# y(t) a of an integrator circuit

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

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 $$t$$.

• 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 🙈 – Elena Martini Feb 6 at 14:04
• 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 $Y$? :) – GKH Feb 6 at 14:40
• 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!!! – Elena Martini Feb 6 at 15:18
• 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 – Elena Martini Feb 6 at 15:27
• – GKH Feb 6 at 16:34