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!