Chapter 2 solutions
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2.2
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2.5
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2.6
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2.7
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2.8
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(c) xo(0)=-xo(-0)= -xo(0). The only number with a=-a is a=0 so this implies xo(0)=0.
x(0)=xe(0)+xo(0)=xe(0).
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2.11
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2.17
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2.18
2.19
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(continued)…
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2.29
i) not memoryless unless t0=0
ii) invertible: x(t)=y(t+t0)
2.30
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2.31
(parts c,d on next page)
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Chapter 3 Solutions
3.7
Parts c,d on next pageÆ
3.12, continued
parts d,e next pageÆ
3.22, continued
3.26
ContinuedÆ
3.28
3.29
3.31
3.32
3.33
3.34
3.35
3.36
3.37
3.38
Chapter 4 solutions
ContinuedÆ
4.3
(a)
(i)
(ii)
(iii)
(iv)
ContinuedÆ
ContinuedÆ
4.12, continued
ContinuedÆ
ContinuedÆ
4.19, continued
ContinuedÆ
4.19, continued
ContinuedÆ
4.19, continued
ContinuedÆ
4.19 continued
ContinuedÆ
4.20, continued
ContinuedÆ
4.25, continued
ContinuedÆ
4.27, continued
Chapter 5 solutions
ContinuedÆ
5.2, continued
ContinuedÆ
5.3, continued
ContinuedÆ
5.4, continued
ContinuedÆ
5.4, continued
ContinuedÆ
5.5, continued
5.6 on next page
ContinuedÆ
5.6, continued
ContinuedÆ
5.9, continued
5.10
(a)
ContinuedÆ
5.10, continued
5.11
(a)
ContinuedÆ
5.14, continued
ContinuedÆ
note the time axis is w/(500pi)
(a)
Chapter 6 solutions
See figures of output signals, next pageÆ
6.3, continued
a
b
1.5
1.5
1
1
0.5
0.5
0
0
-0.5
-0.04
-0.02
0
t
c
0.02
0.04
-0.5
-0.04
1.5
1.5
1
1
0.5
0.5
0
0
-0.5
-0.04
-0.02
0
t
e
0.02
0.04
-0.5
-0.04
1.5
1.5
1
1
0.5
0.5
0
0
-0.5
-0.04
-0.02
0
t
0.02
0.04
-0.5
-0.04
-0.02
0
t
d
0.02
0.04
-0.02
0
t
f
0.02
0.04
-0.02
0
t
0.02
0.04
1
0.0016 s+1
Pulse
Generator
Transfer Fcn
Scope 1
Scope
Part (a)
Part (b)
Part (c)
Part (d)
Part (e)
Part (f)
6.9
ContinuedÆ
6.9(a), continued
(b)
(c)
6.11
(a) (Note that you don’t need the “Analog Butterworth LP Filter” block; just use a Transfer Function block
with the coefficients derived from the ‘butter(N, Wn, ‘s’)’ command.)
We should select a cutoff frequency for the low-pass filter so that the oscillations in the signal are eliminated
as much as possible. This doesn’t specify a precise criterion, however. Here is the signal before and after filtering
with a 2nd order Butterworth low-pass filter with ωc =100π :
The next output plot uses ωc =20π, giving a smoother result, although it takes longer to get there:
(b)
Here is the signal after filtering with a 4th order Butterworth filter with ωc =20π:
6.11, (c)
[b, a] = butter(2, 20*pi, ‘s’);
freqs(b, a);
Frequency Response for 2nd order Butterworth, ωc = 20π
0
10
Magnitude
-1
10
-2
10
-3
10
1
10
2
10
Frequency (rad/s)
3
10
Phase (degrees)
0
-50
-100
-150
-200
1
10
2
10
Frequency (rad/s)
3
10
[b, a] = butter(4, 20*pi, ‘s’);
freqs(b, a);
Frequency Response for 4th order Butterworth, ωc = 20π
0
Magnitude
10
-5
10
1
10
2
10
Frequency (rad/s)
3
10
Phase (degrees)
200
100
0
-100
-200
1
10
2
10
Frequency (rad/s)
3
10
6.11, (d) For the 2nd order filter:
[b, a]=butter(2, 20*pi, ‘s’);
h = freqs(b, a, [377:378]);
abs(h(1));
angle(h(1));
Gives: |H(377)| = 0.0278, θ(377) = -2.9.
For the 4th order filter: |H(377)| =7.715e-4 , θ(377) =0.44
6.13
(a) Filter A is a high-pass filter since the DC component of the signal was removed and the highfrequency components remain
(b) Filter B is a low-pass filter since the signal was smoothed
6.14
6.15 (a) Frequency spectra:
ContinuedÆ
6.15(a), continued
ContinuedÆ
6.15, continued
(c) (a)
6.16
6.17
6.18
6.19
6.20
6.21
6.22
(b)
6.25
6.27
6.28
6.29
6.31
6.32
6.33
6.34
CHAPTER 7
ContinuedÆ
7.1, continued
7.2
ContinuedÆ
7.2, continued
ContinuedÆ
7.2, continued
(g)
(h)
7.5
(a)
ContinuedÆ
7.6(a), continued
ContinuedÆ
7.6, continued
ContinuedÆ
7.7, continued
ContinuedÆ
7.13, continued
ContinuedÆ
7.14, continued
ContinuedÆ
7.17, continued
ContinuedÆ
7.17(b), continued
7.18 (Note that these are just possible answers; any other answer that satisfies the conditions is correct)
(a)
(b)
(c)
continuedÆ
7.18, continued
(d)
(e)
(f)
h(t ) = δ (t ) + Ce − t cos(t − Θ)
(g)
7.20
(a)
7.21
(a)
(c)
, ROC: Re(s) < 2
(e)
ContinuedÆ
7.21, continued
ContinuedÆ
7.23, continued
Part (b) continuedÆ
7.30(b), continued
Chapter 8 Solutions
ContinuedÆ
8.4, continued
ContinuedÆ
8.5, continued
(d)
>> A=[0 1; -24 10]; B=[0; 1]; C=[64 0]; D=0;
>> [n d] = ss2tf(A, B, C, D)
ContinuedÆ
(d)
>> A=[0 1 0; 0 0 1; -3 -10 -4]; B=[0; 0; 1]; C=[10 0 0]; D=0;
>> [n d] = ss2tf(A, B, C, D)
(b)
sI − A = s + 3
H ( s ) = C ( sI − A) −1 B = 4
continuedÆ
1
24
(6) =
s+3
s+3
continuedÆ
8.7(c)
>> A=[-5 3; -6 1]; B=[1; 2]; C=[5 4]; D=0;
>> [n d] = ss2tf(A, B, C, D)
ContinuedÆ
(g)
>> A=[0 1; -13 -4]; B=[0; 1]; C=[41 13]; D=0;
>> [n d] = ss2tf(A, B, C, D);
ContinuedÆ
8.8, continued
ContinuedÆ
>> syms s;
>> M=[s -1 0; 5 s+2 -4; 3 4 s+3];
>> inv(M)
(j)
>> A=[0 1 0; -5 -2 4; -3 -4 -3]; B=[0; 0; 1]; C=[3 4 0]; D=0;
>> [n d] = ss2tf(A, B, C, D)
ContinuedÆ
ContinuedÆ
8.9, continued
(i)
d2y
dy
du
+ 10 + 11 y (t ) = 6u (t ) + 8
2
dt
dt
dt
(j) >> A=[0 1; -11 -10]; B=[0; 2]; C=[3 4]; D=0;
>> [n d] = ss2tf(A, B, C, D);
(d)
>> A=[0 1; 4 -3]; B=[0; 1]; C=[9 1]; D=0;
>> [n d] = ss2tf(A, B, C, D)
ContinuedÆ
(c)
>> A=-2; B=4; C=1; D=0;
>> [n d] = ss2tf(A, B, C, D)
(c)
>> A = [0 1 0; 0 0 1; 1 1 -1]; B = [2 0 0]; C=[1 0 0]; D=0;
>> [n d] = ss2tf(A, B, C, D)
ContinuedÆ
ContinuedÆ
ContinuedÆ
Note: part (b) can be different for each student; parts (c)-(f) are self-checking.
Note: part (b) can be different for each student; parts (c)-(g) are self-checking.
(c), (f)
>> A = [0 1 0; 0 0 1; 1 1 -1]; B=[2; 0; 0];C = [1 0 0]; D=0;
>> P = [1 1 0; 0 0 1; 1 0 0];
>> Q=inv(P)
>> Av = Q*A*P
>> Bv = Q*B
>> Cv = C*P
>> Dv = D
>> [n d] = ss2tf(Av, Bv, Cv, Dv)
(d) Show that H(s)=Cv (sI-A)-1 Bv gives the same result as in part (a)
(c) >>A = [-4 5; 0 1]; eig(A)
(c)
>> A=[0 1; -5 -4];
>> eig(A)
(c) >>A = [0 1 0; 0 0 1; 1 1 -1];
>> eig(A)
CHAPTER 9 solutions
9.3 (a)
2-3x a[n]
2x a[-n]
8
2
6
0
4
-2
2
0
-2
-6
-4
-4
-2
0
n
3x a[n-2]
2
4
6
4
-6
-4
-2
0
n
3-x a[n]
2
4
6
-4
-2
0
n
2x a[-n]-4
2
4
6
-4
-2
0
n
2
4
6
6
2
4
0
-2
2
-4
-6
-4
-2
0
2
4
n
1+2x a[n-2]
6
8
0
-6
4
0
2
-2
-4
0
-6
-2
-8
-4
-4
-2
0
2
n
4
6
8
-6
9.3 (b)
2-3x b[n]
2x b[-n]
5
5
0
0
-5
-6
-5
-4
-2
0
n
3x b[n-2]
2
4
6
-6
-4
-2
0
n
3-x b[n]
2
4
6
-4
-2
0
n
2x b[-n]-4
2
4
6
-4
-2
0
n
2
4
6
6
5
4
0
2
-5
-4
-2
0
2
4
n
1+2x b[n-2]
6
8
0
-6
6
0
4
-2
2
-4
0
-6
-2
-8
-4
-4
-2
0
2
n
4
6
8
-6
9.3 (c)
2-3x c [n]
2x c [-n]
5
5
0
0
-5
-10
-6
-4
-2
0
n
3x c [n-2]
2
4
6
-5
-6
-4
-2
0
n
3-x c [n]
2
4
6
-4
-2
0
n
2x c [-n]-4
2
4
6
-4
-2
0
n
2
4
6
6
10
4
5
2
0
0
-5
-4
-2
0
2
n
1+2x c [n-2]
4
6
8
-2
-6
10
5
5
0
0
-4
-5
-2
0
2
n
4
6
8
-6
9.3 (d)
2-3x d[n]
2x d[-n]
2
6
0
4
-2
-4
2
-6
-8
-6
-4
-2
0
n
3x d[n-2]
2
4
6
10
0
-6
-4
-2
0
n
3-x d[n]
2
4
6
-4
-2
0
n
2x d[-n]-4
2
4
6
-4
-2
0
n
2
4
6
4
3
5
2
1
0
-4
-2
0
2
4
n
1+2x d[n-2]
6
8
0
-6
8
2
6
0
4
-2
2
-4
0
-4
-2
0
2
n
4
6
8
-6
9.4 (a)
x a[-n]u[n]
x a[n]u[-n]
2
2
1
1
0
0
-1
-1
-2
-2
-3
-6
-4
-2
0
n
2
4
6
-3
-6
-4
-2
x a[n]u[n+2]
2
1
1
0
0
-1
-1
-2
-2
-4
-2
0
n
2
4
6
-3
-6
-4
x a[n] δ [n-2]
2
1
1
0
0
-1
-1
-2
-2
-4
-2
0
n
4
6
-2
0
n
2
4
6
4
6
x a[n](δ [n+1]-δ [n-1])
2
-3
-6
2
x a[-n]u[-2-n]
2
-3
-6
0
n
2
4
6
-3
-6
-4
-2
0
n
2
9.4 (b)
x b[-n]u[n]
x b[n]u[-n]
3
3
2
2
1
1
0
0
-1
-1
-2
-2
-3
-6
-4
-2
0
n
2
4
6
-3
-6
-4
-2
x b[n]u[n+2]
3
2
2
1
1
0
0
-1
-1
-2
-2
-4
-2
0
n
2
4
6
-3
-6
-4
x b[n] δ [n-2]
3
2
2
1
1
0
0
-1
-1
-2
-2
-4
-2
0
n
4
6
-2
0
n
2
4
6
4
6
x b[n](δ [n+1]-δ [n-1])
3
-3
-6
2
x b[-n]u[-2-n]
3
-3
-6
0
n
2
4
6
-3
-6
-4
-2
0
n
2
9.4 (c)
x c [-n]u[n]
x c [n]u[-n]
4
4
2
2
0
0
-2
-2
-6
-4
-2
0
n
2
4
6
-6
-4
-2
x c [n]u[n+2]
4
2
2
0
0
-2
-2
-4
-2
0
n
2
4
6
-6
-4
x c [n] δ [n-2]
4
2
2
0
0
-2
-2
-4
-2
0
n
4
6
-2
0
n
2
4
6
4
6
x c [n](δ [n+1]-δ [n-1])
4
-6
2
x c [-n]u[-2-n]
4
-6
0
n
2
4
6
-6
-4
-2
0
n
2
9.4 (d)
x d[-n]u[n]
x d[n]u[-n]
4
4
3
3
2
2
1
1
0
-6
-4
-2
0
n
2
4
6
0
-6
-4
-2
x d[n]u[n+2]
4
3
3
2
2
1
1
-4
-2
0
n
2
4
6
0
-6
-4
x d[n] δ [n-2]
4
3
3
2
2
1
1
-4
-2
0
n
4
6
-2
0
n
2
4
6
4
6
x d[n](δ [n+1]-δ [n-1])
4
0
-6
2
x d[-n]u[-2-n]
4
0
-6
0
n
2
4
6
0
-6
-4
-2
0
n
2
9.5
9.8
x a,even[n]
x a,odd[n]
2
2
1.5
1.5
1
1
0.5
0.5
0
0
-0.5
-0.5
-1
-1
-1.5
-1.5
-2
-6
-4
-2
0
n
2
4
6
-2
-6
-4
-2
x b,even[n]
2
1.5
1.5
1
1
0.5
0.5
0
0
-0.5
-0.5
-1
-1
-1.5
-1.5
ContinuedÆ
-4
-2
0
n
2
4
6
2
4
6
x b,odd[n]
2
-2
-6
0
n
2
4
6
-2
-6
-4
-2
0
n
9.8, continued
x c ,even[n]
x c,odd[n]
5
5
4
4
3
3
2
2
1
1
0
0
-1
-1
-2
-2
-3
-6
-4
-2
0
n
2
4
6
-3
-6
-4
-2
x d,even[n]
4
3
3
2
2
1
1
0
0
-4
-2
ContinuedÆ
0
n
2
4
6
2
4
6
x d,odd[n]
4
-1
-6
0
n
2
4
6
-1
-6
-4
-2
0
n
9.9, continued
(b)
x[n]=6u[n-3]: neither even nor odd
x[n]=-n: odd
8
5
6
4
0
2
0
-6
-5
-4
-2
0
2
4
6
-6
-4
x[n]=0.2|n|:even
-2
0
2
4
6
x[n]=6+0.2n+0.2-n:even
1
150
0.8
100
0.6
0.4
50
0.2
0
-6
-4
-2
0
2
4
6
-3
1.5
1
1
0.5
0.5
0
0
-0.5
-0.5
-1
-1
ContinuedÆ
-4
-2
0
2
-1
0
1
2
3
x[n]=sin(n-π/6):neither even nor odd
x[n]=sin(2n):odd
1.5
-1.5
-6
-2
4
6
-1.5
-6
-4
-2
0
2
4
6
9.9, continued
even part of x[n]=6u[n-3]
odd part of x[n]=6u[n-3]
4
4
3
3
2
2
1
1
0
0
-1
-1
-2
-2
-3
-3
-4
-6
-4
-2
0
2
4
6
-4
-6
-4
even part of x[n]=sin(n-π/6)
2
1.5
1.5
1
1
0.5
0.5
0
0
-0.5
-0.5
-1
-1
-1.5
-1.5
-4
-2
0
2
0
2
4
6
4
6
odd part of x[n]=sin(n-π/6)
2
-2
-6
-2
4
6
-2
-6
-4
-2
0
2
ContinuedÆ
9.23, continued
ContinuedÆ
9.23, continued
ContinuedÆ
9.25, continued
ContinuedÆ
9.27, continued
continuedÆ
9.28, continued
Chapter 10 Solutions
ContinuedÆ
10.3d, continued
ContinuedÆ
10.5(d), continued
See plot next pageÆ
10.5e plot
ContinuedÆ
10.9, continued
ContinuedÆ
10.9e, continued
ContinuedÆ
10.9, continued
ContinuedÆ
10.10b, continued
continuedÆ
10.11, continued
(d)
ContinuedÆ
10.12, continued
ContinuedÆ
10.14, continued
ContinuedÆ
10.19, continued
Chapter 11 solutions
ContinuedÆ
11.2, continued
ContinuedÆ
11.10a, continued
ContinuedÆ
11.10,continued
ContinuedÆ
11.10 continued
ContinuedÆ
11.10 continued
continuedÆ
11.10 continued
ContinuedÆ
11.13b,d next pageÆ
11.13, continued
11.19
11.21
ContinuedÆ
11.23 (c), continued
11.25
11.28
11.29
11.30
Chapter 12 solutions
12.18
12.20
12.21
(a)
(b) To have resolution of 1 rad/sec, at ωs=300rad/sec, need 300 samples.
12.22
12.25
12.26
12.27
12.28
12.29
12.30
12.31
function compressimage(percentzero)
inputimage=imread('filename','pgm');
s=size(inputimage);
height=s(1);
width=s(2);
INPUTIMAGE=dct2(inputimage);
numbercoefficients=height*width*percentzero/100
side_percentzero=sqrt(numbercoefficients)
tpic=zeros(height,width);
for i=[1:round(side_percentzero)]
for j=[1:round(side_percentzero)]
tpic(i,j)=INPUTIMAGE(i,j);
end
end
iinputimage=idct2(tpic);
figure
imshow(iinputimage, [ 0 255])