信号与系统陈后金版答案

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)]
y g (t )
1 0 1 2 3
∵h[1] = 0, h[2] = 0
2 3 h 等 效 初 始 条 件 1]/6 代 入 [1] = 5h[0]/6 h[: +δ[1] = 5/6
∴h[0]C 1 5h[)1]/6 (h12]/6k ]δ[0] =1 = ( 1 k + C 2 [ ) k ]u [ + h[ k ] = [
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
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
6
0
↓ ∴ y [ k ] = 3,1, 7 , 7 , 9 , 5 , 1, 1
0
4
t
来自百度文库
2-2: (2)
x(t) = sin(πt / 2)u(t 1)
0
1
1
…
2-3: (2)
t
(t + 2t + 3)δ (t 2) = (8 +8 + 3)δ (t 2) =19δ (t 2)
3 2
2-4: (4)
∫
2-4: (5) ∞
∞
∞
sin(t)δ (t π / 4)dt = sin(t)|t=π /4
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 )
∫
∞
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 2 4
t
2 -3 0
x(t/3+1)
3 6 12
t
2-9:
x(t) = et [u(t 1) u(t 2)] + tδ (t 3), 求 (1) (t), x '(t) x ∵x(t) = et [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
3-31:
5 1 y[k ] y[k 1] + y[k 2] = x[k ], y(1) = 0, y(2) = 1, 6 6 x[k ] = u[k ]
根据单位脉冲响应的定义,应满足方程 解: (1) 根据单位脉冲响应的定义 应满足方程: 应满足方程 5 1 h[k ] h[k 1] + h[k 2] = δ [k ] 6 6 第一步:求等效初始条件 第一步 求等效初始条件 :
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
t
3-4 已知离散时间 系统,输入 x1[k ] = δ [k 1] 时,输出 已知离散时间LTI系统 输入 输出; 系统 输出
1 k 1 y1[k ] = ( ) u[k 1], 求当输入x2 [k ] = 2δ [k ] + u[k ]时系统响应y2 [k ]. 2
x2 [k ] = 2 x1[k + 1] +
(3) 计算固有响应与强迫响应 计算固有响应与强迫响应:
1 7 1 k 4 1 k y[k ] = [ ( ) + ( ) ]u[k ] 完全响应: 完全响应 2 2 2 3 3 7 1 k 4 1 k 固有响应: yh [k ] = [ ( ) + ( ) ]u[ k ] 固有响应 2 2 3 3 1 强迫响应: 强迫响应 y p [k ] = u[k ] 2 (4) 计算瞬态响应与稳态响应 计算瞬态响应与稳态响应:
t
x1(t)
A -T0/2 A -T0/2 T0/2 T0
x '(t) = e δ (t 1) e δ (t 2) e [u(t 1) u(t 2)] + 3δ '(t 3)
1 2 t
=et [u(t 1) u(t 2)] + e1δ (t 1) e2δ (t 2) + 3δ '(t 3)
2-11:(3)
x[k ] = 0.9k {u[k ] u[k 5]} = 0.9k ,0 ≤ k ≤ 4
∴x(1) (t) = (e1 et )[u(t 1) u(t 2)] + (e1 e2 )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) = et [δ (t 1) δ (t 2)]et [u(t 1) u(t 2)]+3δ '(t 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
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
∞
1 k n k 1 k n = ∑ 3( ) ∑ 2( ) 2 3 n =0 n=0 1 k 1 k = [ 3 3( ) + ( ) ]u[k ] 2 3
完全响应: 完全响应
k
y[k ] = y zi [k ] + yzs [k ] 1 7 1 k 4 1 k = [ ( ) + ( ) ]u[k ] 2 2 2 3 3
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) = e2t [u(t) u(t 4)]
k 1 k 1 n 1 k ∴ y 2 [ k ] = 2( ) u [ k ] + ∑ ( ) =[2+( ) ]u [ k ] 2 2 n=0 2
3-14(1)
[δ (t +1) + 2δ (t 1)][δ (t 1) δ (t 3)] = δ (t) + 2δ (t 2) δ (t 2) 2δ (t 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)
∵ x ( 1) (t ) = r (t ) r (t 1)
∴ 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
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
(2) :(a)计算零输入响应 计算零输入响应: 计算零输入响应 特征方程为: 特征方程为
r2 5r /6 +1/ 6 = 0 ∴r =1/ 2, r2 =1/3 1 1 k 1 k 则 y zi [ k ] = C1 ( ) + C 2 ( ) , k ≥ 0 2 3 代入初始条件,有: 代入初始条件 有 y[ 1] = 2C1 + 3C 2 = 0
瞬态响应: 瞬态响应 稳态响应: 稳态响应
7 1 k 4 1 k yt [k ] = [ ( ) + ( ) ]u[k ] 2 2 3 3 1 ys [k ] = u[k ] 2
第四章
4-5(d)波形如图 波形如图: 波形如图 x(t)
A -T0/2 T0/2 T0 -A
x(t) = x1(t) + x2 (t)
∴x
x
(1)
(t) = ∫ { τ [u(τ 1) u(τ 2)] + 3δ (τ 3)}dτ e
∞
t ∞
t
(1)
(t) = ∫
eτ [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 eτ 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
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