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Homework Statement
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]
Homework 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]).
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?
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