Filter Pole Locations

See

The classic method is just to resonate all the reactive elements to the geometric mean frequency of the bandpass. All series elements to series resonance and all parallel elements to parallel resonance.

For a pi low-pass this means adding parallel inductors to the capacitors and series capacitors to the inductors. Pretty often the series element values are impractical with very small capacitors and hefty inductors. The remedy is to tap them down into the parallel elements. The taps can be taken from either capacitive or inductive legs of the parallel elements.

No "elements", just math.

What I want is an algorithm that transforms...

-A + jB (from a lowpass with the desired TOTAL bandwidth) into a

pair of complex poles around a center frequency OmegaN

...Jim Thompson

Take the LP prototype, and resonate all the elements at the centre frequency.

Cheers

Phil Hobbs

I'm not sure, For a two pole filter, the white noise bandwidth of the low pass is the same as for the band pass. (wow, I think that's right, but it also seems like the band pass should be more.. I can check.)

George H. have to check tomorrow.

That's right, the BW of the BP is the same as the LP, at least in the high- Q limit where the transformation applies. In the LC case, the reactance cha nges twice as fast with frequency, but you have the low- and high-frequency sides of resonance, so the net is the same BW.

The impulse response is twice as wide, though.

Cheers

Phil Hobbs

That's right, the BW of the BP is the same as the LP, at least in the high- Q limit where the transformation applies. In the LC case, the reactance cha nges twice as fast with frequency, but you have the low- and high-frequency sides of resonance, so the net is the same BW.

The impulse response is twice as wide, though.

Cheers

Phil Hobbs

Not sure of this Jim, because I don't/haven't used it. But, I think octave will give you the poles/pole pairs.

For example, take a LP polynomial, stick in the resonnant pairs to move everything to BP, then apply some of octave's built-in polynomial math functions and out pops a string of new poles/pole pairs, so you don't have to do the 'noodling'.

Yes. It is covered in, literally, every filter synthesis text.

For example, see if you can download Omar Wing's /Classical Circiut Theory/ . The transformation is on pp.234-237. Each LP pair of +/-omega is mapped to four BP Omega. If you know omega, then you know two Omega.

Honestly, I don't get your "factoring" question. Who "factors" these thing s analytically? There is no analytical solution beyond 4th degree anyway, and 4th is impractically messy. The most anyone ever does is use a root fi nder (numerical) and say "good enough" for the factorization. BTW, these p olynomials are notoriously ill-conditioned. For high order equations, spec ial techniques must sometimes be used to maintain accuracy in the realizati on.

Seems like you are, of late, paying a heavy price for skipping Ernst Guillemin's classes. :-)

h-Q limit where the transformation applies. In the LC case, the reactance c hanges twice as fast with frequency, but you have the low- and high-frequen cy sides of resonance, so the net is the same BW.

Thanks Phil, It's true for any Q. (A colleague did the mathematical proof .) What I forgot was that at low Q the bandpass has a maximum gain of Q.

George H.

The usual low-pass to band-pass transformation (preserving BW) is

S -> (s^2-wo^2)/s

giving two s points for each point in the original S plane.

Pere

Isn't it '+' instead of '-'?

It produces, from a second-order pole, s^/(s^4...) which is a pain to factor into 2 second order pieces. ...Jim Thompson

I replied to this but did not see any comment from you yay, or nay.

old reply: Not sure of this Jim, because I don't/haven't used it. But, I think octave will give you the poles/pole pairs.

For example, take a LP polynomial, stick in the resonnant pairs to move everything to BP, then apply some of octave's built-in polynomial math functions and out pops a string of new poles/pole pairs, so you don't have to do the 'noodling'.

I haven't used Octave. Maybe I should give it a spin? ...Jim Thompson

Do you run Linux on one of your computers?

My i7 boot windows 8.1, but a simple mSATA drive in a USB enclosure allows me to boot my favorite Linux distro, and put whatever I want on it.

I have a few. One is Ubuntu Studio, and I use it to grab high res, huge files from my bluray discs, and then downconvert those into files I can then view on whatever media device I want. No more discs.

I have another running Debian Wheezy and my math and PCB apps.

I have another Debian Wheezy install for my Cubox-i ARM computer, and I can compile for Android and ARM machines on that.

The cubox is nice because all I need is a microSD swap to boot a new OS on it. and it is only a $135 computer!

it is

s = (S^2 + Omega_0^2)/(B*S)

s := lowpass complex freq S := bandpass complex freq Omega_0 := bandpass (geometric) center frequency B := bandpass bandwidth

The assumption is a normalized lowpass. (corner freq is 1 rad/s)

Whatever the lowpass function is "s" is, you replace it by the right side o f the equation to get a bandpass. That means if you know a pole at the the low pass frequency, you know two at the bandpass, just by plain transforma tion (solve a quadratic).

S^2 - s_k*B*S + Omega_0^2 = 0

s_k := lowpass pole

Can you solve a quadratric? "No one" bothers to do so analytically, btw.

You will be constrained by geometric symmetry with this transform method of producing a bandpass. (There are some tricky ways around that, but that i s another topic.)

Re. Octave: It is a very powerful free math package. It is easier to hack code in "m" (m-code) than something like Python with numpy and scipy (pyth onxy). Of course, Python is free and "better" for broader reasons. So yea h, it is nice to have math package to do root finding, and many other wondr ous things.

Good luck with your impossible active filter.

Sounds like you've quite the bitcoin mining operation going on there.

Wow