Problems 13–61
10 ≤t≤40:
[c] The expressions are
0≤t≤1: y(t)=Zt
0400 dλ= 400λ
t
0
= 400t;
13–62 CHAPTER 13. The Laplace Transform in Circuit Analysis
[d]
[e] Yes, note that h(t) is approaching 40δ(t), therefore y(t) must approach
40x(t), i.e.
P 13.64 [a] h(λ)=2
5λ0≤λ≤5;
0≤t≤5:
5≤t≤10:
vo= 10 Z5
2
Problems 13–63
10 ≤t≤∞:
2
[b]
[c] Area = 1
2(10)(2) = 10 ·
.. 1
2(4)h= 10 so h= 5;
13–64 CHAPTER 13. The Laplace Transform in Circuit Analysis
0≤t≤2:
2≤t≤4:
vo= 10 Z2
0
5
2λdλ+10Zt
2✓10 −5
2λ◆dλ
4≤t≤∞:
vo= 10 Z2
0
5
2λdλ+10Z4
2✓10 −5
2λ◆dλ
P 13.65 [a]
Problems 13–65
[b]
y(t)=0 t<0;
0≤t≤10 : y(t)=Zt
13–66 CHAPTER 13. The Laplace Transform in Circuit Analysis
[c]
y(t)=0 t<0;
0≤t≤1: y(t)=Zt
0625 dλ= 625t;
P 13.66 [a] From Problem 13.51(a)
H(s)= 200
s+ 200.
h(λ) = 200e−200λ.
[b]
P 13.67 [a] H(s)= 2000
s+ 2000
·
.. h(λ) = 2000e−2000λ.
0≤t≤5 ms:
P 13.68 [a] −1≤t≤4:
vo=Zt+1
t+1
=5t2+10t+ 5 V;
13–68 CHAPTER 13. The Laplace Transform in Circuit Analysis
=5λ2
10
t−4
+100λ
t+1
10
=−5t2+ 140t−480 V;
14 ≤t≤19:
24 ≤t≤29:
vo= 10 Zt+1
t−4(30 −λ)dλ= 300λ
t+1
t−4
−5λ2
t+1
t−4
Summary:
vo=0 −∞≤t≤−1;
vo=5t2+10t+5V −1≤t≤4;
Problems 13–69
[b]
P 13.69 H(s)= 16s
40 + 4s+16s=0.8s
s+2 =0.8✓1−2
s+2◆=0.8−1.6
s+2;
P 13.70 [a] H(s)=Vo
Vi
=1/LC
s2+(R/L)s+ (1/LC)
13–70 CHAPTER 13. The Laplace Transform in Circuit Analysis
0≤t≤0.5:
0.5≤t≤∞:
[b]
P 13.71 [a] Vo=16
20Vg;
Problems 13–71
[b]
0.5s≤t≤1.0 s:
[c]
P 13.72 [a]
13–72 CHAPTER 13. The Laplace Transform in Circuit Analysis
[b]
0≤t≤0.5 s;
0.5s≤t≤1 s:
vo=Zt−0.5
0(−75)(0.8e−λ)dλ+Zt
t−0.575(0.8e−λ)dλ
Problems 13–73
1s≤t≤∞:
[c]
[d] No, the circuit has memory because of the capacitive storage element.
P 13.73 [a]
vo=Zt
010(10e−4λ)dλ
13–74 CHAPTER 13. The Laplace Transform in Circuit Analysis
[b]
0≤t≤0.5:
0.5≤t≤∞:
[c]
P 13.74 [a] Ig=Vo
105+Vos
5×106=Vo(s+ 50)
5×106;
Problems 13–75
0.1s≤t≤0.2 s:
vo=Zt−0.1
0(−50 ×10−6)(5 ×106e−50λdλ)
0.2s≤t≤∞:
vo=Zt−0.1
t−0.2
−250e−50λdλ+Zt
t−0.1250e−50λdλ
[b] Io=Vos
5×106=s
5×106
·
5×106Ig
s+50 ;
0<t<0.1 s:
Problems 13–77
0.1s<t<0.2 s:
io=Zt−0.1
0(−50 ×10−6)[δ(λ)−50e−50λ]dλ
0.2s<t<∞:
io=Zt−0.1
t−0.2(−50 ×10−6)(−50e−50λ)dλ
[c] At t=0.1−:
From the solution for iowe have
io(0.1−)=50e−5=0.34 µA.(checks)
At t=0.1+:
Problems 13–79
From the solution for io,
From the solution for io,
P 13.75 H(s)=Vo
Vi
=5
5+2.5s=2
s+2;
(π/2) s ≤t≤∞:
13–80 CHAPTER 13. The Laplace Transform in Circuit Analysis
= 40e−2t“e2λ
= 10e−2t[eπ(sin π−cos π)−1(0 −1)]
P 13.76
Vo
Ig
=H(s)= 4000s
s+ 100;