Determining if the given system is LTI

Question

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).

• 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!

Answer 1

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.