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How to get (frequency response) from this [difference equation]?

  1. Dec 7, 2013 #1
    1. The problem statement, all variables and given/known data
    Given this difference equation [itex]y(k)[/itex] (of a bandpass FIR filter) ...
    [tex]y(k) = \frac{1}{K^2} \sum_{m = k-K+1}^k \; \sum_{n = m-K+1}^m x(n) - \frac{1}{L^2} \sum_{m = k-L+1}^k \; \sum_{n = m-L+1}^m x(n)[/tex]

    ... how does one derive this frequency response [itex]H(f)[/itex]?
    [tex]H(f) = \frac{1}{K^2} \left( \frac{\sin{\pi f K}}{\sin{\pi f}} \right)^2 e^{-2j\pi f(K-1)} - \frac{1}{L^2} \left( \frac{\sin{\pi f L}}{\sin{\pi f}} \right)^2 e^{-2j\pi f(L-1)}[/tex]


    2. Relevant equations
    Usually, I'd do Z-transform on (disregarding general case of IIR filters) [tex]y(n) = \sum_{k=0}^K b_l x(n-k)[/tex] to get the transfer function [tex]H(z) = \frac{Y(z)}{X(z)} = \sum_{k=0}^K b_l z^{-k}X(z)[/tex] and then evaluate on unit circle (i.e. change [itex]z[/itex] with [itex]e^{jw}[/itex]).


    3. The attempt at a solution
    Don't know how to handle those double-sums in [itex]y(k)[/itex]. :shy:

    Since the issue is "symmetrical", I've tried solving the 1st half of [itex]y(k)[/itex] (the lowpass filter with a higher cutoff frequency, [itex]K < L[/itex]). I've tried expanding the sum, hoping a pattern would emerge:
    [tex]\sum_{m = k-K+1}^k \; \sum_{n = m-K+1}^m x(n) = \\
    \left[x(k-2K+2) + x(k-2K+3) + \dots + x(k-K+1) \right] +\\
    \left[x(k-2K+3) + x(k-2K+4) + \dots + x(k-K+2) \right] +\\
    \cdots +\\
    \left[x(k-K+1) + x(k-K+2) + \dots + x(k) \right][/tex]

    So, there's some kind of "triangular/diagonal pattern" since [itex]x(k-2K+2)[/itex] and [itex]x(k)[/itex] (the very first and last x-element) are the only one to appear only once. Now what? Is this even the right approach? :confused:
     
    Last edited: Dec 7, 2013
  2. jcsd
  3. Dec 11, 2013 #2
    I'd appreciate (any) help. Continuing from 1st post:
    [tex]\cdots = \left[ x(k-2K+2) + 2x(k-2K+3) + \cdots + (K-1)x(k-K+1) + \cdots + x(k) \right][/tex]

    Applying [itex]Z[/itex]-transform, we get
    [tex]X(z) \left[ z^{-2K+2} + 2z^{-2K+3} + \cdots + (K-1)z^{-K+1} + \cdots + 1 \right][/tex]

    So, the transfer function [itex]H(z)[/itex] of the whole [itex]y(k)[/itex] (from the 1st post) can be written as
    [tex]H(z)=\frac{Y(z)}{X(z)}= \frac{1}{K^2}\left[ z^{-2K+2} + 2z^{-2K+3} + \cdots + (K-1)z^{-K+1} + \cdots + 1 \right] - \frac{1}{L^2}\left[ z^{-2L+2} + 2z^{-2L+3} + \cdots + (L-1)z^{-L+1} + \cdots + 1 \right][/tex]

    If the above is correct, then how can this
    [tex]\left[ z^{-2K+2} + 2z^{-2K+3} + \cdots + (K-1)z^{-K+1} + \cdots + 1 \right][/tex]
    be equal to this, when evaluated on the unit circle [itex]z=e^{j2\pi f}[/itex]?
    [tex]\left( \frac{\sin{\pi f K}}{\sin{\pi f}} \right)^2 e^{-2j\pi f(K-1)}[/tex]
     
    Last edited: Dec 11, 2013
  4. Dec 15, 2013 #3
    I haven't made any progress on this ... anybody help me out, please?
     
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