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

Question

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!

Answer 1

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, $\delta[n-k]=\delta[k-n]$.

Answer 2

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 :

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

and

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

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...