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

[courteous]

Homework Statement

Given this difference equation y(k) (of a bandpass FIR filter) ...
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)

... how does one derive this frequency response H(f)?
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)}

Homework Equations

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

The Attempt at a Solution

Don't know how to handle those double-sums in y(k). :shy:

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

So, there's some kind of "triangular/diagonal pattern" since x(k-2K+2) and x(k) (the very first and last x-element) are the only one to appear only once. Now what? Is this even the right approach?

 

[courteous]

I'd appreciate (any) help. Continuing from 1st post:
\cdots = \left[ x(k-2K+2) + 2x(k-2K+3) + \cdots + (K-1)x(k-K+1) + \cdots + x(k) \right]

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

So, the transfer function H(z) of the whole y(k) (from the 1st post) can be written as
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]

If the above is correct, then how can this
\left[ z^{-2K+2} + 2z^{-2K+3} + \cdots + (K-1)z^{-K+1} + \cdots + 1 \right]
be equal to this, when evaluated on the unit circle z=e^{j2\pi f}?
\left( \frac{\sin{\pi f K}}{\sin{\pi f}} \right)^2 e^{-2j\pi f(K-1)}

 

[courteous]

I haven't made any progress on this ... anybody help me out, please?