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I came across a concept that says we can design LPH and HPF from APF when the transfer function of the two APFs given as

$$\begin{align}A_0(z)&=1\\ A_1(z)&=\frac{-a+z^{-1}}{1-az^{-1}}\\\\ H_{LP}(z)&=A_0(z)+A_1(z)\\ H_{HP}(z)&=A_0(z)-A_1(z)\end{align}$$

i tried to implement this in matlab but i did not get the desired frequency response..kindly help me with the matlab code for the same..

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  • $\begingroup$ Your A1 formula is ambiguous. Please add more () to make clear what you divide by which (come on, this is basic "how to write down a formula"). $\endgroup$ – Marcus Müller Jan 16 '16 at 14:27
  • $\begingroup$ i didnt understand what is the ambiguity in A1..are you asking me the value for a?? $\endgroup$ – pavan sunder Jan 16 '16 at 18:13
  • $\begingroup$ With Gilles edit, it's now clear that you meant A1 = (-a+z^(-1))/(1-az^(-1)) with your formula, and not A1= -a + (z^(-1))/(1-az^(-1)), which obviously is something completely different! $\endgroup$ – Marcus Müller Jan 17 '16 at 10:15
  • $\begingroup$ but i did not get the desired frequency response: What did you get instead? Where's your Matlab code? Please make it easy to answer your question! $\endgroup$ – Marcus Müller Jan 17 '16 at 10:24
  • $\begingroup$ Does your title how to design LPH and HPF from APF really reflect your question well? I feel your question is about understanding and analyzing the resulting filter, not designing it. $\endgroup$ – Marcus Müller Jan 17 '16 at 10:54
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Since you don't say what response you're actually getting from Matlab, this is all guesswork.

So let's do the basics: A little formula juggling.

$$\begin{align}A_0(z)&=1\\ A_1(z)&=\frac{-a+z^{-1}}{1-az^{-1}}\\[4em] H_{LP}(z)&=A_0(z)+A_1(z)\\ &= 1 + \frac{-a+z^{-1}}{1-az^{-1}}\\ &= \frac{{1-az^{-1}}-a+z^{-1}}{1-az^{-1}}\\ &= \frac{(1-a)\quad +\quad (1-a)z^{-1}}{1-az^{-1}}\\[4em] H_{HP}(z)&=A_0(z)-A_1(z)\\ &= 1 - \frac{-a+z^{-1}}{1-az^{-1}}\\ &= \frac{{1-az^{-1}}+a-z^{-1}}{1-az^{-1}}\\ &= \frac{(1+a)\quad +\quad (-1-a)z^{-1}}{1-az^{-1}} \end{align}$$

So, taken from these formulas, the filter coefficients for $H_{LP}$ are $b=[1-a,1-a]$, $a=[1,-a]$, and for $H_{HP}$ it's $b=[1+a,-1-a]$, $a=[1,-a]$ for the canonical form of recursive filters

$$H= \frac{\sum\limits_{i=0}^N b_i z^{-i}}{\sum\limits_{i=0}^M a_i z^{-i}}\text.$$

Plugging that into scipy's scipy.signal.freqz for different $a$ yielded (abscissa: normalized frequency $\in[0,\pi[$, ordinate: magnitude):

$H_{LP,a}$ for $a\in\{0,0.1,\dots,0.9,1\}$:

LPFs

$H_{HP,a}$ for $ a\in\{0,0.1,\dots,0.9,1\}$:

HPFs

which both very nicely match the terms low and high pass filters, respectively.

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  • $\begingroup$ what is that scipy.signal.freqz..i didnt find that function in matlab documentation.. $\endgroup$ – pavan sunder Jan 19 '16 at 14:40
  • $\begingroup$ unfortunately i am not able to copy the matlab code which i wrote..i have used the coefficients which you suggested but though i am getting the response. its not varying when i change the parameter a.. $\endgroup$ – pavan sunder Jan 19 '16 at 14:49
  • $\begingroup$ here is my mail id: pavansunder@gmail.com..Would you kindly send me the matlab code..thanks :) $\endgroup$ – pavan sunder Jan 19 '16 at 14:50
  • $\begingroup$ @pavansunder: why can't you share your Matlab code? This sounds like the lamest excuse to get someone else to do your homework ever. $\endgroup$ – Marcus Müller Jan 19 '16 at 16:32
  • $\begingroup$ i tried to copy and paste my matlab code....but was unable to do..can u share your mail id so that i ll send my code.. $\endgroup$ – pavan sunder Jan 19 '16 at 17:48

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