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?



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. $\endgroup$ – Jeff isaacs Oct 26 '17 at 16:43

Yes it's a time invariant system.

Even though you could try to formulate it as a convolution integral to deduce the LTI property of the system, you don't have to do so, as the direct 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 $y_d(t)$; $$y_d(t) = \mathcal{T} \{x_d(t)\} = \int_{-\infty}^{t} e^{-9(t-\tau)} x_d(\tau) d\tau = \int_{-\infty}^{t} e^{-9(t-\tau)} x(\tau-d) d\tau $$

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

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


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