Analog filter design metgod questions

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Could some analog filter design guru please
clarify the following ?  
Canonical analog filter design specifications
for e.g., a normamlized low-pass Butterworths  
filter are:
1. pass band edge frequency
2. stop band edge frequency
3. maximum pass band attenuation
4. maximum allowable syop band attenuation  
Given these, the filter order, selectivity  
parameter, discrimination factor and then  
the poles etc., may be computed. Also, there
exist standard tricks to transform a low-pass
filter to a high-pass filter and so on.  
Given that the pass band for a low pass filter
is the stop band for the high pass filter, how
do the above parameters (1 -> 4) map in case of the high pass filter. That is if I start with a
low-pass filter with a given set of values for  
1 - 4, how would they map to the corresponding  
high pass filter. Any hints, suggestions would
be of immense hekp. Tganks in advance.


Analog filter design metgod questions
Make the reactances at f_-3dB the same magnitude and opposite sign as the low pass prototype.  

For instance, a 50-ohm filter with f_-3dB = 1 MHz might have a 5-uH inductor someplace. Replace it with C=1/((2 pi*1MHz)**2)* 5uH) ~5 nF, if my mental arithmetic is holding up today.  

Cheers

Phil Hobbs

Re: Analog filter design metgod questions
you need THE BOOK

(Amazon.com product link shortened)

Mark




Re: Analog filter design metgod questions
On Mon, 22 Jun 2015 07:43:08 -0700 (PDT), snipped-for-privacy@yahoo.com wrote:

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Gag me with a spoon >:-}
        
                                        ...Jim Thompson
--  
| James E.Thompson                                 |    mens     |
| Analog Innovations                               |     et      |
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Re: Analog filter design metgod questions
On 6/22/2015 10:49 AM, Jim Thompson wrote:
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Nah, I like Zverev.  (Of course I've been using it on and off since  
1981-ish, so it's an old friend.)

Cheers

Phil Hobbs

--  
Dr Philip C D Hobbs
Principal Consultant
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Re: Analog filter design metgod questions
On Mon, 22 Jun 2015 17:06:21 -0400, Phil Hobbs

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I've used it, thus the gag ;-)
        
                                        ...Jim Thompson
--  
| James E.Thompson                                 |    mens     |
| Analog Innovations                               |     et      |
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Re: Analog filter design metgod questions
On Monday, June 22, 2015 at 10:43:12 AM UTC-4, snipped-for-privacy@yahoo.com wrote:
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Thanks. I have access to this book, but ploughing
through it is quite an exercise.

Re: Analog filter design metgod questions
On Sunday, June 21, 2015 at 1:08:48 PM UTC-4, Phil Hobbs wrote:
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Thanks, I have been using these tricks for years.

Re: Analog filter design metgod questions
On 06/23/2015 12:23 AM, snipped-for-privacy@gmail.com wrote:
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Oh, okay.  From a matho's POV, then, the usual filter transformations  
are based on a conformal mapping in the s plane:

LP -> HP     s' = s_0**2/s    (swap inductors and capacitors, keeping  
the magnitude of the reactance at s_0 the same)

LP -> BR     s' = s_0/(s_0/s -s/s_0)
Maps s_0 to infinity, zero to zero, and infinity to zero. (Keep the LP  
components, but series-resonate every C at s_0, and parallel-resonate  
all the inductors.)

LP -> BP     s' = s_0**2/s - s**2/s_0
  Maps s_0 to the origin, and both 0 and infinity to infinity.  (Keep  
the LP components, but parallel-resonate all the capacitors and  
series-resonate the inductors)

(I think those are right.  I just did it on the hoof.)

The BR and BP filters have the same FWHM as the lowpass prototype,  
because although the response is two-sided, the reactances change twice  
as fast, so the BW nets out the same.

These transformations are good starting points when you mostly care  
about amplitude, but they're not much use with linear-phase filters.  
Instead of the phase being linear with frequency, it's nicely hyperbolic  
instead.

Zverev has tables with pre-distorted coefficients for good bandpasses.

Cheers

Phil Hobbs

--  
Dr Philip C D Hobbs
Principal Consultant
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Re: Analog filter design metgod questions
On 06/23/2015 03:51 PM, Phil Hobbs wrote:
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Just to add on that, the LP -> BP transformation only works for  
sufficiently high percentual bandwidths or equivalently, low filter Q.  
Otherwise element values become impractical. For moderate to high filter  
Q, the coupled resonator approach is the way to go and you may obtain  
several flavors of frequency (and phase) response. Furthermore,  
pre-distorted (coupling) coefficients may even give a nice response for  
finite Q rsonators.

Pere



Re: Analog filter design metgod questions
On Tuesday, June 23, 2015 at 7:05:44 AM UTC-7, o pere o wrote:
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Predistortion wrecks the return loss, and reduces sensitivity.  I have never found a modern case where I thought it was worth it.  Digital equalization post-ADC seems to be a better approach.

Re: Analog filter design metgod questions
On 6/23/2015 7:47 PM, Simon S Aysdie wrote:
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Hmm, interesting.  Could you say more about that?

Cheers

Phil Hobbs

--  
Dr Philip C D Hobbs
Principal Consultant
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Re: Analog filter design metgod questions
On Tuesday, June 23, 2015 at 5:00:55 PM UTC-7, Phil Hobbs wrote:
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Simplistically:  
It is usually the corners (BPF) that droop due to finite Q. So the "middle,
" where response "peaks," is pushed down to match the corners by the predis
tortion. (There is no way to bring the corners up by predistortion techniqu
e; we are assuming a given or fixed component Q.) Since the elements are mo
stly reactive, the pushing down of the middle area response amounts to refl
ecting that energy back to the source rather than transmitting it to the lo
ad.  "Reflecting" is the same thing as saying the return loss is degraded.  
After all, the return loss approaches 0 dB in the stopband.  

A passive filter works based on the principle of reflection.  Ideally (pure
 reactive elements), what is not transmitted is reflected, and vice versa.  
 This amounts to the famous Feldtkeller energy equation for filters. It is  
at the root of why return loss gets degraded for predistorted filters.

If the filter is pre-LNA (or post PA), then the amount of middle "push-down
" will come straight off the noise figure and link budget.  It won't hurt t
he corner sensitivity because it is already down.  It will probably not mat
ter if, for example, it is an IF filter.  If the equalization is done post  
ADC, then the noise profile will look bathtub-like (without predistortion).
  


Re: Analog filter design metgod questions
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," where response  
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is no way to bring  
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d component Q.)  

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a response  
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ing it to the load.  


fter all, the return  
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ure reactive

mounts to the  
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eturn loss  
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n" will come  
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will probably not  
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bathtub-like  


Thanks, that makes sense. Loudspeaker efficiency behaves the same way, for  
much the same reason--you have to knock down the peaks becaose it's hard to
 fill in the valleys.  

Cheers

Phil Hobbs

Re: Analog filter design metgod questions
On Sat, 20 Jun 2015 22:19:16 -0700 (PDT), snipped-for-privacy@gmail.com wrote:

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Get a copy of my favorite filter book...

"Synthesis of Filters"
Jose Luis Herrero & Gideon Willoner


Library of Congress Number: 66-27547

(Peruse the old/used bookstores... http://tinyurl.com/cezq5kq )

Math intensive, transforms required, not for those who can't already
do a hand-math filter design with Laplace (and need to fall back on
FilterPro :-)

But excellent for handling depth of stop bands, etc.
        
                                        ...Jim Thompson
--  
| James E.Thompson                                 |    mens     |
| Analog Innovations                               |     et      |
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Re: Analog filter design metgod questions
On Monday, June 22, 2015 at 10:52:07 AM UTC-4, Jim Thompson wrote:
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at is if I start with a
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Thanks. Will try it out. Math does not bother me.
I started out in my career as a physicist.

Re: Analog filter design metgod questions
On 06/21/2015 07:19 AM, snipped-for-privacy@gmail.com wrote:
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If you have a low pass transfer function H(S), the change S->1/s gives  
you a high pass transfer function G(s)=H(1/s). (Note upper -LP- and  
lower cases -HP-)

Attenuations are unchanged and frequencies are related by jW=1/jw, i.e.  
w=-1/W.

After this LP to HP transformation you (often) have to perform a  
frequency scaling, i.e. translate G(s) to G(s/wo). Now, whatever was  
happening at w=1 will now be happening at w=wo

Others have suggested some textbooks, and this is the way to go if you  
want to understand. Once you know what you are doing, even wikipedia  
(and perhaps my explanation) looks useful  
https://en.wikipedia.org/wiki/Prototype_filter#Lowpass_to_highpass

HTH, Pere

Re: Analog filter design metgod questions
On a sunny day (Tue, 23 Jun 2015 10:20:05 +0200) it happened o pere o

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FYI
if you use an opamp to subtract the output from the lowpass
you are left with the highpass.
That reduces the problem in half :-)
Same for bandpass-bandstop.
:-)

Re: Analog filter design metgod questions
On 06/23/2015 02:26 PM, Jan Panteltje wrote:
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Right! But you have to be able to do the subtraction with enough  
precision to get significant stopband attenuation and this may not be so  
easy...

Pere

Re: Analog filter design metgod questions
On 6/23/2015 8:26 AM, Jan Panteltje wrote:
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The problem is that you generally wind up with only a single-pole  
characteristic when you do this, on account of the phase shifts.

Cheers

Phil Hobbs

--  
Dr Philip C D Hobbs
Principal Consultant
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