I have to find the Fourier transform , and $y(t)$ of an $ x(t) = e^{- \frac {t}{T} } u(t) $ that passes into a integrator filter. I know that $ Y(f) = X(f) H(f) $ so I first calculate the Fourier transform of $x(t)$ and I found that $ X(f)= \frac{T}{1+ i2\pi fT} $. Now I don’t know how to calculate the step response of the integrator filter. After, using inverse transform I should obtain $y(t)$.

  • $\begingroup$ Why would you need the step response of an integrator in order to solve the problem? $\endgroup$
    – Matt L.
    Jun 13 '20 at 10:37

As you know, we don't solve homework problems here, but you can get some hints. I don't know if you're really required to solve the problem in the frequency domain, but I would suggest to compute the output of the integrator in the time domain:


This integral turns into something very simple after noting how the infinite integration limits change to finite values when taking into account the two step functions. I'm sure you can continue from here.

If you really want to know the Fourier transform of the unit step $u(t)$, it can be found in most text books on signal processing, and in most Fourier transform tables:

$$\mathcal{F}\{u(t)\}=U(f)=\frac12\delta(f)+\frac{1}{j2\pi f}\tag{2}$$

It's not straightforward to derive this result yourself, so I don't think that this was expected of you.


If you solve the problem in the frequency domain, then with $(2)$ and with $X(f)=\frac{T}{1+j2\pi fT}$ you obtain

$$\begin{align}Y(f)&=X(f)U(f)\\&=\frac{T}{1+j2\pi fT}\left[\frac12\delta(f)+\frac{1}{j2\pi f}\right]\\&=\frac{T}{2}\delta(f)+\frac{T}{1+j2\pi fT}\cdot\frac{1}{j2\pi f}\tag{3}\end{align}$$

where the last equality follows from the fact that $X(f)\delta(f)=X(0)\delta(f)$ is $X(f)$ is continuous at $f=0$.

  • $\begingroup$ Thank you , I really don’t want someone who solve my exercises. First I have the solution on my book, second ( I most important ) I want to understand :) after studying better my book I understand that ( if the signal area is not equal to 0 ) the step response of an integrator circuit is $ Y(f) = X(f) H(f) $ where $ H(f) = \frac{1}{2} \delta (f) + \frac {1}{i 2 \pi f} $ so I finally obtained that $ Y(f) = \frac { T}{1+ iwT} [\frac{1}{2} \delta (f) + \frac {1}{i 2 \pi f}]$ But book obtained $ Y(f) = \frac { T}{1+ iwT} [\frac{T}{2} \delta (f) + \frac {1}{i 2 \pi f}]$ $\endgroup$ Jun 13 '20 at 12:47
  • $\begingroup$ @ElenaMartini: Please check my updated answer concerning the frequency domain solution. $\endgroup$
    – Matt L.
    Jun 13 '20 at 15:56
  • $\begingroup$ Thank you so much ! now i found my error and I solved it :) $\endgroup$ Jun 13 '20 at 16:59

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