Re: Single pole lowpass --> highpass - z transform?

On Aug 3, 2:55 am, Tim Wescott <t...@xxxxxxxxxxxxxxxx> wrote:
dbell wrote:
On Aug 2, 6:09 pm, Tim Wescott <t...@xxxxxxxxxxxxxxxx> wrote:
dbell wrote:
On Aug 2, 2:24 pm, Tim Wescott <t...@xxxxxxxxxxxxxxxx> wrote:
bharat pathak wrote:
(top posting fixed)
//-------------------------- in response to ------------------------
Transfer function
    (1+beta) z
H = ----------
     z - beta
Oops -- that's not a high-pass filter at all, it's just a low-pass
filter with a non-unity gain at DC for all beta != 0.
Dangit.
//------------------------------------------------------------------
Hello Tim,
    Before passing out your judgement, I think it is better
    to verify things first, specially when matlab code is
    provided it won't take more than a minutes time to run
    it and check the results. Even if you don't have matlab
    octave is freely available.
    As OP found, both approaches lead to HPF. If you are still
    not convinced read page 329 330 and 331 of proakis, manolakis
    4th Edition.
Regards
Bharat
The day a person thinks he knows everything, is already brain dead.
I was going from your equations, I generally don't bother looking at
code.  If your code is correct then more power to you, but that doesn't
change the fact that your published difference equation is in error.
If you think that you can find mathematical truths with Matlab when you
can't even do the basics with pencil and paper then I have only pity for
you.
--
Tim Wescott
Wescott Design Serviceshttp://www.wescottdesign.com
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"Applied Control Theory for Embedded Systems" gives you just what it says.
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Tim,
His HP equations are right. I think the confusion rests on him
changing the sign of beta between the LP and HP equations. For HP, he
has -1<beta<=0. Plug in z=1 for DC gain, z=-1 for max HP gain. It's a
HP.
Dirk
Extracting his equations from his text:

LP:   y(n) = (1-beta)*x(n) + beta*y(n-1)
"HP": y(n) = (1+beta)*x(n) + beta*y(n-1)

The only difference is the gain of the filter, both are low-pass with a
zero at z = 0 and a pole at z = 1-beta.  You can test the DC gain of the
high-pass case by assuming that y and x are constant, and you get

y = (1+beta)x + beta*y
=> (1-beta)y = (1+beta)x
=> y = x (1+beta)/(1-beta).

And that ain't zero, so that ain't a high pass filter.

--

Tim Wescott
Wescott Design Serviceshttp://www.wescottdesign.com

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"Applied Control Theory for Embedded Systems" gives you just what it says.
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Tim,

Who said the gain has to be zero at DC to be a HP? Look at the gains
at DC and Fs/2 to see the types of filters, and that they are not
scaled versions of each other.

"HP": y(n) = (1+beta)*x(n) + beta*y(n-1)

HHP(Z)=(1+beta)/(1-beta*Z^-1), (for beta in ( -1, 0] as Bharat
specified)

At DC   Z=  1, so HHP(Z= 1) = (1+beta)/(1-beta)
At Fs/2 Z= -1, so HHP(Z=-1) = (1+beta)/(1+beta) = 1.0

Plug in a value for beta, say beta= - 0.9 (negative as Bharat said)

At DC   HHP(Z= 1) = (.1)/(1.9) = 0.0526
At Fs/2 HHP(Z=-1) = (.1)/(.1)   =1.0

Gain going from DC to Fs/2 goes from 0.0526 to 1.0, almost an increase
by a factor of 20.
Alternately you could say the low frequencies are attenuated to as
little as ~1/20 relative to the highest frequencies.

Sounds like a high-pass to me.  Sure isn't a lowpass.

"LP":   y(n) = (1-beta)*x(n) + beta*y(n-1), (for beta in [0, 1) as
Bharat specified)

HLP(Z)=(1-beta)/1-beta*Z^-1)

At DC   Z= 1, HLP(Z= 1) = (1-beta)/(1-beta) =1.0
At Fs/2 Z=-1, HLP(Z=-1) = (1-beta)/(1+beta)

Plug in a value for beta, say beta=0.9 (now positive as bharat said)

At DC   HLP(Z= 1)=(.1)/(.1)  = 1.0
At Fs/2 HLP(Z=-1)=(.1)/(1.9) = 0.0526

Sounds like a low-pass to me.  Sure isn't a high-pass.

The two filters are clearly not the same with the exception of a scale
factor. The equations do look the same except for a scale factor, but
only until you see that the betas are defined differently.

Dirk

BTW, for the HP, I used beta= -0.9, but making beta closer to -1 will
get the DC gain closer to 0, with the Fs/2 gain fixed at 1.0 .

Where I come from (analog circuit design), that's a lowpass with a
highly resonant peak -- which you can get with Bharat's first equation
just by setting beta < 0.  Just because the resonance happens to be
around Fs/2, and allows one to express the filter as a 1st-order, is a
mathematical curiosity of sampled-time systems.

_Real_ high pass filters do, indeed, have zero gain at DC (at least
assuming ideal components and no quantization, etc).  Real low pass
filters have zero gain at infinite frequencies, if you're working in a
domain that has such (and yes, you could argue that this implies that
there are no 'real' sampled-time low-pass filters).

Did I miss that exceptionally screwy bit of mathematical legerdemain?

Why yes,  I did.  My bad.  I'll have to remember in future just how
profoundly impractical Bharat's posts really are.

Do I think that there is any wide practical implementation of this filter?

Why no.  I don't.  Narrow ones, yes.  Wide ones, no.

He certainly went far afield of the OP's question, which involved making
high-pass filters from low-pass prototypes.  Since you're so hot on this
filter, perhaps you could say how to set beta if I want to have a high
pass filter with a cutoff of Fs/100?

--

Tim Wescott
Wescott Design Serviceshttp://www.wescottdesign.com

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Tim,

In digital filters that exhibit characteristics of significant
attenuation on a band from DC to F1 and approximately unity gain from
some F2 to Fs/2, that would no doubt be called highpass by most DSP
people, there is no requirement that gain at DC be 0.

Try

plot(20*log10(abs(fft(firpm(62,[0 .1 .2 .5]/.5, [ 0 0 1 1],[50,1]),
8000))))

120 dB stopband attenuation, ripple +-4*10^-4 dB passband ripple about
0 dB, nonzero at DC (actually a peak in the ripple at DC).

I do not mistake "real" to mean "analog", so not having 0 at DC
doesn't cause me any problem when designing useful digital filters. If
I did I would be over at comp.asp instead of here :-).

Dirk
.

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