文档介绍：第十章线性电路的复频域分析
习题解答
10-1 求下列各函数的象函数
⑴ sin(ωt+θ) ⑵ cos(ωt+θ)
⑶ e–2 tcos(3t) ⑷ e–2 tsin(4t)
⑸ tcos(ωt) ⑹ te– tsin(2t)
⑺ 10ε(t-2)+2δ(t-1) ⑻(t+4)ε(t)+2e– tε(t-4)
解根据拉普拉斯变换的性质可得
⑴ L[sin(ωt+θ)]= L[sin(ωt)cosθ+cos(ωt)sinθ]
=
=
⑵ L[cos(ωt+θ)] = L[cos(ωt)cosθ– sin(ωt)sinθ]
=
⑶ L[e –2 tcos(3t)] =
⑷ L[e–2 tsin(4t)] =
⑸ L[t cos(ωt)] = L
= =
⑹先求出
L[t sin(2t)]= L
= =
则
L[t e– tsin(2t)]=
⑺ L[10ε(t-2)+2δ(t-1)]=
⑻ L[(t+4)ε(t)+2e– tε(t-4)]=
=
10-2 用两种不同的方法求的拉普拉斯变换。
解方法一:利用题10-1中第(5)小题的结论
L[t cos(ωt)] =
可求出
L[te– tcost] =
再利用拉氏变换的导数性质可求出
L= s– f (0–)
=
方法二:将改写为
f (t) = (te– tcos t)
= cost + te– t(–sint)
= (e– t–t e– t) cost–te– t sint
= e– tcost – t e– tcost – t e– tsint
利用题10-1第⑸、⑹小题的结论,可求出
L[f (t)] = L[e– tcost– t e– tcost–t e– tsint]
= ––
=
10-3 计算图示函数的拉普拉斯变换。
1
2
0
5
f (t)
t
题10-3图
解由图示波形可得出
5t 0≤t≤1
f (t) =
–5t+10 1≤t≤2
或写成
f (t) = 5t [ε(t) –ε(t–1)] + ( –5t + 10)[ ε(t –1) –ε(t –2)]
= 5t ε(t) –5t ε(t–1)–5t ε(t –1) + 10ε(t–1) + 5t ε(t-2)–10ε(t –2)
= 5t ε(t) –10t ε(t–1) + 10ε(t –1) + 5t ε(t-2)–10ε(t –2)
= 5t ε(t) –10(t –1)ε(t –1) + 5(t–2)ε(t –2)
则
L[f (t)] = L[5t ε(t)–10(t –1)ε(t –1) + 5(t–2)ε(t –2)]
=
或用定义式求出
L[f (t)] =
=
=
10-4 计算图示函数的拉普拉斯变换。
1
2
0
10
f (t)
t
题10-4图
-10
解先写出f (t)的时域表达式
10 0≤t≤1
f (t) =
–10 1≤t≤2
或写成
f (t) = 10[ε(t)–ε(t – 1)] –10[ε(t – 1)–ε(t – 2)]
则
L[f (t)]= L[10ε(t) –20ε(t – 1) +10ε(t – 2)]
=
或
L[f (t)] ==
=
=
10-5 求下列象函数的原函数
⑴ F(s) = ⑵ F(s) =
⑶ F(s) = ⑷ F(s) =
解应用部分分式展开法可求出
⑴ L–1[F(s)] = = 1+2e– t
⑵ L–1[F(s)] = L–1 = L–1
= 2e– t – 2e–3 t
⑶ L–1[F(s)]= L–1 = L–1 = 3δ(t) –11e–4 tε(t)
⑷ F(s) =
其中
K11 =
K12 =
K2=
则有
[F(s)] =
10-6 对每个F(s),求其f (t):
⑴ F(s) = ⑵ F(s) =
⑶ F(s) = ⑷ F(s) =
解 ⑴ F(s) =
其中
K11 =
K12=
K13=
K 2 =
则有
(2)
其中
则有
(3)
其中
则有
⑷
10-7 用拉普拉斯变换法求图(a)所示电路中的i(t)、和。
(a)
Ω
1Ω
1H
Ω
1F
Ω
ε(t)V
Ω
i(t)
Ω
uC(t)
Ω
+
-
+
-
-
+
uL(t)