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?


1 Answer 1


I was hoping someone could have a look through what I've done so far, and also help me with the final integration.

You don't need to perform that integration.

In the s-domain the unit impulse is unity hence, you have calculated the response already and did the inverse transform correctly. The answer is \$t\cdot e^{-t}\$. The great thing about transferring to the s-domain is that if you can also transfer the time domain input signal to the s-domain, you have a fairly simple calculation and there are on-line resources (such as this one) that can do the inverse manipulation for you.

  • \$\begingroup\$ So I don't need to bother with convolution at all? Damn talk about making it harder for myself! Just to check I'm understanding you correctly: \$V_i(s) = 1\$, so \$V_o(s)=G(s)\$? And so the impulse response is just \$v_o(t) = L^{-1}\{G(s)\}\$? \$\endgroup\$
    – MendelumS
    Jul 27, 2020 at 14:46
  • \$\begingroup\$ Correct and (a little bit of extra info) some loudspeaker manufacturers use this for a quality check on their speakers. They inject a very thin pulse (an impulse) and have a calibrated microphone pick up the response - the response in the time domain is the actual spectral response of the speaker. Well, they used to do this back in the 80s. \$\endgroup\$
    – Andy aka
    Jul 27, 2020 at 14:58

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