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Solutions t o B Problems

CHAPTER 2
B-2-1.
Note that

&amp;rtj =

+S

R e f erring to Egua tion (2-2 1, we obtain

Note t h a t

3 sin (5t

+

45 &quot;) = 3 sin 5t cos 45'

-- - 3

Jz

So we have

fl

-+ 3

cos

s i n 5t + 3 cos 5t

5t sin 45'

Note that

Referring to Equation (2-21, we obtain

f(t) =

B-24.

t &lt;U

0

=cos2Wtros3LLIt

t 1 0

Noting that
cos 2 W t cos 3wt = + ( m s 5 d t + c Q s &amp; J t )
w e have

F(S)

B-2-5.

=

L [ f ( t =) ~

~ + r tn s r ~ t ) ]

[%ms 5

The function f I t ) can be written as

f(t) = (h

- a)

l(t

- a)

The Laplace transform of f(t) is
~ ( s =)

[~(t)] =

J [(t - a )

~ ( -t a)] =

+
-as

6

B-2-6.
-

- a)

f[t) = c l[t

- c I(t

- b)

The Laplace transform oE f ( t ) is
~ ( a =
) c

e-a S
- ce-bs - 2 (,-as S
S

S

B-2-7.

The Function f ( t ) can be written as

So the Laplace transform of f ( t ) becomes

As

a approacl~eszero, t h e l i m i t i n g value of F(s ) becomes as follows:
F ( s ) = lim
a+O

lim

= lim

= lirn
a+O

B-2-8.

10

- 12.5

e-(a/5)s

+

2 5 =-as

a 2s

da

( 2 . 5 e-(a/5)s

-0.5 s e-(a/5)s
2

The function f(t) can be w r i t t e n as

So the Laplace transform of f(t) becomes

+

2 . 5 .-as)

2 . 5 s e-as

The limiting ~ l u 05
e F(s) as a approaches zero is
l i m F(s) = l i r n

a30

24(1

- as

e-*=
a3s2

3-2-11.

Define
y = x

Thw
y(O+) = \$(O+)

The i n i t i a l value of y can be obtained by use of the initial value theorem
as follows:

Since
y(s1

we obtain

B-2-12.

Note that

= J'+[y!t)l

= d;rGct,1

= sX(S)

- x((U)

Referring to Problm A-2-12,

we have

B-2-14.
f ( t ~emst dt =

Referring to Problem A-2-12,

e-st at = 1
0-

we obtain

where

%(s) can thus be written as

and the inverse Laplace transform of Fl(s) is

Cbl

where

FZ(s) can thus be written as

and the inverse Laplace transform of F Is) is
2

me inverse Laplace transform of F (s) is
1

(bl

where

E (s) can thus be written as
2

and the inverse Laplace transform of F,(s)

.

is

L-

i p ) = -3 .-t

+

3 .-2t

B t .-2t

--2 + - + - -2 - - 2 - 5
s + l
s

5
s + L

-

3
s + l

Rre inverse Laplam transform of f ( s ) is

The inverse Lapla-

transform of F(S) is

f { t )=

S ( t ) 'F

2

Hence
f(t) = e-t cos 3t

4-

4t

1 e-t
--

3

sin 3t

+- 5

s

Hence

The inverse Laplace transform of F(s) is

Hence

The inverse Laplace transform of F l s ) is