# Is $y[n] = n x[n]$ an LTI system?

How can I test if $y[n] = n x[n]$ is an LTI system? And any other system for that matter?

For example, how come $y[n] = \left( \frac{1}{2} \right)^n u[n]$, where $u[n]$ is the unit step, is an LTI system?

• There's no magic here. You just go into your textbook, write down the very definition of "L", and "TI", and check your system against that. Basically, you go ahead and just do your homework :) – Marcus Müller Dec 3 '17 at 13:29
• @MattL. Thank you! I was confused why that is an LTI. Apparently its not. What about the first one? – anon Dec 3 '17 at 13:33
• $$y[n] = \left( \frac{1}{2} \right)^n u[n]$$ is not an LTI system. It is the output of a causal LTI system. – Rodrigo de Azevedo Dec 4 '17 at 4:44

Here instead of a general proof method, let us use the method known as the counter example which relies on the fact that a single counter-example that violates the claim is sufficient (enough) to prove that the claim is wrong.

Assume that the system was LTI,(the claim is that system is LTI). Then consider the following inputs $x_1[n] = \delta[n]$ and $x_2[n] = \delta[n-1]$ with corresponding outputs $y_1[n] = 0 \cdot \delta[n] = 0$ for all $n$ and $y_2[n] = 1.\delta[n-1]=\delta[n-1]$.

Now if the system is actually LTI, then its output should satisfy $y_2[n] = y_1[n-1]$. However one can see that this is not the case, as $$y_2[n]= \delta[n-1] \neq y_1[n-1] = 0.$$ Hence this counter-example has proved that the claim was wrong; i.e., the system is not LTI. In fact, what we have actually shown is that the system is not time-invariant, but then the fact that it's also not LTI follows as a consequence.

• Thanks for the awesome explanation. However isn't y1[n]=n⋅δ[n] ? Why is it 0⋅δ[n]? – anon Dec 3 '17 at 14:06
• @anon It's the sifting property of the impulse function : $$f[n] \delta[n-a] = f[a]\delta[n-a]$$ which becomes specifically as $$n \delta[n] = 0 \delta[n] = 0 , \forall n$$ – Fat32 Dec 3 '17 at 14:19