例题5

转移导纳函数
例3、试求图示电路的冲激 响应 u 1(t )。
2
+
L 2H
R1
SL
R1
+
2 R2 Is (s) U 1( s)
i (s )
u1( s)
1F
C
1 CS
R2
解：H ( s ) =
R( s ) U 1( s) = — —策动点阻抗 E ( s) Is ( s ) 1 ( LS + R 2) CS = 2 + S + 1 H ( s ) = R1 + 1 1 LS + R 2 + S2 + S + CS 2 1 1 1 2 令D ( s ) = S + S + S 1、 = ± j 2 2 2 2 k1 k2 + H ( s) = 2 + S S1 S S 2
联立求解得 R1CSUs ( s ) H ( s )与Us ( s )无关， 2 R1LCS + ( R1R 2C + L) S + R1 + R 2 由网络结构和参数决定 I 2( s ) R1CS ∴ H ( s) = = U 2( s ) R1LCS 2 + ( R1R 2C + L) S + R1 + R 2 I 2( s ) =
1
β
+ β
2
β
(cos β t )′]
s β = (s 2 1) = 2 2 β s +β s +β2
例5、求图示RL串联电路的电流响应。
2H
i (t )
3ε (t)V
+
解：列回路方程 2
di + 4i = 3ε (t ) dt
4 方程两边取拉氏变换 2[ sI ( s) i (0)] + 4 I ( s) = 3 s 3 3 + 10s i (0 ) = 5 A 整理 I ( s )(2s + 4) = + 10 = s s
2
s+4 5 2t 1 t ∴L [ 2 ] = 3e e 2 s + 5s + 3 2 se 2 s 例2、求L1[ 2 ] s + 2s + 5 解：令D( s ) = s 2 + 2 s + 5 = 0
3
t≥0
θ = tg 1
S 1、 = 1 ± jα 2
1 2 可以证明 k 2 = k 1 = 5 jθ e 4
例2：试求图示电路的网络函数H ( s )，设I 2( s )为响应象函数， 电路为单电源作用，初始条件为零。
I 1( s )
R 2 I 2( s ) R1
解：列回路方程 回路1： (
+ Us (s )
1 sc
1
2
1 + R1) I 1( s) R1I 2( s ) = Us ( s) (1) sc SL 回路2： R1I 1( s) + ( R1 + R 2 + SL) I 2( s ) = 0 (2)
T
L[ f (t )] =
t
E 1 E L[(t T )ε (t T ) + Tε (t T )] 2 T s T E 1 E st 1 1 = e E e st T s2 T s2 s E = 2 [1 (Ts + 1)e st ] Ts
注意 f (t )ε (t t 0) ≠ f (t t 0)ε (t t 0)
s k1 k2 = + s 2 + 2 s + 5 s s1 s s 2 N (s) s 1 + j2 5 jθ k1 = = = = e D′( s ) s = s1 2 s + 2 s = 1+ j 2 2 + 4 j + 2 4
s 5 t L1[ 2 e cos(2t + θ ) ] = 2× s + 2s + 5 4 = 2 k 1 eαt cos(ωt + θ )
f (0+ ) = lim sF 0( s ) = 5
s →∞ 3 2
s + s + 2s + 1 f (∞) = lim s =0 s →0 ( s + 1)( s + 2)( s + 3) (3) f (0+ ) = lim sF ( s ) = 1
s →∞
F ( s ) 不为真分式， 先化为真分式
1 2s
+
S
6 s
+
+
U 0( s )
3V
2
=
+
1 2s 6 s
S
+ 3V
U 0 zi ( s )
2
+
(b)
S
U 0 zs ( s )
(a )
1
+
6 s
1 2s
2
(c)
(4)求响应象函数 图(b)列节点方程 1 1 3 ( + 2s + )U 0 zi ( s ) = 12 1 s+2 s+2 12 s + 21 9 3 解得：U 0 zi ( s ) = = ( s + 1)(2 s + 3) s + 1 s + 3 2 图(c)列节点方程 1 6 (1 + 2 s + )U 0 zs ( s ) = s+2 s 6( s + 2) 4 6 2 解得：U 0 zs ( s ) = = + s ( s + 1)(2 s + 3) s s + 1 s + 3 2
t≥0
s 3 + 6s 2 + 15s + 11 例4、求L1[ ] 2 s + 5s + 6 解 : ∵ n < m 不能直接用部分分式法。 将F ( s )中分子多项式与分母多项式相除，得 4s + 5 3 7 F (s) = s + 1 + 2 = s +1+ + s + 5s + 6 s+2 s+3
δ n (t ) s n
1 ε (t ) 例6、求L[tε (t )] s t 1 解：L[tε (t )] = L[ ∫ ε (t )dt ] = 2 0 s
例8、试求下列F ( s )所对应的f (t )的初值和终值。
s 2 + 2s + 1 (1) F ( s ) = ( s 1)( s + 2)( s + 3) s 2 + 2s + 3 (3) F ( s ) = ( s + 1)( s 2 + w0 2 )
图(a )列节点方程 1 6 3 )U 0( s ) = + 12 s+2 s s+2 12 s 2 + 27 s + 12 4 3 1 解得：U 0( s ) = = + s ( s + 1)(2 s + 3) s s + 1 s + 3 2 (5)求拉氏逆变换，得 (1 + 2 s +
u 0 zi (t ) = L1[U 0 zi ( s )] = [9e t 3e 1.5t ]ε (t ) u 0 zi (t ) = L1[U 0 zs ( s)] = [4 6e t + 2e 1.5t ]ε (t ) 全响应 = u 0(t ) = u 0 zi (t ) + u 0 zs (t ) = [4 + 3e t e 1.5t ]ε (t )
由时域平移性质
s+2 例3、求L [ ] 2 s ( s + 1) ( s + 3)
1
解：D ( s ) = 0, 得s1 = 0, s 2 = 3, s 3、 = 1, 4 （重根）
s+2 k1 k2 k 31 k 32 = + + + 2 2 s ( s + 1) ( s + 3 ) s s + 3 ( s + 1) ( s + 1) s+2 k1 = ( s + 1) 2 ( s + 3 ) k2 =
例 3、求 L [ e α t cos β t ] 解 L[e
α t
s +α cos β t ] = (s + α )2 + β
2
同理由 sin β t 得 L[e
α t
β
s2 + β
2
sin β t ] =
例 4、求 L[sin β t ] 解 : L[sin β t ] = L[ 1
(s + α )2
例1、求 L [ ε ( t )] 解： L [ ε ( t )] =
∫
∞
0
ε (t ) e
st
dt = 1 = s
∫
∞
0
e st dt
1 st = e s
∞
0
例 2、求 L [δ ( t )] 解： L [δ ( t )] =
∫
∞
0
δ ( t ) e st dt = 1
例 3、求 L [ e + α t ε ( t )]，并讨论收敛域。 解： L [ e
k 31 =
s=0
2 = 3
s+2 s ( s + 1) 2
=
s = 3
1 12
s+2 1 × ( s + 1) 2 = s = 1 s ( s + 1) 2 ( s + 3) 2