Q4.) F(s) L' = 45+93 252+85+40 } 45+9 252 = = 2 = +85+40 =2 2 L" { -1 L+ تو 2/2-1 -2t ? (pat し S+ 2.25 52+45+20 5+2 2-25-2 (5+2)²+(4)2 5+2 ((S+2)²+(4)² Cos (47) + = 2 (e-²+ cos (4t) + } Учи } 1.4 (5+2)²+(432} //ext sin (ut) = 20-2+ cos (ut) + 1/2 e 2+ sin (ut). 82 +48 +20 (5+2)²+20-4 (5+2)²+16
Q4.) F(s) L' = 45+93 252+85+40 } 45+9 252 = = 2 = +85+40 =2 2 L" { -1 L+ تو 2/2-1 -2t ? (pat し S+ 2.25 52+45+20 5+2 2-25-2 (5+2)²+(4)2 5+2 ((S+2)²+(4)² Cos (47) + = 2 (e-²+ cos (4t) + } Учи } 1.4 (5+2)²+(432} //ext sin (ut) = 20-2+ cos (ut) + 1/2 e 2+ sin (ut). 82 +48 +20 (5+2)²+20-4 (5+2)²+16
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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The question asks to find the inverse Laplace transform function of : (4s+9)/(2s2+8s+40)
I have included my working out of the questions in hopes that along with providing me with the answer to the question you can also tell me where I went wrong.
Thanks
Transcribed Image Text:64.) F(s) =
L-1 { 43 + 93
252+88+40
45+9
282
=
+85+40
2
むす。
= 2 " {
S+2.25
52+45+20
5+42-25-2
}
82 +45 +20
(5 + 2)² +20-4
(5+2)²+16
(5+2)²+(4)2
(+)
(5+2)²+(4)2
= 2 ( e^2+ cos (4+) + 1/4 4
+ {
1.4
(5+ 2)²+14)²)
(5+2)²+(4)2
})
= 2 (e-²+ cos (ut) + 1/4 ext sin cut)
ть
= 2 e²²+ cos (ut) + ke-2+ sin (ut).
e-2+ +
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