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Problem 3.08PP
Find the time function corresponding to each of the following Laplace transforms:
(b) =
(0 =
(d) =
(f) F (s ) =
W i W
(g) F(.s) = taa ' ( i )
Step-by-step solution
step 1 of 12
(a)
Consider the Laplace transform equation.
Take partial fraction expansion and rewrite F (.r)- 
r^/ \ O b C
Use the cover up method and find a , ^and c.
' tU='( , + i y
c = i u = - i
- 7 U .
* - l
Step 2 Of 12
Substitute 1 for a. • ] for ^and . ] for c in equation (1). 
Take Inverse Laplace Transform.
/ ( 0 = ( i - e - ' - « - ' ) i ( 0
Thus, the time function c
1
Step 3 of 12
(b )
Consider the Laplace transform equation.
Take partial fraction expansion and rewrite F (.r)-
Ff$)=— fl±£±l—
1
^ “ 1
Take Inverse Laplace Transform.
r'(F (.))= r'(^ ]
/ ( ' ) = ^ ' W
Thus, the time function of is
Step 4 of 12
(c)
Consider the Laplace transform equation.
. . 2 ( j * + j + l )
--------- 5 - i
Take partial fraction expansion and rewrite F ( 4)- 
r^/ \ O b C
' " W 'T + T T n T + T T T T T (2)s (»+l) (i+l)’
j + i y
2^5* + lV
2(5^+5 + 1),
V . = - 2
2 ( j * + j + l )
2(2j + 1 )5 -2 ( j *+ « + 1)
1
u .
step 5 of 12
Substitute 2 for 0 1.280
Find the value of
d ^ ( s - 2 j ) ^ F ( s ) \ ^ ,
^ -8 3 -3 9 y
20
*-4,150-;i.950
Step 9 of 12
Find the value of e.
e ^ d r
- -4 .1 5 0 + y l.9 5 0
Find the value of J}.
^ -1 2 8 -5 7 9 j
200
= -0 .64-y2 .895
Find the value of c. 
c - b *
= -0 .6 4 + y 2.895
Substitute 1,280 for a, -0 .64 -> 2 .895 for 6 , -0.64+>2.895 for c . -4.150->1.950 for
d and -4.150+>1.950 for e in equation (3).
F ( 1-280 ̂-0.64->2.895 ̂-0.64+>2.895 ̂ .̂150->1.950 ̂-4.150+>1.950 
s + \ * S -2 J ^ S + 2J { s - 2 j y * { s * 2 j ) ‘
Step 10 Of 12
Take Inverse Laplace Transform.
/ ( ' ) ■
1.28«-' + /[(-4 .1 5 0 + y i.9 5 )« -^ -(4 .1 5 0 + y i.9 5 )e ^ ]1 
-(0 .6 4 -;2 .8 9 5 )e -^“ -(0 .6 4 + ;2 .895)e '” |
fl.28«-' + /[(-4.150e--'“ + yl.95«-^“ )- (4 .I5 0 « '’' + yl.95e'“ )]l 
|- ( 0 .6 4 e - ^ -y 2 .8 9 5 e -^ ) - (0 .6 4 « ^ + J 2 M S e ‘ ^ ) J
_ |l.2 8 « - ' -/4.150e-^“ + -«4 .150e^ -yrt.O Se'” !
~ |-0 .6 4 « -^ + y 2 .8 9 5 e -F '-0 .6 4 e ^ -y 2 .8 9 5 e '” J
Rearrange the equation.
[l.28«^ -/4.150(e-^“ + « '“ )+ y/1.95(«-^
”|-0.64(«-F' +*>“)+y2.895(«-''“ -e^)
'^“ )1
r
>(')
1.28*'' -/4.150(2cos(2/))+yrt.95(-y2sin(2/)) 
“|-0.64(2cos(2/))+;2.895(-y2sin(2/))
1.28*- -/4.1S0(2cos(2/))-/rt.95(2sin(2/)) 
”|-0.64(2cos(2/))-/2.895(2sin(2/))
= {l.28e-'-»8.3cos(2»)+3.9/sin(2/)-1.28cos(2/)+5.79(sin(2/))}l(/)
Rearrange the equation.
/(/) = [l.28e-'+3.9/sin(2/)+5.79siii(2/)-8.3/cos(2/)-1.28cos(2/)]l(l) 
2(i + 2)(j+5)'
Thus, the time function of -------- ^ is(*+ !)(*’+4)
|[l.28g-* + 3.9)|.
Step 11 of 12
(f)
Consider the Laplace transform equation.
Take partial fraction expansion and rewrite F (s )- 
5* 1
( * + 0 ( » + ' )
Take Inverse Laplace Transform.
r'(F(*))=r
W - ' f (.=+■)'
/ ( 0 = ( / c o s » ) l( / )
Thus, the time function of -------- ^ is ( /c o s /) l( /) •
15* +11 *----------------- *
Step 12 Of 12
(g)
Consider the Laplace transform equation.
F(.)=tan-[i)
Rewrite F (5) in terms of series.
^W=7-37^57--
Take Inverse Laplace Transform.
r'(F(*))=r'[t»-g]]
°'^'(7"37‘"57""]
/ ( 0 = i - 3 ( - ^ -
sin(/)
• w
Thus, the time function of tan"' - is
■ 0 '
s in (/) >(0