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The transform function of a digital FIR filter can be characterized by $$ H(z) = \sum_{n=0}^{2N} h(n)z^{-n}\quad\text{where $2N$ is the order of the filter.} $$

  • But when we use $2N$ or $N-1$ , $2N+1, \ldots$, on any basis chosen, is it related to linear phase or filter characters , or related to various configurations (low pass, high pass, band pass and band cut).
  • On the other hand , is $h(n)$ the coefficients or weights ?
  • Can you give me a realistic example of bringing the idea closer? Are they converted to electrical signals later? Or what?

I Need Clarification Please.

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First, we call $H(z)$ the transfer function (not transform).

Next, I believe you should have a look at Wikipedia first, many of your questions shows no preliminary research.

Still, I'll try to clarify few things for you.

I am not sure why your summation uses $2N$, but it should be $N$.

$$H(z)=\frac{Y(z)}{X(z)}=\sum_{n=0}^{N}{h(n)z^{-n}}$$ Where N is the order of the filter.

Looking at a block diagram from Wikipedia should help you understand what this equation means: (Note that $b(n)$ is the same as $h(n)$ here)

enter image description here

To your questions:

  • $N$ represents the "oldest" delay, which is the same as the filter order. 5 delays makes a fifth order filter. With bigger order, you can achieve more steep rolloff (better attenuation between passband and stopband). All other characteristics are defined by the value of $h(n)$

  • Coefficients or weights? Aren't they the same thing? Weighting a signal comes down to multiplying it with a coefficients. $h(n)$ are constants value with which you multiply the delayed version of your signal.

Then, your last question is hard to understand, what would be converted to electric signal? In any case, what we consider, is a system with an input ($x[n]$) and an output ($y[n]$). The signal that is fed to that system could be anything, including a electric signal that is converted to digital values.

The coefficients $h(n)$ are merely constants that have no units. DSP theory tells us how to properly select these constants in order to achieve a given behaviour.

Let's make a very easy example with a moving average. The moving average is easy to understand. We could define the moving average like this:

$$y[n]=\frac{x[n]+x[n-1]}{2}$$

If we use this moving average on a noisy signal, we can obtain the following image. Yellow is the noisy signal, blue is the averaged signal.

enter image description here

If we take our moving average equation from above and do a little math we find.

$$Y(z)=\frac{X(z)+X(z)z^{-1}}{2}$$ $$Y(z)=\frac{X(z)(1+z^{-1})}{2}$$ $$\frac{Y(z)}{X(x)}=H(z)=\frac{1z^0}{2}+\frac{z^{-1}}{2}$$ So we got our transfer function. The $h(n)$ coefficients are $$h(n)=[\frac{1}{2}, \frac{1}{2}]$$

These constants values are what define the moving averager, which is itself a type of low-pass filter. Select other constants and you will find other type of filters.

Last point, phase linearity will happen when the $h(n)$ coefficients are symmetrical. For example : $h(n) = [0, 0.2, 0.4, 0.2, 0]$

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  • $\begingroup$ that was beautiful. @N.na91: Hamming's digital filters book is a nice somewhat broader introduction to what Pier described. then, after that, I recommends Lyons DPS text. I'm new at all this also. $\endgroup$ – mark leeds Nov 19 '17 at 7:12

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