Question

Find the inverse place transform

Answer 1

As you mentioned in your other question, I don't think this needs to be done with the convolution relation of the transform.

Note that you can rewrite the transform as


`(7s+66-(9(s+5))/((s+5)^2+64))/(s^2+10s+74)`

`=(7(s+5))/((s+5)^2+49)+(31)/((s+5)^2+49)-(9(s+5))/(((s^2+49)(s^2+64))`

`=7(s+5)/((s+5)^2+7^2)+\frac{31}7\frac7{(s+5)^2+7^2}+\frac35(s+5)/((s+5)^2+8^2)-\frac35(s+5)/((s+5)^2+7^2)`

Now, recall the following transforms:


`\mathcal L_s\{\cos at\}=\frac s{s^2+a^2}`

`\mathcal L_s\{\sin at\}=\frac a{s^2+a^2}`

`\mathcal L_s\{e^(ct)f(t)\}=F(s-c)`

With
`c=-5`, you can see that


`\mathcal L_s\{f(t)\}=7(s+5)/((s+5)^2+7^2)+\frac{31}7\frac7{(s+5)^2+7^2}+\frac35(s+5)/((s+5)^2+8^2)-\frac35(s+5)/((s+5)^2+7^2)`

`\implies\mathcal L_s\{e^(-5t)f(t)\}=7\frac s{s^2+7^2}+\frac{31}7\frac7{s^2+7^2}+\frac35\frac s{s^2+8^2}-\frac35\frac s{s^2+7^2}`

Taking the inverse transform of both sides gives


`e^(-5t)f(t)=7\cos7t+\frac{31}7\sin7t+\frac35\cos8t-\frac35\cos7t`

`\implies f(t)=e^(5t)\left(\frac{32}5\cos7t+\frac35\cos8t+\frac{31}7\sin7t\right)`