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Ideal Filter - Windowed - DTFT/Highpass

  1. Apr 2, 2014 #1
    1. The problem statement, all variables and given/known data
    Given:
    [itex]H_{dd}\left(e^{j\omega}\right)=j\omega e^{\frac{-j\omega}{2}}, \left|\omega\right|\le\pi[/itex]

    Find: [itex]H_{3}\left(e^{j\omega}\right)[/itex] where
    [itex]H_{3}\left(e^{j\omega}\right)[/itex] is the spectrum of [itex] h_{dd}\left(n\right)\left(W_N\left(n\right)\right) [/itex] and [itex] W_N\left(n\right)=1 for \frac{-N}{2}\le n \le \frac{N}{2} , [/itex] 0 else


    2. Relevant equations
    DTFT Synthesis: [itex] \frac{1}{2\pi}\int_{-\pi}^{\pi} H\left(e^{j\omega}\right)e^{j\omega n}d\omega [/itex]
    [itex] DTFT Analysis: \sum_{n=\frac{-N}{2}}^{\frac{N}{2}} h(n)e^{-j\omega n} [/itex]

    3. The attempt at a solution

    Step 1: Synthesis [itex]h_{dd}\left(n\right)[/itex]. This is done through integration by parts and my result is:

    [itex] h_{dd}(n)=-sin\left(\pi\left(n-\frac{1}{2}\right)\right) [/itex]

    Step 2: Window - ok. [itex] -\frac{N}{2}\le n \le \frac{N}{2} [/itex] is the new range.

    Step 3: DTFT windowed function result:
    [itex] H_3\left(e^{j\omega}\right) = -\sum_{n=-N/2}^{N/2} \frac{sin\left(\pi n - \frac{\pi}{2}\right)}{\pi \left(n-1/2\right)^2}e^{-j\omega n} [/itex]

    Really cool - but when plotted versus frequency, I get a constant, regardless of the size of N.
    The goal of this problem was to prove that as N--> big that the filter approaches the ideal high-pass differentiator. I must have made a mistake somewhere but I'm not sure where, if anyone could see if they receive a different result I would be much obliged. Thanks!

    -DR
     
  2. jcsd
  3. Apr 3, 2014 #2
    Found my mistake - it turns out that the above is actually correct and corresponds to Sum( (-1)^(n-1/2)/(denom) * e^-jwn).

    There was an error in my MATLAB code (ridiculous error) where my for loop looked like:

    for i=length(n)

    instead of

    for i=1:length(n)

    hence the reason I was only getting one constant value... You think it's some important mistake in your math and it turns out to be a typo.

    Oh well, Go Bears.
     
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