I'm trying to calculate the output given an impulse response $h(t)=e^{-t}u(t)$ and input $x(t)=\cos(2\pi t)$.

I know I need to use the convolution but given that cosine is periodic, I don't see how I would get an answer to converge. Am I thinking about this the right way or missing something?

  • $\begingroup$ Duplicate: math.stackexchange.com/questions/1935230/… $\endgroup$
    – Unknown123
    Feb 1, 2018 at 23:51
  • $\begingroup$ Have you tried using the Fourier transform? $\endgroup$
    – Tendero
    Feb 2, 2018 at 0:04
  • $\begingroup$ Hint 1. cos(2 pi t) = ( exp( -j 2 pi t) + exp( j 2pi t))/2 $\endgroup$
    – user28715
    Feb 2, 2018 at 1:24

1 Answer 1


Eventhough you could solve this problem using other means, such as frequency domain methods, you could also follow a direct time domain path as the following. Given the impulse response $h(t) = e^{-t} u(t)$ of an LTI system and the applied excitation input $x(t) = \cos(2\pi t)$ (which is of infinite extend from $t=-\infty$ to $t=\infty$), the output can be written as the convolution integral:

$$y(t) = x(t)\star h(t) = \int_{-\infty}^{\infty} h(t-\tau) x(\tau)d\tau = \int_{-\infty}^{\infty} e^{-(t-\tau)}u(t-\tau) \cos(2\pi \tau)d\tau$$

$$y(t) = e^{-t} \int_{-\infty}^{t} e^{\tau} \cos(2\pi \tau)d\tau$$

At this point we have to evaluate the integral which can be found on most calculus texts or in an integral table. I will use MATLAB to evaluate it which results in $$\boxed{ y(t) = \frac{\cos(2\pi t) }{1 + 4\pi^2} + \frac{2\pi \sin(2\pi t) }{1 + 4\pi^2} }$$

You can simplify the result using $a\cos(wt) + b\sin(wt) = \sqrt{a^2+b^2} \cos(wt - \tan^{-1}(b/a) )$ into $$\boxed{ y(t) = \frac{1 }{\sqrt{1 + 4\pi^2}} \cos(2\pi t)}$$

as the solution. The solution has no transient part as the input was applied beginning from $t=-\infty$.


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