第 第 第
第二二二二章習題章習題章習題章習題解答解答解答 解答
1. 計算以下的捲加演算,
] [ ] [ ]
[n x n hn
y = ∗
2 ) cos(
]
[n n
x = π
] 3 [ ] [ 2 ] 5 [ ]
[n =u n+ − u n +u n− h
<解答解答解答解答>
] 3 [ ] [ 2 ] 5 [ ]
[n =u n+ − u n +u n−
h h[ m− ]
-5 0 3 n 1
n m 1
] [m x
m 1
n m 1
2
)) 1 2( cos(
2 ] [ ] [ ]
[n =x n ∗h n = n+
y π
2. 計算以下的捲積演算,
) ( ) ( )
(t x t h t
y = ∗
) ( )
(t e 2u t x = − t
) 3 ( )
(t =u t+ h
<解答解答解答解答>
) 3 (
)
(t−τ =u t−τ + h
τ τ
τ τ
τ
τ h t d e τu u t d
x t
y( )= ( ) ( − ) = ∞ −2 ( ) ( − +3)
∞
−
∞
∞
−
∫
∫
0 ) 3 (
) (
3
0 2
0 − + =
=
−
<
∫
e−τu t τ dτt y t
2 1 2
) 1
| ( 2 ) 1
( 3
) 3 ( 2 )
3 ( 2 3
0 2 3 2
0
+
− +
+ −
−
+ − = − = − − = −
=
−
>
∫
t e d e t e t e tt y t
τ τ τ
3. 一個 LTI 系統的脈衝響應h(t)如下圖所示,
0
∆
∆
1
−∆1
如果輸入為x(t),請寫出其輸出y(t)。在∆→0的情形下,h(t)具有什麼功能?
<解答解答解答解答>
)}
( ) ( 1{ )
( − −∆
= ∆ t t
t
h δ δ
)}
( ) ( 1{ )}
( ) ( ) ( ) ( 1{ ) ( ) ( )
( − −∆
= ∆
∗
∆
−
−
∆ ∗
=
∗
=h t x t t x t t x t x t x t
t
y δ δ
) ( )
( )
( lim )
( (1)
0 t t
dt t d h t
g = = δ =δ
→
∆ .
dt t t dx
x t x t
y ( )
)}
( ) ( 1{ lim )
( 0 − −∆ =
= ∆
→
∆
It is a differentiator.
4. 一個 LTI 系統的脈衝響應為 h(t)=e−|at|
請導出此系統的步進響應。
<解答解答解答解答>
τ τ
τ d e τd
h t
s( ) t ( ) t −|a |
∞
−
∞
−
∫
∫
==
at t
a t a
ae ae
d e t
s t
for 1
1 | )
( ,
0 = = =
< −∞
∞
∫
− τ τ τa e a
e a
ae ae
d e d
e t
s t
for
at at
t a a
t a a
−
−
∞ −
− −
∞
−
= −
− −
= −
−
= +
=
≥
∫ ∫
2 1 0
1
1 | 1 |
) ( ,
0 0 0
0
0 τ τ τ τ τ τ
5. 一個系統的輸入x(t)與輸出y(t)如下,
x(t)=u(t)−u(t−1)
≤ <
= otherwise t t t
y 0,
1 0 ) ,
(
請問此系統是否符合因果律?是否非時變?是否具有記憶性?
<解答解答解答解答>
≤ <
=
∫
otherwise t d
t x y
t
, 0
1 0 , ) ) (
( 0 τ τ
= ≤ <
=
∫
otherwise t t d
h
t
, 0
1 0 , 1 ) ) (
( 0 δ τ τ
τ τ τ h t d x
t
y( ) t ( ) ( )
0 −
=
∫
It is a causal system.
It is not a memoryless system Let x(t) → x(t−t0)
) ( )
( ) ( )
( )
( 0 0
0 0
0t x −t h t− d =
∫
t x −t h t− d ≠ y t−t∫
τ τ τ τ τ τThe system is time-varying.
6. 一個符合因果律的離散時間系統以下式表示,
] [ ] 1 [ 6 . 0 ]
[n yn xn
y = − +
] [n
x 為輸入訊號,y[n]為輸出訊號。
(a) 請計算其單位脈衝響應h[n] (b) 若輸入訊號為
≤ ≤
= otherwise n n
x 0,
4 0
, ] 1 [
請計算其輸出訊號。
<解答解答解答解答>
(a)
] [ ] 1 [ 6 . 0 ]
[n yn xn
y = − +
1 1 0 ] 0 [ ] 1 [ 6 . 0 ] 0
[ = y − +x = + =
y
0 6 .
0 =
−
r r=0.6 0 , ) 6 . 0 ( ]
[n =c n≥
yh n
Let x[n]=u[n] 0 ]
[n =d n≥
yp
1 6 .
0 +
= d
d d =1/0.4=5/2 0 , 2 / 5 ) 6 . 0 ( ]
[n =c + n≥
y n
1 2 / 5 ) 6 . 0 ( ] 0
[ =c 0 + =
y c=1−5/2=−3/2
2 ) 5 6 . 0 2( ] 3
[n =− n + h
(b)
2) ) 5 6 . 0 2( ( 3 ]
[ ]
[ ] [ ]
[ ] [ ] [
4
0 4
0
+
−
=
−
=
−
=
∗
= −
=
=
∞
−∞
=
∑ ∑
∑
n kk k
k
k n h k
n h k x n
h n x n y
n n
n k
k n k
n k
n y
) 6 . 0 ( 4589 . 3 5 . 4 12 . 0 92224 . ) 0 6 . 0 2( 3 2 25
6 . 0 1
) 6 . 0 ( ) 1 6 . 0 2( 3 2 ) 25 6 . 0 ( ) 6 . 0 2( 3 2 ) 25 6 . 0 2( 3 2
] 25 [
4 5
0 4
0
×
−
=
−
=
−
− −
=
−
=
−
= −
=
−
=
∑
∑
7. 一個 LTI 系統之差分方程式表示如下
[ 1]
2 ] 1 [ ] 2 8 [
] 1 1 4 [
] 1
[n + y n− − y n− =x n − x n− y
此系統的初始條件為y[−1]=1,y[−2]=2,若輸入為x[n]=2u[n]時,請計算其 輸出y[n]。
<解答解答解答解答>
] 1 2 [
] 1 [ ] 2 8 [
] 1 1 4 [
] 1
[n + y n− − y n− =x n − x n− y
先求系統方程式的同質解,
8 0 1 4
2 + 1r− =
r
2 , 1
4 1
2 1
= −
= r
r
0 , 2 ) ( 1 4)
(1 ]
[n =c1 +c2 − n≥
yh n n
再求系統方程式的特別解,
] [ 2 ] [n u n
x =
0 ]
[n =d n≥
yp
1 22 2 1 8 1 4
1 − = − =
+ d d
d
9
= 8 d
最後求系統方程式的完全解,
0 9,
) 8 2 ( 1 4)
(1 ]
[n =c1 +c2 − + n≥
y n n
1 ] 1 [− =
y y[−2]=2 2 ] 2 8 [ ] 1 1 4 [ ] 1 0
[ + y − − y − =
y 2
8 2 4 ] 1 0
[ + − =
y y[0]=2
1 22 2 1 ] 1 8 [ ] 1 0 4 [ ] 1 1
[ + y − y − = − =
y 1
8 2 1 4 ] 1 1
[ + − =
y
8 ] 5 1 [ = y
9 2 ] 8
0
[ =c1+c2 + =
y
9 10
2
1 +c =
c
8 5 9 ) 8 2 ( 1 4) (1 ] 1
[ =c1 +c2 − + =
y
72 ) 19 2 (1 4) (1 2
1 −c =
c
54 59
1 =
c
54 1
2 = c
0 9,
) 8 2 ( 1 54 ) 1 4 (1 54 ] 59
[n = + − + n≥
y n n
8. 一個 LTI 系統的微分方程式為 2 ( ) 3 ( ) 2 ( ) ( )
2
t x t y t dt y t d dt y
d + + =
其初始條件為
5 ) 3 0 ( − =−
y , 與
5 2 ) 0
( =
= − t t
dt y d
若輸入為x(t)=cos(t)u(t),請計算其輸出結果。
<解答解答解答解答>
) ( ) ( 2 ) ( 3 )
( 2
2
t x t dt y t d dt y t d
y + + =
先求系統方程式的同質解,
0 2 3
1+ r+ r2 = 2 1
, 1 2
1 =− r =−
r
0 , )
( 2
1
2
1 + >
=c e− c e− t t
yh t t
再求系統方程式的特別解,
) ( ) cos(
)
(t t u t
x =
0 ),
sin(
) cos(
)
(t =d1 t +d2 t t>
yp
) cos(
)) sin(
) cos(
( 2
)) cos(
) sin(
( 3 ) sin(
) cos(
2 1
2 1
2 1
t t
d t d
t d t d t
d t d
=
−
− +
+
− + +
1 3 2
1+ =
−d d −3d1−d2 =0
10 3
10 1
˙
2
1 = − d =
d
最後求系統方程式的完全解,
0 ),
10sin(
) 3 10cos(
) 1
( 2
1
2
1 + − + >
=ce− c e− t t t
t
y t t
0 ),
10cos(
) 3 10sin(
1 2
) 1
( 2
1
2
1 − + + >
−
= ce− c e− t t t
t dt y
d t t
5 ) 3 0 ( + =−
y
5 2 ) 0
( =
= + t t
dt y d
5 3 10
1
2 1
= −
− +c
c
5 2 10
3 2
1
2
1 − + =
−c c 1
2
2c1+ c2 =− −10c1 −5c2 =1
5 4 10
3
2 1
= −
= c
c
0 ),
10sin(
) 3 10cos(
1 5
4 10
) 3
( 2
1
>
+
−
−
= e− e− t t t
t
y t t
9. 一個 LTI 系統的微分方程式如下
) ( 3 ) ( 2 ) ( 3 ) ( 2 )
2 (
2
t dtx t d x t y t dt y t d dt y
d + + = +
請繪其直接型第一式與第二式的方塊圖。
<解答解答解答解答>
) ( 3 ) ( 2 ) ( ) ( 2 ) (
3y(−2) t + y(−1) t + y t = x(−2) t + x(−1) t 直接型第一式
) (t
x y(t)
3
2
− 2
− 3
∫
∫
∫
∫
直接型第二式 )
(t
x y(t)
∫
∫
− 2
− 3
3
2
10. 驗證以下的捲積加法演算具有互換性,
] [ ] [ ]
[n x n y n
w = ∗
<解答解答解答解答>
] [ ] [ ]
[n x k y n k
w
k
−
=
∑
∞−∞
=
令n−k =m,則
] [ ] [ ] [ ] [ ]
[ ] [ ]
[n x n m y m y m x n m y n x n
w
m m
∗
=
−
=
−
=
∑ ∑
∞−∞
=
−∞
∞
=