There is this following question.

Consider the transformation H {x}[n] = n x [n]. Does H define an LTI system?

What I understood from the question is:

x[n] -> H -> y[n] or according to the question, x[n] -> H -> n x[n]

to prove the time invariance, if I put n-no instead of n, it should be,

x[n-no] -> H -> y[n-no] = n x[n-no] but it turns out to be, x[n-no] -> H -> n x[n - no] - no x[n-no] which means the output depends on extra term no x[n-n0] and is not time invariant.

Incase, the trasformation was any constant * x[n] the it would have conserve the time invariance property, if I am not wrong.

But the proof mentioned in the notes is, enter image description here

I have difficulty understanding the proof. If anyone could explain it to me or if I understood the question wrong, I'd appreciate it.

  • 1
    $\begingroup$ Your approach and conclusion are correct. There are different ways to solve the same problem and the one shown in the solution is different from yours. Yours is actually "better" since it's more general. $\endgroup$
    – Hilmar
    Commented Jan 30, 2021 at 18:51
  • $\begingroup$ @Hilmar Thank you. But can you explain to me why is there H{x}[n] = 0? I think I am lacking the mathematical concept behind it. $\endgroup$
    – Rima
    Commented Jan 31, 2021 at 9:45

1 Answer 1


Your notes gave a counter example, you just proved it in the general case. Your note used your x[n] as delta and showed that y[n-1] unequal to H{x[n-1]} which is exactly what you showed.

In the notes example H{x}=0 for all n, as discrit delta equals zero foll all n but n=0, which means H is zero every where for all input of the kind delta[n-n0] (becouae of the n factor which zeros y at its only non zero point) Than you have zero output for delta input. But as you showed time varied output isn't zeros every where.


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