2024年4月7日发(作者:郑笑雯)
第二章
2-1
解:
(1):
F(S)L[(4t)
(t)]L[5
(t)]L[t1(t)]L[21(t)]
05
(2):
F(S)
1212
5
S
2
S
S
2
S
3s5
2
2(s25)
1e
s
(3):
F(S)
2
s1
(4):
F(S)L{[4cos2(t
s
6
)]1(t)e
5t
1(t)}
66
s
6
4Se14Se1
s
2
2
2
S5
s
2
4
S5
e
2s
e
2s
6
(5):
F(S)006
SS
(6):
F(S)L[6cos(3t4590)1(t
S
4
4
)]
S
4
6Se6Se
L[6cos3(t)1(t)]
2
44
S3
2
S
2
9
(7):
F(S)L[e
6t
cos8t1(t)0.25e
6t
sin8t1(t)]
S62S8
22222
(S6)8(S6)8S12S100
(8):
F(S)2
2-2
解:
(1):
f(t)L
1
(
259e
s20
(s20)
2
s
2
9
s
6
12
)(e
2t
2e
3t
)1(t)
S2S3
1
(2):
f(t)sin2t1(t)
2
1
(3):
f(t)e
t
(cos2tsin2t)1(t)
2
e
s
)e
t1
1(t1)
(4):
f(t)L(
S1
1
(5):
f(t)(te
t
2e
t
2e
2t
)1(t)
81515
t
(6):
f(t)L
1
(
152
)
815
e
2
sin
15
t1(t)
(S
1
2
)
2
(
15
2
152
2
)
(7):
f(t)(cos3t
1
3
sin3t)1(t)
2-3
解:
(1) 对原方程取拉氏变换,得:
S
2
X(S)Sx(0)x
(0)6[SX(S)x(0)]8X(S)
1
S
将初始条件代入,得:
S
2
X(S)S6SX(S)68X(S)
1
S
(S
2
6S8)X(S)
1
S
S6
1
7
7
X(S)
S
2
6S1
S(S
2
6S8)
8
S
4
S2
8
S4
取拉氏反变换,得:
x(t)
1
7
4
e
2t
7
8
e
4t
8
(2) 当t=0时,将初始条件
x
(0)50
代入方程,得:
50+100x(0)=300
则x(0)=2.5
对原方程取拉氏变换,得:
sx(s)-x(0)+100x(s)=300/s
将x(0)=2.5代入,得:
SX(S)-2.5100X(S)
300
S
X(S)
2.5S30030.
S(S100)
s
5
s100
取拉氏反变换,得:
x(t)3-0.5e
-100t
2-4
解:该曲线表示的函数为:
u(t)61(t0.0002)
则其拉氏变换为:
6e
0.0002s
U(s)
s
- 1 -
2024年4月7日发(作者:郑笑雯)
第二章
2-1
解:
(1):
F(S)L[(4t)
(t)]L[5
(t)]L[t1(t)]L[21(t)]
05
(2):
F(S)
1212
5
S
2
S
S
2
S
3s5
2
2(s25)
1e
s
(3):
F(S)
2
s1
(4):
F(S)L{[4cos2(t
s
6
)]1(t)e
5t
1(t)}
66
s
6
4Se14Se1
s
2
2
2
S5
s
2
4
S5
e
2s
e
2s
6
(5):
F(S)006
SS
(6):
F(S)L[6cos(3t4590)1(t
S
4
4
)]
S
4
6Se6Se
L[6cos3(t)1(t)]
2
44
S3
2
S
2
9
(7):
F(S)L[e
6t
cos8t1(t)0.25e
6t
sin8t1(t)]
S62S8
22222
(S6)8(S6)8S12S100
(8):
F(S)2
2-2
解:
(1):
f(t)L
1
(
259e
s20
(s20)
2
s
2
9
s
6
12
)(e
2t
2e
3t
)1(t)
S2S3
1
(2):
f(t)sin2t1(t)
2
1
(3):
f(t)e
t
(cos2tsin2t)1(t)
2
e
s
)e
t1
1(t1)
(4):
f(t)L(
S1
1
(5):
f(t)(te
t
2e
t
2e
2t
)1(t)
81515
t
(6):
f(t)L
1
(
152
)
815
e
2
sin
15
t1(t)
(S
1
2
)
2
(
15
2
152
2
)
(7):
f(t)(cos3t
1
3
sin3t)1(t)
2-3
解:
(1) 对原方程取拉氏变换,得:
S
2
X(S)Sx(0)x
(0)6[SX(S)x(0)]8X(S)
1
S
将初始条件代入,得:
S
2
X(S)S6SX(S)68X(S)
1
S
(S
2
6S8)X(S)
1
S
S6
1
7
7
X(S)
S
2
6S1
S(S
2
6S8)
8
S
4
S2
8
S4
取拉氏反变换,得:
x(t)
1
7
4
e
2t
7
8
e
4t
8
(2) 当t=0时,将初始条件
x
(0)50
代入方程,得:
50+100x(0)=300
则x(0)=2.5
对原方程取拉氏变换,得:
sx(s)-x(0)+100x(s)=300/s
将x(0)=2.5代入,得:
SX(S)-2.5100X(S)
300
S
X(S)
2.5S30030.
S(S100)
s
5
s100
取拉氏反变换,得:
x(t)3-0.5e
-100t
2-4
解:该曲线表示的函数为:
u(t)61(t0.0002)
则其拉氏变换为:
6e
0.0002s
U(s)
s
- 1 -