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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 \lbrack n \rbrack = \delta\lbrack n + 3\rbrack\$ and \$h_2\lbrack n \rbrack = \delta\lbrack 5 − n\rbrack\$. What is the output when the input is \$x\lbrack n \rbrack = n\$, i.e., find \$y \lbrack n \rbrack = \mathcal{T}_2 \{ \mathcal{T}_1 \{ x \lbrack n \rbrack\}\}\$.

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.

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!

EDIT: While answering @jramsay's comments, 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, technically, not the exact description of the underlying system?

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  • \$\begingroup\$ Being LTI doesn't imply that the input and output of the second impulse response should be the same. Or did you mean something else? \$\endgroup\$ – jramsay42 Apr 13 at 1:40
  • \$\begingroup\$ I tried to get the input-output relationship using the impulse response, but it gave me a different relationship because it is not LTI, at least according to me (since there is scaling by -1 in the index of the impulse response). \$\endgroup\$ – Pranshu Malik Apr 13 at 1:54
  • \$\begingroup\$ All you showed was that convolving a signal x[k] with the impulse response gives a different output signal, that doesn't mean the system isn't LTI \$\endgroup\$ – jramsay42 Apr 13 at 2:11
  • \$\begingroup\$ No, but given that I'm convolving with the impulse response, I should get the same output. I don't, which would, I'm assuming, only happen if the system is not LTI. Also, 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). Also, since \$\delta[n] \$ is even, the non-LTI system's impulse response, for example \$\delta[1-n]\$, will equal \$\delta[n-1]\$ which corresponds to an LTI system. This explains why I am getting LTI characteristics after the convolution. \$\endgroup\$ – Pranshu Malik Apr 13 at 2:20
  • \$\begingroup\$ Why would you get the same output? That depends on what the impulse response of the system is. Also I think you're confusing LTI with Causal. \$\endgroup\$ – jramsay42 Apr 13 at 3:06
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This question has been answered on Signal Processing StackExchange here. (Thanks to @Matt L.)

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