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Homework Help: Determining if a System is Time-Invariant and/or Linear

  1. May 6, 2012 #1
    1. The problem statement, all variables and given/known data
    For each of the following systems, determine whether or not the system is Time-Invariant, Linear, and causal.

    a.) y[n] = x[n]cos(0.2*pi*n)
    there are more but if I can figure this out I should be able to get the others

    2. Relevant equations
    Time Invariant ---> if x[n] produces y[n] then x[n - d] produces y[n - d]
    Linear ---> if x1[n] produces y1[n] and x2[n] produces y2[n], then x[n] = ax1[n] + bx2[n] produces y[n]=ay1[n] + by2[n]

    3. The attempt at a solution
    I am able to prove that y[n] = x[n]cos(0.2*pi*n) is linear by saying
    let x[n] = ax1[n] + bx2[n]
    then y[n] = (ax1[n] + bx2[n])cos(0.2*pi*n) = ax1[n]cos(0.2*pi*n) + bx2[n]cos(0.2*pi*n)
    so y[n] = ay1[n] + by2[n] proving that it is linear

    I am not able to prove, at this point, that y[n] is either time-invariant or not.
    All I have is this:
    let g[n] = x[n - d]
    then y[n - d] = g[n]cos(0.2*pi*n) = x[n - d]cos(0.2*pi*n), so it is time-invariant?
    This doesn't seem right though for whatever reason.
  2. jcsd
  3. May 8, 2012 #2


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    Homework Helper

    The system is time invariant if it gives the same output regardless of whether a delay is made to the input signal, or to the output signal.
    So, given
    y[n] = x[n]cos(0.2*pi*n)
    If the output signal is delayed,
    y[n-d] = x[n-d]cos(0.2*pi*(n-d))
    However, if the delay was made to the input signal, before inputing it into the system ... Can you take it from here?
  4. May 8, 2012 #3
    Im following you until you talk about making the delay prior to inputting the signal.

    Let me see if I am understanding. It looks to me like, unless you pick a d that delays it by 2*pi*k where k is an interger, it wont produce the same output?
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