PDF《第8章习题及解答》

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2 T T T T T T T T e z e z e z Te e z Te e T z e T ? ? ? ? ? ? ? ? ? + + + ? ? ? + ? ? ? + + + ? = )

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2 由MATLAB 语言的符号数学工具,拉氏反变换函数 invlaplace 和z变换函数 ztrans 求解. (1) )

10 )(

5 (

50 ) ( + + = s s s X Fs='

50/(s+5)/(s+10)'

;

ft=invlaplace(Fs) ft = 10*exp(-5*t)-10*exp(-10*t) Fz=ztrans(ft) Fz = 10*z*(-exp(-10)+exp(-5))/(z-exp(-5))/(z-exp(-10)) (2) )

10 )(

5 (

50 ) ( + + = ? s s e s X Ts Fz = 10*z*(-exp(-10)+exp(-5))/(z-exp(-5))/(z-exp(-10)) Fi='

z^(-1)'

;

93 Fe=symmul(Fi,Fz) Fe = 10*(-exp(-10)+exp(-5))/(z-exp(-5))/(z-exp(-10)) (3) )

1 (

1 ) (

2 + = s s s X Fs='

1/s^2/(s+1)'

;

ft=invlaplace(Fs) ft = t-1+exp(-t) Fz=ztrans(ft) Fz = z*(-2*exp(-1)+z*exp(-1)+1)/(z-1)^2/(z-exp(-1)) (4) )

1 (

1 ) (

2 + ? = ? s s e s X Ts Fz1=Fz;

Fi='

z^(-1)'

;

Fz2=symmul(Fi,Fz) Fz2 = (-2*exp(-1)+z*exp(-1)+1)/(z-1)^2/(z-exp(-1)) FFz=symsub(Fz1,Fz2) FFz = z*(-2*exp(-1)+z*exp(-1)+1)/(z-1)^2/(z-exp(-1))... -(-2*exp(-1)+z*exp(-1)+1)/(z-1)^2/(z-exp(-1)) 解毕. 8-3 已知采样信号的 z 变换 如下,试求 z 反变换 . ) (z X ) (t x? (1) )

2 )(

1 ( ) ( ? ? = z z z z X (2) )

2 )(

1 (

1 ) ( ? ? = z z z X (3) ) )( ( ) ( 2T T e z e z z z X ? ? ? ? = (4) )

2 ( )

1 ( ) (

2 ? ? = z z z z X 题解: (1) )

2 )(

1 ( ) ( ? ? = z z z z X 由于

2 1

1 1 )

2 )(

1 (

1 ) ( ? + ? ? = ? ? = z z z z z z X

2 1 ) ( ? + ? ? = z z z z z X

94 查z变换表,由nT a a z z Z = ? ? ] [

1 得到 nT nT x

2 1 ) ( + ? = (2) )

2 )(

1 (

1 ) ( ? ? = z z z X 由于

2 1

1 1 )

2 )(

1 (

1 ) ( ? + ? ? = ? ? = z z z z z X

2 1 ) ( ? + ? ? = ? z z z z z X z nT nT z z z z Z z X z Z ? ? ? + ? = ? + ? ? = ?

2 ) (

1 ]

2 1 [ )] ( [

1 1 所以 nT nT z X z Z T n x ? ? + ? = ? = +

2 ) (

1 )] ( [ ] )

1 [(

1 T n T n nT x )

1 (

2 ] )

1 [(

1 ) ( ? ? + ? ? = (3) ) )( ( ) ( 2T T e z e z z z X ? ? ? ? = 由于 ]

1 1 [

1 ) )( (

1 ) (

2 2

2 T T T T T T e z e z e e e z e z z z X ? ? ? ? ? ? ? ? ? ? = ? ? = ] [

1 ) (

2 2 T T T T e z z e z z e e z X ? ? ? ? ? ? ? ? = 查z变换表,由anT aT e e z z Z ? ? ? = ? ] [

1 得到 ] [

1 ) (

2 2 nT nT T T e e e e nT x ? ? ? ? ? ? = (4) )

2 ( )

1 ( ) (

2 ? ? = z z z z X 由于

2 1

1 1 )

1 (

1 )

2 ( )

1 (

1 ) (

2 2 ? + ? ? ? ? = ? ? = z z z z z z z X

2 1 )

1 ( ) (

2 ? + ? ? ? ? = z z z z z z z X 查z变换表 nT z Tz Z = ? ? ] )

1 ( [

2 1 , nT a a z z Z = ? ? ] [

1 得到 nT nT T nT X

2 1 ) (

1 ) ( + ? =

95 由MATLAB 语言的符号运算工具,z 反变换函数 invztrans 求解. (1) )

2 )(

1 ( ) ( ? ? = z z z z X F='

z/(z-1)/(z-2)'

;

f=invztrans(F) f = -1+2^n (2) )

2 )(

1 (

1 ) ( ? ? = z z z X F='

1/(z-1)/(z-2)'

;

f=invztrans(F) f = 1/2*Delta(n)-1+1/2*2^n (3) ) )( ( ) ( 2T T e z e z z z X ? ? ? ? = F='

z/(z-exp(-1))/(z-exp(-2))'

;

f=invztrans(F);

vpa(f,4) ans = 4.300*.3679^n-4.300*.1353^n (4) )

2 ( )

1 ( ) (

2 ? ? = z z z z X F='

z/(z-1)^2/(z-2)'

;

f=invztrans(F) f = -1-n+2^n 解毕. 8-4 分别用 z 变换法和迭代法(解出

5 项以上)求解如下差分方程. (1) ) ( ) (

2 )

1 (

3 )

2 ( k u k x k x k x = + + ? + 输入信号: ) (

1 ) ( k k u = 初始条件:

0 )

1 ( ,