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Discrete Time question

  1. Dec 2, 2005 #1
    w[n] = exp(j*pi*n)

    w[n] = (−1)^n

    Hi, can anyone off hand verify why these two exressions are equal? Thanks
     
  2. jcsd
  3. Dec 2, 2005 #2
    exp(i * pi * n) = cos(pi * n) + isin(pi * n).

    You figure out the rest.
     
  4. Dec 2, 2005 #3
    I am assuming n takes on integer values. In which case and expansion of exp(i*pi*n)=cos(pi*n)+i*sin(pi*n). Now for n in Z, sin(pi*n)=0, since: sin(0)=sin(pi)=sin(2pi)=sin(3pi)=...=0. Similarily, cos(pi*n)=(-1)^n, since cos(0)=cos(2pi)=cos(4pi)=..cos(2kpi)=1, and cos(pi)=cos(3*pi)=...cos((2k+1)pi)= -1. Therefore exp(i*pi*n)=(-1)^n for any n in Z.
     
  5. Dec 2, 2005 #4

    uart

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    Science Advisor

    The easiest way is just to write [tex]e^{j \pi n} = (e^{j \pi})^n[/tex] and note that [tex]e^{j \pi} = -1[/tex]
     
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