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第二章
2—1 试证明图 P2—1 中周期性信号可以展开为 (图略)
0
4 ( 1)
( ) cos(2 1)
2 1
n
n
s t n t
n
p
p
¥
=
-
= +
+
å
证明:因为
所以
0 0 0
0
2 2
( ) cos cos cos
2
k k k
k k k
kt kt
s t c c c kt
T
p p
p
¥ ¥ ¥
= = =
= = =
å å å
1
0
1
( ) 0 0s t dt c
-
= Þ =
ò
1 1
1 1
2 2
1
1
1 1
2
2
4
( ) cos ( ) cos cos sin
2
k
k
c s t k tdt k tdt k tdt
k
p
p p p
p
-
- -
-
= = - + + =
ò ò ò ò
0, 2
4
( 1) 2 1
(2 1)
n
k n
k n
n
p
=
ì
ï
=
í
- = +
ï
+
î
所以
0
4 ( 1)
( ) cos(2 1)
2 1
n
n
s t n t
n
p
p
¥
=
-
= +
+
å
2—2 设一个信号
可以表示成
( ) 2cos(2 )s t t t
p q
= + -¥ < < ¥
试问它是功率信号还是能量信号,并求出其功率谱密度或能量谱密度。
解:功率信号.
2
2
2
( ) cos(2 )
sin ( 1) sin ( 1)
[ ]
2 ( 1) ( 1)
j ft
j j
s f t e dt
f f
e e
f f
t
p
t
t
q q
p q
t p t p t
p t p t
-
-
-
= +
- +
= +
- +
ò
2 2
2 2 2 2 2 2 2 2
sin ( 1) sin ( 1) sin ( 1) sin ( 1)
lim 2 cos 2
4 ( 1) ( 1) ( 1)( 1)
f f f f
f f f f
t
t p t p t p t p t
q
p t p t p t
® ¥
- + - +
= + +
- + - +
由公式