1
$\begingroup$

Professor asked me what filter is better -1/3 2/3 -1/3 or -1/4 1/2 -1/4, the answer was second. But what those coefficients means? It was during discussion of linear filter but i'm not sure was it about it.

$\endgroup$
6
$\begingroup$

Both filters can be re-written in the form: \begin{equation} H(z) = \alpha ( 1 -2z^{-1} +z^{-2} ) \end{equation} Clearly, in the first case, $\alpha=-\frac{1}{3}$ and in the second $\alpha=-\frac{1}{4}$.

In other words, the shape of the filter's frequency response is the same, the only difference is the scaling.

I can think of two reasons of why the second filter may be preferred.

  • In the case of the second filter, the maximum gain of the filter (occuring at the Nyquist frequency) is normalized to $1$, whereas the first filter has a maximum gain of $\frac{4}{3}$.

  • The other more compelling reason however, is probably that the coefficients of the second filter can be expressed as powers of two ($\frac{1}{4}=2^{-2}$, $\frac{1}{2}=2^{-1}$) and therefore multiplication by these coefficients can be quickly implemented as bitwise operations on a computer or digital signal processor.

$\endgroup$
  • 1
    $\begingroup$ Good answer. I'd add that the recurring decimal co-efficients required to represent $\frac{1}{3}$ will be truncated in any real implementation of the filter leading to inaccuracies in the response. $\endgroup$ – tobassist Oct 29 '13 at 14:08
  • $\begingroup$ What is H(z)? Derived from this? postimg.org/image/sbgs9a27z $\endgroup$ – Олег Кривцов Nov 1 '13 at 14:26
  • $\begingroup$ $H(z)$ is the transfer function which describes the difference equation that you link to. Specifically, $H(z) = \frac{Y(z)}{X(z)}$, where $X(z)$ and $Y(z)$ are the Z-transforms of $y[n]$ and $x[n]$. Check out: en.wikipedia.org/wiki/Z-transform $\endgroup$ – Kenneide Nov 1 '13 at 16:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.