信号与系统陈后金版答案

1 k 1 k yzs [k ] = ∑ x[n]h[k − n] = u[k ] ∗ 3( ) − 2( ) u[k ] 3 2 n =−∞ ∞ 1 1 = ∑ u[n] ⋅ 3( ) k − n − 2( ) k − n u[k - n] 3 2 n =−∞ k 1 k −n 1 k −n = ∑ 3( ) − 2( ) 2 3 n=0
∴x
x
(−1)
(t) = ∫ { −τ [u(τ −1) −u(τ − 2)] + 3δ (τ −3)}dτ e
−∞
t −∞
t
(−1)
(t) = ∫
e−τ [u(τ −1) −u(τ − 2)]dτ + 3u(t −3)
0, t <1 t t −τ −τ ∫−∞ e [u(τ −1) −u(τ − 2)]dτ = ∫1 e dτ ,1< t < 2 2 e−τ dτ , t > 2 ∫1 0, t <1 t −1 −t −τ ,1 ∫−∞ e [u(τ −1) −u(τ − 2)]dτ = e −−e −2 < t < 2 e 1 −e ,t > 2
特征根为 s1 = -2, s2 = -5, 又因为 n > m , 所以： 则 h ( t ) = K 1e − 2 t u ( t ) + K 2 e − 5 t u ( t )
h '(t ) = − 2 K 1e −2 t u (t ) + K 1δ (t ) − 5 K 2 e −5 t u (t ) + K 2δ (t ) = − 2 K 1e −2 t u (t ) − 5 K 2 e −5 t u (t ) + ( K 1 + K 2 )δ (t ) h ''(t ) = 4 K 1e −2 t u (t ) − 2 K 1δ (t ) + 25 K 2 e −5 t u (t ) − 5 K 2δ (t ) + ( K 1 + K 2 )δ '(t ) 代入方程有： = K 1 + K 2 = '( t ) = 2 K 2δ ( t ) + 5 K∴K2 + (7/3; K1 )δ −1/3; 2δ '( t ) + 3δ ( t ) 1δ ( t )
x[k ] = 0.9k {u[k ] − u[k − 5]} = 0.9k ,0 ≤ k ≤ 4
2-13:(3)
x[3k ]
2 1 1
2
k
-1 0 1 2
2-13:(4)
3-2
g (t ) = r (t ) − 2r (t − 1) + r (t − 2)
Q x ( −1) (t ) = r (t ) − r (t − 1)
第二步:求差分方程的齐次 解： 2 求差分方程的齐次 第二步 h [ 0 ] = C 1 + C 2 r −5r /6 +1/6 = 0 1 k1 1 k 1 特征方程为: [ ( + 特征方程为=hCk1 ] = )[3 (C 2) ( −) 2 ( 求 ] u [ C ] = 3, C 2 = − 2 h [1] ⇒ ) 出 k1 ∴r =1/ 2, r2 =1/3 2 3 3 1 2
∫
−∞
e− jω0t [δ (t +T) −δ (t −T)]dt = e− jω0 (−T ) − e− jω0 (T )
= 2 j sin(ω0T)
2-5: (4)
x(t)
2 0 2 3 5
t
2 -1 0
x(t+1)1源自2 4t2 -3 0
x(t/3+1)
3 6 12
t
2-9:
x(t) = e−t [u(t −1) −u(t − 2)] + tδ (t −3), 求 (−1) (t), x '(t) x Qx(t) = e−t [u(t −1) −u(t − 2)] + 3δ (t −3)
−
∴ y zi (t ) = 2e −2 t − e −5t , t ≥ 0 −
3:求零状态响应： 求零状态响应： 求零状态响应
yzs (t ) = x(t ) ∗ h(t ) = ∫ x(τ )h(t − τ )dτ
−∞
∞
4:求全响应： 求全响应： 求全响应
1 −2τ 7 −5τ = ∫ [− e u (τ ) + e u (τ )]e− (t −τ )u (t − τ )dτ −∞ 3 3 ∞ 1 −2τ 7 −5τ − (t −τ ) = ∫ [− e + e ]e dτ 0 3 3 1 −t 1 −2t 7 −5t = ( e + e − e )u (t ) 4 3 12
x '(t) = e δ (t −1) − e δ (t − 2) − e [u(t −1) −u(t − 2)] + 3δ '(t −3)
−1 −2 −t
=−e−t [u(t −1) −u(t − 2)] + e−1δ (t −1) − e−2δ (t − 2) + 3δ '(t −3)
2-11:(3)
[u(t) −u(t −1)]∗[u(t − 2) −u(t −3)] = r(t − 2) − r(t −3) − r(t −3) + r(t − 4) = r(t − 2) − 2r(t −3) + r(t − 4)
3-14(2)
3-20: y ''( t ) + 7 y '( t ) + 10 y ( t ) = 2 x '( t ) + 3 x ( t ); x ( t ) = e − t u ( t ); y (0 − ) = 1, y '(0 − ) = 1 解: 求冲激响应h(t)：输入x(t ) = δ(t),有： １:求冲激响应 求冲激响应 ： h ''( t ) + 7 h '( t ) + 10 h ( t ) = 2δ '( t ) + 3δ ( t ), t ≥ 0
y[ − 2] = 4C1 + 9C 2 = 1 ⇒ C1 = − 1 / 2, C 2 = 1 / 3
1 1 k 1 1 k 则 y zi [ k ] = − ( ) + ( ) ， k ≥ 0 2 2 3 3 1 k + 1 1 k +1 = − ( ) + ( ) ，k ≥ 0 2 3
(2) (b)计算零状态响应 计算零状态响应: 计算零状态响应
x2 [k ] = 2 x1[k + 1] +
x2[k ] = 2x1[k +1] + ∑ x1[n +1]
∴ y2 [k ] = 2 y1[k + 1] +
n=−∞ n=−∞ k
n =−∞ k
n =−∞
∑ δ [ n] k
1
k
∑ y [n + 1]
1 k 1 n ∴ y2 [k ] = 2( ) u[k ] + ∑ ( ) u[n] 2 n =−∞ 2
求零输入响应： ２:求零输入响应： 求零输入响应
特征根为s1 = −2, s2 = −5; 所以： 则 y zi (t ) = K1e −2t + K 2 e −5t , t ≥ 0 −
利用初始条件,有 利用初始条件 有:
y(0− ) = K1 + K 2 = 1 y '(0 ) = −2 K1 − 5K 2 = 1 ⇒ K1 = 2, K 2 = −1
2-1: (1) x(t) = [u(t +1) −u(t)] −[u(t) −u(t −1)]
x(t)
1
2-1: (2)
x(t) = r(t +1) − r(t −1) − u(t −1) +δ (t +1)
-1
0
1
t
−1
x(t)
(1)
-1 0 0 1 t
t
x(t )
1
2-1: (7)
x(t) = e−2t [u(t) −u(t − 4)]
y (t ) = yzi (t ) + yzs (t )
∞
3-28:
x(−1) =−3 x (0) = 4 h(0) = 1 −3 4
−3 −3 −3
4
x (1) = 6 x (2) = 0 x (3) = −1 6 0 −1 6
6
h (1) = 1
h(2) = 1
0
0
−1 −1 −1
4
4
h(3) = 1
∴x(−1) (t) = (e−1 − e−t )[u(t −1) − u(t − 2)] + (e−1 − e−2 )u(t − 2) + 3u(t −3)
2-9:
x(t) = e [u(t −1) −u(t − 2)]+tδ (t −3),求x (t), x'(t)
−t (−1)
∴x'(t) = e−t [δ (t −1) −δ (t − 2)]−e−t [u(t −1) −u(t − 2)]+3δ '(t −3)
∴ g (t ) = x
( −1)
(t ) − x
( −1)
(t − 1)
根据系统积分特性:输入信号积分 输出也积分,有: 根据系统积分特性 输入信号积分,输出也积分 有 输入信号积分 输出也积分
yg (t ) = y ( −1) (t ) − y ( −1) (t − 1) = [u (t + 1) + u (t + 2)] − [u (t − 1) + u (t − 3)] = [u (t + 1) − [u (t − 1)] + [u (t − 2) − u (t − 3)]