Solution Manual for Signals and Systems

634 Pages·4.393 MB·English

by  Simon Haykin

CHAPTER 1
1.1 to 1.41 - part of text
1.42 (a) Periodic:
Fundamental period = 0.5s
(b) Nonperiodic
(c) Periodic
Fundamental period = 3s
(d) Periodic
Fundamental period = 2 samples
(e) Nonperiodic
(f) Periodic:
Fundamental period = 10 samples
(g) Nonperiodic
(h) Nonperiodic
(i) Periodic:
Fundamental period = 1 sample
(cid:230) (cid:230) p(cid:246) (cid:246) 2
l.43 y(t) = 3cos 200t + ---
Ł Ł ł ł
6
2(cid:230) p(cid:246)
= 9cos 200t + ---
Ł ł
6
9 (cid:230) p(cid:246)
= --- cos 400t + --- 1
Ł ł
2 3
9
(a) DC component = ---
2
9 (cid:230) p(cid:246)
(b) Sinusoidal component = ---cos 400t + ---
Ł ł
2 3
9
Amplitude = ---
2
1
200
Fundamental frequency = ---------Hz
p
1.44 The RMS value of sinusoidal x(t) is A⁄ 2. Hence, the average power of x(t) in a 1-ohm
resistor is (A⁄ 2)2 = A2/2.
1.45 LetN denote the fundamental period ofx[N]. which is defined by
2p
N = ------
W
The average power ofx[n] is therefore
N-1
P = -1---(cid:229) x2[n]
N
n=0
N-1
= -1--- (cid:229) A2cos2(cid:230) 2----p---n--- + f(cid:246)
Ł ł
N N
n=0
= -A----2- N(cid:229) -1cos2(cid:230) 2----p---n--- + f(cid:246)
Ł ł
N N
n=0
1.46 The energy of the raised cosine pulse is
E = (cid:242) p w⁄ 1---(cos(w t) +1)2dt
–p w⁄ 4
= 1---(cid:242) p w⁄ (cos2(w t)+ 2cos(w t) +1)dt
2
0
= 1---(cid:242) p w⁄ (cid:230) 1---cos(2w t)+ 1--- +2cos(w t) +1(cid:246) dt
Ł ł
2 2 2
0
1(cid:230) 3(cid:246) (cid:230) p(cid:246)
= --- --- ---- = 3p ⁄4w
2Ł 2ł Ł w ł
1.47 The signal x(t) is even; its total energy is therefore
E = 2(cid:242) 5x2(t)dt
0
2
= 2(cid:242) 4(1)2dt +2(cid:242) 5(5 –t)2dt
0 4
= 2[t]4 +2 –1---(5–t)3 5
t=0 3
t=4
2 26
= 8+ --- = ------
3 3
1.48 (a) The differentiator output is
(cid:236)
1 for –5< t < –4
(cid:239)
y(t) = (cid:237) –1 for 4 <t < 5
(cid:239)
(cid:238) 0 otherwise
(b) The energy of y(t) is
E = (cid:242) –4(1)2dt +(cid:242) 5(–1)2dt
–5 4
= 1+1 = 2
1.49 The output of the integrator is
t
y(t) = A(cid:242) t dt = At for 0£ £t T
0
Hence the energy ofy(t) is
2 3
(cid:242) T 2 2 A T
E = A t dt = -------------
3
0
1.50 (a)
x(5t)
1.0
-1 -0.8 0 0.8 1 t
(b)
x(0.2t)
1.0
-25 -20 0 20 25 t
3
1.51
x(10t - 5)
1.0
t
0 0.1 0.5 0.9 1.0
1.52 (a)
x(t)
1
t
-1 1 2 3
-1
y(t- 1)
t
-1 1 2 3
-1
x(t)y(t- 1)
1
1
t
-1 2 3
-1
4
1.52 (b)
x(t+ 1)
x(t- 1)
1
1
t
-1 1 2 3 4
-1
y(-t)y(-t)
1
t
-2 -1 1 2 3 4
-1
x(t- 1)y(-t)
1
t
-2 -1 1 2 3 4
-1
1.52 (c)
-2
t
-1 1 2 3
-1
-2 -1
t
1 2 3 4
x(t+ 1)y(t - 2)
-2 -1 1 2 3 4
t
5
1.52 (d)
x(t)
1
t
-3 -2 -1 1 2 3
-1
y(1/2t+ 1)
6 -5 -4 -3 -2 -1
t
1 2 4 6
-1.0
x(t- 1)y(-t)
1
t
-3 -2 -1 1 2 3
-1
1.52 (e)
x(t)
1
-4 -3 -2 -1
t
1 2 3
-1
y(2 - t)
1 2 3
t
-4 -3 -2 -1
x(t)y(2 - t)
-1
t
1 2 3
-1
6
1.52 (f)
x(t)
1
t
-2 -1 1 2
-1
y(t/2+ 1)
1.0
-5 -3 -2 -1
t
-6 1 1 2 3
-1.0
x(2t)y(1/2t+ 1)
+1
-0.5
t
-1 1 2
-1
1.52 (g)
x(4 -t)
1
t
-7 -6 -5 -4 -3 -2
-1
y(t)
t
-2 -1 1 2 4
x(4 -t)y(t) = 0
t
-3 -2 -1 1 2 3
7
1.53 We may represent x(t) as the superposition of 4 rectangular pulses as follows:
g (t)
1
1
1 2 3 4
t
g (t)
2
1
11 2 3 4
t
g (t)
3
1
1 2 3 4
t
g (t)
4
1
1 2 3 4
t
0
To generateg (t) from the prescribedg(t), we let
1
g (t) = g(at –b)
1
wherea and b are to be determined. The width of pulse g(t) is 2, whereas the width of
pulseg (t) is 4. We therefore need to expand g(t) by a factor of 2, which, in turn, requires
1
that we choose
1
a = ---
2
Themid-pointofg(t)isatt=0,whereasthemid-pointofg (t)isatt=2.Hence,wemust
1
chooseb to satisfy the condition
a(2) –b = 0
or
(cid:230) 1(cid:246)
b = 2a = 2 --- = 1
Ł ł
2
(cid:230) 1 (cid:246)
Hence, g (t) = g ---t –1
1 Ł 2 ł
Proceeding in a similar manner, we find that
(cid:230) 2 5(cid:246)
g (t) = g ---t – ---
2 Ł 3 3ł
g (t) = g(t –3)
3
g (t) = g(2t –7)
4
Accordingly,wemayexpressthestaircasesignalx(t)intermsoftherectangularpulseg(t)
as follows:
8
(cid:230) 1 (cid:246) (cid:230) 2 5(cid:246)
x(t) = g ---t –1 + g ---t – --- +g(t –3) +g(2t –7)
Ł ł Ł ł
2 3 3
1.54 (a)
x(t) =u(t) -u(t - 2)
t
0 1 2
(b)
x(t) =u(t + 1) - 2u(t) +u(t - 1)
-2 0 1 2
t
-1 3
-1
(c)
x(t) =-u(t + 3) + 2u(t +1) -2u(t - 1) +u(t - 3)
-3 1 2 3
t
0
-1
(d)
x(t) =r(t + 1) -r(t) +r(t - 2)
1
t
-2 -1 0 1 2 3
(e)
x(t) =r(t + 2) -r(t + 1) -r(t - 1)+r(t - 2)
1
t
-3 -2 -1 0 1 2
9
1.55 We may generatex(t) as the superposition of 3 rectangular pulses as follows:
g (t)
1
1
t
-4 -2 0 2 4
g (t)
2
1
t
-4 -2 0 2 4
g (t)
3
1
t
-4 - 2 0 2 4
All three pulses,g (t), g (t), and g (t), are symmetrically positioned around the origin:
1 2 3
1. g (t) is exactly the same asg(t).
1
2. g (t) is an expanded version of g(t) by a factor of 3.
2
3. g (t) is an expanded version of g(t) by a factor of 4.
3
Hence, it follows that
g (t) = g(t)
1
(cid:230) 1(cid:246)
g (t) = g ---t
2 Ł 3ł
(cid:230) 1(cid:246)
g (t) = g ---t
3 Ł 4ł
That is,
(cid:230) 1(cid:246) (cid:230) 1(cid:246)
x(t) = g(t)+ g ---t +g ---t
Ł ł Ł ł
3 4
1.56 (a)
x[2n]
o 2 o
o n
-1 0 1
(b)
x[3n- 1]
2 o
o1
o n
-1 0 1
10