# Contradiction while using the convolution sum for a non-LTI system

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?

• 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? – jramsay42 Apr 13 at 1:40
• 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). – Pranshu Malik Apr 13 at 1:54
• 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 – jramsay42 Apr 13 at 2:11
• 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. – Pranshu Malik Apr 13 at 2:20
• 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. – jramsay42 Apr 13 at 3:06