Applied Digital Signal Processing 1st Edition Manolakis Solutions Manual

CHAPTER 2
Discrete-Time Signals and Systems
Tutorial Problems
1. (a) MATLAB script:
% P0201a: Generate and plot unit sample
close all; clc
n = -20:40; % specifiy support of signal
deltan = zeros(1,length(n)); % define signal
deltan(n==0)=1;
% Plot:
hf = figconfg(’P0201a’,’small’);
stem(n,deltan,’fill’)
axis([min(n)-1,max(n)+1,min(deltan)-0.2,max(deltan)+0.2])
xlabel(’n’,’fontsize’,LFS); ylabel(’delta[n]’,’fontsize’,LFS);
title(’Unit Sample delta[n]’,’fontsize’,TFS)
−20 −10 0 10 20 30 40
−0.2
0
0.2
0.4
0.6
0.8
1
n
δ[n]
Unit Sample δ[n]
FIGURE 2.1: unit sample δ[n].
1
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CHAPTER 2. Discrete-Time Signals and Systems 2
(b) MATLAB script:
% P0201b: Generate and plot unit step sequence
close all; clc
un = zeros(1,length(n)); % define signal
un(n>=0)=1;
% Plot:
hf = figconfg(’P0201b’,’small’);
stem(n,un,’fill’)
axis([min(n)-1,max(n)+1,min(un)-0.2,max(un)+0.2])
xlabel(’n’,’fontsize’,LFS); ylabel(’u[n]’,’fontsize’,LFS);
title(’Unit Step u[n]’,’fontsize’,TFS)
−20 −10 0 10 20 30 40
−0.2
0
0.2
0.4
0.6
0.8
1
n
u[n]
Unit Step u[n]
FIGURE 2.2: unit step u[n].
(c) MATLAB script:
% P0201c: Generate and plot real exponential sequence
close all; clc
x1n = 0.8.^n; % define signal
% Plot:
hf = figconfg(’P0201c’,’small’);
stem(n,x1n,’fill’)
axis([min(n)-1,max(n)+1,min(x1n)-5,max(x1n)+5])
xlabel(’n’,’fontsize’,LFS); ylabel(’x_1[n]’,’fontsize’,LFS);
title(’Real Exponential Sequence x_1[n]’,’fontsize’,TFS)
(d) MATLAB script:

−20 −10 0 10 20 30 40
0
20
40
60
80
n
x
1
[n]
Real Exponential Sequence x
1
[n]
FIGURE 2.3: real exponential signal x1[n] = (0.80)n.
% P0201d: Generate and plot complex exponential sequence
close all; clc
x2n = (0.9*exp(j*pi/10)).^n; % define signal
x2n_r = real(x2n); % real part
x2n_i = imag(x2n); % imaginary part
x2n_m = abs(x2n); % magnitude part
x2n_p = angle(x2n); % phase part
% Plot:
hf = figconfg(’P0201d’);
subplot(2,2,1)
stem(n,x2n_r,’fill’)
axis([min(n)-1,max(n)+1,min(x2n_r)-1,max(x2n_r)+1])
xlabel(’n’,’fontsize’,LFS); ylabel(’Re{x_2[n]}’,’fontsize’,LFS);
title(’Real Part of Sequence x_2[n]’,’fontsize’,TFS)
subplot(2,2,2)
stem(n,x2n_i,’fill’)
axis([min(n)-1,max(n)+1,min(x2n_i)-1,max(x2n_i)+1])
xlabel(’n’,’fontsize’,LFS); ylabel(’Im{x_2[n]}’,’fontsize’,LFS);
title(’Imaginary Part of Sequence x_2[n]’,’fontsize’,TFS)
subplot(2,2,3)
stem(n,x2n_m,’fill’)
axis([min(n)-1,max(n)+1,min(x2n_m)-1,max(x2n_m)+1])
xlabel(’n’,’fontsize’,LFS); ylabel(’|x_2[n]|’,’fontsize’,LFS);
title(’Magnitude of Sequence x_2[n]’,’fontsize’,TFS)
subplot(2,2,4)

stem(n,x2n_p,’fill’)
axis([min(n)-1,max(n)+1,min(x2n_p)-1,max(x2n_p)+1])
xlabel(’n’,’fontsize’,LFS); ylabel(’phi(x_2[n])’,’fontsize’,LFS);
title(’Phase of Sequence x_2[n]’,’fontsize’,TFS)
−20 −10 0 10 20 30 40
−4
−2
0
2
4
6
8
n
Re{x2
[n]}
Real Part of Sequence x
2
[n]
−20 −10 0 10 20 30 40
−2
0
2
4
6
n
Im{x2
[n]}
Imaginary Part of Sequence x
2
[n]
−20 −10 0 10 20 30 40
0
2
4
6
8
n
|x2
[n]|
Magnitude of Sequence x
2
[n]
−20 −10 0 10 20 30 40
−4
−2
0
2
4
n
φ(x2
[n])
Phase of Sequence x
2
[n]
FIGURE 2.4: complex exponential signal x2[n] = (0.9ejπ/10)n.
(e) MATLAB script:
% P0201e: Generate and plot real sinusoidal sequence
close all; clc
x3n = 2*cos(2*pi*0.3*n+pi/3); % define signal
% Plot:
hf = figconfg(’P0201e’,’small’);
stem(n,x3n,’fill’)
axis([min(n)-1,max(n)+1,min(x3n)-0.5,max(x3n)+0.5])
title(’Real Sinusoidal Sequence x_3[n]’,’fontsize’,TFS)

−20 −10 0 10 20 30 40
−2
−1
0
1
2
n
x
3
[n]
Real Sinusoidal Sequence x
3
[n]
FIGURE 2.5: sinusoidal sequence x3[n] = 2 cos[2π(0.3)n + π/3].
2. MATLAB script:
% P0202: Illustrate the noncommutativity of folding and shifting
close all; clc
nx = 0:4; % specify the support
x = 5:-1:1; % specify sequence
n0 = 2;
% (a) First folding, then shifting
[y1 ny1] = fold(x,nx);
[y1 ny1] = shift(y1,ny1,-n0);
% (b) First shifting, then folding
[y2 ny2] = shift(x,nx,-n0);
[y2 ny2] = fold(y2,ny2);
% Plot
hf = figconfg(’P0202’);
xylimit = [min([nx(1),ny1(1),ny2(1)])-1,max([nx(end),ny1(end)...
,ny2(end)])+1,min(x)-1,max(x)+1];
subplot(3,1,1)
stem(nx,x,’fill’)
axis(xylimit)
ylabel(’x[n]’,’fontsize’,LFS); title(’x[n]’,’fontsize’,TFS);
set(gca,’Xtick’,xylimit(1):xylimit(2))
subplot(3,1,2)
stem(ny1,y1,’fill’)
axis(xylimit)
ylabel(’y_1[n]’,’fontsize’,LFS);

title(’y_1[n]: Folding and Shifting’,’fontsize’,TFS)
subplot(3,1,3)
axis(xylimit)
xlabel(’n’,’fontsize’,LFS); ylabel(’y_2[n]’,’fontsize’,LFS);
title(’y_2[n]: Shifting and Folding’,’fontsize’,TFS)
−7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5
0
5
x[n]
x[n]
−7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5
0
5
y1
[n]
y
1
[n]: Folding and Shifting
−7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5
0
5
n
y2
[n]
y
2
[n]: Shifting and Folding
FIGURE 2.6: Illustrating noncommunativity of folding and shifting operations.
Comments:
From the plot, we can see y1[n] and y2[n] are different. Indeed, y1[n] repre-
sents the correct x[2 − n] signal while y2[n] represents signal x[−n − 2].
3. (a) x[−n] = {4, 4, 4, 4, 4
↑
, 3, 2, 1, 0, −1}
x[n − 3] = {−1, 0, 1
↑
, 2, 3, 4, 4, 4, 4, 4}
x[n + 2] = {−1, 0, 1, 2, 3, 4, 4, 4
↑
, 4, 4}

(b) see part (c)
(c) MATLAB script:
% P0203bc: Illustrate the folding and shifting effect
close all; clc
nx = -5:4; % specify support
x = [-1:4,4*ones(1,4)]; % define sequence
[y1 ny1] = fold(x,nx); % folding
[y2 ny2] = shift(x,nx,-3); % right-shifting
[y3 ny3] = shift(x,nx,2); % left-shifting
% Plot
xylimit = [min([nx(1),ny1(1),ny2(1),ny3(1)])-1,max([nx(end),...
ny1(end),ny2(end),ny2(end)])+1,min(x)-1,max(x)+1];
subplot(4,1,1)
stem(nx,x,’fill’); axis(xylimit)
ylabel(’x[n]’,’fontsize’,LFS); title(’x[n]’,’fontsize’,TFS);
subplot(4,1,2)
stem(ny1,y1,’fill’); axis(xylimit)
ylabel(’x[-n]’,’fontsize’,LFS); title(’x[-n]’,’fontsize’,TFS)
subplot(4,1,3)
ylabel(’x[n-3]’,’fontsize’,LFS); title(’x[n-3]’,’fontsize’,TFS);
subplot(4,1,4)
xlabel(’n’,’fontsize’,LFS); ylabel(’x[n+2]’,’fontsize’,LFS);
title(’x[n+2]’,’fontsize’,TFS)
4. MATLAB script:
% P0204: Illustrate the using of repmat, persegen and pulstran
% to generate periodic signal
close all; clc
n = 0:9; % specify support
x = [ones(1,4),zeros(1,6)]; % sequence 1
% x = cos(0.1*pi*n); % sequence 2

−8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 8
−2
0
2
4x[n]
x[n]
−8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 8
−2
0
2
4
x[−n]
x[−n]
−8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 8
−2
0
2
4
x[n−3]
x[n−3]
−8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 8
−2
0
2
4
n
x[n+2]
x[n+2]
FIGURE 2.7: Illustrating folding and shifting operations.
% x = 0.8.^n; % sequence 3
Np = 5; % number of periods
xp1 = repmat(x,1,Np);
nxp1 = n(1):Np*length(x)-1;
[xp2 nxp2] = persegen(x,length(x),Np*length(x),n(1));
xp3 = pulstran(nxp1,(0:Np-1)’*length(x),x);
%Plot
xylimit = [-1,nxp1(end)+1,min(x)-1,max(x)+1];
subplot(3,1,1)
stem(nxp1,xp1,’fill’); axis(xylimit)
ylabel(’x_p[n]’,’fontsize’,LFS);
title(’Function ’’repmat’’’,’fontsize’,TFS);
subplot(3,1,2)
ylabel(’x_p[n]’,’fontsize’,LFS);

title(’Function ’’persegen’’’,’fontsize’,TFS)
subplot(3,1,3)
xlabel(’n’,’fontsize’,LFS); ylabel(’x_p[n]’,’fontsize’,LFS);
title(’Function ’’pulstran’’’,’fontsize’,TFS)
0 5 10 15 20 25 30 35 40 45 50
−1
0
1
2
xp
[n]
Function ’repmat’
0 5 10 15 20 25 30 35 40 45 50
−1
0
1
2
xp
[n]
Function ’persegen’
0 5 10 15 20 25 30 35 40 45 50
−1
0
1
2
n
xp
[n]
Function ’pulstran’
FIGURE 2.8: Periodically expanding sequence {1 1 1 1 0 0 0 0 0 0}.

0 5 10 15 20 25 30 35 40 45 50
−1
0
1
2
x
p
[n]
0 5 10 15 20 25 30 35 40 45 50
−1
0
1
2
x
p
[n]
0 5 10 15 20 25 30 35 40 45 50
−1
0
1
2
n
x
p
[n]
FIGURE 2.9: Periodically expanding sequence cos(0.1πn), 0 ≤ n ≤ 9.
0 5 10 15 20 25 30 35 40 45 50
0
1
2
xp
[n]
0 5 10 15 20 25 30 35 40 45 50
0
1
2
xp
[n]
0 5 10 15 20 25 30 35 40 45 50
0
1
2
n
xp
[n]
FIGURE 2.10: Periodically expanding sequence 0.8n, 0 ≤ n ≤ 9.

5. (a) Proof: If the sinusoidal signal cos(ω0n + θ0) is periodic in n, we need
to ﬁnd a period Np that satisfy cos(ω0n+θ0) = cos(ω0n+ω0Np +θ0)
for every n. Since f0
ω0
2π is a rational number, we can substitute ω0 =
2πf0 = 2πM
N into the previous periodic condition to have cos(2πM
N n+
θ0) = cos(2πM
N n + 2πM
N Np + θ0). No matter what integers M and N
take, Np = N is a period of the sinusoidal signal.
(b) The sequence is NOT periodic.
(c) The sequence is periodic with fundamental period N = 10. N can
be interpreted as period and M is the number of repetitions the corre-
sponding continuous signal repeats itself.
−20 −15 −10 −5 0 5 10 15 20
−1
−0.5
0
0.5
1
n
x1
[n]
Nonperiodic Sequence
−20 −15 −10 −5 0 5 10 15 20
−1
−0.5
0
0.5
1
n
x2
[n]
Periodic Sequence
FIGURE 2.11: Illustrating the periodicity condition of sinusoidal signals.
MATLAB script:
% P0205: Illustrates the condition for periodicity of discrete
% sinusoidal sequence
close all; clc
% Part (b): Nonperiodic

n = -20:20; % support
w1 = 0.1; % angular frequency
x1 = cos(w1*n-pi/5);
% Part (c): Periodic
w2 = 0.1*pi; % angular frequency
x2 = cos(w2*n-pi/5);
%Plot
xylimit = [n(1)-1,n(end)+1,min(x1)-0.5,max(x1)+0.5];
subplot(2,1,1)
stem(n,x1,’fill’); axis(xylimit)
title(’Nonperiodic Sequence’,’fontsize’,TFS);
% set(gca,’Xtick’,xylimit(1):xylimit(2))
subplot(2,1,2)
stem(n,x2,’fill’); axis(xylimit)
title(’Periodic Sequence’,’fontsize’,TFS)
6. MATLAB script:
% P0206: Investigates the effect of downsampling using
% audio file ’handel’
close all; clc
load(’handel.mat’)
n = 1:length(y);
% Part (a): original sampling rate
sound(y,Fs); pause(1)
% Part (b): downsampling by a factor of two
y_ds2_ind = mod(n,2)==1;
sound(y(y_ds2_ind),Fs/2); pause(1)
% Part (c): downsampling by a factor of four
y_ds4_ind = mod(n,4)==1;
sound(y(y_ds4_ind),Fs/4)
% save the sound file
wavwrite(y(y_ds4_ind),Fs/4,’handel_ds4’)
7. Comments: The ﬁrst system is NOT time-invariant but the second system is
time invariant.
MATLAB script:

0 2 4 6 8 10 12 14 16 18 20
0
0.2
0.4
0.6
0.8
1
n
y1
[n]
y
1
[n]
0 2 4 6 8 10 12 14 16 18 20
0
0.2
0.4
0.6
0.8
1
n
y2
[n]
y
2
[n]
FIGURE 2.12: System responses with respect to input signal x[n] = δ[n].
% P0207: Compute and plot sequence defined by difference equations
close all; clc
n = 0:20; % define support
yi = 0; % zero initial condition
xn = delta(n(1),0,n(end))’; % input 1
% xn = delta(n(1),5,n(end))’; % input 2
% Compute sequence 1:
yn1 = zeros(1,length(n));
yn1(1) = n(1)/(n(1)+1)*yi+xn(1);
for ii = 2:length(n)
yn1(ii) = n(ii)/(n(ii)+1)*yn1(ii-1)+xn(ii);
end
% Compute sequence 2:
yn2 = filter(1,[1,-0.9],xn);
%Plot

0 2 4 6 8 10 12 14 16 18 20
0
0.5
1
n
y1
[n]
y
1
[n]
0 2 4 6 8 10 12 14 16 18 20
0
0.5
1
n
y2
[n]
y
2
[n]
FIGURE 2.13: System responses with respect to input signal x[n] = δ[n − 5].
xylimit = [n(1)-1,n(end)+1,min(yn1)-0.2,max(yn1)+0.2];
subplot(2,1,1)
stem(n,yn1,’fill’); axis(xylimit)
title(’y_1[n]’,’fontsize’,TFS);
subplot(2,1,2)
stem(n,yn2,’fill’); axis(xylimit)
title(’y_2[n]’,’fontsize’,TFS)
8. (a)
y[n] =
1
5
(x[n] + x[n − 1] + x[n − 2] + x[n − 3] + x[n − 4])
(b)
h[n] = u[n] − u[n − 5]

0 5 10 15 20
−0.1
0
0.1
0.2
0.3
n
h[n]
5−Point Moving Average Filter Impulse Response
FIGURE 2.14: Impulse response of a 5-point moving average ﬁlter.
(c) Block diagram.
x[n]
y[n]
0.2 0.2
z
−1
0.2
z
−1
0.2
z
−1
0.2
z
−1
FIGURE 2.15: Block diagram of a 5-point moving average ﬁlter.
MATLAB script:
% P0208: Plot the 5-point moving average filter
% y[n] = 1/5*(x[n]+x[n-1]+x[n-2]+x[n-3]+x[n-4]);
close all; clc
n = 0:20;
xn = delta(n(1),0,n(end))’;
hn = filter(ones(1,5)/5,1,xn);
%Plot
hf = figconfg(’P0208’,’small’);
xylimit = [n(1)-1,n(end)+1,min(hn)-0.1,max(hn)+0.1];
stem(n,hn,’fill’); axis(xylimit)

xlabel(’n’,’fontsize’,LFS); ylabel(’h[n]’,’fontsize’,LFS);
title(’5-Point Moving Average Filter Impulse Response’,...
’fontsize’,TFS);
9. (a) Proof:
∞
n=0
an
= 1 + a + a2
+ · · ·
a
∞
n=0
an
= a + a2
+ a3
+ · · ·
(1 − a)
∞
n=0
an
= 1 + (a − a) + (a2
− a2
) + · · · + (a∞
− a∞
)
(1 − a)
∞
n=0
an
= 1 + 0 + 0 + · · · + 0
∞
n=0
an
=
1
1 − a
(b) Proof:
N−1
n=0
an
=
∞
n=0
an
−
∞
n=N
an
=
∞
n=0
an
− aN
∞
n=0
an
Substituting the result in part (a), we have
N−1
n=0
an
= (1 − aN
)
∞
n=0
an
=
1 − aN
1 − a
10. (a) Solution:
x[−m] = {−1, 2, 3, 1
↑
}
x[3 − m] = {−1
↑
, 2, 3, , 1}
h[m] = {2
↑
, 2(0.8)1
, 2(0.8)2
, 2(0.8)3
, 2(0.8)4
, 2(0.8)5
, 2(0.8)6
}
x[3 − m] ∗ h[m] = {−2
↑
, 4(0.8)1
, 6(0.8)2
, 2(0.8)3
}
y[3] =
3
m=0
x[3 − m] ∗ h[m] = 6.064

(b) MATLAB script:
% P0210: Graphically illustrate the convolution sum
close all; clc
nx = 0:3;
x = [1,3,2,-1]; % input sequence
nh = 0:6;
h = 2*(0.8).^nh; % impulse response
nxf = fliplr(-nx); xf = fliplr(x); %folding
nxfs = nxf+3; % left shifting
[y1 y2 n] = timealign(xf,nxfs,h,nh);
y = y1.*y2;
y3 = sum(y);
%Plot
subplot(5,1,1)
axis([-4 7 min(x)-1 max(x)+1])
ylabel(’x[k]’,’fontsize’,LFS);
subplot(5,1,2)
stem(nh,h,’fill’)
axis([-4 7 min(h)-1 max(h)+1])
ylabel(’h[k]’,’fontsize’,LFS);
subplot(5,1,3)
stem(nxf,xf,’fill’)
ylabel(’x[-k]’,’fontsize’,LFS);
subplot(5,1,4)
stem(nxfs,xf,’fill’)
ylabel(’x[-k+3]’,’fontsize’,LFS);
subplot(5,1,5)
stem(n,y,’fill’)
axis([-4 7 min(y)-1 max(y)+1])
xlabel(’k’,’fontsize’,LFS);
ylabel(’h[k]*x[-k+3]’,’fontsize’,LFS);

−4 −3 −2 −1 0 1 2 3 4 5 6 7
−2
0
2
4
x[k]
−4 −3 −2 −1 0 1 2 3 4 5 6 7
0
1
2
3
h[k]
−4 −3 −2 −1 0 1 2 3 4 5 6 7
−2
0
2
4
x[−k]
−4 −3 −2 −1 0 1 2 3 4 5 6 7
−2
0
2
4
x[−k+3]
−4 −3 −2 −1 0 1 2 3 4 5 6 7
−2
0
2
4
k
h[k]*x[−k+3]
FIGURE 2.16: Graphically illustration of convolution as a superposition of scaled
and scaled replicas.

11. Comments: The step responses of the two equivalent system representations
are equal.
0 2 4 6
−1
0
1
2
3
n
h[n]
System h[n]
−1 0 1 2 3 4 5
−1
0
1
2
3
n
u[n]
Unit Step u[n]
0 2 4 6 8 10
2
4
6
8
n
y1
[n]
Step Response of System I
0 2 4 6 8 10
2
4
6
8
n
y2
[n]
Step Response of System II
FIGURE 2.17: Illustrating equivalent system representation.
MATLAB script:
% P0211: Illustrating the combination of parallel and
% series systems
close all; clc
n1 = 0:4;
h1 = ones(1,5);
h2 = [1 -1 -1 -1 1];
n2 = 0:2;
h3 = ones(1,3);
[h n] = conv0(h1+h2,n1,h3,n2);
un = unitstep(n1(1),0,n1(end));
[ytemp1 nyt] = conv0(h1,n1,un,n1);
ytemp2 = conv(h2,un);

[y1 ny1] = conv0(h3,n2,ytemp1+ytemp2,nyt);
[y2 ny2] = conv0(h,n,un,n1);
%Plot
subplot(2,2,1)
stem(n,h,’fill’)
axis([n(1)-1 n(end)+1 min(h)-1 max(h)+1])
xlabel(’n’,’fontsize’,LFS);
ylabel(’h[n]’,’fontsize’,LFS);
title(’System h[n]’,’fontsize’,TFS);
subplot(2,2,2)
stem(n1,un,’fill’)
axis([n1(1)-1 n1(end)+1 min(h)-1 max(h)+1])
ylabel(’u[n]’,’fontsize’,LFS);
title(’Unit Step u[n]’,’fontsize’,TFS);
subplot(2,2,3)
axis([ny1(1)-1 ny1(end)+1 min(y1)-1 max(y1)+1])
ylabel(’y_1[n]’,’fontsize’,LFS);
title(’Step Response of System I’,’fontsize’,TFS);
subplot(2,2,4)
axis([ny1(1)-1 ny1(end)+1 min(y2)-1 max(y2)+1])
title(’Step Response of System II’,’fontsize’,TFS);
12. MATLAB script:
% P0212: Illustrating the usage of function ’convmtx’
close all; clc
nx = 0:5; nh = 0:3;
x = ones(1,6); h = 0.5.^(0:3);
A = convmtx(x,length(h));
y = h*A; % compute convolution
ny = (nx(1)+nh(1)):(nx(end)+nh(end)); % compute support
% [y2 ny2] = conv0(x,nx,h,nh);
%Plot

stem(ny,y,’fill’)
axis([ny(1)-1 ny(end)+1 min(y)-1 max(y)+1])
xlabel(’n’,’fontsize’,LFS); ylabel(’y[n]’,’fontsize’,LFS);
title(’Convolution y[n]’,’fontsize’,TFS);
set(gca,’XTick’,ny(1):ny(end))
−1 0 1 2 3 4 5
0
5
10
15
n
y[n]
Convolution y[n]
FIGURE 2.18: Compute the convolution of the ﬁnite length sequences in (2.38)
using convmtx.
0 1 2 3 4 5 6 7 8
−0.5
0
0.5
1
1.5
2
2.5
n
y[n]
Convolution y[n]
FIGURE 2.19: Compute the convolution of the ﬁnite length sequences in (2.39)
using convmtx.

13. Proof:
Since the linear time-invariant system is stable, we have
∞
n=−∞
|h[n]| < ∞
∞
n=−∞
|h[n]| = lim
N→∞
N
n=−∞
|h[n]| +
∞
n=N+1
|h[n]|
lim
N→∞
∞
n=N+1
|h[n]| = 0
y[n] = x[n] ∗ h[n] =
∞
m=−∞
h[m]x[n − m] =
∞
m=n−n0
h[m]x[n − m]
lim
n→∞
|y[n]| = lim
n→∞
|
∞
m=n−n0
h[m]x[n−m]| ≤ lim
n→∞
∞
m=n−n0
|h[m]||x[n−m]| = 0
Hence, we proved
lim
n→∞
y[n] = 0
14. MATLAB script:
% P0214: Use function ’conv(h,x)’ to compute noncausal
% h convolves causal x
close all; clc
nh = -4:4;
nx = 0:5;
h = ones(1,9);
x = 1:6;
y1 = conv(h,x); % compute convolution
ny1 = (nh(1)+nx(1)):(nh(end)+nx(end)); % define support
[y2 ny2] = conv0(h,nh,x,nx); % verification

15. Comments: The image is blurred by both ﬁlters and the larger the ﬁlter is the
more blurred the image is.
MATLAB script:
% P0215: Filtering 2D image lena.jpg using 2D filter
close all; clc
x = imread(’lena.jpg’);
% Part (a): image show
hfs = figconfg(’P0215a’,’small’);
imshow(x,[])
% Part (b):
hmn = ones(3,3)/9;
y1 = filter2(hmn,x);
% hmn is symmetric and no change if rotated by 180 degrees
% we can use 2d correlation instead of 2d convolution
hfs1 = figconfg(’P0215b’,’small’);
imshow(y1,[])
% Part (c):
hmn2 = ones(5,5)/25;
y2 = filter2(hmn2,x);
hfs2 = figconfg(’P0215c’,’small’);
imshow(y2,[])

(a)
(b) (c)
FIGURE 2.20: (a) Original image. (b) Output image processed by 3 × 3 impulse
response h[m, n] given in (2.75). (c) Output image processed by 5 × 5 impulse
response h[m, n] deﬁned in part (c).

16. (a) See plots.
(b) Comments: The resulting image is horizontally blurred.
(c) Comments: The resulting image is vertically blurred.
(d) Comments: The resulting image is blurred the same way as the one in
part (c) in Problem 16.
MATLAB script:
% P0216: Filtering 2D image lena.jpg using 1D filter
[nx ny] = size(x);
hfs = figconfg(’0216a’,’small’);
imshow(x,[])
n = -2:2;
h = ones(1,5)/5;
% Part (b): horizontal filtering
yh = zeros(nx,ny);
for ii = 1:ny
temp = conv(h,double(x(ii,:)));
yh(ii,:) = temp(3:end-2);
end
hfs1 = figconfg(’0216b’,’small’);
imshow(yh,[])
% Part (c): vertical filtering
yv = zeros(nx,ny);
for ii = 1:nx
temp = conv(h,double(x(:,ii)));
yv(:,ii) = temp(3:end-2);
end
hfs2 = figconfg(’0216c’,’small’);
imshow(yv,[])
% Part (d): horizontal and vertical filtering
yhv = zeros(nx,ny);
for ii = 1:nx
temp = conv(h,yh(:,ii));
yhv(:,ii) = temp(3:end-2);
end
hfs3 = figconfg(’0216d’,’small’);
imshow(yhv,[])

(a) (b)
(c) (d)
FIGURE 2.21: (a) Original image. (b) Output image obtained by row processing.
(c) Output image obtained by column processing. (d) Output image obtained by
row and column processing.

17. (a) Impulse response.
0 10 20 30 40 50 60 70 80 90 100
−0.2
0
0.2
0.4
0.6
n
h[n] Impulse Response h[n]
FIGURE 2.22: Impulse response h[n].
(b) Output using y=filter(b,a,x).
0 10 20 30 40 50 60 70 80 90 100
0
0.5
1
1.5
n
y
1
[n]
Unit Step Response: filter(b,a,x)
FIGURE 2.23: System step output y[n] computed using the function
y=filter(b,a,x).
(c) Output using y=conv(h,x).
(d) Output using y=filter(h,1,x).
MATLAB script:
% P0217: Illustrating the usage of functions ’impz’,’filter,’conv’
close all; clc
n = 0:100;
b = [0.18 0.1 0.3 0.1 0.18];
a = [1 -1.15 1.5 -0.7 0.25];
% Part (a):

0 20 40 60 80 100 120 140 160 180 200
−0.5
0
0.5
1
1.5
n
y
2
[n]
Unit Step Response: conv(h,x)
y=conv(h,x).
0 10 20 30 40 50 60 70 80 90 100
0
0.5
1
1.5
n
y
3
[n]
Unit Step Response: filter(h,1,x)
y=filter(h,1,x).
h = impz(b,a,length(n));
% Part (b):
u = unitstep(n(1),0,n(end));
y1 = filter(b,a,u);
% Part (c):
y2 = conv(h,u);
% Part (d):
y3 = filter(h,1,u);
%Plot
hf = figconfg(’P0217a’,’long’);
stem(n,h,’fill’)
xlabel(’n’,’fontsize’,LFS); ylabel(’h[n]’,’fontsize’,LFS);
title(’Impulse Response h[n]’,’fontsize’,TFS);

hf2 = figconfg(’P0217b’,’long’);
stem(n,y1,’fill’)
title(’Unit Step Response: filter(b,a,x)’,’fontsize’,TFS);
hf3 = figconfg(’P0217c’,’long’);
stem(0:2*n(end),y2,’fill’)
title(’Unit Step Response: conv(h,x)’,’fontsize’,TFS);
hf4 = figconfg(’P0217d’,’long’);
title(’Unit Step Response: filter(h,1,x)’,’fontsize’,TFS);
18. (a) Block diagrams.
x[n]
y[n]
0.1667 0.1667
z
−1
0.1667
z
−1
0.1667
z
−1
0.1667
z
−1
0.1667
z
−1
FIGURE 2.26: Block diagram representations of the nonrecursive implementation
of M = 5 moving average ﬁlter.
(b) MATLAB script:
% P0218: Implement nonrecursive and recursive implementations
% of moving average filter
close all; clc
M = 5;
n = 0:M;
un = unitstep(n(1),0,n(end));
% Nonrecursive implementation:
y_nr = filter(ones(1,M+1)/(M+1),1,un);
% Recursive implementation:
y_re = filter([1 zeros(1,M) -1]/(M+1),[1 -1],un);
subplot(2,1,1)

x[n] y[n]
0.1667
0z
−1
0z−1
0z−1
0z−1
0
z−1
−0.1667
z−1
1
FIGURE 2.27: Block diagram representations of the recursive implementation of
M = 5 moving average ﬁlter.
stem(n,y_nr,’fill’)
axis([n(1)-1 n(end)+1 min(y_nr)-0.5 max(y_nr)+0.5])
title(’Nonrecursive Implementation’,’fontsize’,TFS);
subplot(2,1,2)
stem(n,y_re,’fill’)
axis([n(1)-1 n(end)+1 min(y_re)-0.5 max(y_re)+0.5])
title(’Recursive Implementation’,’fontsize’,TFS);
19. MATLAB script:
% P0219: Generate digital reverberation using audio file ’handel’
close all; clc
n = 1:length(y);
a = 0.7; % specify attenuation factor
tau = 50e-3; % Part (a)

−1 0 1 2 3 4 5 6
0
0.5
1
1.5
n
y1
[n]
Nonrecursive Implementation
−1 0 1 2 3 4 5 6
0
0.5
1
1.5
n
y2
[n]
Recursive Implementation
FIGURE 2.28: Step response computed by nonrecursive and recursive implemen-
tations.
% tau = 100e-3; % Part (b)
% tau = 500e-3; % Part (c)
D = floor(tau*Fs); % compute delay
yd = filter(1,[1 zeros(1,length(D)-1),-a],y);
sound(yd,Fs)
20. (a) Solution:
y1(t) = x1(t) ∗ h(t) =
∞
−∞
h(τ)x1(t − τ)dτ =
∞
−∞
e−τ/2
u(τ)u(t − τ)dτ
= u(t)
t
0
e−τ/2
dτ = u(t)(−2)e−τ/2
|t
0= 2(1 − e−t/2
)u(t)

y2(t) = x2(t) ∗ h(t) =
∞
−∞
h(t − τ)x2(τ)dτ = 2
3
0
e−(t−τ)/2
u(t − τ)dτ
= (u(t) − u(t − 3))2
t
0
e−(t−τ)/2
dτ + u(t − 3)2
3
0
e−(t−τ)/2
dτ
= (u(t) − u(t − 3))4e−(t−τ)/2
|t
0+u(t − 3)4e−(t−τ)/2
|3
0
= 4(1 − e−t/2
)u(t) − 4(1 − e−(t−3)/2
)u(t − 3)
(b) Proof:
x2(t) = 2x1(t) − 2x1(t − 3)
y2(t) = 2y1(t) − 2y1(t − 3)
= 4(1 − e−t/2
)u(t) − 4(1 − e−(t−3)/2
)u(t − 3)

Basic Problems
21. See book companion toolbox for the function.
22. (a) x[n] versus n.
−30 −20 −10 0 10 20 30
−1.5
−1
−0.5
0
0.5
1
1.5
n
x[n]
x[n]
FIGURE 2.29: x[n] versus n.
(b) A down sampled signal y[n] for M = 5.
−30 −20 −10 0 10 20 30
−1.5
−1
−0.5
0
0.5
1
1.5
n
y[n]
Downsampling y[n]= x[nM]
FIGURE 2.30: A down sampled signal y[n] for M = 5.
(c) A down sampled signal y[n] for M = 20.
(d) Comments: The downsampled signal is compressed.
MATLAB script:
% P0222: Illustrate downsampling: y[n] = x[nM]
close all; clc
nx = -30:30;
x = cos(0.1*pi*nx);

−30 −20 −10 0 10 20 30
−0.5
0
0.5
1
1.5
n
y[n]
Downsampling y[n]= x[nM]
FIGURE 2.31: A down sampled signal y[n] for M = 20.
% M = 5; % Part (b)
M = 20; % Part (c)
yind = mod(nx,M)==0;
y = x(yind);
ny = nx(yind)/M;
[x y n] = timealign(x,nx,y,ny);
stem(n,x,’fill’)
axis([n(1)-1 n(end)+1 min(x)-0.5 max(x)+0.5])
xlabel(’n’,’fontsize’,LFS); ylabel(’x[n]’,’fontsize’,LFS);
title(’x[n]’,’fontsize’,TFS);
axis([n(1)-1 n(end)+1 min(y)-0.5 max(y)+0.5])
title(’Downsampling y[n]= x[nM]’,’fontsize’,TFS);
23. (a) y[n] = x[−n] (Time-ﬂip)
linear, time-variant, noncausal, and stable
(b) y[n] = log(|x[n]|) (Log-magnitude )
nonlinear, time-invariant, causal, and unstable
(c) y[n] = x[n] − x[n − 1] (First-difference)
linear, time-invariant, causal, and stable
(d) y[n] = round{x[n]} (Quantizer)
nonlinear, time-invariant, causal, and stable

24. Comments: The ﬁltered data are smoother and y1[n] is 25 samples delayed
than y2[n].
0 100 200 300 400 500 600
0
200
400
600
800
1000
1200
time index n
DowJonesIndustrialIndex
x[n]
y1
[n]
y2
[n]
FIGURE 2.32: Dow Jones Industrial Average weekly opening value x[n] and its
moving averages.
MATLAB script:
% P0224: Write MATLAB script to compute moving averages
close all; clc
x = load(’djw6576.txt’);
N = length(x);
nx = 0:N-1;
xepd1 = [zeros(50,1);x];
y1 = zeros(N,1);
for ii = 1:N
y1(ii) = sum(xepd1(ii:ii+50))/51;
end
xepd2 = [zeros(25,1);x;zeros(25,1)];

y2 = zeros(N,1);
for ii = 1:N
y2(ii) = sum(xepd2(ii:ii+50))/51;
end
% Plot:
plot(nx,x,’.’,nx,y1,’.’,nx,y2,’.’)
xlabel(’time index n’,’fontsize’,LFS)
ylabel(’Dow Jones Industrial Index’,’fontsize’,LFS)
legend(’x[n]’,’y_1[n]’,’y_2[n]’,’fontsize’,LFS,’location’,’best’)
25. (a) Solution:
y[n] = h[n] ∗ x[n] =
∞
m=−∞
h[m]x[n − m]
=
∞
m=−∞
m(u[m] − u[m − M])(u[n − m] − u[n − M − N])
if n ∈[0, M − 1]
y[n] =
n
m=0
m =
n(n + 1)
2
if n ∈[M − 1, N − 1]
y[n] =
M−1
m=0
m =
M(M − 1)
2
if n ∈[N − 1, M + N − 3]
y[n] =
M−1
m=n−(N−1)
m =
M−1
m=0
m −
n−N
m=0
m =
M(M − 1)
2
−
(n − N + 1)(n − N)
2
(b) Comments: The analytical solution can be veriﬁed.
MATLAB script:
% P0225: Verify the analytical expression
close all; clc
N = 10; M = 5;
n = 0:N-1;

0 5 10 15
0
2
4
6
8
10
n
y[n]
y[n] = h[n]*x[n]
FIGURE 2.33: MATLAB veriﬁcation of analytical expression for the sequence
y[n] = h[n] ∗ x[n].
x = unitpulse(0,0,N-1,N-1)’;
h = n.*unitpulse(0,0,M-1,N-1)’;
[y ny] = conv0(h,n,x,n);
% Plot:
stem(ny,y,’fill’)
axis([ny(1)-1,ny(end)+1,min(y)-1,max(y)+1])
title(’y[n] = h[n]*x[n]’,’fontsize’,TFS)
26. Solution:
y[n] = x[n] ∗ h[n] =
∞
k=−∞
x[n − k]h[k]
if n ∈[0, N − 1]
y[n] =
n
k=0
ak
=
1 − an+1
1 − a
if n ∈[N − 1, M − 1]
y[n] =
n
k=n−N+1
ak
=
n
k=0
ak
−
n−N
k=0
ak
=
an+1(a−N − 1)
1 − a

if n ∈[M − 1, M + N − 2]
y[n] =
M−1
k=n−N+1
ak
=
M−1
k=0
ak
−
n−N
k=0
ak
=
an−N+1 − aM
1 − a
y[n] = 0, otherwise
27. Solution:
y[n] = h[n] ∗ x[n] = an
u[n] ∗ bn
u[n]
=
∞
m=−∞
am
u[m]bn−m
u[n − m] = u[n]
n
m=0
am
bn−m
= u[n]bn
n
m=0
am
b−m
=
bn+1 − an+1
b − a
u[n]
0 5 10 15 20
−0.2
0
0.2
0.4
0.6
0.8
1
n
y[n]
y[n] = h[n]*x[n]
FIGURE 2.34: MATLAB veriﬁcation of analytical expression for the sequence
y[n] = h[n] ∗ x[n].
MATLAB script:
close all; clc
a = 1/4; b = 1/3;
N = 20;
n = 0:N-1;
x = a.^n;
h = b.^n;
y = conv(h,x);

% Plot:
stem(n,y(1:N),’fill’)
axis([n(1)-1,n(end)+1,min(y)-0.2,max(y)+0.2])
title(’y[n] = h[n]*x[n]’,’fontsize’,TFS)
28. (a) Solution:
y[n] = x[n] ∗ h[n] =
∞
m=−∞
(0.9)m
u[m](0.9)n−m
u[n − m]
= u[n]
n
m=0
(0.9)n
= (n + 1)(0.9)n
u[n]
0 10 20 30 40 50 60 70 80 90
0
1
2
3
4
n
y[n]
y[n] = h[n]*x[n] Analytical Expression
FIGURE 2.35: y[n] plot determined analytically.
(b) y[n] computed by conv function.
(c) y[n] computed by filter function.
(d) Comments: (c) comes closer to (a). Because in (b) the tail parts (sam-
ples from n = 50) of both x[n] and h[n] are curtailed, the second part
samples (samples from n = 50) of (b) differ from the ones in (a).
MATLAB script:
close all; clc
a = 0.9;
% Part (a): Analytical Result:

0 10 20 30 40 50 60 70 80 90
0
1
2
3
4
n
y
2
[n]
y[n] = h[n]*x[n] Computed by ’conv’
FIGURE 2.36: y[n] plot determined by conv function.
0 10 20 30 40 50 60 70 80 90
0
1
2
3
4
n
y
3
[n]
y[n] = h[n]*x[n] Computed by ’filter’
FIGURE 2.37: y[n] plot determined by filter function.
n = 0:98;
y1 = (n+1).*a.^n;
% Plot:
hf1 = figconfg(’P0228a’,’long’);
axis([n(1)-1,n(end)+1,min(y1)-0.2,max(y1)+0.2])
title(’y[n] = h[n]*x[n] Analytical Expression’,’fontsize’,TFS)
% Part (b): Using ’conv’
N = 50;
n = 0:N-1;
x = a.^n;
h = a.^n;
y2 = conv(h,x);
ny = 0:length(y2)-1;

% Plot:
stem(ny,y2,’fill’)
axis([ny(1)-1,ny(end)+1,min(y2)-0.2,max(y2)+0.2])
title(’y[n] = h[n]*x[n] Computed by ’’conv’’’,’fontsize’,TFS)
% Part (c): Using ’filter’
N = 99;
n = 0:N-1;
x = a.^n;
h = a.^n;
y3 = filter(h,1,x);
ny = 0:length(y2)-1;
% Plot:
hf3 = figconfg(’P0228c’,’long’);
stem(ny,y3,’fill’)
axis([ny(1)-1,ny(end)+1,min(y3)-0.2,max(y3)+0.2])
title(’y[n] = h[n]*x[n] Computed by ’’filter’’’,’fontsize’,TFS)

29. MATLAB script:
% P0229: Verify the properites of convolution summarized
% in Table 2.3 on page 54
close all; clc
%% Specify signals:
nx = -15:9;
x = nx*(-1);
nh = 0:9;
h = 0.5.^nh;
nh1 = 0:20;
h1 = cos(0.05*pi*nh1);
nh2 = -3:5;
h2 = [2 0 0 0 2 0 0 0 -3];
[d nd] = delta(0,0,0); % unit impulse
n0 = 3;
[dd ndd] = delta(n0,n0,n0); % unit delay
%% Verify Identity Property:
y = conv(x,d);
% Plot:
hf1 = figconfg(’P0229a’);
subplot(2,1,1)
axis([nx(1)-1,nx(end)+1,min(x)-1,max(x)+1])
xlabel(’n’,’fontsize’,LFS); ylabel(’x[n]’,’fontsize’,LFS)
title(’x[n]’,’fontsize’,TFS)
subplot(2,1,2)
stem(nx,y,’fill’)
axis([nx(1)-1,nx(end)+1,min(y)-1,max(y)+1])
title(’y[n] = x[n]times delta[n]’,’fontsize’,TFS)
%% Verify Delay Property:
[y1 ny1] = shift(x,nx,-n0);
y2 = conv(x,dd);
ny2 = nx + n0;
% Plot:
hf2 = figconfg(’P0229b’);
subplot(2,1,1)
axis([ny1(1)-1,ny1(end)+1,min(y1)-1,max(y1)+1])

title(’x[n-n_0]’,’fontsize’,TFS)
subplot(2,1,2)
xlabel(’n’,’fontsize’,LFS)
title(’x[n]times delta[n-n_0]’,’fontsize’,TFS)
%% Verify Commutative Property:
y1 = conv(x,h);
ny1 = nx(1)+nh(1):nx(end)+nh(end);
y2 = conv(h,x);
ny2 = nx(1)+nh(1):nx(end)+nh(end);
% Plot:
hf3 = figconfg(’P0229c’);
subplot(2,1,1)
title(’x[n]times h[n]’,’fontsize’,TFS)
subplot(2,1,2)
title(’h[n]times x[n]’,’fontsize’,TFS)
%% Verify Associative Property:
[y1 ny1] = conv0(x,nx,h1,nh1);
[y1 ny1] = conv0(y1,ny1,h2,nh2);
[y2 ny2] = conv0(h1,nh1,h2,nh2);
[y2 ny2] = conv0(x,nx,y2,ny2);
% Plot:
hf4 = figconfg(’P0229d’);
subplot(2,1,1)
title(’(x[n]times h_1[n])times h_2[n]’,’fontsize’,TFS)
subplot(2,1,2)

title(’x[n]times (h_1[n]times h_2[n])’,’fontsize’,TFS)
%% Verify Distributive Property:
[hh1 hh2 nh12] = timealign(h1,nh1,h2,nh2);
[y1 ny1] = conv0(x,nx,hh1+hh2,nh12);
[y2a ny2a] = conv0(x,nx,h1,nh1);
[y2b ny2b] = conv0(x,nx,h2,nh2);
[y2a y2b ny2] = timealign(y2a,ny2a,y2b,ny2b);
y2 = y2a + y2b;
% Plot:
hf5 = figconfg(’P0229e’);
subplot(2,1,1)
title(’x[n]times (h_1[n]+h_2[n])’,’fontsize’,TFS)
subplot(2,1,2)
title(’x[n]times h_1[n]+x[n]times h_2[n]’,’fontsize’,TFS)

−15 −10 −5 0 5 10
−10
−5
0
5
10
15
n
x[n]
x[n]
−15 −10 −5 0 5 10
−10
−5
0
5
10
15
n
y[n]
y[n] = x[n]× δ[n]
FIGURE 2.38: Verify identity property.
−10 −5 0 5 10
−10
−5
0
5
10
15
n
x[n−n
0
]
−10 −5 0 5 10
−10
−5
0
5
10
15
n
x[n]× δ[n−n
0
]
FIGURE 2.39: Verify delay property.

−15 −10 −5 0 5 10 15
−10
0
10
20
n
x[n]× h[n]
−15 −10 −5 0 5 10 15
−10
0
10
20
n
h[n]× x[n]
FIGURE 2.40: Verify commutative property.
−15 −10 −5 0 5 10 15 20 25 30 35
−300
−200
−100
0
100
200
n
(x[n]× h
1
[n])× h
2
[n]
−15 −10 −5 0 5 10 15 20 25 30 35
−300
−200
−100
0
100
200
n
x[n]× (h
1
[n]× h
2
[n])
FIGURE 2.41: Verify associative property.

−15 −10 −5 0 5 10 15 20 25 30
−100
−50
0
50
100
n
x[n]× (h
1
[n]+h
2
[n])
−15 −10 −5 0 5 10 15 20 25 30
−100
−50
0
50
100
n
x[n]× h1
[n]+x[n]× h2
[n]
FIGURE 2.42: Verify distributive property.
30. MATLAB script:
function [y,L1,L2] = convol(h,M1,M2,x,N1,N2)
% P0230: Compute the convolution of two arbitrarily positioned finite
% length sequences using the procedure illustrated in Figure 2.16
L1 = M1+N1; L2 = M2+N2;
ny = L1:L2;
y = zeros(1,length(ny));
nx = N1:N2;
[hf nhf] = fold(h,M1:M2);
for ii = 1:length(ny)
[hfs nhfs] = shift(hf,nhf,-ny(ii));
[y1 y2 ny] = timealign(hfs,nhfs,x,nx);
y(ii) = sum(y1.*y2);
end

31. (a) See below.
0 10 20 30 40 50 60 70 80 90 100
0
1
2
3
4
5
x 10
20
n
y[n] LCCDE: y[n] = y[n−1] + y[n−2] + x[n], y[−1] = y[−2] = 0
FIGURE 2.43: System impulse response for 0 ≤ n ≤ 100, using function filter.
(b) Comments: The system is unstable.
(c) Comments: h[n] is 1 sample left moved Fibonacci sequence.
MATLAB script:
% P0231: Use function ’filter’ to realize LCCDE resting
% at zero initial condition
close all; clc
% Part (a):
n = 0:100;
x = delta(n(1),0,n(end));
y = filter(1,[1 -1 -1],x);
% Plot:
axis([n(1)-1,n(end)+1,min(y)-1,max(y)+1])
xlabel(’n’,’fontsize’,LFS); ylabel(’y[n]’,’fontsize’,LFS)
title(’LCCDE: y[n] = y[n-1] + y[n-2] + x[n], y[-1] = y[-2] = 0’...
,’fontsize’,TFS)

32. MATLAB script:
% P0232: Use function ’filter’ to study the impulse response and
% step response of a system specified by LCCDE
close all; clc
N = 60;
n = 0:N-1;
b = [0.18 0.1 0.3 0.1 0.18];
a = [1 -1.15 1.5 -0.7 0.25];
[d nd] = delta(n(1),0,n(end));
[u nu] = unitstep(n(1),0,n(end));
y1 = filter(b,a,d);
y2 = filter(b,a,u);
% Plot:
subplot(2,1,1)
title(’Impulse Response’,’fontsize’,TFS);
subplot(2,1,2)
title(’Step Response’,’fontsize’,TFS)
33. MATLAB script:
% P0233: Realize a first-order digital differentiator given by
% y[n] = x[n] - x[n-1]
close all; clc
% Part (a):
n = -10:19;
x = 10*ones(1,length(n));
% % Part (b):
% nx1 = 0:9;
% x1 = nx1;
% nx2 = 10:19;
% x2 = 20-nx2;

0 10 20 30 40 50 60
−0.2
0
0.2
0.4
n
Impulse Response
0 10 20 30 40 50 60
0
0.5
1
1.5
n
Step Response
FIGURE 2.44: System impulse response and step response for ﬁrst 60 samples
using function filter.
% [x1 x2 n] = timealign(x1,nx1,x2,nx2);
% x = x1 + x2;
% % Part (c):
% n = 0:39;
% x = cos(0.2*pi*n-pi/2);
% Differentiator:
y = filter([1,-1],1,x);
% Plot:
subplot(2,1,1)
stem(n,x,’fill’)
axis([n(1)-1,n(end)+1,min(x)-1,max(x)+1])
title(’Input Signal x[n]’,’fontsize’,TFS)
subplot(2,1,2)

axis([n(1)-1,n(end)+1,min(y)-1,max(y)+1])
title(’Response y[n] = x[n] - x[n-1]’,’fontsize’,LFS)
−10 −5 0 5 10 15 20
9
9.5
10
10.5
11
n
Input Signal x[n]
−10 −5 0 5 10 15 20
0
2
4
6
8
10
n
Response y[n] = x[n] − x[n−1]
FIGURE 2.45: Differentiator output if input is x[n] = 10{u[n + 10] − u[n − 20]}.
34. MATLAB script:
% P0234: Use function ’filter’ to study the impulse response
% and step response of a system specified by LCCDE
close all; clc
N = 100;
n = 0:N-1;
b = 1;
a = [1 -0.9 0.81];
% a = [1 0.9 -0.81];
[u nu] = unitstep(n(1),0,n(end));
y1 = filter(b,a,d);
y2 = filter(b,a,u);
% Plot:

0 2 4 6 8 10 12 14 16 18 20
0
2
4
6
8
10
n
Input Signal x[n]
0 2 4 6 8 10 12 14 16 18 20
−2
−1
0
1
2
n
FIGURE 2.46: Differentiator output if input is x[n] = n{u[n]−u[n−10]}+(20−
n){u[n − 10] − u[n − 20]}.
axis([n(1)-1,n(end)+1,min(y1)-1,max(y1)+1])
title(’Impulse Response’,’fontsize’,TFS)
axis([n(1)-1,n(end)+1,min(y2)-1,max(y2)+1])
title(’Step Response’,’fontsize’,TFS)
35. (a) y(t) = x(t − 1) + x(2 − t)
linear, time-variant, noncausal, and stable
(b) y(t) = dx(t)/dt
linear, time-invariant, causal, and unstable
(c) y(t) =
3t
−∞ x(τ)dτ
linear, time-variant, noncausal, and unstable

0 5 10 15 20 25 30 35 40
−1
0
1
n
Input Signal x[n]
0 5 10 15 20 25 30 35 40
−1
0
1
n
FIGURE 2.47: Differentiator output if input is x[n] = cos(0.2πn − π/2){u[n] −
u[n − 40]}.
(d) y(t) = 2x(t) + 5
nonlinear, time-invariant, causal, and stable
36. (a) Solution:
y(t) = h(t) ∗ x(t) =
∞
−∞
x(τ)h(t − τ)dτ
if t ∈[−1, 1]
y(t) =
1+t
0
τ/3dτ =
(1 + t)2
6
if t ∈[1, 2]
y(t) =
1+t
−1+t
τ/3dτ =
2t
3
if t ∈[2, 4]
y(t) =
3
−1+t
τ/3dτ =
−t2 + 2t + 8
6
y(t) = 0 otherwise

0 10 20 30 40 50 60 70 80 90 100
−1.5
−1
−0.5
0
0.5
1
1.5
2
n
Impulse Response
FIGURE 2.48: System impulse response.
0 10 20 30 40 50 60 70 80 90 100
0
0.5
1
1.5
2
2.5
n
Step Response
FIGURE 2.49: System step response.
(b) Proof:
y(t) = h(t) ∗ x(t) =
∞
−∞
h(τ)x(t − τ)dτ
=
∞
k=−∞
T
2
− T
2
h(kT + τ)x(t − kT − τ)dτ
≈
∞
k=−∞
[h(kT)x(t − kT)T]
= T
∞
k=−∞
h[k]x[n − k] = ˆy(t)
(c) Comments: When T = 0.01, the error becomes negligible.
MATLAB script:

−1 −0.5 0 0.5 1 1.5 2 2.5 3 3.5 4
0
0.5
1
1.5
MSE = 0.003512
t (∆ T = 0.1)
yhat[n]
y[n]
FIGURE 2.50: Plot of sequences ˆy(nT) and y(nT) for T = 0.1.
% P0236: Compute and plot continuous-time convolution
% using discrete approximation
close all; clc
dT = 0.1;
% % dT = 0.01;
n = -1/dT:4/dT;
t = n*dT;
h = zeros(1,length(n));
ind = (t >= -1 & t <=1);
h(ind) = 1;
x = zeros(1,length(n));
ind = (t >= 0 & t <=3);
x(ind) = t(ind)/3;
% Theoretical continuous y(t):
y = zeros(1,length(n));
ind = (t >= -1 & t <= 1);
y(ind) = (t(ind).^2+2*t(ind)+1)/6;
ind = (t >=1 & t <= 2);
y(ind) = 2*t(ind)/3;
ind = (t >=2 & t <= 4);

−1 −0.5 0 0.5 1 1.5 2 2.5 3 3.5 4
0
0.2
0.4
0.6
0.8
1
1.2
1.4
MSE = 3.4845e−05
t (∆ T = 0.01)
yhat[n]
y[n]
y(ind) = (-t(ind).^2+2*t(ind)+8)/6;
% Approximated y(t):
[yhat nyhat] = conv0(h,n,x,n);
tyhat = dT*nyhat;
yhat = dT*yhat;
ind = (tyhat <-1 | tyhat>4);
yhat(ind) = [];
% Compute mean square error:
mse = mean((y-yhat).^2);
% Plot:
hf1 = figconfg(’P0236’);
plot(t,yhat,’.’,t,y,’-.’)
TB = [’MSE = ’,num2str(mse)];
title(TB,’fontsize’,TFS)
LB = [’t (Delta T = ’,num2str(dT),’)’];
xlabel(LB,’fontsize’,LFS);
legend(’yhat[n]’,’y[n]’,’location’,’best’)

Assessment Problems
37. Comments: y1[n] represents the correct x[−n − 4] signal.
−10 −9 −8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5
−1
0
1
2
3
4
5
6
n
y
1
[n]
First Folding Then Shifting
FIGURE 2.52: y1[n] obtained by ﬁrst folding and then shifting.
−10 −9 −8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5
−1
0
1
2
3
4
5
6
n
y
2
[n]
First Shifting Then Folding
FIGURE 2.53: y2[n] obtained by ﬁrst shifting and then folding.
MATLAB script:
% P0237: Exercise the manipulations of folding and shifting signals
close all; clc
nx = 0:5;
x = 0:5;
n0 = 4;
% Part (a): First folding, then shifting
[y1 ny1] = fold(x,nx);
[y1 ny1] = shift(y1,ny1,n0);

% Part (b): First shifting, then folding
[y2 ny2] = shift(x,nx,n0);
[y2 ny2] = fold(y2,ny2);
% Plot:
hf = figconfg(’P0237a’,’small’);
xlim([min([ny1(1),ny2(1)])-1,max([ny1(end),ny2(end)])+1])
ylim([min(y1)-1,max(y1)+1])
xlabel(’n’,’fontsize’,LFS); ylabel(’y_1[n]’,’fontsize’,LFS)
title(’First Folding Then Shifting’,’fontsize’,TFS)
set(gca,’Xtick’,min([ny1(1),ny2(1)])-1:max([ny1(end),ny2(end)])+1)
hf2 = figconfg(’P0237b’,’small’);
xlim([min([ny1(1),ny2(1)])-1,max([ny1(end),ny2(end)])+1])
ylim([min(y2)-1,max(y2)+1])
xlabel(’n’,’fontsize’,LFS); ylabel(’y_2[n]’,’fontsize’,LFS)
title(’First Shifting Then Folding’,’fontsize’,TFS)
set(gca,’Xtick’,min([ny1(1),ny2(1)])-1:max([ny1(end),ny2(end)])+1)
38. MATLAB script:
% P0238: Generate and plot signals using function ’stem’
close all; clc
%% Part (a):
[x1a nx1a] = delta(-1,-1,-1);
x1a = 5*x1a;
nx1b = -5:3;
x1b = nx1b.^2;
nx1c = 4:7;
x1c = 10*0.5.^nx1c;
[x1a x1b nx1] = timealign(x1a,nx1a,x1b,nx1b);
x1 = x1a + x1b;
[x1 x1c nx1] = timealign(x1,nx1,x1c,nx1c);
x1 = x1 + x1c;
% Plot:
hf1 = figconfg(’P0238a’,’small’);
stem(nx1,x1,’fill’)
axis([min(nx1)-1,max(nx1)+1,min(x1)-1,max(x1)+1])
xlabel(’n’,’fontsize’,LFS); ylabel(’x_1[n]’,’fontsize’,LFS)
title(’x_1[n]’,’fontsize’,TFS)

set(gca,’Xtick’,nx1(1)-1:nx1(end)+1)
%% Part (b):
nx2 = 0:20;
x2 = 0.8.^nx2.*cos(0.2*pi*nx2+pi/4);
% Plot:
axis([min(nx2)-1,max(nx2)+1,min(x2)-0.2,max(x2)+0.2])
%% Part (c):
nx3 = 0:20;
x3 = zeros(1,length(nx3));
for ii = 0:4
[d1 nd1] = delta(nx3(1),ii,nx3(end));
[d2 nd2] = delta(nx3(1),2*ii,nx3(end));
x3 = x3 + (ii+1)*(d1-d2)’;
end
% Plot:
hf3 = figconfg(’P0238c’,’small’);
axis([min(nx3)-1,max(nx3)+1,min(x3)-1,max(x3)+1])
−6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 8
0
5
10
15
20
25
n
x
1
[n]
x
1
[n]
FIGURE 2.54: x1[n] = 5δ[n + 1] + n2(u[n + 5] − u[n − 4]) + 10(0.5)n(u[n −
4] − u[n − 8]).

0 5 10 15 20
−0.5
0
0.5
n
x2
[n]
x
2
[n]
FIGURE 2.55: x2[n] = (0.8)n cos(0.2πn + π/4), 0 ≤ n ≤ 20.
0 5 10 15 20
−6
−4
−2
0
2
4
n
x
3
[n]
x
3
[n]
FIGURE 2.56: x3[n] = 4
m=0(m + 1){δ[n − m] − δ[n − 2m]}, 0 ≤ n ≤ 20.
39. MATLAB script:
% P0239: Generate and plot signals using function ’stem’
close all; clc
nx = -5:14;
x = nx;
x(x<0) = 0;
x(x>10) = 0;
% Part (a):
[x1 nx1] = shift(x,nx,-4);
x1 = 2*x1;
% Part (b):
[x2 nx2] = shift(x,nx,-5);

x2 = 3*x2;
% Part (c):
[x3 nx3] = shift(x,nx,3);
[x3 nx3] = fold(x3,nx3);
% Plot:
xlimit = [min([nx(1) nx1(1) nx2(1) nx3(1)])-1,...
max([nx(end) nx1(end) nx2(end) nx3(end)])+1];
subplot(2,2,1)
xlim(xlimit); ylim([min(x)-1,max(x)+1])
xlabel(’n’,’fontsize’,LFS); title(’x[n]’,’fontsize’,LFS)
subplot(2,2,2)
xlim(xlimit); ylim([min(x1)-1,max(x1)+1])
xlabel(’n’,’fontsize’,LFS); title(’2x[n-4]’,’fontsize’,TFS)
subplot(2,2,3)
xlim(xlimit)
ylim([min(x2)-1,max(x2)+1])
xlabel(’n’);
title(’3x[n-5]’)
subplot(2,2,4)
xlim(xlimit); ylim([min(x3)-1,max(x3)+1])
title(’x[3-n]’,’fontsize’,TFS)
40.
T{a1x1[n] + a2x2[n]} = 10(a1x1[n] + a2x2[n]) cos(0.25πn + θ)
= a1y1[n] + a2y2[n]
The system is linear.
T{x[n − n0]} = 10x[n − n0] cos(0.25πn + θ)
= y[n − n0] = 10x[n − n0] cos(0.25π(n − n0) + θ)
The system is time-variant.
h[n] = 10δ[n] cos(0.25πn + θ)
The system is causal and stable.

−10 −5 0 5 10 15 20
0
2
4
6
8
10
n
x[n]
−10 −5 0 5 10 15 20
0
5
10
15
20
n
2x[n−4]
−10 −5 0 5 10 15 20
0
10
20
30
n
3x[n−5]
−10 −5 0 5 10 15 20
0
2
4
6
8
10
n
x[3−n]
FIGURE 2.57: x[n], 2x[n − 4], 3x[n − 5], and x[3 − n].
41. Comments: The system is unstable.
MATLAB script:
% P0241: Compute and plot the output of the discrete-time system
% y[n] = 5y[n-1]+x[n], y[-1]=0
close all; clc
n = 0:1e3;
x = ones(1,length(n));
y = filter(1,[1 -5],x);
% Plot:
axis([n(1)-1 n(end)+1 min(y)-1 max(y)+1])
title(’y[n] = 5y[n-1]+x[n], y[-1]=0’,’fontsize’,TFS)
42. MATLAB script:

0 200 400 600 800 1000
0
1
2
3
4
x 10
307
n
y[n] = 5y[n−1]+x[n], y[−1]=0
FIGURE 2.58: Step response of system y[n] = 5y[n − 1] + x[n], y[−1] = 0.
% P0242: Compute the outputs of a LTI system defined by
% ’y=angosto(x)’ for different inputs
close all; clc
n = 0:100;
h = agnosto(d);
x1 = ones(1,length(n));
x2 = (1/2).^n;
x3 = cos(2*pi*n/20);
y1 = conv(h,x1);
y2 = conv(h,x2);
y3 = conv(h,x3);
% Reference:
yr1 = agnosto(x1);
yr2 = agnosto(x2);
yr3 = agnosto(x3);
% Plot:
hf1 = figconfg(’P0242a’);
subplot(2,1,1)
stem(n,y1(1:length(n)),’fill’)
axis([n(1)-1 n(end)+1 min(y1(1:length(n)))-1 max(y1(1:length(n)))+1])
title(’y_1[n] Computed by Impulse Response h[n]’,’fontsize’,TFS)
subplot(2,1,2)
stem(n,yr1,’fill’)
axis([n(1)-1 n(end)+1 min(yr1)-1 max(yr1)+1])

title(’y_1[n] Computed by Function ’’agnosto’’ Directly’,’fontsize’,TFS)
% Plot:
hf2 = figconfg(’P0242b’);
subplot(2,1,1)
title(’y_2[n] Computed by Impulse Response h[n]’,’fontsize’,TFS)
subplot(2,1,2)
title(’y_2[n] Computed by Function ’’agnosto’’ Directly’,’fontsize’,TFS)
% Plot:
subplot(2,1,1)
title(’y_3[n] Computed by Impulse Response h[n]’,’fontsize’,LFS)
subplot(2,1,2)
title(’y_3[n] Computed by Function ’’agnosto’’ Directly’,’fontsize’,LFS)

0 10 20 30 40 50 60 70 80 90 100
0
1
2
3
4
n
y
1
[n] Computed by Impulse Response h[n]
0 10 20 30 40 50 60 70 80 90 100
0
1
2
3
4
n
y1
[n] Computed by Function ’agnosto’ Directly
FIGURE 2.59: System response y1[n] of input x1[n] = u[n].
0 10 20 30 40 50 60 70 80 90 100
−2
−1
0
1
2
n
y2
0 10 20 30 40 50 60 70 80 90 100
−2
−1
0
1
2
n
y
2
FIGURE 2.60: System response y2[n] of input x2[n] = (1/2)n.

0 10 20 30 40 50 60 70 80 90 100
−2
0
2
n
y3
0 10 20 30 40 50 60 70 80 90 100
−2
0
2
n
y
3
FIGURE 2.61: System response y3[n] of input x3[n] = cos(2πn/20).
43. (a) Proof:
Ay =
n
y[n] =
n
∞
k=−∞
x[k]h[n − k] =
∞
k=−∞
x[k]
n
h[n − k]
=
∞
k=−∞
x[k]
n
h[n] = AxAh
(b) Comments: Ay = AxAh
(c) Comments: Ay = AxAh
MATLAB script:
% P0243: Compute the sum of the sequence
close all; clc
nx = 0:100;
x = sin(2*pi*0.01*(0:100)) + 0.05*randn(1,101);
h = ones(1,5);
nh = 0:4;
% Part (b):

[y ny] = conv0(h,nh,x,nx);
Ay = sum(y);
Ax = sum(x);
Ah = sum(h);
% Plot:
plot(nx,x,’.’,ny,y,’.’,’markersize’,16)
legend(’x[n]’,’y[n]’,’fontsize’,LFS,’location’,’best’)
% Part (c):
[y2 ny2] = conv0(h/Ah,nh,x,nx);
Ay2 = sum(y2);
% Plot:
plot(nx,x,’.’,ny2,y2,’.’,’markersize’,16)
legend(’x[n]’,’y[n]’,’fontsize’,LFS,’location’,’best’)
0 20 40 60 80 100 120
−6
−4
−2
0
2
4
6
n
x[n]
y[n]
FIGURE 2.62: Plots of x[n] and y[n].

0 20 40 60 80 100 120
−1.5
−1
−0.5
0
0.5
1
1.5
n
x[n]
y[n]
FIGURE 2.63: Plots of x[n] and y[n].
44. Solution:
Constraint that
Ah =
n
h[n] = 1
n
h[n] =
n
ban
u[n] =
∞
n=0
ban
=
b
1 − a
= 1
We choose
b = 1 − a
The step response can be found:
s[n] = h[n] ∗ u[n] =
∞
m=−∞
h[m]u[n − m] =
∞
m=−∞
bam
u[m]u[n − m]
=
n
m=0
ban
= b
1 − an+1
1 − a
= 1 − an+1

45. Solution:
y[n] = x[n] ∗ h[n] =
∞
k=−∞
x[n − k]h[k]
if n ∈ [0, M − 1],
y[n] =
n
k=0
ak
=
1 − an+1
1 − a
if n ∈ [M − 1, N − 1],
y[n] =
M−1
k=0
ak
=
1 − aM
1 − a
if n ∈ [N − 1, M + N − 2],
y[n] =
M−1
k=n−(N−1)
ak
=
M−1
k=0
ak
−
n−N
k=0
ak
=
1 − aM
1 − a
−
1 − an−N+1
1 − a
= an−N+1 (1 − aM−n+N−1)
1 − a
otherwise, y[n] = 0.
46. Comments: The result highlights line feature oriented horizontally.
MATLAB script:
% P0246: Filtering 2D image lena.jpg using 2D Sobel filter
close all; clc
[nx ny] = size(x);
imshow(x,[])
% Part (b): Sobel finding vertical edges
hver = fspecial(’sobel’)’;
yver = filter2(hver,x);
hfver = figconfg(’P0246b’,’small’);
imshow(yver,[])
% Part (c): Sobel finding horizontal edges
hhor = fspecial(’sobel’);
yhor = filter2(hhor,x);
hfhor = figconfg(’P0246c’,’small’);
imshow(yhor,[])

(a)
(b) (c)
FIGURE 2.64: (a) Original image. (b) Filtered image using 3 × 3 vertical Sobel
ﬁlter. (c) Filtered image using 3 × 3 horizontal Sobel ﬁlter.

47. MATLAB script:
function y = lccde(b,a,x)
% P0247: LCCDE stream implementation
% Eq.(2.94) y[n] = -sum_{k=1}^N{a_k*y[n-k]}+sum_{k=0}^M{b_k*x[n-k]}
%% Testing:
% b = [0.18 0.1 0.3 0.1 0.18];
% a = [1 -1.15 1.5 -0.7 0.25];
% [x n] = delta(0,0,59);
%% function:
na = length(a); nb = length(b);
if a(1)~=1
a = a/a(1);
end
nab = max([na,nb]);
nx = length(x);
x = [zeros(1,nab-1),x(:)’];
y = zeros(1,nx+nab-1);
for ii = nab:nx+nab-1
y(ii) = b(1)*x(ii);
for jj = 2:nab
y(ii) = y(ii)-a(jj)*y(ii-jj+1)+b(jj)*x(ii-jj+1);
end
end
y = y(end-nx+1:end);
48. Solution:
y(t) = h(t) ∗ x(t) =
∞
−∞
h(τ)x(t − τ)dτ
=
∞
−∞
[u(τ) − u(τ − 3)] [u(t − τ) − u(t − τ − 2)] dτ
=
3
0
[u(t − τ) − u(t − τ − 2)] dτ
0 ≤ t ≤ 2, y(t) =
t
0
1dτ = t
2 ≤ t ≤ 3, y(t) =
t
t−2
1dτ = 2

3 ≤ t ≤ 5, y(t) =
3
t−2
1dτ = 5 − t
otherwise, y(t) = 0.
49. Solution:
h(t) = e−t/2
u(t), x(t) = x2(t) = 2, 0 ≤ t ≤ 3
y(t) = x(t) ∗ h(t) =
∞
−∞
h(τ)x(t − τ)dτ
0 ≤ t ≤ 3
y(t) = 2
t
0
e−τ/2
dτ = 4 − 4e−t/2
t ≥ 3
y(t) = 2
t
t−3
e−τ/2
dτ = 4e−t/2
(e3/2
− 1)
y(t) = 0, otherwise
0 1 2 3 4 5 6 7 8 9 10
0
0.5
1
1.5
2
2.5
3
3.5
MSE = 0.008718
t (∆ T = 0.1)
yhat[n]
y[n]
MATLAB script:

0 1 2 3 4 5 6 7 8 9 10
0
0.5
1
1.5
2
2.5
3
3.5
MSE = 8.5688e−05
t (∆ T = 0.01)
yhat[n]
y[n]
% P0249: Compute and plot continuous-time convolution
% using discrete approximation
close all; clc
% dT = 0.1;
dT = 0.01;
n = 0:10/dT;
t = n*dT;
h = exp(-t/2); % h in P0220
x = zeros(1,length(n));
ind = (t >= 0 & t <=3);
x(ind) = 2; % x2 in P0220
% Theoretical continuous y(t):
y = zeros(1,length(n));
ind = (t >= 0 & t <= 3);
y(ind) = 4-4*exp(-t(ind)/2);
ind = (t >=3);
y(ind) = 4*(exp(1.5)-1)*exp(-t(ind)/2);
% Approximated y(t):
yhat = conv(h,x);

yhat = dT*yhat;
yhat = yhat(1:length(n));
% Compute mean square error:
mse = mean((y-yhat).^2);
% Plot:
hf1 = figconfg(’P0249’);
plot(t,yhat,’.’,t,y,’-.’)
TB = [’MSE = ’,num2str(mse)];
title(TB,’fontsize’,TFS)
LB = [’t (Delta T = ’,num2str(dT),’)’];
xlabel(LB,’fontsize’,LFS);
legend(’yhat[n]’,’y[n]’,’location’,’best’)

Review Problems
50. MATLAB script:
% P0250: Generate digital reverberation using audio file ’handel’
% and compare it with the one generated in P0219
close all; clc
[d nd] = delta(0,0,1e3);
n = 1:length(y);
a = 0.7; % specify attenuation factor
tau = 50e-3; % Part (a)
% tau = 100e-3; % Part (b)
% tau = 500e-3; % Part (c)
D = floor(tau*Fs); % compute delay
yir = filter([zeros(1,length(D)) 1],[1 zeros(1,length(D)-1),-a],d);
% Plot:
plot(nd,yir)
% axis([n(1)-1 n(end)+1 min(y3(1:length(n)))-1 max(y3(1:length(n)))+1])
title(’Impulse Response’,’fontsize’,LFS)
yd = filter([zeros(1,length(D)) 1],[1 zeros(1,length(D)-1),-a],y);
yd_ref = filter(1,[1 zeros(1,length(D)-1),-a],y);
% sound(y,Fs)
% pause(1)
sound(yd,Fs)
pause(1)
sound(yd_ref,Fs)

0 100 200 300 400 500 600 700 800 900 1000
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
n
Impulse Response
FIGURE 2.67: Impulse response for a = 0.7.
51. (a) Solution:
h[n] ∗ g[n] =
∞
k=0
akδ[n − kD] ∗
∞
l=0
blδ[n − lD]
=
∞
k=0
∞
l=0
akblδ[n − kD] ∗ δ[n − lD]
=
∞
k=0
∞
l=0
akblδ[n − kD − lD]
=
∞
k=0
ckδ[n − kD] = δ[n]
Hence, we can conclude
c0 = a0b0 = 1
ck = k
m=0 ak−mbm = 0, k > 0

(b) Solution:



c0 = a0b0 = 1
c1 = a0b1 + a1b0 = 0
c2 = a0b2 + a1b1 + a2b0 = 0
=⇒



b0 = a−1
0
b1 = −a1a−2
0
b2 = −a2a−2
0 + a2
1a−3
0
(c) Solution:
Combined the conditions and the results of previous parts, we have
bk + 0.5bk−1 + 0.25bk−2 = 0, b0 = 1, b1 = −0.5, b2 = 0
k 0 1 2 3 4 5 6 7 · · ·
bk 1 −0.5 0 0.53 −0.54 0 0.56 −0.57 · · ·
We can conclude that
bk =



0.5k, k = 3l
−0.5k, k = 3l + 1
0, k = 3l + 2
l = 0, 1, 2, . . .
52. (a) Solution:
u[n] =
∞
k=0
δ[n − k] =
n
k=−∞
δ[k]
s[n] = h[n] ∗ u[n] = h[n] ∗
∞
k=0
δ[n − k] =
∞
k=0
h[n] ∗ δ[n − k]
=
∞
k=0
h[n − k]
if ∞
k=0 h[n − k] = 0, for all n < 0
∞
k=0 h[−1 − k] = ∞
k=0 h[−2 − k] + h[−1] = 0
∞
k=0 h[−2 − k] = 0
=⇒ h[−1] = 0
Follow the above procedure, we can prove by mathematical induction
that h[n] = 0, for all n < 0. Hence, we conclude a system is causal if
the step response s[n] is zero for n < 0.
(b) Solution:
y[n] = x[n] ∗ h[n] =
∞
m=−∞
h[n − m]x[m]

Suppose a period of h[n] is N, that is h[n + N] = h[n].
y[n + N] =
∞
m=−∞
h[N + n − m]x[m] =
∞
m=−∞
h[n − m]x[m] = y[n]
Hence, we proved the output is periodic.
(c) Solution: Wrong. A counter example is the output of the stable system
is always zero.
(d) Solution: Wrong: The inverse of the identity system is itself and it is
causal.
(e) Solution: Wrong. A counter example, h[n] = (0.5)nu[n], is both of
inﬁnite-duration and stable.
(f) Solution: Wrong. A counter example is h[n] = u[n] which is unstable.
53. (a) Solution:
h[n] = δ[n + 1] − 2δ[n] + δ[n − 1]
(b) Comments: The resulting image highlights vertical edges.
(c) Comments: The resulting image highlights horizontal edges.
MATLAB script:
% P0253: Filtering 2D image lena.jpg using 1D second
% derivative filter: y[n] = x[n+1]-2x[n]+x[n-1]
close all; clc
[nx ny] = size(x);
% Part (b): row processing
imshow(x,[])
h = [1 -2 1];
y1 = zeros(nx,ny);
for ii = 1:nx
temp = conv(double(x(ii,:)),h);
y1(ii,:) = temp(2:end-1);
end
imshow(y1,[])
% Part (c): column processing
y2 = zeros(nx,ny);

for ii = 1:ny
temp = conv(double(x(:,ii)),h);
y2(:,ii) = temp(2:end-1);
end
imshow(y2,[])

(a)
(b) (c)
FIGURE 2.68: (a) Original image. (b) Filtered image after row-by-row processing.
(c) Filtered image after column-by-column processing.

54. (a) Comments: The resulting image is about the ﬁne edge details.
(b) Solution:
h[m, n] =
0 −1 0
−1 5 −1
0 −1 0
(c) Comments: The edge details are enhanced in the resulting image than
the original one.
MATLAB script:
% P0254: Filtering 2D image lena.jpg using 2D Laplacian filter
% Illustrating edge enhancement
close all; clc
% Part (a):
[nx ny] = size(x);
imshow(x,[])
h = [0 1 0;1 -4 1;0 1 0];
y1 = filter2(h,x);
imshow(y1,[])
% Part (c):
heh = [0 -1 0;-1 5 -1;0 -1 0];
y2 = filter2(heh,x);
imshow(y2,[])

(a)
(b) (c)
FIGURE 2.69: (a) Original image. (b) Filtered image using the impulse re-
sponse (2.125). (c) Filtered image using the edge-enhanced ﬁlter speciﬁed in
part (b).
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