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Inverse Laplace Transform Guide

1. The inverse Laplace transform of 1/s^2 and 1/s^5 is t and 2t^4, respectively. 2. The inverse Laplace transform of (1/s^2)*(1/s^3) is 4t + (2/3!)*(1/5!)*t^3*t^5 = 4t + 2/120*t^3*t^5. 3. The inverse Laplace transform of 1/(4s+1) is e^(-4t)/4.

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ALLEN TANAKA MUZORERA
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130 views5 pages

Inverse Laplace Transform Guide

1. The inverse Laplace transform of 1/s^2 and 1/s^5 is t and 2t^4, respectively. 2. The inverse Laplace transform of (1/s^2)*(1/s^3) is 4t + (2/3!)*(1/5!)*t^3*t^5 = 4t + 2/120*t^3*t^5. 3. The inverse Laplace transform of 1/(4s+1) is e^(-4t)/4.

Uploaded by

ALLEN TANAKA MUZORERA
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© © All Rights Reserved
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TWK2A

The Inverse Laplace transform (Section 7.2)


Solutions
1.

1 1 48 1 1 1 4!
L = L 2L
s2 s5 s2 s5
= t 2t4

2.
( )
2
1 2 1 1 4 4 1
L = L + 6
s s3 s2 s 4 s
t3 t5
= 4t 4 +
3! 5!
t3 t5
= 4t 2 +
3 120

3.

1 1 1 1 1
L = L 1
4s + 1 4 s+ 4
1 1
t
= e 4
4

4.

1 10s 1 s
L = 10L
s2+ 16 s2 + 16
= 10 cos 4t

5.

1 2s 6 1 s 1 3
L = 2L 2L
s2 + 9 s2 +9 s2 +9
= 2 cos 3t 2 sin 3t
6.
1 1 1 1
L = L
s2 +s 20 (s + 5)(s 4)
1 1 1 1 1 1
= L + L
9 s+5 9 s 4
1 5t 1 4t
= e + e
9 9
7.
1 (s + 2)2 s2 + 4s + 4
1
L = L
s3 s3
1 4 4
= L 1 + 2+ 3
s s s
1 1 2
= L 1 + 4L 1 + 2L 1
s s2 s3
= 1 + 4t + 2t2

8.
1 s+1 1 s+1
f (t) L =L
s2 4s s(s 4)
Let
s+1 A B A(s 4) + Bs
= + =
s(s 4) s s 4 s(s 4)
Equate the numerators:
A(s 4) + Bs = s + 1
1
Put s = 0 : 4A = 1; A=
4
5
Put s = 4 : 4B = 5; B=
4
We now have
1 5
1 4 4
f (t) = L +
s s4
1 1 5 1
= L 1 + L 1
4 s 4 s 4
1 5 4t
= + e :
4 4
9.
1 6s + 3 1 6s + 3
f (t) L =L
s4 + 5s2 + 4 (s2 + 1)(s2 + 4)
Let
6s + 3 As + B Cs + D
= 2 + 2
(s2
2
+ 1)(s + 4) s +1 s +4
Equating the numerators, we have:
(As + B)(s2 + 4) + (Cs + D)(s2 + 1) = 6s + 3
Compare the coe¢ cients of s :
s3 : A + C = 0
) A = 2; C = 2
s : 4A + C = 6
s2 : B + D = 0
0 ) B = 1; D = 1
s (= 1) : 4B + D = 3
Therefore,
1 2s + 1 2s + 1
f (t) = L
s2 + 1 s2 + 4
1 2s 1 2s 1
= L 2
+ 2
s +1 s +1 s2 +4 s +4 2

1 s 1 s 1 2
= 2L 2
+L 1 2
2L 1 L 1
s +1 s +1 s2 +4 2 s2 +4
1
= 2 cos t + sin t 2 cos 2t sin 2t:
2
10. Assume
s2 + 1 A Bs + C D
= + 2 +
s(s 1) (s + 1) (s 2) s s 1 s 2
| {z }
s2 1
8
>
> A+B+D =0
<
2A 2B + C = 1
)
>
> A 2C D = 0
:
2A = 1
8
>
> A = 12
< 4
B= 3
) 2
>
> C= 3
:
D = 56
s2 + 1 1 1 4 s 2 1 5 1
) = +
s (s 1) (s + 1) (s 2) 2 s 3 s2 1 3 s2 1 6 s 2
Now,

1 1
L = 1
s
1 s ekt + e kt
L = cosh t =
s2 1 2
kt kt
1 1 e e
L = sinh t =
s2 1 2
1 1
L = e2t
s 2
and so

1 s2 + 1 1 e t 5e2t
L = et + :
s (s 1) (s + 1) (s 2) 2 3 6

11. We have
(s + 1)3 1 3 3 1
4
= + 2+ 3+ 4
s s s s s
1 1 3 2 1 6
= +3 2 + +
s s 2 s3 6 s4
and so ( )
1 (s + 1)3 3t2 t3
L = 1 + 3t + + :
s4 2 6

12.

1 1 1 1
L = L
s2 +s s(s + 1)
1 1 1
= L
s s+1
1 1 1
= L L 1
s s+1
t
= 1 e
where we used
1 A B
= + ) A(s + 1) + Bs = 1
s(s + 1) s s+1
) A + B = 0 and A = 1
) B= 1

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