PDF《第8章习题及解答》

第8章习题及解答 8-1 已知时间信号 如下,试求取它们的 z 变换 .

) (t x ) (z X (1) t A t x ω cos ) ( = (2)

2 ) ( t t x = (3) (4) t e t x

5 1 ) ( ? ? = t e t t x

2 2 ) ( ? ? = (5) bt e t x at sin ) ( ? = ? 题解: (1) t A t x ω cos ) ( = 查z变换表,因为

1 cos

2 ) cos ( ] [cos

2 + ? ? ? = z T z T z z t Z ω ω ω ,所以有

1 cos

2 ) cos ( ) (

2 + ? ? ? = z T z T z Az z X ω ω (2)

2 ) ( t t x = 查z变换表有

3 2

2 )

1 ( )

1 ( ] [ ) ( ? + = = z z z T t Z z X (3) t e t x

5 1 ) ( ? ? = 查z变换表,

1 )] (

1 [ ? = z z t Z , aT at e z z e Z ? ? ? = ] [ ,所以 T T T T t e z e z z e e z z z z e Z z X

5 5

2 5

5 5 )

1 ( )

1 (

1 ]

1 [ ) ( ? ? ? ? ? + + ? ? = ? ? ? = ? = (4) t e t t x

2 2 ) ( ? ? = 查z变换表,

2 ) ( ] [ aT aT at e z z Te te Z ? ? ? ? = ,所以

2 2

2 2 ) (

2 ]

2 [ ) ( T T t e z z Te te Z z X ? ? ? ? = = (5) bt e t x at sin ) ( ? = ? 查z变换表有 aT aT aT at e z bT e z z bT e bt e Z z X

2 2 cos

2 sin ] sin [ ) ( ? ? ? ? + ? ? ? = = 由MATLAB 语言的符号运算工具,z 变换函数 ztrans 求解. (1) t A t x ω cos ) ( = (2)

2 ) ( t t x = f='

a*cos(w*t)'

;

f='

t^2'

;

F=ztrans(f) F=ztrans(f) F = F = z*a*(-cos(w)+z)/(1-2*z*cos(w)+z^2) z*(z+1)/(z-1)^3

91 (3) (4) t e t x

5 1 ) ( ? ? = t e t t x

2 2 ) ( ? ? = f='

1-exp(-5*t)'

;

f='

2*t*exp(-2*t)'

;

F=ztrans(f) F=ztrans(f) F = F = -z*(exp(-5)-1)/(z-1)/(z-exp(-5)) 2*z*exp(-2)/(z-exp(-2))^2 (5) bt e t x at sin ) ( ? = ? f='

exp(-a*t)*sin(b*t)'

;

F=ztrans(f) F = exp(-a)*sin(b)*z/(-2*z*exp(-a)*cos(b)+z^2+exp(-2*a)) 解毕. 8-2 已知时间函数 的拉氏变换 ,试求取它们的 z 变换 . ) (t x ) (s X ) (z X (1) )

10 )(

5 (

50 ) ( + + = s s s X (2) )

10 )(

5 (

50 ) ( + + = ? s s e s X Ts (3) )

1 (

1 ) (

2 + = s s s X (4) )

1 (

1 ) (

2 + ? = ? s s e s X Ts 题解: (1) )

10 )(

5 (

50 ) ( + + = s s s X 因为

10 10

5 10 )

10 )(

5 (

50 ) ( + ? + = + + = s s s s s X t t e e t x

10 5

10 10 ) ( ? ? ? = 所以 T T T T T T T T e e z e e z z e e e z z e z z z X

10 5

10 5

2 10

5 10

5 ) ( ) (

10 10

10 ) ( ? ? ? ? ? ? ? ? + + ? ? = ? ? ? = (2) )

10 )(

5 (

50 ) ( + + = ? s s e s X Ts 因为

10 10

5 10 )

10 )(

5 (

50 ) ( + ? + = + + = ′ s s s s s X t t e e t x

10 5

10 10 ) ( ? ? ? = ′ 所以 T T T T T T T T e e z e e z z e e e z z e z z z X

10 5

10 5

2 10

5 10

5 ) ( ) (

10 10

10 ) ( ? ? ? ? ? ? ? ? + + ? ? = ? ? ? = ′ 又 ,所以

1 ] [ ? ? = z e Z Ts T T T T T T e e z e e z e e z z X z X

10 5

10 5

2 10

5 1 ) ( ) (

10 ) ( ) ( ? ? ? ? ? ? ? + + ? ? = ? ′ =

92 (3) )

1 (

1 ) (

2 + = s s s X 因为

1 1

1 1 )

1 (

1 ) (

2 2 + + ? = + = s s s s s s X t e t t t x ? + ? = ) (

1 ) ( 所以 T T T T T T T e z e z e z z Te e z e T e z z z z z Tz z X ? ? ? ? ? ? ? ? + + + ? ? ? + + + ? = ? + ? ? ? = )

2 1 ( )

2 ( )

1 ( )

1 (

1 )

1 ( ) (

2 3

2 2 (4) )

1 (

1 ) (

2 + ? = ? s s e s X Ts 因为

1 1

1 1 )

1 (

1 ) (

2 2 + + ? = + = ′ s s s s s s X t e t t t x ? + ? = ′ ) (

1 ) ( 所以 T T T T T T T e z e z e z z Te e z e T e z z z z z Tz z X ? ? ? ? ? ? ? ? + + + ? ? ? + + + ? = ? + ? ? ? = ′ )

2 1 ( )

2 ( )

1 ( )

1 (

1 )

1 ( ) (

2 3

2 2 又, ,所以

1 ] [ ? ? = z e Z Ts

1 ) ( ) ( ) ( ? ? ′ ? ′ = z z X z X z X T T T T T T T T T T T T e z e z e z Te e z e T e z e z e z z Te e z e T ? ? ? ? ? ? ? ? ? ? ? ? ? + + + ? ? ? + + + ? ? ? + + + ? ? ? + + + ? = )