I have a circuit with the transfer function \$G(s) = \frac{1}{(s+1)^2}\$ and I'm asked to calculate the impulse response of the circuit. After looking into it I know that I need to use convolution with the delta function to solve this.
I've managed to get as far as forming the integration to solve this, but I'm struggling as it seems to be quite a complex integration. I'm not sure if I've gone wrong somewhere leading up to it, so I was hoping someone could have a look through what I've done so far, and also help me with the final integration. I'll post my working first and I'll explain the integration problem at the end.
So the transfer function \$G(s) = \frac{V_o(s)}{V_i(s)} \therefore V_o(s)=G(s)V_i(s)\$, where \$V_o(s)\$ is the output response and \$V_i(s)\$ is the input impulse.
To find the output as a function of time: $$v_o(t) =L^{-1}\{G(s)V_i(s)\}=g(t)\ast v_i(t)\\ g(t) = L^{-1}\{\frac{1}{(s+1)^2}\}=e^{-t}t\\v_i(t) = \delta(t)\\ \therefore v_o(t) = g(t)\ast v_i(t)=e^{-t}t\ast \delta(t)$$ So the convolution integral will be: $$\int_{0}^{t}g(t-\tau)v_i(\tau)d\tau = \int_{0}^{t}e^{-(t-\tau)}(t-\tau)\delta(\tau)d\tau$$
This is where I get stuck as I can't solve this integral. Firstly I'm not 100% sure about the integral of \$\delta(\tau)\$, but I believe it equals 1? Is this correct?
I'm also not sure if I should be including that second \$(t-\tau)\$ bracket, I thought maybe the integral should be \$\int_{0}^{t}e^{-(t-\tau)}\delta(\tau)d\tau\$. This seems much easier to solve.
Can anyone point me in the right direction/correct me if I've made an error?