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For Linear Time-invariant systems mapping any complex sequence $x[n]$ to an output $y[n]$, the question is if there the kernel must be trivial, given that the LTI system does not map everything to the zero sequence.

So , the ideal low-pass filter (which is LTI) has a kernel of any frequencies above the cutoff.

Ok, so then I wanted to find conditions that would make the kernel trivial. The best I could come up with are:

  1. input signal is causal ($x[n] = 0$ for n<0) WLOG assume $x[0] \neq 0$
  2. LTI system is causal ($y[n]$ depends only on values of $x[n-k]$ for positive $k$)
  3. LTI system is relaxed ($y[n] =0$ until $x[n]$ starts)

If these conditions are true, then I argue that the kernel is trivial.

proof: The LTI system can be written as a convolution, $x[n] * h[n]$, where $h[n] $ is the impulse response. Then y[n] at n=0 depends only on x[0]. If $h[0] \neq 0$, we are done. If not, there will be a output caused by x[0] sometime later, and it cannot be canceled out by the impulse response to the signal at time n, because that would be delayed further.


My questions:

A.These conditions appear sufficient but not necessary. Is there a sufficient and necessary condition? Can I remove some of the 3 conditions and still be sufficient to prove the kernel is trivial?

B. Is my proof correct/ suggestions to make it better.

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