Freq. Independent Phase Shifter
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From: John Fields on 11 May 2010 10:17 On Mon, 10 May 2010 07:56:44 -0700 (PDT), George Herold <gherold(a)teachspin.com> wrote: >On May 9, 6:14�pm, Bill Sloman <bill.slo...(a)ieee.org> wrote: >> On May 9, 11:33�pm, billmur...(a)protech.com (Bill Murphy) wrote: >> >> > I am using a commercial stereo amp to output continuous wave test >> > signals in the low audio range, up to about 2KHz. However, I need a >> > third channel with a 120 degree phase shift. Is there a circuit that >> > will do this evenly across this entire frequency range? >> >> > Is it possible to do same using an off-the-shelf transformer and >> > current subtraction? >> >> > Any advice would be appreciated. >> >> Check out Horowitz and Hill's "The art of Electronics". Section 5.16 >> talks about phase-sequence filters, which give a constant 90 degree >> shift over a range of frequencies. They consist of strings of equal >> value resistors with cross-connected capacitors whose values decrease >> in constant proportion per stage, halving in the example given, which >> isn't all that practical to set up. The bottom line is that it isn't >> trivial, and if you need to ask, you probably don't know enough to put >> together a circuit that will work. >> >> -- >> Bill Sloman, Nijmegen > >Yeah, I've used this circuit. It gives you nice quadrature output >sine/cosine waves that can be used to make any particular phase >shift. But phase shift relative to the input signal changes with >frequency... The phase tends to keep wrapping around. Mixing the two >signals with a potentiometer also causes amplitude variations if that >would be a problem. >That said I used a 10 section filter that has less than 1 degree of >pahse ripple from 3 Hz to 3kHz. Caps are standard 1, 2.2, 4.7, 10... >values. The ratio between sections is not as important as keeping the >same value in each section. You can also change the order of the >sections with no change at the output. > >If the OP wants only a few known frequincies then he could make a few >dedicated RC sections... or some dedicated All-pass opamp filters. >and switch each in when testing at that frequency. --- Have you both lost your minds? What Jan's talking about is an old system which looks like this: (View in Courier) FIN>---[COUNT]-+-[LUT 0�]---[DAC 0�]--->OUT 0� | | +-[LUT n�]---[DAC n�]--->OUT n� How it works is that for any given output from the counter, the LUTs will have outputs which, after being run through the DACs, will differ from each other by the difference in voltage/current caused by the difference in phase between them, that difference being programmed into the LUTs.
From: George Herold on 11 May 2010 12:01 On May 11, 10:17 am, John Fields <jfie...(a)austininstruments.com> wrote: > On Mon, 10 May 2010 07:56:44 -0700 (PDT), George Herold > > > > > > <gher...(a)teachspin.com> wrote: > >On May 9, 6:14 pm, Bill Sloman <bill.slo...(a)ieee.org> wrote: > >> On May 9, 11:33 pm, billmur...(a)protech.com (Bill Murphy) wrote: > > >> > I am using a commercial stereo amp to output continuous wave test > >> > signals in the low audio range, up to about 2KHz. However, I need a > >> > third channel with a 120 degree phase shift. Is there a circuit that > >> > will do this evenly across this entire frequency range? > > >> > Is it possible to do same using an off-the-shelf transformer and > >> > current subtraction? > > >> > Any advice would be appreciated. > > >> Check out Horowitz and Hill's "The art of Electronics". Section 5.16 > >> talks about phase-sequence filters, which give a constant 90 degree > >> shift over a range of frequencies. They consist of strings of equal > >> value resistors with cross-connected capacitors whose values decrease > >> in constant proportion per stage, halving in the example given, which > >> isn't all that practical to set up. The bottom line is that it isn't > >> trivial, and if you need to ask, you probably don't know enough to put > >> together a circuit that will work. > > >> -- > >> Bill Sloman, Nijmegen > > >Yeah, I've used this circuit. It gives you nice quadrature output > >sine/cosine waves that can be used to make any particular phase > >shift. But phase shift relative to the input signal changes with > >frequency... The phase tends to keep wrapping around. Mixing the two > >signals with a potentiometer also causes amplitude variations if that > >would be a problem. > >That said I used a 10 section filter that has less than 1 degree of > >pahse ripple from 3 Hz to 3kHz. Caps are standard 1, 2.2, 4.7, 10... > >values. The ratio between sections is not as important as keeping the > >same value in each section. You can also change the order of the > >sections with no change at the output. > > >If the OP wants only a few known frequincies then he could make a few > >dedicated RC sections... or some dedicated All-pass opamp filters. > >and switch each in when testing at that frequency. > > --- > Have you both lost your minds? Dang! has my mind gone missing again? It's always wandering off. (shuffles through papers on desk) Ahh here it is. > > What Jan's talking about is an old system which looks like this: I thought we were responding to Bill (the OP) and not to Jan. Certainly with DDS and a couple of look up tables you can make sine waves with any phase shift you want. But the phase sequence filter would do what the OP wanted. Perhaps too many parts for your taste? When I first played with this I air wired the R's and C's together... It had a certain beauty to it... Opps there goes the mind again, George H. > > (View in Courier) > > FIN>---[COUNT]-+-[LUT 0°]---[DAC 0°]--->OUT 0° > | > | > +-[LUT n°]---[DAC n°]--->OUT n° > > How it works is that for any given output from the counter, the LUTs > will have outputs which, after being run through the DACs, will differ > from each other by the difference in voltage/current caused by the > difference in phase between them, that difference being programmed > into the LUTs.- Hide quoted text - > > - Show quoted text -
From: whit3rd on 11 May 2010 13:27 On May 11, 6:47 am, George Herold <gher...(a)teachspin.com> wrote: > On May 11, 3:21 am, billmur...(a)protech.com (Bill Murphy) wrote: > > > On Mon, 10 May 2010 18:24:52 -0700 (PDT), MooseFET > > > <kensm...(a)rahul.net> wrote: > > >What are the two other channels making? > > > Same signal, all spearated by 120 degrees. Done in CoolEdit. > > > >If you have two signals at 90 degrees, you can get any other phase > > >and the same amplitude. > > The simple way is to mix them in a pot. > > Sine----+ > | > P > O<----output > T > | > Cosine--+ > > But this has an amplitude variation as you move the pot wiper around, No, it doesn't! Just buffer that output and it's good to go. It has output impedance variation, not amplitude. That'll get phase shifts from 0 to 90 degrees ("0" shift -> sine, "90" shift -> cosine). sin( w*t + phi) = sin(w*t) cos(phi) + cos(w*t) sin(phi) so your weighted sum of sine and cosine just needs the phase sine and cosine for its coefficients; this in general requires negative coefficients so the potentiometer is joined with an inverting amplifier or something in the way of a coupling transformer.
From: Fred Bartoli on 11 May 2010 13:49 whit3rd a �crit : > On May 11, 6:47 am, George Herold <gher...(a)teachspin.com> wrote: >> On May 11, 3:21 am, billmur...(a)protech.com (Bill Murphy) wrote: >> >>> On Mon, 10 May 2010 18:24:52 -0700 (PDT), MooseFET >>> <kensm...(a)rahul.net> wrote: >>>> What are the two other channels making? >>> Same signal, all spearated by 120 degrees. Done in CoolEdit. >>>> If you have two signals at 90 degrees, you can get any other phase >>>> and the same amplitude. > >> The simple way is to mix them in a pot. >> >> Sine----+ >> | >> P >> O<----output >> T >> | >> Cosine--+ >> >> But this has an amplitude variation as you move the pot wiper around, > > No, it doesn't! Just buffer that output and it's good to go. > It has output impedance variation, not amplitude. > But it does! > That'll get phase shifts from 0 to 90 degrees ("0" shift -> sine, "90" > shift -> cosine). > > sin( w*t + phi) = sin(w*t) cos(phi) + cos(w*t) sin(phi) > Pb is that with a pot you're mixing proportionally to the pot angle, not the sin and cos of the pot angle. IOW output = phi sin(w*t) + (1-phi) cos(w*t) (instead of sin(w*t) cos(phi) + cos(w*t) sin(phi) ) which has a min sqrt(1/2) amplitude for phi=0.5 > so your weighted sum of sine and cosine just needs the phase sine > and cosine for its coefficients; this in general requires negative > coefficients > so the potentiometer is joined with an inverting amplifier or > something > in the way of a coupling transformer. -- Thanks, Fred.
From: whit3rd on 11 May 2010 14:04
On May 11, 10:27 am, whit3rd <whit...(a)gmail.com> wrote: > On May 11, 6:47 am, George Herold <gher...(a)teachspin.com> wrote: > > > > > > > On May 11, 3:21 am, billmur...(a)protech.com (Bill Murphy) wrote: > > > > On Mon, 10 May 2010 18:24:52 -0700 (PDT), MooseFET > > > > <kensm...(a)rahul.net> wrote: > > > >What are the two other channels making? > > > > Same signal, all spearated by 120 degrees. Done in CoolEdit. > > > > >If you have two signals at 90 degrees, you can get any other phase > > > >and the same amplitude. > > > The simple way is to mix them in a pot. > > > Sine----+ > > | > > P > > O<----output > > T > > | > > Cosine--+ > > > But this has an amplitude variation as you move the pot wiper around, > > No, it doesn't! Just buffer that output and it's good to go. Oops, got that wrong; there IS amplitude variation, up to about 30 percent; there's another familiar phase-shifter with potentiometer/capacitor that does get the amplitude constant... |