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Problem

Given the compound system below, with the input $x(t)=\operatorname{sinc(t)}$, the output of A is $y(t)=\operatorname{sinc(2t)}$ and the output of B is $z(t)=\operatorname{sinc(t)}$, determine which of the A, B systems are Linear time invariant (LTI).
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  • a) A
  • b) B
  • c) Both
  • d) Neither

My approach:

Using two facts below (if I got them right):

  • A and B are inverse of each other (because the overall output is equal to the input)

  • If an LTI system is reversible, then the reverse is also LTI.

after a quick scanning I crossed a, b as the answered. I concluded that either both systems are LTI or neither of them are.

For the next step, using the given information I derived the system A:

\begin{align} x(t)&=\frac {\sin(\pi t)}{\pi t}\\ y(t)&=\frac {\sin(2\pi t)}{\pi t}=\frac{2\sin(\pi t)\cos(\pi t)}{\pi t}=\frac{2\sin(\pi t)\sin(\pi t+\frac{\pi}{2})}{\pi t}=2\pi t x(t)x\left(t+\frac{\pi}{2}\right) \end{align}

now with the $t$ as the coefficient of the system we can say $y$ is not time invariant and therfore not LTI. so I chose d as the correct answer.
But the text book I'm reading proves that b is the answer, using another method.

Questions:

So my questions are:

  • Is my approach/facts flawed?
  • If so, what is the correct approach to solve this problem.
  • OR possibly did the book get it wrong?

Any help would be highly apprecieted!

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    $\begingroup$ What's quite flawed about your approach is the idea that a single such simple pair of input and output signals would give you conclusive information about whether a system is LTI. Really, you can never prove something is LTI by simply looking at a given input and output, only disprove that a system is LTI. (Though in practice, if you compare the power spectrums for many different signals and notice that the output for each frequence is always same same frequency-wise multiple of the same frequency in the input, this is a good indication that the system may well be LTI). $\endgroup$
    – leftaroundabout
    Mar 20, 2017 at 15:57
  • $\begingroup$ @Amen could you say the name of the book you are learning from? $\endgroup$
    – arash javan
    Mar 22, 2017 at 13:44

1 Answer 1

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This is how I would approach the problem.

System $A$ cannot be LTI because its output has frequencies not present in its input. The input spectrum is a brickwall from $f=-1/2$ to $f=1/2$, but the output spectrum is a brickwall from $f=-1$ to $f=1$.

It's impossible to arrive at a conclusion for system B. It could be an ideal LTI low-pass filter, but it could also be something else (for example, it could be a signal generator).

Note that we cannot say that A and B are inverses of each other. That statement requires $x(t)=B[A[x(t)]]$ to be true for all $x(t)$. Here, we only know that is true for one specific input.

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  • $\begingroup$ Can we use my two facts to conclude that system B is not LTI as well? $\endgroup$
    – Amen
    Mar 19, 2017 at 16:03
  • $\begingroup$ by the way the book exactly used your approach for the system A and reversed this logic to state B is LTI. $\endgroup$
    – Amen
    Mar 19, 2017 at 16:04
  • $\begingroup$ I don't think so -- B could be LTI. However, given the information we have, there's no way to tell. $\endgroup$
    – MBaz
    Mar 19, 2017 at 16:04
  • $\begingroup$ @Amen, please see my edit. We can't be sure A and B are inverses. $\endgroup$
    – MBaz
    Mar 19, 2017 at 16:08

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