Doepfer

hi group,

feeding the various outputs of the a108 filter to its own cv and resonance inputs, it becomes immediately obvious that there is more difference here than just slope.

i wonder if anybody can tell me what the phases of the outputs are on this incredibly versatile module.

thanks,

stu

Hi Stu,

> i wonder if anybody can tell me what the phases of the outputs are on this incredibly versatile module.

According to my simulation, as you go up the outputs: 6dB, 12dB, 18dB, etc, the phase of each lags that of the previous output, by an amount which increases with frequency, and which depends on the resonance setting. I've created a folder 'A-108 phase' in the files section, and uploaded 3 gifs. (And I hope I got all my signs/inversions right - willing to be proved wrong by Dieter/whoever...!)

Tim

Ack! I should add that these are all at a cut-off of about 1kHz, which is where the big 'wobble' is!

Tim

> I've created a folder 'A-108 phase' in the files section, and 
> uploaded 3 gifs.

wow. 

thanks tim, that's really helpful. of course, like all the best answers, it raises more questions :)

like this one: what's so special about a cut off of 1kz and a slope of 30db that the filtered signal stays in phase with the original up the cut off frequency?

stu

Hi Stu,

> thanks tim, that's really helpful. of course, like all the best answers, it raises more questions :)
> 
> like this one: what's so special about a cut off of 1kz and a slope of 30db that the filtered signal stays in phase with the original up the cut off frequency?

Indeed, that is an interesting observation! And the short answer is I have no idea. However, having run a few more simulations, at frequencies above and below 1kHz, and stepping the resonance from min to max, it appears to be something special about the fifth stage, and doesn't appear to be particularly frequency related, nor related too much to the resonance (though it is more pronounced at higher res settings): for the stage below (24dB), below the cut-off it peaks above zero, and for above (36dB), it falls away from zero before reaching the cut-off.

I'm guessing it is just some funny facet of the algebra concerned. The (normalized) transfer function for 'stage n', n=1 to 8 (so 30dB is n=5), appears to be

H(s)=1/((s+1)^n+k*(s+1)^(2n-8))

where k is the gain in the feedback loop. If one were to be bothered to work out the phase from this, tan^-1 im(H(wj))/re(H(wj)), I dare say it might be possible to see what it is about n=5 that keeps it nearly zero for w<1 (for 'w', read 'omega'). I have to say I'm not even remotely tempted, far too busy doing other things!

Tim

> I'm guessing it is just some funny facet of the algebra concerned. The (normalized) transfer function for 'stage n', n=1 to 8 (so 30dB is n=5), appears to be
> 
> H(s)=1/((s+1)^n+k*(s+1)^(2n-8))
> 
> where k is the gain in the feedback loop. If one were to be bothered to work out the phase from this, tan^-1 im(H(wj))/re(H(wj)), I dare say it might be possible to see what it is about n=5 that keeps it nearly zero for w<1 (for 'w', read 'omega'). I have to say I'm not even remotely tempted, far too busy doing other things!

this is the best post i've ever read, may i put it on a t-shirt?

Hi Stu, Hi Tim,

I hope you are doing well.

The observations you have made about the changing phase shifts in dependence on the frequency and resonance value can be explain by Fourier-transformations which is the mathematical theory behind every kind of frequency filtering. 

It says something like that every manipulation of a periodic function in the frequency space results in a manipulation of the signal in the phase space and vice versa.

That's why a signal filtered with different filter slopes, has different phase shifts along the slop. Because of the different gradients among the slopes, the phase shifts between the different frequencies alter as well.
If you change the resonance of a filter, the slope is changing again and you will get again changing phases between the different frequencies of your audio signal.

This is perfectly consistent with the phase shift behavior you have measured, Tim! 
The phase shift "... increases with frequency... " due to the heavier manipulation of higher frequencies to obtain low pass filtering.

I'm not sure, but I would say that the reason why you don't here a phase shift at 1 Khz with a 30dB slope is, that the phase shift is coincidentally exactly 2*pi or 360 degrees with these settings. So that one could think that there is no phase shift, but the phase is already shifted with 2 periods.

I hope I could help with my knowledge of my physics studies.

Ollie

--- In Doepfer_a100@yahoogroups.com, "grimshaw_stuart" <grimshaw@...> wrote:

>
> wow. 
> 
> thanks tim, that's really helpful. of course, like all the best answers, it raises more questions :)
> 
> like this one: what's so special about a cut off of 1kz and a slope of 30db that the filtered signal stays in phase with the original up the cut off frequency?
> 
> stu
>

Hi Stu

> this is the best post i've ever read, may i put it on a t-shirt?
Me want this shirt too ;-)

Florian

OK, so, now before anyone goes sticking this on a T-shirt, believing it to be the answer to life, the universe and everything, I think it only fair to warn you that this:

> > I'm guessing it is just some funny facet of the algebra concerned. The (normalized) transfer function for 'stage n', n=1 to 8 (so 30dB is n=5), appears to be
> > 
> > H(s)=1/((s+1)^n+k*(s+1)^(2n-8))

is complete bollocks (i.e. it is _wrong_)!! I finished that 'other thing' I was working on (a new webpage), and so couldn't help trying the short-cut route of banging this expression into Mathematica to see how the phase from it looked, and in the process discovered the most elementary of schoolboy mistakes in my working (bad, bad boy!). So I now think it is:

H(s)=(s+1)^(8-n)/((s+1)^8+k)

which at least gives the correct expression for n=8 (the 48dB output). The phase plots look similar to the simulation output, but I didn't get any insight as to what might be 'special' about n=5 (maybe some property of the odd 16th roots of unity or something - make a nice little homework for some budding student!).

Tim
__________________________________________________________
Tim Stinchcombe 

Cheltenham, Glos, UK
www.timstinchcombe.co.uk

Hi Tim

> H(s)=(s+1)^(8-n)/((s+1)^8+k)
it is ok, that someone makes me feel again like that stupid 15 year old 
boy back 35 years ago, sitting in the class room and understanding 
nothing. But posting it twice? Sorry, this is too much... ;-)

Florian

Hi Tim

I enjoyed your web site - particularly the old brochures.
I think I was at the same, fine ELP show in Cardiff.

Cheers, Nick

I see, you want to get an explicit expression of the phase-frequency relation of your filter.
I doubt that you can derive this expression directly from the transfer function of a filter.

Look, since s = iw + d, where w is frequency, i is imaginary unity, d is phase factor. The phase factor is just a parameter which were set as initial condition. It doesn't depend on the frequency again and is constant for all frequencies.

To derive such a phase-frequency relation, one needs a formula which relates the phase of the "incoming" frequency to the phase of the "outgoing" frequency. So you could think of a system where a harmonic oscillator with frequency w1 and phase d1 is coupled to another harmonic oscillator with freq. w2 and phase d2. The exact value of the coupling strength of these oscillators and the value of the inertia of each oscillator would specify, which kind of filter this system should describe.  
An appropriate model for such a system is the driven harmonic oscillator with damping:

http://upload.wikimedia.org/math/a/1/0/a1030f6cb947558b4fd472723ad2b059.png
Also see:
http://en.wikipedia.org/wiki/Harmonic_oscillator#Driven_harmonic_oscillators

It's possible to derive the phase shift of the oscillation with w1 relative to the driving force with w2:

http://upload.wikimedia.org/math/d/2/0/d20d6b00980a90b649899fb6b402e04e.png
Also see the paragraph in wikipedia above.

Which would represent the wanted expression in the most general form. 

The next step would be to iterate the parameter zeta, m1 and m2 in the differential equation in that way, that the amplitude-frequency-responce of the DEQ fits the transfer function of your filter of interest.

Here you go,
Ollie

PS: This was my last replay to this thread.

--- In Doepfer_a100@yahoogroups.com, "Tim" <timothy@...> wrote:

>
> OK, so, now before anyone goes sticking this on a T-shirt, believing it to be the answer to life, the universe and everything, I think it only fair to warn you that this:
> 
> > > I'm guessing it is just some funny facet of the algebra concerned. The (normalized) transfer function for 'stage n', n=1 to 8 (so 30dB is n=5), appears to be
> > > 
> > > H(s)=1/((s+1)^n+k*(s+1)^(2n-8))
> 
> is complete bollocks (i.e. it is _wrong_)!! I finished that 'other thing' I was working on (a new webpage), and so couldn't help trying the short-cut route of banging this expression into Mathematica to see how the phase from it looked, and in the process discovered the most elementary of schoolboy mistakes in my working (bad, bad boy!). So I now think it is:
> 
> H(s)=(s+1)^(8-n)/((s+1)^8+k)
> 
> which at least gives the correct expression for n=8 (the 48dB output). The phase plots look similar to the simulation output, but I didn't get any insight as to what might be 'special' about n=5 (maybe some property of the odd 16th roots of unity or something - make a nice little homework for some budding student!).

Hi Ollie,

> I see, you want to get an explicit expression of the phase-
> frequency relation of your filter.
> I doubt that you can derive this expression directly from the 
> transfer function of a filter.

Yes you can, as I posted before: tan^-1 im(H(wj))/re(H(wj)). (Aaargh, I see I'm being forced into it!) - it is pretty nasty, here it is for the 5th stage o/p:

tan^-1(-5*w + 3*k*w - 5*w^3 - k*w^3 + 14*w^5 + 22*w^7 + 7*w^9 - w^11)/
(1 + k - 7*w^2 - 3*k*w^2 - 22*w^4 - 14*w^6 + 5*w^8 + 5*w^10)

So plug in frequency, w, and k (resonance), and get the phase back.

[Mostly via Mathematica + time courtesy of not being able to get into work due to snow!]

> To derive such a phase-frequency relation, one needs a formula
> which relates the phase of the "incoming" frequency to the phase
> of the "outgoing" frequency.

This is given by the transfer function, see above!

> So you could think of a system where a harmonic oscillator with

<<snip lots of stuff I have no intention of trying to understand, as I don't see the relevance>>

There are lots of good books out there on active filters if you want to learn more: _my_ particular favourite is 'Design of Analog Filters', Schaumann & van Valkenburg, OUP

(Sorry everyone else for the algebra overload - especially
Florian :-) - back to what you were doing!)

Tim
__________________________________________________________
Tim Stinchcombe 

Cheltenham, Glos, UK
www.timstinchcombe.co.uk

FS: 5000 t-shirts, various colours, with caption 

"The (normalized) transfer function for 'stage n', n=1 to 8 (so 30dB is n=5), appears to be H(s)=1/((s+1)^n+k*(s+1)^(2n-8))"

and photo of tim poking out his tongue.

diavolo pizza or nearest offer.

stu

Hi Tim, Hi everybody,

I want to apologize for continuing this thread. I don't want to do any math here. I hope I can clarify this interesting theoretical discussion, where I have learned a lot and I hope some others as well.

> tan^-1(-5*w + 3*k*w - 5*w^3 - k*w^3 + 14*w^5 + 22*w^7 + 7*w^9 - w^11)/
> (1 + k - 7*w^2 - 3*k*w^2 - 22*w^4 - 14*w^6 + 5*w^8 + 5*w^10)

This is not the the function which tells you how the phase changes, when you changes the filter! 
The transfer function and the phase function above describes how the filter affects the input signal in a _steady state_.

I think you mean the phase delay, which is:
http://homepages.physik.uni-muenchen.de/~oliver.gretz/a-100/phase.pdf
(taken from "Analog Filters Using MATLAB" from Wanhammer)

Cheers,
Ollie