I have a signal $$ x(t)= \frac{1}{T} e^{-\frac{t}{T}} u(t) - \frac{1}{T} e^{\frac{t}{T}} u(-t) $$

and I know that it transits in a integrator circuit and I have to find y(t) in time and frequency domain. From theory on my book i know that

$$ y(t) = \int_{-\infty}^{t} x(\tau) d \tau $$

So I started to calculate this integral with my $x(t)$:

$$ \frac{1}{-RCT} \left[ \int_{0}^{+\infty} e^{\frac{-\tau}{T}} d\tau - \int_{—\infty}^{0} e^{\frac{\tau}{T}}d\tau \right] $$

but this gave me

$$ \frac{1}{-RCT} \left[ \dfrac{e^{-\infty } -1}{-\frac{1}{T}} - \dfrac{-e^{-\infty } +1}{\frac{1}{T}} \right] $$

and this gave me 0.

The correct result should be $$ -e^{ -\frac{|t|}{T} } $$

Thank you so much


1 Answer 1


The input and the integrator output are shown below:

enter image description here

So, you should consider two cases: $t<0$ and $t>0$.

In the first case ($t<0$), you have $$\begin{align} y(t) &= \int_{-\infty}^t x(\tau)d\tau = -\frac{1}{T}\int_{-\infty}^t e^{\frac{\tau}{T}}u(-\tau)d\tau \\ &= -\frac{1}{T}\int_{-\infty}^t e^{\frac{\tau}{T}}d\tau = -\frac{1}{T}Te^{\frac{\tau}{T}}\Big]_{-\infty}^t \\ &= -e^{\frac{\tau}{T}}\Big]_{-\infty}^{t} = -(e^{\frac{t}{T}} - 0) = -e^{\frac{t}{T}} \end{align}$$ again, for $t<0$. So now you have $$y(t) = \left\{\begin{array}{ll} -e^{\frac{t}{T}}, & t < 0 \\ ? \quad \:\:\:, & t > 0 \end{array}\right.$$ and you're looking for the other branch of the output. Can you show that it is $$y(t) = -e^{-\frac{t}{T}}, \quad t > 0?$$

If you do, then you'll get $$y(t) = \left\{\begin{array}{ll} -e^{\frac{t}{T}}, & t < 0 \\ -e^{-\frac{t}{T}}, & t > 0 \end{array}\right.= -e^{-\frac{|t|}{T}}$$ for all $t$.

  • $\begingroup$ I though I understood and for t>0 I wrote $$ y(t) = \int_{t}^{+ \infty} \frac{1}{T} e^{\frac{\tau}{T}} u(\tau) d\tau $$ but this gave me $$ \frac{1}{T} [ \frac{0 - e^{\frac{-t}{T} }}{-\frac{1}{T}} ]$$ that is $$ e^{-\frac{t}{T} } $$ And I have also another question.. what about $$ \frac{-1}{RC} $$ ?? thank you !! You’re better than Wikipedia 🙈 $\endgroup$ Feb 6, 2020 at 14:04
  • 1
    $\begingroup$ No, that's not the correct integral. Check the figure. You have to integrate from $-\infty$ up to $t$. According to the figure you should split the integral from $-\infty$ to $0$, which includes the left part of the signal $x(\tau)$, and from $X$ to $Y$ to include the right part of the signal, up to $t$. What are $X$ and $Y$? :) $\endgroup$
    – GKH
    Feb 6, 2020 at 14:40
  • 1
    $\begingroup$ Checking the figure ( thank for that !) I should have integrate the left part of the graph from $$ -\infty $$ to $$ 0 $$ and the right part from $$ 0 $$ to $$ t $$. The sum of these two integrals give me the result of y(t) for t>0. Now the result is correct!! thanks!!! $\endgroup$ Feb 6, 2020 at 15:18
  • $\begingroup$ The question about the constant 1/RC is about the fact that I study on my book that if x(t) enters on a integrator circuit , so $$ y(t) = \frac{1}{RC} \int_{- \infty}^{t} x(\tau) d\tau $$ but when we studied the signal for t<0 e t>0 we ignore the this constant. This formula $$ y(t) = \frac{1}{RC} \int_{- \infty}^{t} x(\tau) d\tau $$ is what I study in my book $\endgroup$ Feb 6, 2020 at 15:27
  • 1
    $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – GKH
    Feb 6, 2020 at 16:34

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