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Subtraction of step functions

  1. Apr 7, 2015 #1
    Hi. Here is one example from my book.
    Calculate Fourier transform of signal:

    p1.png

    Here is solution:
    We can write x(n) as:

    p2.png ,

    where x1(n) is u(n+N)-u(n-N-1). We can write:

    p3.png

    (we used that cos(n)=(1/2)*(exp(j*n)+exp(-j*n)).
    Using properties of Fourier transform of discrete signal:

    p4.png ,
    Fourier transform of our signal will be:

    p5.png

    We will find Fourier transform of x1(n):

    p7.png


    How they calculated that u(n+N)-u(n-N-1) equals 1 for -N<=n<=N and 0 for other values of N?
     
  2. jcsd
  3. Apr 8, 2015 #2
    Try setting up a number line with -N and N + 1 on it, and then mark the intervals on that number line where u(n + N) and u(n - N - 1) are either 0 or 1.

    You should be able to use that to say something about where u(n + N) - u(n - N - 1) is then 0 or 1.
     
  4. Apr 8, 2015 #3
    Here is how I tried to solve this algebraically.
    We have function u(n+N)-u(n-(N+1)). Let's say N is some positive number, for example N=1.
    1. When n is less then -N, for example n=-2, we have u(n+N)-u(n-(N+1))=0-0=0.
    2. When -N<n<N+1 , for example n=-0.5, we have u(n+N)-u(n-(N+1))=1-0=1.
    3. When n>N+1, for example n=3, we have u(n+N)-u(n-(N+1))=1-1=0.
    So our function equals 1 for n between -N and N+1 but solution from my book doesn't agree with it?
     
  5. Apr 8, 2015 #4
    Take care to consider the endpoints of your intervals. What happened to -N?

    Also, n and N are integers.

    What happened to N + 1?
     
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