
0 0
1
H(ejω) = = 0.9343e−j0.2517π
1−0.8e−j2π/5
Hence y (n)= 2(0.9343)cos(2πn/5+π/3−0.2517π),or
ss
yc,ss(t) = 1.8686cos(40πt +0.585π)
(c) Any x (t) that has the same digital frequency ω after sampling and the same phase shift as above will
c 0
havethesamesteadystateresponse. Since F = 100sam/sec,thetwootherfrequenciesare120and220
s
Hz.
2.2 Thediscrete-time signalis
x(n) = A cos(ω n)w (n)
0 R
wherew (n)isan N-point rectangular window.
R
(a) TheDTFTofx(n)isdeterminedas
X(ejω) = F[A cos(ω n)w (n)]
(cid:3) 0 R (cid:4) (cid:3) (cid:4)
= (A/2)F ejω0w (n) +(A/2)F e−jω0w (n) (1)
R R
UsingtheDTFTofw (n)as
R
(cid:5)N−1 sin(ωN/2)
F[w (n)]= e−jωn = e−jω(N−1)/2 (2)
R sin(ω/2)
n=0
and the fact that complex exponential causes a translation in the frequency domain (1) can be written
afterafairamountofalgebraandtrigonometry as
X(ejω) = X (ejω)+ j X (ejω)
R I
1
2 StatisticalandAdaptiveSignalProcessing-SolutionManual
32−point DTFT (Real)
32−point DTFT (Imaginary)
20
15
15 10
5
10
0
5
−5
0
−10
−5 −15
−4 −2 0 2 4 −4 −2 0 2 4
Figure2.2bc: RealandImaginaryDTFTandDFTPlots(ω = π/4)
0
where
A sin[(ω−ω )N/2]
X (ejω) = cos[(ω−ω )(N −1)/2] 0
R 2 0 sin[(ω−ω )/2]
0
A sin{[ω−(2π −ω )]N/2}
+ cos[(ω+ω )(N −1)/2] 0 (3)
2 0 sin{[ω−(2π −ω )]/2}
0
and
A sin[(ω−ω )N/2]
X (ejω) = − sin[(ω−ω )(N −1)/2] 0
R 2 0 sin[(ω−ω )/2]
0
A sin{[ω−(2π −ω )]N/2}
− sin[(ω+ω )(N −1)/2] 0 (4)
2 0 sin{[ω−(2π −ω )]/2}
0
(b) N = 32andω = π/4.TheDTFTplotsareshowninFigure2.2bc.
0
(c) TheDFTsamplesareshowninFigure2.2bc.
(d) N = 32andω = 1.1π/4. TheplotsareshowninFigure2.2d.
0
Theaddedspectrumforthesecondcaseabove(ω = 1.1π/4)isaresultoftheperiodicextensionofthe
0
DFT.Fora32-point sequence, theendofeachextension doesnotlineupwiththebeginning ofthenext
extension. Thisresultsinsharpedgesintheperiodicextension, andaddedfrequencies inthespectrum.
2.3 Thesequence isx(n) = cos(πn/4),0 ≤ n ≤15.
(a) The16-point DFTisshowninthetop-left plotofFigure2.3.
(b) The32-point DFTisshowninthetop-right plotofFigure2.3.
(c) The64-point DFTisshowninthebottom plotofFigure2.3.
(d) Thezeropaddingresultsinalowerfrequencysamplinginterval. HencetherearemoretermsintheDFT
representation. TheshapeoftheDTFTcontinues tofillinas N increases from16to64.
2.4 x(n) = {1,2,3,4,3,2,1};h(n) = {−1,0,1}
StatisticalandAdaptiveSignalProcessing-SolutionManual 3
32−point DTFT (Real)
32−point DTFT (Imaginary)
20
15
15 10
5
10
0
5
−5
0
−10
−5 −15
−4 −2 0 2 4 −4 −2 0 2 4
Figure2.2d: RealandImaginaryDTFTPlots(ω = 1.1π/4)
0
16−point DFT 32−point DFT
7 7
6 6
5 5
4 4
3 3
2 2
1 1
0 0
0 π/ 4 π/ 2 3π /4 π 5π /4 3π /2 7π /4 0 π/ 4 π/ 2 3π /4 π 5π /4 3π /2 7π /4
64−point DFT
8
6
4
2
0
0 π/ 4 π/ 2 3π /4 π 5π /4 3π /2 7π /4
Figure2.3: The16,32,and64-point DFTsof x(n) = cos(πn/4)
4 StatisticalandAdaptiveSignalProcessing-SolutionManual
(a) Convolution usingmatrix-vector multiplication approach.
y = Xh
   
−1 1 0 0
 −2   2 1 0 
 −2   3 2 1  
 −2   4 3 2  −1
 0  =  3 4 3  0 
   
 2   2 3 4  1
   
 2   1 2 3 
   
2 0 1 2
1 0 0 1
(b) TheMatlabfunction convtoep
function [y]=convtoep(x,h)
% Convolution using Toeplitz Arrays
% y = convtoep(x,h)
nx = length(x); x = reshape(x,nx,1);
nh = length(h); h = reshape(h,nh,1);
X = toeplitz([x; zeros(nh-1,1)],[x(1) zeros(1,nh-1)]);
y = X*h;
(c) Verification:
2.5 x(n) = 0.9nu(n)
(a) Analytical evaluation ofx(n)∗x(n) :
(cid:5)∞
y(n) = x(n)∗x(n) = x(k)x(n −k)
k=−∞
(cid:5)∞
= (0.9)ku(k)(0.9)n−ku(n−k)
k=−∞
(cid:5)∞
= (0.9)k(0.9)n−ku(n−k)
k=0
y(n) = (n+1)(0.9)n
Thissequence isshownintheleftmostplotinFigure2.5.
(b) Convolutionusingtheconvfunction: Thesequencex(n)istruncatedto51samples. Thisconvolutionis
doneusing
n = 0:50; x =(0.9).^n; y = conv(x,x);
Thissequence isinthecenterplotininFigure2.5.
(c) Convolution using the filter function: To use this function, we have to represent one of the x(n) by
coefficients inanequivalent difference equation. Thisdifference equation isgivenby
x(n) = δ(n)+0.9x(n −1)
whichmeansthatthefiltercoefficientsareb = 1,a = [1,-0.9]. Thusthisconvolution isdoneusing
y = filter(1,[1,-0.9],x);
Thissequence isintherightmostplotinFigure2.5.
StatisticalandAdaptiveSignalProcessing-SolutionManual 5
(d) The three plots appear to be identical. However, the conv function gives the largest error since both
sequences aretruncated. Thefilterfunction wouldbebestsuitedforinfinitelengthsequences.
y(n)=(n+1)(0.9)n Using Conv Using filter
4 4 4
3.5 3.5 3.5
3 3 3
2.5 2.5 2.5
2 2 2
1.5 1.5 1.5
1 1 1
0.5 0.5 0.5
0 0 0
0 20 40 60 80 100 0 20 40 60 80 100 0 20 40 60 80 100
Figure2.5: x(n) = (0.9)nu(n)convolved withitself
2.6 H (n)isacausalandstableall-passsystem withinput x(n)andoutput y(n). Thenwehave
ap
(cid:5)∞ (cid:5)∞
|y(n)|2 = |x(n)|2
n=0 n=0
Consider
x(n) = x (n)+x (n) and y (n) = h(n)∗x (n)
0 1 i i
Then
(cid:5)∞ (cid:5)∞
|y (n)+ y (n)|2 = |x (n)+x (n)|2
0 1 0 1
n=0 n=0
(cid:5)∞ (cid:5)∞ (cid:5)∞ (cid:5)∞ (cid:5)∞
|y (n)|2+ |y (n)|2+ |y (n)y (n)| = |x (n)|2+ |x (n)|2
0 1 0 1 0 1
n=0 n=0 n=0 n=0 n=0
(cid:5)∞ (cid:5)∞ (cid:5)∞
|y (n)|2+ |y (n)y (n)| = |x (n)|2
0 0 1 0
n=0 n=0 n=0
Hence
(cid:5)∞ (cid:5)∞
|y (n)|2 ≤ |x (n)|2
0 0
n=0 n=0
Definex (n)as0aboven ,then
0 0
(cid:5)n0 (cid:5)n0
|y(n)|2 ≤ |x(n)|2
n=0 n=0
2.7 Monotone phase-response property ofacausalandstablePZ-APsystem:
(a) Arealfirst-ordersystem: Consider
p−z−1
H(z)= , |p| < 1
1− pz−1
6 StatisticalandAdaptiveSignalProcessing-SolutionManual
Then
(cid:1) (cid:2) (cid:1) (cid:2)
sinω sinω
H(ejω) = arctan −arctan
] cosω−1/p cosω− p
Clearly, H(ejω) decreases from H(ej0) = π to H(ej2π) = −π.Toshow that it decreases mono-
] ] ]
tonically, consider
(cid:1) (cid:1) (cid:2) (cid:1) (cid:2)(cid:2)
d (cid:12) (cid:13) d sinω sinω
H(ejω) = arctan −arctan
dω ] dω cosω−1/p cosω− p
p2 −1
=
1+ p2 −2(cosω) p
which isnegative for |p| < 1. Thisproves that H(ejω)decreases monotonically from H(ej0) = π
to H(ej2π) = −π. ] ]
]
(b) Arealsecond-order (complex-conjugate pole-pair) system: Thesystemfunction is
(cid:14) (cid:15)(cid:14) (cid:15)
(r θ)−z−1 (r θ)∗ −z−1
H(z)= 1−](r θ)∗z−1 1−](r θ)z−1 , 0 <r < 1
] ]
Considertheterm
(r θ)−z−1
H1(z), 1−](r θ)z−1
]
Thenanglecalculationsaresimilartothosefortherealfirst-ordercaseifwerotatethecoordinatesystem
bytheangleθ,thatis,
(cid:14) (cid:1) (cid:2) (cid:15) (cid:14) (cid:1) (cid:2) (cid:15)
sinω sinω
H (ejω) = arctan +θ − arctan +θ
] 1 cosω−1/r cosω−r
(cid:1) (cid:2) (cid:1) (cid:2)
sinω sinω
= arctan −arctan
cosω−1/r cosω−r
Thusfollowingtheargumentsin(a),weconcludethat H (ejω)decreasesmonotonicallyfrom H (ej0)
1 1
to H (ej0)−π. Similarly,consider ] ]
1
]
(r θ)−z−1
H2(z) , 1−](r θ)z−1
]
Then H (ejω)decreases monotonically from H (ej0)to H (ej0)−π. Finally,since
2 2 2
] ] ]
H (ejω) = H (ejω)+ H (ejω)
1 1 2
] ] ]
weconclude that H(ejω)decreases monotonically from H(ej0)to H(ej0)−2π.
] ] ]
(c) Generalizingtheaboveresultforareal,causal,andstablePZ-APsystemwith P pole-zeropairs,wecan
showthat H(ejω)decreases monotonically from H(ej0)to H(ej0)−2πP.
] ] ]
2.8 Minimumgroupdelayproperty: Considerthesystems
(cid:12) (cid:13)(cid:12) (cid:13)
H (z) = 1−0.25z−1 1+0.5z−1
min (cid:12) (cid:13)(cid:12) (cid:13)
H (z) = 0.25−z−1 0.5+z−1
max (cid:12) (cid:13)(cid:12) (cid:13)
H (z) = 1−0.25z−1 0.5+z−1
mix