I've been working at this problem for a while now, and can't seem to come to a solid conclusion - is this system time invariant?

$y(t) = \int_{-\infty}^{t} e^{-9(t-\tau)} x(\tau)d\tau $

My reasoning for the system being time invariant is that it is clearly a convolution integral. Time shifting the input will shift the output.

But, it is not a normal convolution integral. The top bound is only to $t$ rather than to $\infty$. This is all well and good if the system is causal, because for any $\tau$ greater than $t$, the output would be zero. But because there is no unit step $u(t)$ clearly defined in the transfer function (like $e^{-9(t-\tau)}u^\tau(t)$), one cannot make a conclusion on the causality of the system. So the question at hand is, does shifting the input also shift the bounds of the integral?

So what do you think? Can anyone prove if the system is time invariant as given?


2 Answers 2



You are on the right track. You note that the integral is

not a normal convolution integral


there is no unit step $u(t)$ clearly defined in the transfer function.

So, can you introduce a unit step yourself and change the limits of integration to make it look like a normal convolution integral?

STOP! Full answer below.

Let $\mathbf{1}_{A}$ be the indicator function which is 1 on the set $A$ and 0 everywhere else. \begin{eqnarray} y(t) &=& \int_{-\infty}^t e^{-9(t-\tau)} x(\tau) d\tau \\ &=& \int_{-\infty}^{\infty} e^{-9(t-\tau)} x(\tau) \mathbf{1}_{\{\tau \leq t\}} d\tau \\ &=& \int_{-\infty}^{\infty} e^{-9(t-\tau)} u(t-\tau) x(\tau) d\tau \end{eqnarray}

This is an LTI system with transfer function $e^{-9t}u(t)$.

  • $\begingroup$ Thanks a lot! That's what I was working towards, but was unsure whether the problem as given was always a LTI. $\endgroup$ Commented Oct 26, 2017 at 16:43

Yes it's a time invariant system. Derivation is easy.

Let the output $y(t)$ of the system to the input $x(t)$ be: $$y(t) = \mathcal{T} \{x(t)\} = \int_{-\infty}^{t} e^{-9(t-\tau)} x(\tau) d\tau $$

Let's apply a shifted input $x_d(t) = x(t-d)$ such that its output is $y_d(t)$: $$ \begin{align} y_d(t) &= \mathcal{T} \{x_d(t)\} \\ \\ &= \int_{-\infty}^{t} e^{-9(t-\tau)} x_d(\tau) d\tau \\ \\ &= \int_{-\infty}^{t} e^{-9(t-\tau)} x(\tau-d) d\tau \end{align} $$

Make a change of varibles such that $\tau-d = \beta$: $$y_d(t) = \int_{-\infty}^{t-d} e^{-9(t-d-\beta)} x(\beta) d\beta $$

Now, to see whether $y_d(t)$ equals the shifted output $y(t-d)$, apply the shift on the definition of $y(t)$ as: $$y(t-d) = \int_{-\infty}^{t-d} e^{-9(t-d-\tau)} x_d(\tau) d\tau $$

Then you can deduce that $y_d(t) = y(t-d)$ for all $d$; hence the system is time-invariant.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.