y(t) a of an integrator circuit

Question

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

Answer 1

The input and the integrator output are shown below:

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 $t$.