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The output system is:

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

Recall that the system is causal if the output at $t$ depends only on input before $t$, or if the impulse response $h(t)=0,\forall t <0$

I guess the system is not causal, but I am not sure how to obtain the impulse response if we are given the output.

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  • $\begingroup$ You might want to consider posting questions like this on electronics.stackexchange $\endgroup$ – Math1000 Feb 4 '16 at 7:05
  • $\begingroup$ @Math1000: Rather on Signal Processing SE. $\endgroup$ – Matt L. Feb 4 '16 at 8:12
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Let $ (Lx)(t) = \int_{-\infty}^t x(\tau)d \tau$.

You want to show that if $x_1,x_2$ are such that $x_1(s) = x_2(s)$ for $s \le t$, then $(Lx_1)(t) = (L x_2)(t)$.

You have $(Lx_1)(t) = \int_{-\infty}^t x_1(\tau)d \tau = \int_{-\infty}^t x_2(\tau)d \tau = (L x_2)(t)$.

Hence the system $L$ is causal.

Alternatively, albeit the use of distributions is unnecessary, we can look at the response of the system when subjected to an impulse input. You need to verify that the system is linear and time invariant first, then note that if $t < 0$, we have $h(t) = (L \delta)(t) = 0$.

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  • $\begingroup$ So, that does not depend on the upper and lower bounds of the integral? So if I changed the bounds to be from $t-2$ to $t$, is it still causal? $\endgroup$ – El Qanas Feb 4 '16 at 6:46
  • $\begingroup$ I don't understand your question. $\endgroup$ – copper.hat Feb 4 '16 at 6:47
  • $\begingroup$ I have edited my comment, making change the bounds of the integral of the output $\endgroup$ – El Qanas Feb 4 '16 at 6:51
  • $\begingroup$ Yes, just follow exactly the same steps above. If the upper bound was $>t$ then you might have an issue. $\endgroup$ – copper.hat Feb 4 '16 at 6:52
  • $\begingroup$ Oh, OK. So, how if using the impulse response $h(t)$? If suppose I have shown that the system is LTI, and would like to show that $h(t)=0,\forall t<0$, is it correct to calculate $h(t)=\int^{t}_{-\infty} \delta (\tau) d\tau$ $\endgroup$ – El Qanas Feb 4 '16 at 6:57
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Note that the input/output relation of your system can be written in terms of a convolution:

$$y(t)=\int_{-\infty}^{\infty}x(\tau)u(t-\tau)d\tau\tag{1}$$

where $u(t)$ is the unit step function, which is zero for $t<0$ and equals $1$ for $t>0$. Eq. (1) proves that the system is linear and time-invariant. Furthermore, it proves that the system is causal because the system's impulse response equals the unit step function, which is zero for $t<0$.

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  • $\begingroup$ Oh I see. But is it correct that the impulse response is $h(t)=\int^t_{-\infty}\delta(\tau)d\tau$? $\endgroup$ – El Qanas Feb 4 '16 at 8:42
  • $\begingroup$ @ElQanas: Yes, that integral equals $u(t)$ because as soon as $t>0$ you integrate over the Dirac delta and the result is $1$. $\endgroup$ – Matt L. Feb 4 '16 at 8:44

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