Allpass filter & (locally) negative delay

In a causal system there is a relationship between the amplitude variation and the phase delay -- you can get some phase lead with a differentiator, but only at the cost of an increasing gain vs. frequency characteristic. This of course causes _other_ problems, but you find a lot of differentiators in control loops just for this predictive capability.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
Reply to
Tim Wescott
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Unfortunately you can't go backwards in time. All noncausal networks, i.e. those which have negative group delay in any frequency band whatever, are unstable. The reason is that when you compute the impulse response from the transfer function, you have to close the contour at

-infinity for t < 0 and +infinity for t > 0. (Modulo your choice of Fourier transform sign convention.) Getting a response before t=0 means that there have to be poles in the unstable half plane.

Cheers,

Phil Hobbs

Cheers,

Phil Hobbs

Reply to
Phil Hobbs

Dear colleagues,

If I'm not mistaken, the standard all-pass filter sections, the first-order T(s)=(s - a0) / (s + a0) and the second-order T(s)=(s^2 - wr/Q s + wr^2)/(s^2 + wr/Q s + wr^2) both yield a monotonously increasing phase shift as a function of frequency.

Does an all-pass transfer function exist, which would result in a decreasing phase shift within a limited range of frequencies? This would correspond to a negative group delay, but I don't immediately see why such a phenomenon would be forbidden by the Kramers-Kronig relation (a.k.a the Bode relation, or causality). Provided that outside of the (limited) frequency range the group delay would be positive, of course.

Such an effect takes place in a notch filter, when one moves from the low-pass slope to the hi-pass slope with the associated switch from a lagging phase shift into a leading phase shift. I wonder if it would be possible to construct a filter function with (locally) similar phase behaviour, but a constant magnitude.

Regards, Mikko

Reply to
Mikko Kiviranta

I read in sci.electronics.design that Tim Wescott wrote (in ) about 'Allpass filter & (locally) negative delay', on Wed, 9 Nov 2005:

Use AGC to combat the change in gain.

--
Regards, John Woodgate, OOO - Own Opinions Only.
If everything has been designed, a god designed evolution by natural selection.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk
Reply to
John Woodgate

Yes I think you can...

but you don't do it by reducing delay...you do it by increasing delay using an all pass network...

you increase the delay for a range of frequencies, then (relativley speaking) the other frequencies are advanced.

All pass group delay equalizers work by INCREASING the delay at those frequencies where there is insufficent delay.

Got it? You can only add delay but you can choose the frequencies where you do it so the result is flat or whatever shape you want...

Mark

Mark

Reply to
Mark

Yes, this is exactly the (Bode version of) the Kronig-Kramers relation. Its rule-of-thumb version says that on a log-log plot, at each frequency the phase shift is proportional to the slope of the gain, and the constant of proportionality is 90 degrees for a 20 dB/decade slope. Lagging phase shift for a downhill slope and leading for an uphill slope.

The more careful version, however, says that the phase shift is not proportional to the slope at the same frequency point, but to the weighted average of the slope at surrounding frequencies. The weighing function log{coth(u/2)) has 90% of its weight within the f/10...10*f range. So there is some room to fiddle with contributions which cancel out within this frequency range. The ordinary allpass filters succeed in generating the phase shift without any gain slope, after all.

This line of thinking makes me wonder why would it be impossible to obtain a brief region of negative dphi/df in the transfer function.

Regards, Mikko

Reply to
Mikko Kiviranta

Agreed.

This 'any frequency band' detail sounds surprising and hence interesting! Do you have a reference on this?

I agree that getting a finite impulse response at t

Reply to
Mikko Kiviranta

I been thinkin bout this. The first and second all-pass sections you originally described are, I think, the only ones available. (and combinations of them...)

Draw the complex plane and plot those two as pole-zero pairs (quads for the second order). Poles always on left half plane, zeros always on right half. (pole-zero pairs always symmetric about the axis, too.)

No way to get phase advance from these in any combination.

The negative sign on deltaphase/deltafrequency that's achievable with bandlimiting filters must always come from one or more zeros in the left half plane. (causal filters only -- poles in right half plane are a Bad Idea).

Regards PN2222A "Of course I'm biased!"

Reply to
PN2222A

OK Mark

please, give an example of a linear circuit (or other system) with frequency response flat over the frequency range of interest, in which there are two frequencies, F0 and F1, F1>F0 and for which the phase shift of the transfer function at F1 is less than the phase shift at F0.

You may freely choose the frequencies.

Regards PN2222A NPN (Is = 14.34f Xti = 3 Eg = 1.11 Vaf = 74.03 Bf = 255.9 Ne = 1.307 Ise =

14.34 Ikf = .2847 Xtb = 1.5 Br = 6.092 Isc = 0

Ikr = 0 Rc = 1 Cjc = 7.306p Mjc = .3416 Vjc = .75 Fc = .5 Cje = 22.01p Mje = .377 Vje = .75 Tr = 46.91n Tf = 411.1p Itf = .6

Vtf = 1.7 Xtf = 3 Rb = 10)

Reply to
PN2222A

A network with a couple of all pass delay poles tuned to 1 MHz such that the total group delay is 21 uS at 1 MHz and 10 us at 2 MHz.

Mark

Reply to
Mark

(s^2 - a*s + b)/(s^2 + a*s + b)

Left half-plane complex pole mirrored by right half-plane zero ;-)

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
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Reply to
Jim Thompson

Sorry, I didn't get enough detail from your post to discover how to actually build this circuit.

(Note: 10 uS at 2 MHz is 628 radians -- perhaps you had a different delay TIME in mind?)

Would you kindly post one of: ..a more detailed explanation of your "all pass delay poles" ..an s-plane sketch showing pole and zero locations ..the laplace-domain polynomial describing the filter ..a schematic diagram or netlist ..a photograph of your breadboard with 'scope photos or another bit of information that would 'allow one normally skilled in the state of the art' to duplicate the design

Regards PN2222A

Reply to
PN2222A

Sounds like Mark is groping for a circuit resembling the one I considered before posting here in s.e.d.

(1) Take a bank of bandpass filters, with center frequencies at, say,

10MHz, 12MHz, 14MHz, 16MHz ... (2) Couple lengths of cable at outputs of the filters, giving delays of, say, 100ns, 80ns, 60ns, 40ns ... (3) Sum the output from the cables to get the phase shift of 360deg, 346deg, 302deg, 230deg...

The idea is that the signal at an increasing frequency would get coupled to shorter and shorter cables. If passbands of the filters are separated from each other, the phase oscillates wildly between -90deg and +90deg as alternating uphill and downhill slopes of the bandpass filters kick in. When the passbands get closer to each other, the behaviour gets nicer.

This was a clumsy way to construct such a filter, and I wondered that perhaps there is a body of theory (going beyond elementary filter books which I know of) on how to construct a minimal complexity filter given the constraints.

Regards, Mikko

Reply to
Kiviranta, Mikko

Problem: You can't match the phases of the filters at the crossover points. Thus there will be plenty of amplitude ripple at 11, 13 and 15 MHz.

Phase and amplitude response are tightly coupled. TANSTAAFL.

Regards PN2222A

Reply to
PN2222A

read up on all pass filters and group delay equalizers, they work by creating delay where needed

here are a few links

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It's an interesting question, but there is an answer to your problem.

Mark

Reply to
Mark

Mark

You have shown a number of websites where the all-pass function (pole-zero pairs symmetric about the Y-Axis, zeros in RHP, poles in RHP) is described and used. I'm choosing not to consider the Discrete-time versions.

All of these articles and all of the functions and circuits are topologically the same. All of the functions have a monotonically increasing phase as function of frequency. And there are no others, except the case of poles in RHP, zeros in LHP which I claim are not of practical use.

Mikko proposes a decreasing phase-with-frequency implementation by splitting the signal and delaying the low-frequency section. In the simplest case,

V1 = Vin * (s/s+1), V2 = (1/s+1). Delay V1 with, say, a long piece of coax. Recombine the signals with an adder. At the cutoff frequencies, there is an ugly phase shift characteristic ADDED BY THE COAX. So the two signals recombine with amplitude ripples whose number is proportional to the amount of delay.

So, says Mark. We'll throw an all-pass function into V2, to equalize the phase as needed. Indeed, this can be done. BY ADDING DELAY TO V2. The all pass function to remove the ripple in the cutoff band duplicates the function of the coax.

Net result, there is still no example given of a linear circuit (or other system)with frequency response flat over the frequency range of interest, in which there are two frequencies, F0 and F1, F1>F0 and for which the phase shift of the transfer function at F1 is less than the phase shift at F0.

TANSTAAFL.

PN2222A Now lead free!

Reply to
PN2222A

How about a number of lower-rated reed switches in series, in the same coil..?

Reply to
Mikko Kiviranta

Sounds to me that what you want is an inverse Chebysev, which places the amplitude ripple and phase irregularities in the stop band. Notably twichier in design and implementation though.

--
JosephKK
Reply to
JosephKK

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