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

In summary, the homework statement is saying that to derive the frequency response for a bandpass FIR filter, one first does a z-transform on the y(n) values, and then uses the H(z) equation to find Y(z). However, the author says that they have not been able to solve for H(z) for the double-sum x(k-2K+2) + x(k-2K+3) + ... + x(k-K+1) + x(k-K+2), and asks for help.
  • #1
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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? :confused:
 
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  • #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]
 
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  • #3
I haven't made any progress on this ... anybody help me out, please?
 

1. What is a difference equation?

A difference equation is a mathematical representation of a system in discrete time, where the output of the system depends on its previous inputs and outputs.

2. How do I obtain the frequency response from a difference equation?

To obtain the frequency response from a difference equation, you can use the Z-transform, which converts the difference equation into a transfer function in the complex frequency domain. The frequency response can then be obtained by evaluating the transfer function at different frequencies.

3. Can I use the difference equation to analyze any type of signal?

Yes, the difference equation can be used to analyze any type of discrete signal, including audio signals, image signals, and control signals.

4. What is the significance of the frequency response in signal processing?

The frequency response provides information about how the system responds to different frequencies in the input signal. This is important in signal processing because it allows us to understand and manipulate the characteristics of the system, such as its filtering properties.

5. Are there any limitations to using a difference equation for frequency response analysis?

One limitation of using a difference equation for frequency response analysis is that it assumes the system is linear and time-invariant. In real-world systems, this may not always be the case and other methods may need to be used for more accurate analysis.

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