Solution manual fundamentals of electric circuits 3rd edition chapter15

Chapter 15, Problem 1.
Find the Laplace transform of:
(a) cosh at

(b) sinh at

(

)

(

)

1 x
1
e + e − x , sinh x = e x − e − x .]
2
2

[Hint: cosh x =

Chapter 15, Solution 1.

(a)

(b)

e at + e - at
cosh(at ) =
2

1⎡ 1
1 ⎤
s
L [ cosh(at ) ] = ⎢
+
=
2 ⎣ s − a s + a ⎥⎦ s 2 − a 2
e at − e - at
sinh(at ) =
2
a
1⎡ 1
1 ⎤
L [ sinh(at ) ] = ⎢
−
= 2
⎥
2 ⎣ s − a s + a ⎦ s − a2

Chapter 15, Problem 2.

Determine the Laplace transform of:
(a) cos( ωt + θ )

(b) sin( ωt + θ )

Chapter 15, Solution 2.

(a)

f ( t ) = cos(ωt ) cos(θ) − sin(ωt ) sin(θ)
F(s) = cos(θ) L [ cos(ωt ) ] − sin(θ) L [ sin(ωt ) ]
s cos(θ) − ω sin(θ)
F(s) =
s 2 + ω2

(b)

f ( t ) = sin(ωt ) cos(θ) + cos(ωt ) sin(θ)
F(s) = sin(θ) L [ cos(ωt ) ] + cos(θ) L [ sin(ωt ) ]
s sin(θ) − ω cos(θ)
F(s) =
s 2 + ω2

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Chapter 15, Problem 3.

Obtain the Laplace transform of each of the following functions:
(a) e −2t cos 3tu (t )

(c) e −3t cosh 2tu (t )
(e) te − t sin 2tu (t )

(b) e −2t sin 4tu (t )

(d) e −4t sinh tu (t )

Chapter 15, Solution 3.

(a)

L [ e -2t cos(3t ) u ( t ) ] =

s+2
(s + 2 ) 2 + 9

(b)

L [ e -2t sin(4 t ) u ( t ) ] =

4
(s + 2) 2 + 16

(c)

Since L [ cosh(at ) ] =

(d)

Since L [ sinh(at ) ] =

(e)

L [ e - t sin( 2t ) ] =

s
s − a2
s+3
L [ e -3t cosh(2 t ) u ( t ) ] =
(s + 3 ) 2 − 4
2

a
s − a2
1
L [ e -4t sinh( t ) u ( t ) ] =
(s + 4) 2 − 1
2

2
(s + 1) 2 + 4

f (t) ←
⎯→ F(s)
-d
F(s)
t f (t) ←
⎯→
ds
-d
-1
2 ( (s + 1) 2 + 4)
Thus, L [ t e - t sin(2 t ) ] =
ds
2

=
⋅ 2 (s + 1)
((s + 1) 2 + 4) 2
4 (s + 1)
L [ t e -t sin( 2t ) ] =
((s + 1) 2 + 4) 2
If

[

]

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Chapter 15, Problem 4.
Find the Laplace transforms of the following:
(a) g (t ) = 6 cos(4t − 1)
(b) f (t ) = 2tu (t ) + 5e −3(t − 2 )u (t − 2 )

Chapter 15, Solution 4.
s

e −s =

(a)

G (s) = 6

(b)

e −2s
F(s) =
+5
s+3
s2

s2 + 42

6se −s
s 2 + 16

2

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Chapter 15, Problem 5.
Find the Laplace transform of each of the following functions:
(b) 3t 4 e −2t u (t )
(a) t 2 cos(2t + 30°)u (t )
d

(c) 2tu (t ) − 4 δ (t )
(d) 2e − (t −1)u (t )
dt
(f) 6e − t 3 u (t )
(e) 5u (t 2 )
dn
(g) n δ (t )
dt

Chapter 15, Solution 5.
(a)

s cos(30°) − 2 sin(30°)
s2 + 4
d 2 ⎡ s cos(30°) − 1 ⎤
L [ t 2 cos(2t + 30°) ] = 2 ⎢
ds ⎣ s 2 + 4 ⎥⎦
L [ cos(2t + 30°) ] =

=

⎞
-1 ⎤
d d ⎡⎛ 3
s − 1⎟⎟ (s 2 + 4) ⎥
⎢⎜⎜
ds ds ⎢⎣⎝ 2
⎥⎦
⎠

=

⎛ 3
⎞
-1
-2 ⎤
d ⎡ 3 2
s − 1⎟⎟ (s 2 + 4) ⎥
⎢ (s + 4) − 2s ⎜⎜
ds ⎣⎢ 2
⎝ 2
⎠
⎦⎥

⎛ 3
⎞
⎛ 3⎞
⎛ 3
⎞
2
3
(- 2s ) 2 ⎜⎜ 2 s − 1⎟⎟ 2s ⎜⎜ 2 ⎟⎟ (8s ) ⎜⎜ 2 s − 1⎟⎟
⎠− ⎝
⎠+
⎝
⎠
= 2
− ⎝
3
2

2
2
2
2
2
2
s +4
s +4
s +4
s +4

(

)

(

)

(

)

(

)

⎛ 3
⎞
(8s 2 ) ⎜⎜

s − 1⎟⎟
- 3s − 3s + 2 − 3s
⎝ 2
⎠
=
+
2
3
2
2
s +4
s +4

(

=

)

(-3 3 s + 2)(s 2 + 4)

(s

2

+4

L [ t 2 cos(2t + 30°) ] =

)

3

(

+

)

4 3 s3 − 8 s 2

(s

2

+4

)

3

8 − 12 3 s − 6s 2 + 3s 3

( s 2 + 4) 3

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[

]

4!

72

(b)

L 3 t 4 e - 2t = 3 ⋅

(c)

⎡
⎤ 2
2
d
L ⎢ 2t u ( t ) − 4 δ( t ) ⎥ = 2 − 4(s ⋅ 1 − 0) = 2 − 4s
⎣
⎦ s
s
dt

(s + 2) 5

=

(s + 2) 5

(d)

2 e -(t-1) u ( t ) = 2 e -t u ( t )
2e
L [ 2 e -(t-1) u ( t ) ] =
s+1

(e)

Using the scaling property,
1
1
1
5
L [ 5 u ( t 2) ] = 5 ⋅
⋅
= 5⋅ 2⋅ =
1 2 s (1 2)
2s s

(f)
(g)

L [ 6 e -t 3 u ( t ) ] =

6
18
=

s + 1 3 3s + 1

Let f ( t ) = δ( t ) . Then, F(s) = 1 .
⎡ dn
⎤
⎡ dn
⎤
L ⎢ n δ( t ) ⎥ = L ⎢ n f ( t ) ⎥ = s n F(s) − s n −1 f (0) − s n − 2 f ′(0) − L
⎣ dt
⎦
⎣ dt
⎦

⎡ dn
⎤
⎡ dn
⎤
L ⎢ n δ( t ) ⎥ = L ⎢ n f ( t ) ⎥ = s n ⋅ 1 − s n −1 ⋅ 0 − s n − 2 ⋅ 0 − L
⎣ dt
⎦
⎣ dt
⎦
n
⎡ d
⎤
L ⎢ n δ( t ) ⎥ = s n
⎣ dt
⎦

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Chapter 15, Problem 6.

Find F(s) given that
⎧2t , 0 < t < 1
⎪
f (t ) = ⎨t , 1 < t < 2
⎪0, otherwise
⎩
Chapter 15, Solution 6.
∞

1

2

0

1

F ( s ) = ∫ f (t )e − st dt = ∫ 2te− st dt + ∫ 2e− st dt
0

− st

2

− st

1
e
e 2 2
(
1)
2
st
−
−
+
= 2 (1 − e − s − se −2 s )
2
0
s
−s 1 s

Chapter 15, Problem 7.

Find the Laplace transform of the following signals:
(a) f (t ) = (2t + 4 )u (t )
(b) g (t ) = (4 + 3e −2t )u (t )
(c) h(t ) = (6 sin (3t ) + 8 cos(3t ))u (t )
(d) x(t ) = (e −2t cosh (4t ))u (t )

Chapter 15, Solution 7.
2 4

(a) F ( s ) = 2 +
s
s

(b) G ( s ) =

4
3
+
s s+2

(c ) H(s) = 6

3
2

s +9

+8

s
2

s +9

=

8s + 18
s2 + 9

(d) From Problem 15.1,
s
L{cosh at} = 2
s − a2
s+2
s+2
= 2
X (s) =
2
2
s + 4s − 12
( s + 2) − 4
(a )

4
4
3
8s + 18
s+2
, (c )
, (d )
+ , (b) +
s s+2
s2 s
s2 + 9
s 2 + 4s − 12
2

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Chapter 15, Problem 8.

Find the Laplace transform F(s), given that f(t) is:
(a)
(b)
(c)
(d)

2tu (t − 4 )
5 cos(t )δ (t − 2 )

e − t u (t − t )
sin (2t )u (t − τ )

Chapter 15, Solution 8.

(a) 2t=2(t-4) + 8
f(t) = 2tu(t-4) = 2(t-4)u(t-4) + 8u(t-4)
2
8
⎛ 2 8⎞
F ( s ) = 2 e −4 s + e−4 s = ⎜ 2 + ⎟ e−4 s
s
s
s⎠

⎝s
∞

∞

0

0

(b) F ( s ) = ∫ f (t )e − st dt = ∫ 5cos tδ (t − 2)e− st dt =5cos te− st
(c)

t=2

5cos(2)e
= 5cos
2e−2 s –2s

e − t = e− ( t −τ ) e−τ
f (t ) = e −τ e− (t −τ )u (t − τ )
−τ

F (s) = e e

−τ s

1
e −τ ( s +1)
=
s +1

s +1

(d) sin 2t = sin[2(t − τ ) + 2τ ] = sin 2(t − τ ) cos 2τ + cos 2(t − τ )sin 2τ
f (t ) = cos 2τ sin 2(t − τ )u (t − τ ) + sin 2τ cos 2(t − τ )u (t − τ )
2
s
F ( s ) = cos 2τ e −τ s 2
+ sin 2τ e−τ s 2
s +4
s +4

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Chapter 15, Problem 9.

Determine the Laplace transforms of these functions:
(a) f (t ) = (t − 4 )u (t − 2 )

(b) g (t ) = 2e −4t u (t − 1)
(c) h(t ) = 5 cos(2t − 1)u (t )
(d) p (t ) = 6[u (t − 2 ) − u (t − 4 )]

Chapter 15, Solution 9.

(a)

f ( t ) = ( t − 4) u ( t − 2) = ( t − 2) u ( t − 2) − 2 u ( t − 2)
e -2s 2 e -2s
F(s) = 2 − 2
s
s

(b)

g( t ) = 2 e -4t u ( t − 1) = 2 e -4 e -4(t -1) u ( t − 1)
2 e -s
G (s) = 4
e (s + 4)

(c)

h ( t ) = 5 cos(2 t − 1) u ( t )
cos(A − B) = cos(A) cos(B) + sin(A) sin(B)
cos(2t − 1) = cos(2t ) cos(1) + sin(2t ) sin(1)
h ( t ) = 5 cos(1) cos(2 t ) u ( t ) + 5 sin(1) sin(2t ) u ( t )
s
2
+ 5 sin(1) ⋅ 2
s +4
s +4
2.702 s 8.415
H(s) = 2
+
s + 4 s2 + 4
H(s) = 5 cos(1) ⋅

(d)

2

p( t ) = 6u ( t − 2) − 6u ( t − 4)
P(s) =

6 - 2s 6 -4s
e − e
s
s

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Chapter 15, Problem 10.

In two different ways, find the Laplace transform of
d −t
g (t ) =
te cos t
dt

(

)

Chapter 15, Solution 10.

(a)

(b)

By taking the derivative in the time domain,
g( t ) = (-t e -t + e -t ) cos( t ) − t e -t sin( t )
g( t ) = e -t cos( t ) − t e -t cos( t ) − t e -t sin( t )
G (s) =

⎤
s +1
d ⎡ s +1 ⎤ d ⎡
1
+ ⎢
⎥+ ⎢
⎥
2
2
2
(s + 1) + 1 ds ⎣ (s + 1) + 1⎦ ds ⎣ (s + 1) + 1⎦

G (s) =

s +1
s 2 + 2s
2s + 2

s 2 (s + 2)
−
−
=
s 2 + 2s + 2 (s 2 + 2s + 2) 2 (s 2 + 2s + 2) 2 (s 2 + 2s + 2) 2

By applying the time differentiation property,
G (s) = sF(s) − f (0)
where f ( t ) = t e -t cos( t ) , f (0) = 0
- d ⎡ s +1 ⎤
(s)(s 2 + 2s)
s 2 (s + 2)
=
G (s) = (s) ⋅ ⎢
=
ds ⎣ (s + 1) 2 + 1 ⎥⎦ (s 2 + 2s + 2) 2 (s 2 + 2s + 2) 2

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Chapter 15, Problem 11.

Find F(s) if:
(b) f (t ) = 3te −2t sinh 4t
(a) f (t ) = 6e − t cosh 2t
(c) f (t ) = 8e −3t cosh tu (t − 2)

Chapter 15, Solution 11.

s
s − a2
6 (s + 1)
6 (s + 1)
F(s) =
= 2
2
(s + 1) − 4 s + 2s − 3

(a)

Since L [ cosh(at ) ] =

(b)

Since L [ sinh(at ) ] =

2

a
s − a2
(3)(4)
12
L [ 3 e -2t sinh(4t ) ] =
= 2
2
(s + 2) − 16 s + 4s − 12
2

-d
[ 12 (s 2 + 4s − 12) -1 ]
ds
24 (s + 2)
F(s) = (12)(2s + 4)(s 2 + 4s − 12) -2 = 2
(s + 4s − 12) 2
F(s) = L [ t ⋅ 3 e -2t sinh(4t ) ] =

(c)

1
⋅ (e t + e - t )
2
1
f ( t ) = 8 e -3t ⋅ ⋅ (e t + e - t ) u ( t − 2)
2
-2t
= 4 e u ( t − 2) + 4 e-4t u ( t − 2)
= 4 e-4 e-2(t - 2) u ( t − 2) + 4 e-8 e-4(t - 2) u ( t − 2)
cosh( t ) =

L [ 4 e -4 e -2(t -2) u ( t − 2)] = 4 e -4 e -2s ⋅ L [ e -2 u ( t )]
4 e -(2s+ 4)
L [ 4 e -4 e -2(t -2) u ( t − 2)] =
s+2

Similarly, L [ 4 e

-8

e

- 4(t - 2)

4e
u ( t − 2) ] =

-(2s+ 8)

s+4

Therefore,
4 e -(2s+ 4) 4 e -(2s+8) e -(2s+ 6) [ (4 e 2 + 4 e -2 ) s + (16 e 2 + 8 e -2 )]
+
=
F(s) =
s+2
s+4
s 2 + 6s + 8
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Chapter 15, Problem 12.

If g (t ) = e −2t cos 4t find G(s).

Chapter 15, Solution 12.
G(s) =

s+2
s+2
= 2
2
2
( s + 2) + 4
s + 4s + 20

Chapter 15, Problem 13.

Find the Laplace transform of the following functions:
(b) e − t t sin t u (t )

(a) t cos t u (t )

(c)

sin βt
u (t )
t

Chapter 15, Solution 13.

←
⎯→

(a) tf (t )

−

d
F (s)
ds

If f(t) = cost, then F(s)=

s
s2 + 1

and -

L ( t cos t ) =

d
(s 2 + 1)(1) − s(2s)
F(s)= −
ds
(s 2 + 1) 2
s2 −1
(s 2 + 1) 2

(b) Let f(t) = e-t sin t.
1
1
= 2
F (s) =
2

( s + 1) + 1 s + 2s + 2
dF ( s 2 + 2s + 2)(0) − (1)(2s + 2)
=
ds
( s 2 + 2s + 2) 2
dF
2( s + 1)
L (e −t t sin t ) = −
= 2
ds ( s + 2s + 2) 2

(c )

f (t )
t

∞

←
⎯→

∫ F (s)ds
s

Let f (t ) = sin βt , then F ( s ) =
∞

β
s +β2
2

β
1
s
⎡ sin βt ⎤
L⎢
=∫ 2
ds = β tan −1
2
⎥
β
β
⎣ t ⎦ s s +β

∞
s

=

π
2

− tan −1

s

β

= tan −1

β
s

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Chapter 15, Problem 14.

Find the Laplace transform of the signal in Fig. 15.26.

Figure 15.26

For Prob. 15.14.

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Chapter 15, Solution 14.

Taking the derivative of f(t) twice, we obtain the figures below.
f’(t)
5

0

t
2

4

6

-2.5

f’’(t)
5 δ (t)

0

2.5δ(t-6)

2

6

-7.5δ(t-2)
f” = 5δ(t) – 7.5δ(t–2) + 2.5δ(t–6)
Taking the Laplace transform of each term,
s2F(s) = 5 – 7.5e–2s + 2.5e–6s or F(s) =

5
e −2s
e −6s

− 7.5
+ 2.5
s
s2
s2

Please note that we can obtain the same answer by representing the function as,
f(t) = 5tu(t) – 7.5u(t–2) + 2.5u(t–6).

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Chapter 15, Problem 15.

Determine the Laplace transform of the function in Fig. 15.27.

Figure 15.27

For Prob. 15.15.
Chapter 15, Solution 15.

This is a periodic function with T=3.
F ( s)
F ( s ) = 1 −3 s
1− e
To get F1(s), we consider f(t) over one period.

f1(t)

f1’(t)

5

f1’’(t)

5
5δ(t)

0

1

t

0

1

t

–5δ(t-1)

0

1

t

–5δ(t-1)
–5δ’(t-1)

f1” = 5δ(t) –5δ(t–1) – 5δ’(t–1)
Taking the Laplace transform of each term,
s2F1(s) = 5 –5e–s – 5se–s or F1(s) = 5(1 – e–s – se–s)/s2
Hence,
F(s) = 5

1 − e −s − se −s
s 2 (1 − e − 3s )

Alternatively, we can obtain the same answer by noting that f1(t) = 5tu(t) – 5tu(t–1) –
5u(t–1).
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Chapter 15, Problem 16.

Obtain the Laplace transform of f(t) in Fig. 15.28.

Figure 15.28

For Prob. 15.16.

Chapter 15, Solution 16.

f ( t ) = 5 u ( t ) − 3 u ( t − 1) + 3 u ( t − 3) − 5 u ( t − 4)

F(s) =

1
[ 5 − 3 e -s + 3 e - 3 s − 5 e - 4 s ]
s

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Chapter 15, Problem 17.

Find the Laplace transform of f(t) shown in Fig. 15.29.

Figure 15.29

For Prob. 15.17.
Chapter 15, Solution 17.

Taking the derivative of f(t) gives f’(t) as shown below.
f’(t)
2δ(t)

t
-δ(t-1) – δ(t-2)
f’(t) = 2δ(t) – δ(t–1) – δ(t–2)
Taking the Laplace transform of each term,
sF(s) = 2 – e–s – e–2s which leads to
F(s) = [2 – e–s – e–2s]/s
We can also obtain the same answer noting that f(t) = 2u(t) – u(t–1) – u(t–2).

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Chapter 15, Problem 18.

Obtain the Laplace transforms of the functions in Fig. 15.30.

Figure 15.30

For Prob. 15.18.

Chapter 15, Solution 18.

(a)

g ( t ) = u ( t ) − u ( t − 1) + 2 [ u ( t − 1) − u ( t − 2)] + 3 [ u ( t − 2) − u ( t − 3)]
= u ( t ) + u ( t − 1) + u ( t − 2) − 3 u ( t − 3)
1

G (s) = (1 + e -s + e - 2s − 3 e - 3s )
s

(b)

h ( t ) = 2 t [ u ( t ) − u ( t − 1)] + 2 [ u ( t − 1) − u ( t − 3)]
+ (8 − 2 t ) [ u ( t − 3) − u ( t − 4)]

= 2t u ( t ) − 2 ( t − 1) u ( t − 1) − 2 u ( t − 1) + 2 u ( t − 1) − 2 u ( t − 3)
− 2 ( t − 3) u ( t − 3) + 2 u ( t − 3) + 2 ( t − 4) u ( t − 4)
= 2t u ( t ) − 2 ( t − 1) u ( t − 1) − 2 ( t − 3) u ( t − 3) + 2 ( t − 4) u ( t − 4)
H(s) =

2
2 - 3s 2 - 4 s
2
-s
+ 2 e = 2 (1 − e -s − e - 3s + e -4s )
2 (1 − e ) − 2 e
s
s
s
s

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Chapter 15, Problem 19.

Calculate the Laplace transform of the train of unit impulses in Fig. 15.31.

Figure 15.31

For Prob. 15.19.
Chapter 15, Solution 19.

Since L[ δ( t )] = 1 and T = 2 , F(s) =

1
1 − e - 2s

Chapter 15, Problem 20.

The periodic function shown in Fig. 15.32 is defined over its period as
⎧sin π t , 0 < t < 1
g (t )⎨
1< t < 2
⎩0,
Find G(s)

Figure 15.32

For Prob. 15.20.
Chapter 15, Solution 20.
Let
g 1 ( t ) = sin(πt ), 0 < t < 1

= sin( πt ) [ u ( t ) − u ( t − 1)]
= sin(πt ) u ( t ) − sin(πt ) u ( t − 1)

Note that sin(π( t − 1)) = sin(πt − π) = - sin(πt ) .
g1 ( t ) = sin( πt) u(t) + sin( π( t - 1)) u(t - 1)
So,
G 1 (s) =

π
(1 + e -s )
s + π2
2

G 1 (s)
π (1 + e -s )
G (s) =
=
1 − e -2s (s 2 + π 2 )(1 − e - 2s )
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Chapter 15, Problem 21.

Obtain the Laplace transform of the periodic waveform in Fig. 15.33.

Figure 15.33

For Prob. 15.21.

Chapter 15, Solution 21.

T = 2π
Let

t ⎞
⎛
f1 ( t ) = ⎜1 − ⎟ [ u ( t ) − u ( t − 2π)]
⎝ 2π ⎠
t
1
f1 ( t ) = u ( t ) −
u(t) +
( t − 2 π) u ( t − 2 π)
2π
2π

[

1
1
e - 2πs 2π s + - 1 + e -2πs
+
=
F1 (s) = −
s 2πs 2 2πs 2
2πs 2

F(s) =

]

F1 (s)
2πs − 1 + e −2πs
=
1 − e -Ts 2πs 2 (1 − e - 2πs )

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Chapter 15, Problem 22.

Find the Laplace transforms of the functions in Fig. 15.34.

Figure 15.34

For Prob. 15.22.
Chapter 15, Solution 22.
(a)
Let
g1 ( t ) = 2t, 0 < t < 1
= 2 t [ u ( t ) − u ( t − 1)]
= 2t u ( t ) − 2 ( t − 1) u ( t − 1) + 2 u ( t − 1)
2 2 e -s 2

G 1 (s) = 2 − 2 + e -s
s
s
s
G 1 (s)
G (s) =
, T =1
1 − e -sT
2 (1 − e -s + s e -s )
G (s) =
s 2 (1 − e -s )

(b)

Let h = h 0 + u ( t ) , where h 0 is the periodic triangular wave.
Let h 1 be h 0 within its first period, i.e.
⎧ 2t
0 < t <1
h 1 (t) = ⎨
⎩ 4 − 2t 1 < t < 2
h 1 ( t ) = 2 t u ( t ) − 2 t u ( t − 1) + 4u ( t − 1) − 2 t u ( t − 1) − 2 ( t − 2) u ( t − 2)
h 1 ( t ) = 2 t u ( t ) − 4 ( t − 1) u ( t − 1) − 2 ( t − 2) u ( t − 2)
2
2 4 -s 2 e -2s
H 1 (s) = 2 − 2 e − 2 = 2 (1 − e -s ) 2
s
s
s
s
-s 2

2 (1 − e )
H 0 (s) = 2
s (1 − e -2s )

1 2 (1 − e -s ) 2
H(s) = + 2
s s (1 − e - 2s )
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Chapter 15, Problem 23.
Determine the Laplace transforms of the periodic functions in Fig. 15.35.

Figure 15.35
For Prob. 15.23.

Chapter 15, Solution 23.
(a)

Let

⎧1 0 < t <1
f1 ( t ) = ⎨
⎩- 1 1 < t < 2

f 1 ( t ) = [ u ( t ) − u ( t − 1)] − [ u ( t − 1) − u ( t − 2)]

f 1 ( t ) = u ( t ) − 2 u ( t − 1) + u ( t − 2)

1
1
F1 (s) = (1 − 2 e -s + e -2s ) = (1 − e -s ) 2
s
s
F1 (s)
, T=2
(1 − e -sT )
(1 − e -s ) 2
F(s) =
s (1 − e - 2s )
F(s) =

(b)

Let
h 1 ( t ) = t 2 [ u ( t ) − u ( t − 2)] = t 2 u ( t ) − t 2 u ( t − 2)
h 1 ( t ) = t 2 u ( t ) − ( t − 2) 2 u ( t − 2) − 4 ( t − 2) u ( t − 2) − 4 u ( t − 2)
2
4
4
H 1 (s) = 3 (1 − e -2s ) − 2 e -2s − e -2s
s
s
s
H 1 (s)
, T=2
(1 − e -Ts )

2 (1 − e -2s ) − 4s e -2s (s + s 2 )
H(s) =
s 3 (1 − e - 2s )
H(s) =

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Chapter 15, Problem 24.
Given that
F (s ) =

s 2 + 10 s + 6
2
s (s + 1) (s + 3)

Evaluate f(0) and f (∞ ) if they exist.

Chapter 15, Solution 24.
s 2 + 10 s + 6
1/ s + 10 / s 2 + 6 / s 3 0
=
lim
= =0
s →∞ ( s + 1) 2 ( s + 2)
s →∞ (1 + 1/ s )(1 + 2 / s )

1

f (0) = lim sF ( s ) = lim
s →∞

s 2 + 10 s + 6
6
=
=3
2
s → 0 ( s + 1) ( s + 2)
(1)(2)

f (∞) = lim sF ( s ) = lim
s →0

Chapter 15, Problem 25.
Let

F (s ) =

5(s + 1)
(s + 2)(s + 3)

(a) Use the initial and final value theorems to find f(0) and f (∞ ) .
(b) Verify your answer in part (a) by finding f(t), using partial fractions.

Chapter 15, Solution 25.
5s ( s + 1)
5(1 + 1/ s )

= lim
=5
s
→∞
( s + 2)( s + 3)
(1 + 2 / s )(1 + 3 / s )
5s ( s + 1)
f (∞) = lim sF ( s ) = lim
=0
s →0
s → 0 ( s + 2)( s + 3)

(a)

f (0) = lim sF ( s ) = lim

(b) F ( s ) =

s →∞

s →∞

5( s + 1)
A
B
=
+
( s + 2)( s + 3) s + 2 s + 3
5(−1)
5(−2)

B=
= −5,
= 10
1
−1
−5
10
F (s) =
+
⎯⎯
→ f (t ) = −5e −2t + 10e −3t
s+2 s+3

A=

f(0) = -5 + 10 = 5
f( ∞ )= -0 + 0 = 0.
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Chapter 15, Problem 26.
Determine the initial and final values of f(t), if they exist, given that:

s2 + 3
s 3 + 4s 2 + 6
s 2 − 2s + 1

(b) F (s ) =
(s − 2) s 2 + 2s + 4

(a) F (s ) =

(

)

Chapter 15, Solution 26.
(a)

s 3 + 3s
=1
3
2
s →∞ s + 4s + 6

f (0) = lim sF(s) = lim
s →∞

Two poles are not in the left-half plane.
f (∞) does not exist
(b)

s 3 − 2s 2 + s
2
s →∞
s →∞ (s − 2)(s + 2s + 4)
2 1

1− + 2
s s
= lim
=1
s →∞ ⎛
2 ⎞⎛ 2 4 ⎞
⎜1 − ⎟ ⎜1 + + 2 ⎟
⎝ s ⎠⎝ s s ⎠

f (0) = lim sF(s) = lim

One pole is not in the left-half plane.
f (∞) does not exist

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Chapter 15, Problem 27.
Determine the inverse Laplace transform of each of the following functions:
1
2
(a) F (s ) = +
s s +1
3s + 1
(b) G (s ) =
s+4

4
(c) H (s ) =
(s + 1)(s + 3)
12
(d) J (s ) =
(s + 2 )2 (s + 4 )

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Chapter 15, Solution 27.
(a)

f ( t ) = u ( t ) + 2 e -t u ( t )

(b)

G (s) =

3 (s + 4) − 11
11
= 3−
s+4
s+4

g( t ) = 3 δ( t ) − 11e -4t u ( t )

(c)

4
A
B
=
+
(s + 1)(s + 3) s + 1 s + 3
A = 2,
B = -2
2
2
H(s) =
−
s +1 s + 3
H(s) =

h ( t ) = 2 e -t − 2 e -3t u(t)
(d)

12
A
B
C
=
+
2
2 +
(s + 2) (s + 4) s + 2 (s + 2)
s+4

12
12
B=
= 6, C=
=3
2
(-2) 2
12 = A (s + 2) (s + 4) + B (s + 4) + C (s + 2) 2
J (s) =

Equating coefficients :
0= A+C ⎯
⎯→ A = -C = -3
s2 :
s1 :
s0 :

0 = 6A + B + 4C = 2A + B ⎯
⎯→ B = -2A = 6
12 = 8A + 4B + 4C = -24 + 24 + 12 = 12

J (s) =

-3
6
3
+
2 +
s + 2 (s + 2)
s+4

j( t ) = 3 e -4t − 3 e -2t + 6 t e -2t u(t)

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