In a recent quiz, we were given the following problem:

The cascaded LTI systems $\mathcal{T}_1$ and $\mathcal{T}_2$ respectively have impulse responses $h_1[n]=\delta[n+3]$ and $h_2[n]=\delta[5−n]$. What is the output when the input is $x[n]=n$, i.e., find $y[n]=\mathcal{T}_2\{\mathcal{T}_1\{x[n]\}\}$.

I am primarily confused about the "LTI-ness" of the system and at a contradiction; given the second stage to be LTI (assuming that for a system to be LTI as a whole, all its stages/subsystems must also be LTI). Thus, we can say that a sub-system with $h \lbrack n \rbrack = \delta \lbrack 5 - n\rbrack$, i.e. $y \lbrack n \rbrack = x \lbrack 5 - n\rbrack$ should be LTI, which already seems to be false. To prove that, I considered the following convolution sum: $$\tilde{y} \lbrack n \rbrack = \displaystyle{\sum_{k = -\infty}^{\infty} x \lbrack k \rbrack \delta \lbrack 5 - n + k \rbrack} = x \lbrack n-5 \rbrack \ne x \lbrack 5-n \rbrack = y \lbrack n \rbrack$$

And this is a contradiction since, after the convolution, the output doesn't match the original output with which we started.

Also, I realized that interestingly, the convolution always gives a result that corresponds to an LTI system (as in the case above too: $y[n]=x[n−5]$ is LTI whereas $y[n] = x[5-n]$ is not).

Also, since $\delta[n]$ is even, any non-LTI system's impulse response, for example, $\delta[1−n]$, will equal $\delta[n−1]$ which corresponds the impulse response of an LTI system. This explains why I am getting LTI characteristics after the convolution. This is interesting too since in either way the impulse response implies delaying the signal by 1 (in the current example).

And so did the impulse response $\delta[5-n]$ stated in the quiz question just qualitatively imply a delay of 5, and is, technically, not the exact description of the underlying system?

So, in summary, I have the following two doubts:

  1. Is the convolution sum only true for the output relation of an LTI system? If so, then can it be used to prove non-LTI-ness of a system in the same manner as above?
  2. Is the quiz question incorrect?

Any help will be greatly appreciated!


2 Answers 2


I think there are two misunderstandings here. First, you seem to think that if a system's impulse response is $h[n]=\delta[5-n]$, then its response to an input sequence $x[n]$ equals $x[5-n]$. This is not the case. And second, you seem to think - or at least you're not sure about it - that a system with a response $x[5-n]$ to an input is LTI, and hence is completely specified by its impulse response. That's also not true.

The system $y[n]=x[5-n]$ is linear but it is not time-invariant. In order to prove this you just have to show that the response to $x[n-k]$ for some integer $k$ does not equal $y[n-k]$, where $y[n]$ is the response to $x[n]$. This is straightforward to show, and I leave this proof up to you.

An LTI system with impulse response $h[n]=\delta[5-n]$ delays its input by $5$ samples: $y[n]=x[n-5]$, as you've shown yourself in your question. This is also immediately obvious from the fact that $\delta[n]$ is an even sequence, i.e., $\delta[n]=\delta[-n]$, and, consequently, $\delta[n-k]=\delta[k-n]$.

  • $\begingroup$ Thanks a lot for pointing out my misconceptions! I still have one doubt: why would the system $y[n] = x[5-n]$ not have an impulse response of $\delta[5-n]$, or do we just avoid using the term "impulse response" for non-LTI systems? Because this is what is still confusing me. $\endgroup$ Commented Apr 13, 2019 at 16:53
  • 1
    $\begingroup$ @PranshuMalik: The response of that system to an impulse $x[n]=\delta[n]$ is indeed $\delta[5-n]$. So in that sense this is its impulse response. But this impulse response does not characterize the system in such a way that the responses to all possible input signals can be computed from it (via the convolution sum). But that's what we usually mean by referring to the impulse response of a (LTI) system. $\endgroup$
    – Matt L.
    Commented Apr 13, 2019 at 16:56
  • $\begingroup$ Ahh. I see my mistake completely now. Thanks again :) $\endgroup$ Commented Apr 13, 2019 at 17:06
  • $\begingroup$ if $y[n] = x[5-n]$ does not obey convolution sum so please show the convolution sum that $y[n] = h[n]$ produce what ? $\endgroup$
    – Fat32
    Commented Apr 13, 2019 at 18:51
  • 1
    $\begingroup$ (produces) a joke may be ;-))) $\endgroup$
    – Fat32
    Commented Apr 13, 2019 at 19:22

You've some troubles about the LTI property and the impulse response. First of all let me give you the answer to your quiz question.

Given two LTI systems $\mathcal{T}_1$ and $\mathcal{T}_2$ with impulse responses :

$$h_1[n] = \delta[n+3]$$ $$h_2[n] = \delta[5-n] = \delta[n-5]$$

you can compute the output of the cascade $$y[n] = \mathcal{T}_2 \{ \mathcal{T}_1 \{ x[n] \} \} $$ for the input signal $x[n] = n$ as :

$$ y_1[n] = \mathcal{T}_1 \{ x[n] \} \implies y_1[n] = n \star \delta[n+3] = (n+3) $$


$$ y_2[n] = \mathcal{T}_2 \{ y_1[n] \} \implies y_2[n] = (n+3) \star \delta[n-5] = (n-2) $$

so your output is $$y[n] = n-2$$

Note that $h_2[n] = \delta[5-n]$ and $h_2[n] = \delta[n-5]$ are the same systems, and yield the same output.

Note again that both systems are declared to be LTI and are given their respective impulse responses $h_1[n]$ and $h_2[n]$ , therefore their cascade will also be LTI and furthermore the output does not depend on the ordering of the systems; i.e.

$$y[n] = \mathcal{T}_2 \{ \mathcal{T}_1 \{ x[n] \} \} = \mathcal{T}_1 \{ \mathcal{T}_2 \{ x[n] \} \} $$

Now coming to your own questions:

1- You cannot use convolution sum to deduce LTI property. You cannot form a convolution sum if the system is not already LTI.

2- The question seems ok:

Note: your claim that $ h[n] = \delta[5-n] $ , signifies a non-LTI system is wrong...

  • $\begingroup$ Thanks a lot for your answer! But I'm still wondering that If $y[n] = x[-n] $ is non-LTI, then why isn't $y[n] = x[5-n]$ also non-LTI? $\endgroup$ Commented Apr 13, 2019 at 11:15
  • $\begingroup$ $y[n] = \mathcal{T} \{x[n]\} = x[-n]$ is a general system input-output definition which is time-varying therefore not LTI and does not have an impulse response $h[n]$. On the other hand, the impulse response $h[n] = \delta[m-n]$ signifies (by definition!) an LTI system whose input output relation is $y[n] = h[n] \star x[n] = \delta[m-n] \star x[n] = x[n-m]$ where there is no conradiction... $\endgroup$
    – Fat32
    Commented Apr 13, 2019 at 12:24
  • $\begingroup$ Hmm, interesting. Why doesn't $y[n] = x[-n] $ not have an impulse response? If your input is $\delta[n] $, then the output will be $h[n] = \delta[-n] $. And I understand that this response will not help us with characterizing the system's output for a general input. But, I don't really understand why $y[n] = \delta[m-n] $ is LTI. This is, I think, the root of my doubts $\endgroup$ Commented Apr 13, 2019 at 12:47
  • $\begingroup$ you have a problem with the concept of system and impulse response. Please consult a book on Signal's and Systems chapter one and two. $\endgroup$
    – Fat32
    Commented Apr 13, 2019 at 13:47
  • $\begingroup$ Yes, I've read through my course notes on these topics. I'll be very grateful if you could redirect me to a specific resource or tell me a particular topic to search for. The notes did cover why time reversal is time-varying, from which I came to the extension that any inversion of index, that is, making $y[n] = x[n-K]$ into $y[n] = x[K-n] $ will result in a time-varying system. And as far as i understand, impulse response of a system is literally an impulse's response. So regardless of the system's linearity or time-invariance, an impulse response has to be defined for any system. $\endgroup$ Commented Apr 13, 2019 at 14:08

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