I have the following input-output relation for a system:

$$y(t) = Odd Part Of [x(t)]$$

My question is: Is the system causal?

What my approach has been:

I expressed $y(t)$ alternatively as: $$y(t) = \frac{x(t) - x(-t)}{2}\tag{1}$$

Here, when I substitute $x(t)$ with the impulse function, I get the impulse response as $0$ because the impulse function is an even function. Its odd part is $0$. This leads me to believe that the system is causal as the impulse response is zero for negative time.

However, when I substitute $t$ with $-t$ in Eq. $(1)$, I find that for negative time, the output depends on the input at a future time. This would lead me to believe that the system is non-causal.

So my question is actually two-fold here:

  1. How do I reconcile the two seemingly contradictory results?

  2. Why is the impulse response of the system 0? What's the difference between this system and having no system at all?

Any help/pointers would be sincerely appreciated.


1 Answer 1


Clearly, for negative values of $t$, the system needs to know the future in order to determine its output. Hence, the system can't be causal.

Since the system is also time-varying (show it!), its response to an impulse doesn't say much about its general behavior, unlike it would be the case for a linear time-invariant (LTI) system. So the given system's response to an impulse being zero for $t<0$ doesn't say anything at all about causality or any other general properties of the system.

Takeaway: don't use a system's impulse response for drawing conclusions about its properties before having verified that the system is LTI and can in fact be characterized by its impulse response.

  • 1
    $\begingroup$ Gosh! I completely forgot to check if the system is LTI or not, and blindly went ahead with computing the impulse response. I, now, see that it is indeed NOT time-invariant. Thanks a lot for the answer! $\endgroup$ Commented Jul 20, 2020 at 16:22

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