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**posted on**

- dakupoto

June 21, 2015, 5:19 am

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.

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

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

On 6/22/2015 10:49 AM, Jim Thompson wrote:

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

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

Dr Philip C D Hobbs

Principal Consultant

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Re: Analog filter design metgod questions

On 06/23/2015 12:23 AM, snipped-for-privacy@gmail.com wrote:

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

the magnitude of the reactance at s_0 the same)

LP -> BR s' = s

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

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

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, keepingthe 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__0******2/s - s******2/s_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

Dr Philip C D Hobbs

Principal Consultant

We've slightly trimmed the long signature. Click to see the full one.

Re: Analog filter design metgod questions

On 06/23/2015 03:51 PM, Phil Hobbs wrote:

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

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 5:00:55 PM UTC-7, Phil Hobbs wrote:

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).

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

," where response

is no way to bring

d component Q.)

a response

ing it to the load.

fter all, the return

ure reactive

mounts to the

eturn loss

n" will come

will probably not

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:

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

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 |

| James E.Thompson | mens |

| Analog Innovations | et |

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Re: Analog filter design metgod questions

On 06/21/2015 07:19 AM, snipped-for-privacy@gmail.com wrote:

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

HTH, Pere

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_highpassHTH, Pere

Re: Analog filter design metgod questions

On 6/23/2015 8:26 AM, Jan Panteltje wrote:

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

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

Dr Philip C D Hobbs

Principal Consultant

We've slightly trimmed the long signature. Click to see the full one.

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