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DTFT with Unit Step Function

  1. Mar 25, 2013 #1
    So I'm trying to find the DTFT of the following; where u(n) is the unit step function.

    [itex]u \left( n \right) =\cases{0&$n<0$\cr 1&$0\leq n$\cr}[/itex]

    I want to find the DTFT of

    [itex]u \left( n \right) -2\,u \left( n-8 \right) +u \left( n-16 \right)[/itex]

    Which ends up being a piecewise defined function looking like

    [itex]u \left( n \right) -2\,u \left( n-8 \right) +u \left( n-16 \right) = \cases{1&$0\leq n$\ and \ $n\leq 7$\cr -1&$8\leq n$\ and \ $n\leq 15$\cr}[/itex]

    With the function zero elsewhere.

    I plug this into the formula for a DTFT and get the following:

    [itex]1+{{\rm e}^{-i\omega}}+{{\rm e}^{-2\,i\omega}}+{{\rm e}^{-3\,i\omega}}
    +{{\rm e}^{-4\,i\omega}}+{{\rm e}^{-5\,i\omega}}+{{\rm e}^{-6\,i\omega
    }}+{{\rm e}^{-7\,i\omega}}-{{\rm e}^{-8\,i\omega}}-{{\rm e}^{-9\,i
    \omega}}-{{\rm e}^{-10\,i\omega}}-{{\rm e}^{-11\,i\omega}}-{{\rm e}^{-
    12\,i\omega}}-{{\rm e}^{-13\,i\omega}}-{{\rm e}^{-14\,i\omega}}-{
    {\rm e}^{-15\,i\omega}}[/itex]

    This should be correct, however, it is very ugly and I'm looking for a better form for my answer. I cannot reduce the summation using a harmonic series because the coefficient |a| = 1.

    I can keep it in summation form but I feel like I'm missing an easy step that can simplify this.

    Thanks for any help you can offer.
     
  2. jcsd
  3. Mar 25, 2013 #2

    jbunniii

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    Try using this fact to simplify your answer:
    $$\sum_{n = 0}^{N-1} x^n = \begin{cases}
    \frac{1-x^{N}}{1-x} & \text{ if }x \neq 1 \\
    \\
    N & \textrm{ if }x = 1
    \end{cases}$$
     
  4. Mar 25, 2013 #3
    Well I feel stupid. I saw this formula but like I said in my original post I thought I couldn't use it since series wasn't approaching zero. I guess I was being stupid and thinking about infinite series. Since this is finite of course I can use that.

    If my math is right I get

    [itex]{\frac { \left( 1-{{\rm e}^{-8\,i\omega}} \right) ^{2}}{1-{{\rm e}^{-i\omega}}}}[/itex]
     
  5. Mar 25, 2013 #4

    jbunniii

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    I haven't checked your math, but I will mention that you can always simplify expressions involving ##1 - e^{-ix}## by factoring out ##e^{-ix/2}## to obtain ##e^{-ix/2}(e^{ix/2} - e^{-ix/2}) = 2i e^{-ix/2} \sin(x/2)##.
     
  6. Mar 26, 2013 #5
    Thanks. I've got it in this form now:

    [itex]{\frac {2\,i \left( \sin \left( 4\,\omega \right) \right) ^{2}{{\rm e}^{-15/2\,i\omega}}}{\sin \left( 1/2\,\omega \right) }}[/itex]

    After all this I found a formula for any equation in the form u(n) - u(n-L). Thanks for the help.
     
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