Question

solve inverse laplace using convolution method question in pic attached to second post 1/(s^2+2s+10)^2......I got up to 1/9 integral e^-t sin3t times e^-(t-tau) sin 3(t-tau)dt ....not sure how to solve it ty....

Answer 1

@ash2326

Answer 2

\[\int_{0}^{t} \frac{1}{9} e^{-\tau}\sin 3\tau \times e^{-(t-\tau)}\sin 3 (t-\tau) d\tau\]
We get
\[\int_{0}^{t} \frac{1}{9} e^{-t}\sin 3\tau \times \sin 3 (t-\tau) d\tau\]
we get
\[\frac{1}{9} e^{-t}\int_{0}^{t} \sin 3\tau \times \sin 3 (t-\tau) d\tau\]
Let's expand the term
\[\sin 3(t-\tau)=\sin 3t \cos 3\tau- \cos 3t \cos 3\tau\]
we get
\[\frac{1}{9} e^{-t}\int_{0}^{t} \sin 3\tau \times (\sin 3t \cos 3\tau- \cos 3t \sin 3\tau)d\tau\]

now we get
\[\frac{1}{9} e^{-t}\int_{0}^{t} (\sin 3t\times \sin 3\tau \cos 3\tau- \cos 3t\times \sin^2 3\tau )d\tau\]
now we have
\[\frac{1}{9} e^{-t}(\sin 3t\int_{0}^{t} \sin 3\tau \cos 3\tau \times d\tau- \cos 3t\int_{0}^{t} \sin^2 3\tau \times d\tau)\]
we know that
\[\sin A \cos A= (\sin 2A) /2\]
and
\[\sin^2 A= (1-\cos 2A)/2\]
so we get
\[\frac{1}{9} e^{-t}(\sin 3t\int_{0}^{t} \frac{\sin 6\tau } {2} d\tau- \cos 3t\int_{0}^{t} (\frac{1-2\cos 6\tau}{2} d\tau))\]
Can you finish the integration now?

Answer 3

hmm i thought the f(t) part is e^-t and sin3t and not tau....i thought is only the second part of g(t-tau) only got tau right?

Answer 4

Covulution of f(t) and g(t)
\[\int_{0}^{t} f(\tau) g(t-\tau) d\tau\]

Answer 5

oh i double check and yup your right no wonder i am stuck cuz both looks quite similar ok so redoing again.......

Answer 6

the last part should be 1-cos6tau and not 1-2cos6tau right>?

Answer 7

@ash2326 i manage to do integration for other parts except the sin 6 tau / 2 part can you guide me for that only ty....

Answer 8

@angalc557 It's simple, check this:
\[\int_{0}^{\tau} \frac{\sin 6\tau}{2} \]
we know integration of sin x is -cos x
so we get
\[ [\frac{-\cos 6\tau}{6\times 2}]_{0}^{t}\]
we get
\[ \frac{1}{12}\times[- \cos 6 t-(-\cos 0)]\]
we get
\[\frac{1}{12}\times(1- \cos 6 t)\]

Answer 9

i see nicely explain i understand that kk so i manage to get up to this line and now is quite long so i got 1/9 e^-t [sin3t * (1-cos6t)/12 - cos3t * (t/2 - sin6t/12) ] right?

Answer 10

Yeah :D

Answer 11

nice then do i expand the terms for each part or how to simplify them? take out constants?

Answer 12

@ash2326 haha can u help me simplify them last question really.....kk

Answer 13

We have
\[\frac{1}{9} e^{-t} (\sin 3t \frac{1-\cos 6t}{12}- \cos 3t(\frac{t}{2}- \frac{\sin 6t}{12})\]
let's simplify this
\[\frac{1}{9} e^{-t} (\sin 3t \frac{1}{12}-\frac{\sin 3t \times \cos 6t}{12}- \frac{tcos 3t}{2}+ \frac{\cos 3t \times\sin 6t}{12})\]
we know that sin(A-B)= sin Acos B-cos A sin B

\[\frac{1}{9} e^{-t} (\sin 3t \frac{1}{12}- \frac{tcos 3t}{2}+ \frac{\cos 3t \times\sin 6t}{12}-\frac{\sin 3t \times \cos 6t}{12})\]
We get
\[\frac{1}{9} e^{-t} (\sin 3t \frac{1}{12}- \frac{tcos 3t}{2}+ \frac{\cos 3t \times\sin 6t-\sin 3t \times \cos 6t}{12})\]
\[\frac{1}{9} e^{-t} (\sin 3t \frac{1}{12}- \frac{tcos 3t}{2}+ \frac{\sin(6t-3t)}{12})\]
\[\frac{1}{9} e^{-t} (\sin 3t \frac{1}{12}- \frac{tcos 3t}{2}+ \frac{\sin(3t)}{12})\]
now combine the two sine (3t)/12
we get
\[\frac{1}{9} e^{-t} (\frac{\sin(3t)}{6}- \frac{tcos 3t}{2})\]
I think this is the simplified form

Answer 14

i understand all the steps wow this long long one kk i really thank you for all you time and effort in replying me and guiding me for the past 4 questions ty so much :) if in future i need your help i will let u know .....take care for now....

Answer 15

Welcome @angalc557 glad to help you:D