信号与系统奥本海姆英文版课后答案chapter10

∞
−n
n =−∞
∑ (1/ 4)
∞
z −n zn
n =−3
∑ (1/ 4)
= ∑ (1/ 4) − n +3 z n −3
n=0
= (1/ 64) z −3 /(1 − 4 z ), z < 1/ 4 = (1/16) z −4 /(1 − (1/ 4) z −1 ), z < 1/ 4
The Fourier transform does not exist because the ROC includes the unit circle. (g) Consider x1 ( z ) = 2n μ[ − n].
X ( z) =
n =−∞
jw
r> 1 . The second summation
2
10.2
∑ ( 5 ) u[n − 3]z
n
∞
1
z −3 ∞ 1 n − n ]∑ ( ) z =[ 125 n = 0 5
∞ 1 = ∑ ( )n z − n n =3 5 z −3 1 ] =[ 125 1 − 1 z −1 5 −n
X ( Z ) = ∑ x[n]z − n
n =∞ ∞
−5
= ∑ (−1) n z − n
n =0
∞
= 1/(1 + z −1 ), z > 1
The Fourier transform does not exist because the ROC does not include the unit circle (d) For x [n]= ( 1 ) n +1 μ[n + 3],
x( z ) =
n =−∞ ∞
n =2 ∞
n−2
μ[n − 2].
∑ x[n]z
∞
−n
= ∑ (1/ 3) n − 2 z − n = ∑ (1/ 3) n z − n − 2
n =0
= z −2 [1/(1 − (1/ 3) z −1 )], z > 1/ 3
The Fourier transform exists because the ROC includes the unit circle.
n=0
∞
= ∑ (0 + σ [n] + σ [n]z − n
n=0
∞
= 2,
All z The unilateral z-transform is
1 n x( z ) = ( ) z − n 2 ∞ 1 n −n = ∑( ) z n=0 2 1 1 = ,z > 2 ⎛ 1 ⎞ −1 1− ⎜ ⎟ z ⎝2⎠
10.20.Applying the unilateral z-transform to given difference equation, we have
z −1 y ( z ) + y[ −1] + 2 y ( z ) = y ( z ).
(a) For the zero-input response, assume that x[n]=0. Since we are given that y[-1]=2,
∞ 1 x( Z ) = ∑ ( ) n u[n + 5] z − n N =0 4
∞ 1 = ∑ ( )n z − n n=0 4 1 1 ,z > 1 −1 4 1− ( )z 4
(b) The unilateral z-transform is
178
x( z ) = ∑ (σ [n + 3] + σ [n] + 2n [−n]) z − n
z −1 y ( z ) + y (−1) + 2 y ( z ) = 0 ⇒ y ( z ) = −1 . 1 1 + ( ) z −1 2
Taking the inverse unilateral z-transform, Y[n]= −(− 1 ) n μ[n].
2
(b) For the zero-state response set y[-1]=0. Also, we have
x( z ) = − 2 z −1 z −1 / 4 + , (1/ 4) < z < 2 (1 − 2 z −1 ) 1 − (1/ 4) z −1
The Fourier transform exist because the ROC include the unit circle.
180
(g) Consider x [n]= (1/ 3)
But if r< function within the summation grows towards infinity with increasing n ,. Also , the summation dose not converge. But if r< 1 , then the summation converges. (a) The summation may be written as ∞ ( 1 r −1 )n cos(π n / 4)e− jwn + 0 ( 1 r )− n cos(π n / 4)e− jwn
n =−1
2 2
2
And the function within the summation grows towards infinity with increasing n ,. Also , the summation dose not converge.
1 , then the summation converges. 2 ∞ r − n + (−r ) − n − jwn jw e . by replacing z with re . If ( c )The summation may be written as ∑ 2 n=0 r> 1 , then the
IM 5n ordu3μ0 (a) IM ( e) 2nd ordu3u0 Re (f) IM
x2 ( z ) =
n =−∞ ∞
n =1 ∞
∑ x [ n]z
2
∞
−n
= ∑ (1/ 4) n z − n = ∑ (1/ 4) n +1 z − n −1 = ( z −1 / 4 ) [1/(1 − (1/ 4) z −1 )], z > 1/ 4
n=0
The z-transform of the overall sequence x [n]= x1[ n] + x2 [n] is
x1 ( z ) =
n =−∞
∑ x [ n] z
1
∞
−n
=
n =−∞ ∞
∑ (2) z
0
n −n
= ∑ (2) − n z n
n=0
= 1/(1 − (1/ 2) z ), z < 2 = −2 z −1 /(1 − 2 z −1 ), z < 2
Consider x2 [ n] = (1/ 4) n μ[ n − 1].
1 1 1 x( z ) = μ z{( ) n μ[n] = ,z > . 1 −1 2 2 1− z 2
Therefore,
1 2 y( z) = ( )( ). 1 −1 2 + z −1 1− z 4
We use partial fraction expansion followed by the inverse unilateral z-transform to obtain
∑ x[n]z
−2 n =−∞ ∞
∞
−n
=
∑ (−1/ 3)
n
z −n
−1 − n n ) z n=2 3 ∞ −1 = ∑ ( )− n−2 z n+2 n =0 3 = ∑( = 9 z 2 /(1 + 3 z ), z < 1/ 3 = 3z , z < 1/ 3 (1 + (1/ 3) −1 )
10.21.the pole –zero plots are all shown in figure S10.21. 5 (a) For x[n] = σ [n + 5], x( z ) = z , all z. The Fourier transform exists because the ROC includes the unit circle. (b) For x [n] = σ [ n − 5], X (z)= z , all z expect 0. The Fourier transform exists because the ROC includes the unit circle. n (c) For x [n]= ( −1) μ[ n],
2
X ( z) =
n =−∞
∑ x[n]z
∞
−n
179
=
n =−3
∑ (2)
∞
1
n +1
z −n
∞ 1 = ∑ ( ) n − 2 z − n +3 n =0 2
=
(e) For x [n]= ( −1) n μ[− n − 2],
3
x( z ) =
n =−∞
4 z3 1 ,z > 1 −1 2 (1 − ( ) z ) 2,来自z >1 5
10.18.
(a) using the analysis of example10.18,we may show that
H (Z ) = 1 − 6 Z −1 + 8Z −2 2 1 1 − Z −2 + −2 3 9Z