0
$\begingroup$

I am trying to see if

y[n] = [cos(πn)]x[n]

is casual, stable, linear and shift-invariant.

I came up with the solution that it is not stable since it is not "summable" (As my professor says). I took the limit of cos(πn) (correct me if that method is wrong)

For the causal part, I learned that for it to be non-causal, the output depends upon future inputs. I believe that this is non-causal since there are no present or past inputs.

for Linearity: I said it was linear, but I am not sure on this one.

for shift-invariance I stated that this system was indeed invariant

$\endgroup$
2
  • 2
    $\begingroup$ For the last two, it'd be very useful if you included more of your analysis for each part; not just what your conclusion is, but how you arrived at it. $\endgroup$ – MBaz Sep 16 '20 at 0:15
  • $\begingroup$ for sure @MBaz, my professor state the following: "The way to prove linearity is to define two inputs sequences and see whether a linear combination of those sequences into the system results in the same linear combination of the corresponding outputs. So if we define x1 and x2 that each have corresponding system outputs of y1 and y2, and form a linear combination of those two input sequences: ax1 + bx2, then the system is linear if the corresponding output is ay1 + by2." So, I did that above where y1 = acos(πn_1)x[n_1] & y2 = bcos(πn_2)x[n_2] and added them together and think its linear. $\endgroup$ – Jorge Juarez Sep 16 '20 at 0:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.