Find the inverse Laplace transform of: 6s2 +8s +3 (a) F₁(s)= ss²+2s+5)

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Author:Robert L. Boylestad
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Where can I find 7/5?
Find the inverse Laplace transform of:
6s² +85 +3
(a) F₁(s)=
s(s²+2s+5)
s²+5s+6
(b) F₂(s)=
(s+1)²(s+4)
(c) F₁₂(s)=
10
2
(s+1)(s²+4s+8)
(a) F₁(s) =
6s² +8s+3
s(s²+2s+5)
=
A
S
Bs +C
+
s² +2s+5
6s²+8s+3=4(s² +2s+5)+ Bs² + Cs
We equate coefficients.
6 = A + B
$²
S:
8=2A+ C
constant: 3=5A or
A=3/5
B=6-A 27/5, C=8-2A 34/5
F₁(s)=
3/5 27s/5+34/5 3/5 27(s+1)/5+7/5
-+
S s²+2s+5
2
=
+
S
(s+1)²+2²
3.27
7
f₁(t) =
+ ·e cos 2t+⋅
e sin 2t
5 5
10
21]u(1)
Transcribed Image Text:Find the inverse Laplace transform of: 6s² +85 +3 (a) F₁(s)= s(s²+2s+5) s²+5s+6 (b) F₂(s)= (s+1)²(s+4) (c) F₁₂(s)= 10 2 (s+1)(s²+4s+8) (a) F₁(s) = 6s² +8s+3 s(s²+2s+5) = A S Bs +C + s² +2s+5 6s²+8s+3=4(s² +2s+5)+ Bs² + Cs We equate coefficients. 6 = A + B $² S: 8=2A+ C constant: 3=5A or A=3/5 B=6-A 27/5, C=8-2A 34/5 F₁(s)= 3/5 27s/5+34/5 3/5 27(s+1)/5+7/5 -+ S s²+2s+5 2 = + S (s+1)²+2² 3.27 7 f₁(t) = + ·e cos 2t+⋅ e sin 2t 5 5 10 21]u(1)
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