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[高清晰版奥本海默信号与系统第二版课后习题解答].solutions.manual.signals.and.systems.(2....pdf
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
l.43
y t( )
=
3
cos
200t
2
---+
6
2
=
9
cos
200t
---+
6
=
9
---
2
cos
400t
1
---+
3
(a) DC component =
9
---
2
(b) Sinusoidal component =
9
---
2
cos
400t
---+
3
Amplitude =
9
---
2
1
p
Ł
ł
Ł
ł
p
Ł
ł
p
Ł
ł
p
Ł
ł
Fundamental frequency =
200
p---------Hz
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
Let N denote the fundamental period of x[N]. which is defined by
N
=
2p
W------
The average power of x[n] is therefore
N -1
x2 n[ ]
n=0
N -1
2
A2
cos
P
=
=
=
1
----
N
1
----
N
A2
------
N
2p n
----------
N
f+
n=0
N -1
n=0
2
cos
2p n
----------
N
f+
1.46
The energy of the raised cosine pulse is
E
w ⁄
w ⁄–
w ⁄
=
=
1
---
2
=
1
---
2
0
--- 3
1
---Ł
2
2
=
0
w ⁄
1
---
4
(
(
(
cos
)
t
1+
)2 td
2
(
cos
)
t
+
2
cos
(
)
t
1+
) td
cos
(
2w
)
t
+ +
1
---
2
2
cos
(
)
t
1+
td
3p
=
⁄
4w
1
---
2
w----
1.47
The signal x(t) is even; its total energy is therefore
E
5=
2
0
x2 t( ) td
2
Ł
ł
Ł
ł
w
p
p
w
w
p
w
Ł
ł
p
ł
p
Ł
ł
4=
2
1( )2 t
0
2 t[ ]
=
4
t=0
+
t–(
5
5+d
2
4
1
t–(
--- 5
3
2
–
)2 td
5
)3
t=4
=
8
2
---+
3
=
26
------
3
1.48
(a) The differentiator output is
y t( )
=
1
1–
0
4–
< <
5
for
–
t
< <
for 4
5
t
otherwise
(b) The energy of y(t) is
5+d
2
4–=
5–
=
1+
1
1( )2 t
E
=
4
1–(
)2 td
1.49
The output of the integrator is
y t( )
for
At
td
=
=
t
A t
0
Hence the energy of y(t) is
E
=
T
0
A2t2 td
=
A2T 3
-------------
3
0
t T
x(5t)
1.0
1.50
(a)
(b)
-1 -0.8 0 0.8 1
x(0.2t)
1.0
-25 -20 0 20 25
t
t
3
£
£
1.51
1.52
(a)
x(10t - 5)
1.0
0 0.1 0.5 0.9 1.0
t
x(t)
1
-1
1 2 3
-1
y(t - 1)
-1 1 2 3
-1
x(t)y(t - 1)
1
1
-1
-1
2 3
t
t
t
4
1.52
(b)
1.52
(c)
x(t + 1)
x(t - 1)
1
1
-1
-1
1 2 3 4
y(-t)
y(-t)
1
-2 -1 1 2 3 4
-1
x(t - 1)y(-t)
1
-2 -1 1 2 3 4
-1
-2
-1 1 2 3
-1
-2 -1
1 2 3 4
x(t + 1)y(t - 2)
-2 -1 1 2 3 4
5
t
t
t
t
t
t
1.52
(d)
x(t)
1
-3 -2 -1 1 2 3
-1
y(1/2t + 1)
6 -5 -4 -3 -2 -1
1 2 4 6
1.52
(e)
-1.0
x(t - 1)y(-t)
1
-3 -2 -1 1 2 3
-1
-4 -3 -2 -1
x(t)
1
-1
1 2 3
y(2 - t)
1 2 3
-4 -3 -2 -1
x(t)y(2 - t)
-1
-1
1 2 3
6
t
t
t
t
t
t
1.52
(f)
x(t)
1
-2 -1 1 2
-1
-5
-3 -2 -1
y(t/2 + 1)
1.0
-6
1 1 2 3
-1.0
x(2t)y(1/2t + 1)
+1
-0.5
-1
1 2
1.52
(g)
-7 -6 -5 -4 -3 -2
-1
x(4 - t)
1
-1
y(t)
-2 -1 1 2 4
x(4 - t)y(t) = 0
-3 -2 -1 1 2 3
7
t
t
t
t
t
t
1.53 We may represent x(t) as the superposition of 4 rectangular pulses as follows:
g1(t)
1
1 2 3 4
g2(t)
1
g3(t)
1
g4(t)
11 2 3 4
1 2 3 4
1
0
1 2 3 4
t
t
t
t
)
=
b–
(
g at
To generate g1(t) from the prescribed g(t), we let
g1 t( )
where a and b are to be determined. The width of pulse g(t) is 2, whereas the width of
pulse g1(t) is 4. We therefore need to expand g(t) by a factor of 2, which, in turn, requires
that we choose
a
1
---=
2
The mid-point of g(t) is at t = 0, whereas the mid-point of g1(t) is at t = 2. Hence, we must
choose b to satisfy the condition
a 2( )
or
0=
b–
b
=
2a
=
2
1
---Ł
2
=
1
Hence,
g1 t( )
=
g
1
---t
2
1–
g
=
Proceeding in a similar manner, we find that
g2 t( )
g3 t( )
g4 t( )
Accordingly, we may express the staircase signal x(t) in terms of the rectangular pulse g(t)
as follows:
2
5
---t
---–
3
3
3–(
)
g t
(
g 2t
7–
=
=
)
8
ł
Ł
ł
Ł
ł
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