frerichs@adsl (dfRERICHS) (05/06/91)
Does anyone know the algorithm that would take a stream of sampled sound ,shift it 180 deg in phase and spit it back out again? ie sound -> a/d -> dsp to shift phase -> d/a -> sound (it would seem from the nature of the problem that a simplistic analog circuit to do this might be possible... some sort of RC net... if anyone know if this is possible, PLEASE make it known.) Thanks djf
karsh@trifolium.esd.sgi.com (Bruce Karsh) (05/06/91)
In article frerichs@adsl.ece.uiuc.edu (dfRERICHS) writes: >Does anyone know the algorithm that would take a stream of sampled sound >,shift it 180 deg in phase and spit it back out again? >ie sound -> a/d -> dsp to shift phase -> d/a -> sound This one is so easy that I wonder if it's not part of a homework problem. Bruce Karsh karsh@sgi.com
cas@media-lab.media.mit.edu.MEDIA.MIT.EDU (Scud) (05/06/91)
In article <1991May5.233533.18783@ux1.cso.uiuc.edu> frerichs@adsl.ece.uiuc.edu (dfRERICHS) writes: >Does anyone know the algorithm that would take a stream of sampled sound >,shift it 180 deg in phase and spit it back out again? Use an op-amp as as inverter (ie with a gain of -1). One chip, one resistor, one pot. cas
wdp@ee.egr.duke.edu (William D. Palmer) (05/06/91)
How about an op-amp inverting amplifier? Dev Palmer wdp@dukee.egr.duke.edu Duke Electrical Engineering Room 309 Old Engineering Bldg. Durham, NC 27706 (919) 660-5282
grayt@Software.Mitel.COM (Tom Gray) (05/07/91)
In article <1991May5.233533.18783@ux1.cso.uiuc.edu> frerichs@adsl.ece.uiuc.edu (dfRERICHS) writes: >Does anyone know the algorithm that would take a stream of sampled sound >,shift it 180 deg in phase and spit it back out again? > >ie sound -> a/d -> dsp to shift phase -> d/a -> sound > Invert the sign bit
jthornto@ee.ubc.ca (Johan Thornton) (05/07/91)
In article <1420@cameron.egr.duke.edu> wdp@ee.egr.duke.edu (William D. Palmer) writes: >How about an op-amp inverting amplifier? Sure... it's simple in the analog domain. In a digital system, the solution is a bit more complex. The transfer function is: X = -X -- Johan Thornton (but my friends call me jthornto@ee.ubc.ca)
mhorne@jomega.rain.com (Michael T. Horne;649-8957;;;jomega) (05/07/91)
In a recent article by frerichs@adsl (dfRERICHS): > Does anyone know the algorithm that would take a stream of sampled sound > ,shift it 180 deg in phase and spit it back out again? > Without giving this much thought :), how about a pair of cascaded Hilbert transform filters? :) Mike
mpurtell@IASTATE.EDU (Purtell Michael J) (05/07/91)
In article <7821@nst>, grayt@Software.Mitel.COM (Tom Gray) writes: > In article <1991May5.233533.18783@ux1.cso.uiuc.edu> frerichs@adsl.ece.uiuc.edu (dfRERICHS) writes: > >Does anyone know the algorithm that would take a stream of sampled sound > >,shift it 180 deg in phase and spit it back out again? > > > >ie sound -> a/d -> dsp to shift phase -> d/a -> sound > > > > Invert the sign bit Not if it's a 2's complement integer!! -- -- Michael Purtell -- | "In a hundred years, | There's an Old Irish Recipe for mpurtell@iastate.edu | we'll all be dead." | Longevity: Leave the Table Iowa State University | -- The January Man | Hungry. Leave the Bed Sleepy. "slow is real" | Leave the Tavern Thirsty.
gibbonsj@iccgcc.decnet.ab.com (05/07/91)
In article <1991May5.233533.18783@ux1.cso.uiuc.edu>, frerichs@adsl (dfRERICHS) writes: > Does anyone know the algorithm that would take a stream of sampled sound > ,shift it 180 deg in phase and spit it back out again? > > ie sound -> a/d -> dsp to shift phase -> d/a -> sound > > Thanks > > djf Its a no-brainer: just reverse your speaker wires! Instant 180 degree phase inversion! 8-] -- John Gibbons N8OBJ Macedonia, Ohio "Welcome My Son, Welcome To The Machine" - Pink Floyd
jfa0522@hertz.njit.edu (john f andrews ece) (05/07/91)
In article <1644@fs1.ee.ubc.ca> jthornto@ee.ubc.ca (Johan Thornton) writes: >In article <1420@cameron.egr.duke.edu> wdp@ee.egr.duke.edu (William D. Palmer) writes: >>How about an op-amp inverting amplifier? > > >Sure... it's simple in the analog domain. In a digital system, the solution >is a bit more complex. The transfer function is: > > X = -X > > >-- >Johan Thornton >(but my friends call me jthornto@ee.ubc.ca) Well, sure, you all can say how easy it is, but you're stopping shy of the hard part. Can't implement a transfer function without getting it into the form of a difference equation! That's the challenge... ----------------------------------------------------------------------------- john f andrews SYSOP The Biomedical Engineering BBS 24 hrs 300/1200/2400/4800 (201) 596-5679 ----------------------------------------------------------------------------- INTERNET jfa0522@hertz.njit.edu CIS 73710,2600 -----------------------------------------------------------------------------
jim@fuji.eng.Yale.edu (James J. Szinger) (05/08/91)
In article <1991May5.233533.18783@ux1.cso.uiuc.edu> frerichs@adsl (dfRERICHS) writes:
Does anyone know the algorithm that would take a stream of sampled sound
,shift it 180 deg in phase and spit it back out again?
ie sound -> a/d -> dsp to shift phase -> d/a -> sound
(it would seem from the nature of the problem that a simplistic analog
circuit to do this might be possible... some sort of RC net... if anyone
know if this is possible, PLEASE make it known.)
Thanks
djf
I think what you are looking for is the Hilbert Transform. A
Hilbert Xform shifts the phase of the signal by 90 degrees. Just
cascade two of them together for a total of 180 degrees.
There might be an easier way but I can't quite think of it.
Jim
--
James J. Szinger Becton Center
jim@fuji.eng.yale.edu 15 Prospect Street
Intelligent Sensors Lab. Box 2157 Yale Station
Electrical Engineering New Haven, Connecticut 06520
Yale University U.S.A.todd@appmag.com (Todd Day) (05/08/91)
%Well, sure, you all can say how easy it is, but you're stopping shy of the %hard part. Can't implement a transfer function without getting it into the %form of a difference equation! That's the challenge... Well, I can see you guys are having fun with this, but let me make a few comments. First of all, 180 deg phase shift is not the same as simply inverting the signal. A phase shift implies a time delay of some sort. Second of all, he *may* have meant a 180 deg phase shift for each of the frequencies that add up to his composite signal. The output signal will definitely NOT look much like the input signal. You can prove this to your self by drawing a time axis and delaying a 100 Hz and 50 Hz signal by 180 degrees. What you can see is similar to what happens in a typical speaker system when an impulse sound like a drum hit is reproduced. Most of the time, it doesn't sound quite right because of the "phase smearing" that results... -- Todd Day | todd@appmag.com
wilf@sce.carleton.ca (Wilf Leblanc) (05/08/91)
todd@appmag.com (Todd Day) writes: >%Well, sure, you all can say how easy it is, but you're stopping shy of the >%hard part. Can't implement a transfer function without getting it into the >%form of a difference equation! That's the challenge... Want a difference equation ?? y(n) = -x(n) >[...] >First of all, 180 deg phase shift is not the same as simply inverting >the signal. A phase shift implies a time delay of some sort. Oh, it does does it ?? Does a transfer function H(z) = -1 imply a time delay ?? [...] >-- >Todd Day | todd@appmag.com -- Wilf LeBlanc, Carleton University, Systems & Comp. Eng. Ottawa, Canada, K1S 5B6 Internet: wilf@sce.carleton.ca UUCP: ...!uunet!mitel!cunews!sce!wilf Oh, cruel fate! Why do you mock me so! (H. Simpson)
karsh@trifolium.esd.sgi.com (Bruce Karsh) (05/09/91)
In article <1991May8.022953.781@appmag.com> todd@appmag.com (Todd Day) writes: >First of all, 180 deg phase shift is not the same as simply inverting >the signal. A phase shift implies a time delay of some sort. Sorry, but a phase shift need not imply a time delay. The amount of delay is called the group delay and it is the slope of the graph of the delay function. It's really only meaningful to think of it as a delay if the delay function is largely a straight line. The problem is that sin waves are periodic, so you can't really say whether on is delayed with respect to another, or advanced. Since the signal is composed of the sum of sin waves, if you delay (or advance) all the sin waves by 180 degrees, then you've just inverted all their signs. When you sum them all together, you get -1 times the original signal. Bruce Karsh karsh@sgi.com
jbuck@janus.Berkeley.EDU (Joe Buck) (05/09/91)
In article <1991May8.022953.781@appmag.com> todd@appmag.com (Todd Day) writes: >First of all, 180 deg phase shift is not the same as simply inverting >the signal. A phase shift implies a time delay of some sort. Todd, you're too smart for your own good, and so are the people talking about Hilbert transforms, even though they are, in a sense, right. >Second of all, he *may* have meant a 180 deg phase shift for each of the >frequencies that add up to his composite signal. OK, this means that |H(j omega)| should be 1 and arg(H (j omega)) should be pi for all omega (H is the transfer function). That means that H(j omega) should be -1 for all omega, so simply inverting the signal really does shift by 180 degrees at all frequencies. Now, for the Hilbert transform people. The Hilber transform is a filter whose response is -j for positive frequencies and j for negative frequencies, and it causes a 90 degree phase shift. Now let's cascade two of them. We square the transfer function to obtain the total transfer function. Note that j^2 = (-j)^2 = -1, so again we get a simple inversion. > The output signal will >definitely NOT look much like the input signal. You can prove this to your >self by drawing a time axis and delaying a 100 Hz and 50 Hz signal by 180 >degrees. You screwed up your plot, Todd. Otherwise you'd find that the right answer is an inversion. Since each individual component is negated by a 180 degree shift, clearly the sum is. The original poster clearly had a homework problem with a trivial answer, and all these "experts" screwed it up. Embarrassing... -- Joe Buck jbuck@janus.berkeley.edu {uunet,ucbvax}!janus.berkeley.edu!jbuck
steveq@syd.dms.CSIRO.AU (Stephen Quigg) (05/09/91)
In article <5781@media-lab.media.mit.edu.MEDIA.MIT.EDU> cas@media-lab.media.mit.edu (Scud) writes: >In article <1991May5.233533.18783@ux1.cso.uiuc.edu> frerichs@adsl.ece.uiuc.edu (dfRERICHS) writes: >>Does anyone know the algorithm that would take a stream of sampled sound >>,shift it 180 deg in phase and spit it back out again? > >Use an op-amp as as inverter (ie with a gain of -1). One chip, one resistor, >one pot. > >cas Inverting is NOT the same as 180 deg phase shift. For a symetric waveform (eg a sine wave) it looks the same, but with something assymetric what you will see is the waveform upside-down, which is not the same as shifted 180 deg. Phase shifting moves a waveform along the time axis: it stays the same way up. Try it on an oscilloscope.
todd@appmag.com (Todd Day) (05/09/91)
I wrote: %Second of all, he *may* have meant a 180 deg phase shift for each of the %frequencies that add up to his composite signal. The output signal will %definitely NOT look much like the input signal. Sorry about this. "Phase smearing" does not occur at multiples of pi/2. I had this visualized totally wrong. After thinking about "the sum of infinite sinewaves", I realized that at 180, this sum does align properly and makes the inverse. I had gotten it confused with an argument I had previously discussing 90 degree phase shift in speakers, which definitely DOES smear the output. Sorry. -- Todd Day | todd@appmag.com
todd@appmag.com (Todd Day) (05/09/91)
steveq@syd.dms.CSIRO.AU (Stephen Quigg) writes:
% Inverting is NOT the same as 180 deg phase shift.
See? I'm not the only one who didn't do his homework... :-)
I gotta start reading my ASSPs again!
--
Todd Day | todd@appmag.commcmahan@netcom.COM (Dave Mc Mahan) (05/09/91)
In a previous article, karsh@trifolium.sgi.com (Bruce Karsh) writes: >In article <1991May8.022953.781@appmag.com> todd@appmag.com (Todd Day) writes: > >>First of all, 180 deg phase shift is not the same as simply inverting >>the signal. A phase shift implies a time delay of some sort. > >Sorry, but a phase shift need not imply a time delay. The amount of delay >is called the group delay and it is the slope of the graph of the delay >function. It's really only meaningful to think of it as a delay if the >delay function is largely a straight line. > >The problem is that sine waves are periodic, so you can't really say whether >on is delayed with respect to another, or advanced. Since the signal >is composed of the sum of sine waves, if you delay (or advance) all the >sin waves by 180 degrees, then you've just inverted all their signs. When >you sum them all together, you get -1 times the original signal. This works great for pure tones, but if you have a signal (such as that from an ECG or a modem that uses phase or frequency shifting) then the input signal will not be a pure tone, but rather a sum of sine waves that vary in amplitude over time. I think (just to add another voice to this mess) that the correct 'theoretical' answer is to find a filter that actually does produce the proper time shift for each frequency. The 'realistic' answer is that this just negating the input signal (Y[n] = -X[n]) is probably what the original poster needs to know. Once again, I guess you have to know the full needs of the application before you can provide the full answer. (Is this another case where we engineers charge off to solve the wrong problem? :-) > Bruce Karsh > karsh@sgi.com -dave -- Dave McMahan mcmahan@netcom.com {apple,amdahl,claris}!netcom!mcmahan
heredia@enel.ucalgary.ca (Edwin Heredia) (05/09/91)
For those who can't believe that inverting the signal produces a 180 deg. phase shift and for those who think that this trick works only with "easy-to-work" signals as pure tones: Given *ANY* signal x(n) with Fourier Transform X(w) Consider the transformation y(n) = (-1) * x(n) in the Fourier domain this gives Y(w) = (-1) * X(w) which is Y(w) = exp(j180) * X(w) result: 180 deg. phase shift! where *ANY* means really *any*, that is ECG's, speech, etc., and (-1)*x(n) means really inverting the signal. --------- If a 180 deg shift produced about 15 articles posted, how many would be produced by a 360 deg shift? Answer: 2*15 = 30
myhui@bnr.ca (Michael Hui) (05/10/91)
In article <1991May9.160115.3494@cpsc.ucalgary.ca> heredia@enel.ucalgary.ca (Edwin Heredia) writes: >For those who can't believe that inverting the signal produces >a 180 deg. phase shift and for those who think that this trick >works only with "easy-to-work" signals as pure tones: > >Given *ANY* signal x(n) with Fourier Transform X(w) ^^^ Really? You surely mean _periodic_ signal. Michael MY Hui Ottawa Canada myhui@bnr.ca
whit@milton.u.washington.edu (John Whitmore) (05/10/91)
In article <1991May8.222501.19572@syd.dms.CSIRO.AU> steveq@syd.dms.CSIRO.AU (Stephen Quigg) writes: >In article <5781@media-lab.media.mit.edu.MEDIA.MIT.EDU> cas@media-lab.media.mit.edu (Scud) writes: >>In article <1991May5.233533.18783@ux1.cso.uiuc.edu> frerichs@adsl.ece.uiuc.edu (dfRERICHS) writes: >>>Does anyone know the algorithm that would take a stream of sampled sound >>>,shift it 180 deg in phase and spit it back out again? >> >>Use an op-amp as as inverter (ie with a gain of -1). > Inverting is NOT the same as 180 deg phase shift. For a symetric waveform >(eg a sine wave) it looks the same, but with something assymetric what you >will see is the waveform upside-down, which is not the same as shifted 180 >deg. Phase shifting moves a waveform along the time axis: it stays the same >way up. Try it on an oscilloscope. That's wrong. The inverter is a perfectly good 180 degree phase shifter, and if you test it at ANY frequency you will see 180 degrees of phase shift; the phase shift versus frequency is EXACTLY what was asked for. Your 'assymmetric waveform' has a lot of Fourier components, and time-shifting it as you seem to be describing is NOT a well-defined operation. You have to find some particular frequency, derive a time delay from THAT ONE FREQUENCY COMPONENT, and apply that time delay to get the time-shift, and that is NOT the correct phase shift for any frequency component except the one you chose as 'most significant'. John Whitmore
grayt@Software.Mitel.COM (Tom Gray) (05/10/91)
In article <1991May10.003817.5593@milton.u.washington.edu> whit@milton.u.washington.edu (John Whitmore) writes: :In article <1991May8.222501.19572@syd.dms.CSIRO.AU> steveq@syd.dms.CSIRO.AU (Stephen Quigg) writes: :>In article <5781@media-lab.media.mit.edu.MEDIA.MIT.EDU> cas@media-lab.media.mit.edu (Scud) writes: :>>In article <1991May5.233533.18783@ux1.cso.uiuc.edu> frerichs@adsl.ece.uiuc.edu (dfRERICHS) writes: :>>>Does anyone know the algorithm that would take a stream of sampled sound :>>>,shift it 180 deg in phase and spit it back out again? :>> :>deg. Phase shifting moves a waveform along the time axis: it stays the same :>way up. Try it on an oscilloscope. : : : That's wrong. The inverter is a perfectly good 180 degree :phase shifter, and if you test it at ANY frequency you will see :180 degrees of phase shift; the phase shift versus frequency is :EXACTLY what was asked for. : Your 'assymmetric waveform' has a lot of Fourier components, :and time-shifting it as you seem to be describing is NOT a :well-defined operation. You have to find some particular frequency, Not trying to be impudent but just trying to expand on the idea which Mr. Whitmore put forward. Time shifting is a well defined operation. A time shift in the time domain is equivalent to multiplying the frequency domain function by e**(-jwT) (where T is the magnitude of the time shift). Remember the requirment in linear systems for a linear phase response. Well this is where it comes from. A time delay will not cause a constant phase change for all frequencies. Instead the it will produce a linear relationship betweem phase and frequency. --------------------------------------------------------------- Tom Gray -------------------------------------------------------------- :derive a time delay from THAT ONE FREQUENCY COMPONENT, and apply :that time delay to get the time-shift, and that is NOT the correct :phase shift for any frequency component except the one you :chose as 'most significant'.
karsh@trifolium.esd.sgi.com (Bruce Karsh) (05/10/91)
>>Given *ANY* signal x(n) with Fourier Transform X(w) >Really? You surely mean _periodic_ signal. A signal doesn't have to be periodic to have a Fourier transform. Any sequence x(n) that is absolutely summable (i.e., the sum of the absolute values is bounded) has a Fourier transform given by: inf X(w) = sum x(n)e^(-jwn) n = -inf Bruce Karsh karsh@sgi.com
jfa0522@hertz.njit.edu (john f andrews ece) (05/10/91)
In article <1991May09.192740.4165@bmerh2.bnr.ca> myhui@bnr.ca (Michael Hui) writes: >In article <1991May9.160115.3494@cpsc.ucalgary.ca> heredia@enel.ucalgary.ca (Edwin Heredia) writes: >>For those who can't believe that inverting the signal produces >>a 180 deg. phase shift and for those who think that this trick >>works only with "easy-to-work" signals as pure tones: >> >>Given *ANY* signal x(n) with Fourier Transform X(w) > ^^^ > >Really? You surely mean _periodic_ signal. > > Michael MY Hui Ottawa Canada myhui@bnr.ca Why do we keep tagging onto this thread? "any signal ..with a fourier ransform" implies any periodic signal....so why the beef? Let it go!
ikw@genrad.UUCP (Isaac K. Wong) (05/11/91)
In article <1991May09.192740.4165@bmerh2.bnr.ca> myhui@bnr.ca (Michael Hui) writes: =>In article <1991May9.160115.3494@cpsc.ucalgary.ca> heredia@enel.ucalgary.ca (Edwin Heredia) writes: =>> =>>Given *ANY* signal x(n) with Fourier Transform X(w) => ^^^ => =>Really? You surely mean _periodic_ signal. => and a linear system too. and I am sure the ANY signal you mentioned doesn't include DC. :) -- Isaac Wong || There's no need to worry about death, GenRad, Milpitas, CA || it will not happen in your lifetime. wongi@topo.genrad.com || wongi@topo.uucp || -Raymond Smullyan
karsh@trifolium.esd.sgi.com (Bruce Karsh) (05/11/91)
In article jfa0522@hertz.njit.edu (john f andrews ece) writes: >Why do we keep tagging onto this thread? "any signal ..with a fourier >ransform" implies any periodic signal....so why the beef? Let it go! Because what you said is wrong. It is simply not the case that the existence of a Fourier transform of a function or a series implies periodicity. The existence of a Fourier sequence implies periodicity, but that is another matter. The Fourier transform exists for non-periodic functions which are absolutely summable. Bruce Karsh karsh@sgi.com
jsorenso@thesis1.med.uth.tmc.edu (JEFFREY MARIUS SORENSON) (05/12/91)
In article <1991May8.222501.19572@syd.dms.CSIRO.AU> steveq@syd.dms.CSIRO.AU (Stephen Quigg) writes: >In article <5781@media-lab.media.mit.edu.MEDIA.MIT.EDU> cas@media-lab.media.mit.edu (Scud) writes: >>In article <1991May5.233533.18783@ux1.cso.uiuc.edu> frerichs@adsl.ece.uiuc.edu (dfRERICHS) writes: >>>Does anyone know the algorithm that would take a stream of sampled sound >>>,shift it 180 deg in phase and spit it back out again? >> >>Use an op-amp as as inverter (ie with a gain of -1). One chip, one resistor, >>one pot. >> >>cas > Inverting is NOT the same as 180 deg phase shift. For a symetric waveform >(eg a sine wave) it looks the same, but with something assymetric what you >will see is the waveform upside-down, which is not the same as shifted 180 >deg. Phase shifting moves a waveform along the time axis: it stays the same >way up. Try it on an oscilloscope. Are we talking about phase shifts or time shifts? Phase Shift: Since the signal can be represented as a sum (or integral) of complex exponetials, phase shifting can be done by simply multiplying the whole sum (integral) by exp( i*p ), where p is the value of the phase shift. This can be visualized by rotating a vector (our signal) by 180 degrees in the complex plane. Since a phase shift of pi (180 degrees) for each component of the Fourier series gives that same component multiplied by -1 (e(i*180) = -1), the sum of the phase shifted (180 degrees) components gives the original signal multiplied by -1. So phase shifting by 180 degrees IS inverting. This is why signals that are 180 degrees out of phase are said to cancel eachother out Time Shift: A time shift is equivalent to phase shifting each component of the signal by its frequency times the value of the time shift, not simply phase shifting each frequency component by the same phase. This is NOT the same thing as performing a simple phase shift on the entire signal. (It just happens that for a signal composed of a single frequency (a sine wave), a time shift of half the period is equivalent to a phase shift of 180 degrees) As I understand it, a phase shift by 180 degrees IS inversion. If I am not using the conventional definitions of the terms "phase shift" and "time shift", please let me know. -- ------------------------------------------------------------------------------- jsorenso@thesis1.med.uth.tmc.edu | | |
mcphail@dataco.UUCP (Alex McPhail) (05/14/91)
>> Inverting is NOT the same as 180 deg phase shift. For a symetric waveform >>(eg a sine wave) it looks the same, but with something assymetric what you >>will see is the waveform upside-down, which is not the same as shifted 180 >>deg. Phase shifting moves a waveform along the time axis: it stays the same >>way up. Try it on an oscilloscope. In article <1991May10.003817.5593@milton.u.washington.edu> whit@milton.u.washington.edu (John Whitmore) writes: > That's wrong. The inverter is a perfectly good 180 degree >phase shifter, and if you test it at ANY frequency you will see >180 degrees of phase shift; the phase shift versus frequency is >EXACTLY what was asked for. > Your 'assymmetric waveform' has a lot of Fourier components, >and time-shifting it as you seem to be describing is NOT a >well-defined operation. You have to find some particular frequency, >derive a time delay from THAT ONE FREQUENCY COMPONENT, and apply >that time delay to get the time-shift, and that is NOT the correct >phase shift for any frequency component except the one you >chose as 'most significant'. > > John Whitmore Sorry, John. I usually don't butt into other people's business here, but you really are dead wrong. The phase shift described above is indeed correct. Since any periodic function is always measured against time, shifting the function's phase is also a time-related operation. In specific response to your article, frequency and Fourier components have nothing to do with phase shifting. The shift of a phase, in degrees, can be expressed by: dt -- x 360 L where: dt is the shift of the function L is the wavelength of the function Diagramatically, the following example shows the difference of a sawtooth wave shifted by 180 degrees, and an inverted sawtooth wave. . . . . /| /| /| /| / | / | / | / | / | / | / | / | / | / | / | / | / | / | / | / | / |/ |/ |/ | Normal sawtooth wave . . . . /| /| /| /| / | / | / | / | / | / | / | / | / | / | / | / | / | / | / | / | / |/ |/ |/ | Shifted sawtooth wave \ |\ |\ |\ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \| \| \| \| ` ` ` ` Inverted sawtooth wave ============================================================================ ________ |\ / Alex McPhail | \ / | \ / mail to mcphail@dataco |---X (uunet!mitel!melair!dataco!mcphail) | / \ | / \ "Wwwwwhat is the air speed of an average swallow?" |/_____\ Alex ***--------------------------------------------------------------*** * DISCLAIMER: * * ==========: * * The opinions expressed are solely of the author and do not * * necessarily reflect the opinions of Canadian Marconi Company. * ***--------------------------------------------------------------***
myhui@bnr.ca (Michael Hui) (05/15/91)
In article <625@fudd.dataco.UUCP> mcphail@taarna.UUCP (Alex McPhail,DC ) writes: > . . . . > /| /| /| /| > / | / | / | / | > / | / | / | / | > / | / | / | / | > / | / | / | / | >/ |/ |/ |/ | Normal sawtooth wave > > . . . . > /| /| /| /| > / | / | / | / | > / | / | / | / | > / | / | / | / | > / | / | / | / | > / |/ |/ |/ | Shifted sawtooth wave > >\ |\ |\ |\ | > \ | \ | \ | \ | > \ | \ | \ | \ | > \ | \ | \ | \ | > \ | \ | \ | \ | > \| \| \| \| > ` ` ` ` Inverted sawtooth wave >============================================================================ >________ >|\ / Alex McPhail >| \ / >| \ / mail to mcphail@dataco >|---X (uunet!mitel!melair!dataco!mcphail) Brilliant. Absolutely brilliant (!). Now for my comments, IMHO: inverting shifts all frequency components by pi degrees. That means, taken time-wise, high frequencies are shifted less in time than low frequencies. That's how you get the inverted sawtooth above. Imagine how the fundamental would look in the original wave. Invert it. Lo and behold, it does fit the inverted sawtooth!!! But the original question was to shift the phase of a signal by 180 degrees, which implicitly means to delay the signal by pi degrees as referred to its fundamental frequency. Michael MY Hui Ottawa Canada myhui@bnr.ca
jthornto@ee.ubc.ca (Johan Thornton) (05/15/91)
In article <1991May15.031706.15559@bmerh2.bnr.ca> myhui@bnr.ca (Michael Hui) writes: >Now for my comments, IMHO: inverting shifts all frequency components by pi >degrees. Pi degrees? Ok, that's easy, pi^2/180 radians. >But the original question was to shift the phase of a signal by 180 degrees, >which implicitly means to delay the signal by pi degrees as referred to its >fundamental frequency. This shouldn't be too difficult. I just finished my thesis on a device that delays a signal by 360 degrees with respect to its fundamental. It was pretty complicated, but your device should be about half as complicated. -- Johan Thornton (but my friends call me jthornto@ee.ubc.ca)
whit@milton.u.washington.edu (John Whitmore) (05/15/91)
In article <625@fudd.dataco.UUCP> mcphail@taarna.UUCP (Alex McPhail,DC ) writes: > Since any periodic function is always measured against time, >shifting the function's phase is also a time-related operation. > > .... frequency and Fourier components >have nothing to do with phase shifting. I disagree. In discussing networks (or filters), one usually refers to the Bode plot of amplitude and of phase shift plotted against frequency. The normal meaning in the context of filters of the phrase 'phase shift' is the phase part of that representation, and 'constant 180 degree phase shift' is clear and unambiguous. It means amplitude inversion. What earthly use are 'degrees' in describing the phase of something which is not a circular function? And, why do you refer to 'any periodic function' in this discussion? To the best of my recollection, there was never any statement made that the function under discussion was periodic. Phase shifts are well-defined in the absence of periodicity, though easier to measure and display in the case of periodic test waveforms. John Whitmore
amr@egr.duke.edu (Anthony M. Richardson) (05/15/91)
I find it interesting that this thread has dragged on as long as
it has. I assume that this is due to a tendency to respond
without taking time to think about the problem.
To phase shift a signal means to multiply its Fourier transform
by exp(j A) where A is the desired phase shift in radians.
The inverse Fourier transform of the signal exp(j A) X(w) is
exp(j A) x(t). (We're just multiplying by a scalar here. There is
no time shift. If we had multiplied by exp(j A w) instead of
exp(j A) the corresponding time signal would be x(t + A).)
If A is equal to pi (180 degrees) then exp(j A) = -1 and so we
are just inverting the signal.
Notice that there is no requirement for the signal to be periodic.
It just has to be Fourier transformable.
I suspect the problem is that many people are thinking in terms
of Fourier series and phase shifting each component of the series.
We still get the same result though. If x(t) is periodic
we can represent it in a Fourier series as
x(t) = Sum a(k) exp(j k w0 t)
where the Sum is over all k and w0 is the fundamental frequency.
Now shift each component by A radians. Call this new signal y(t).
y(t) = Sum a(k) exp[j (k w0 t + A)]
(Notice again that if A is equal to pi, then y(t) = -x(t).)
What is the time shift of each component? Rewrite as
y(t) = Sum a(k) exp[j k w0 (t + A/(k w0))].
The time shift of each component depends upon the frequency of the
component, so the time shift of each component waveform is not
the same. (This is obvious if you sketch two harmonically related
sinusoids and then sketch their 180 degree phase shifted version.)
This has all been said by others. I just that I would throw in my
$.02 because the argument seemed to be shifting to the wrong answer.
Regards,
Tony Richardson
amr@egr.duke.edujefft@phred.UUCP (Jeff Taylor) (05/21/91)
All the postings on 180 deg phase shifts reminds me of standard questions I ask when interviewing for positions which require some knowledge of DSP. "WHEN WOULD YOU USE A FIR FILTER?" "If I wanted a linear phase linear phase characteristics." "HOW DO YOU GET A LINEAR PHASE CHARACTERISTIC? " "Have a symeterical time response." "WHERE ARE THE ZEROS? "On the unit circle." "DOESN'T THAT GIVE PHASE SHIFTS OF 180 DEG'S? (it's kind of fun watching people sort through the dichotomy - everyone KNOWS symeterical fir filters have linear phase, but the zeros are on the unit circle - which does give a 180 deg phase shift) I suppose it depends on how one wants to interperate signal inversion. -- ----------------------------------------------------------------------------- Jeff Taylor Physio Control Corp. -----------------------------------------------------------------------------
wilf@sce.carleton.ca (Wilf Leblanc) (05/24/91)
jefft@phred.UUCP (Jeff Taylor) writes: >: 30 >All the postings on 180 deg phase shifts reminds me of standard questions I >ask when interviewing for positions which require some knowledge of DSP. > "WHEN WOULD YOU USE A FIR FILTER?" > "If I wanted a linear phase linear phase characteristics." > "HOW DO YOU GET A LINEAR PHASE CHARACTERISTIC? " > "Have a symeterical time response." > "WHERE ARE THE ZEROS? > "On the unit circle." > "DOESN'T THAT GIVE PHASE SHIFTS OF 180 DEG'S? >(it's kind of fun watching people sort through the dichotomy - everyone KNOWS > symeterical fir filters have linear phase, but the zeros are on the unit > circle - which does give a 180 deg phase shift) >I suppose it depends on how one wants to interperate signal inversion. "... but the zeros are on the unit circle" ?????? O.K. smartass, I'll play your game: H(z) = (1+0.5 z^-1)(1+2z^-1) = 1 + 2.5z^-1 + z^-2 Looks symmetric, looks too me like either you are wrong or 0.5/1 = 2/1 = 1. Nothing like interviewing someone, trying to screw them up, and being completely wrong, eh ?? The zeros are not on the unit circle. >-- >----------------------------------------------------------------------------- > Jeff Taylor > Physio Control Corp. >----------------------------------------------------------------------------- -- Wilf LeBlanc, Carleton University, Systems & Comp. Eng. Ottawa, Canada, K1S 5B6 Internet: wilf@sce.carleton.ca UUCP: ...!uunet!mitel!cunews!sce!wilf Oh, cruel fate! Why do you mock me so! (H. Simpson)
jefft@phred.UUCP (Jeff Taylor) (05/29/91)
In article <wilf.675023233@rigel.sce.carleton.ca> wilf@sce.carleton.ca (Wilf Leblanc) writes: > >"... but the zeros are on the unit circle" ?????? > >O.K. smartass, I'll play your game: > > H(z) = (1+0.5 z^-1)(1+2z^-1) > = 1 + 2.5z^-1 + z^-2 > >Looks symmetric, looks too me like either you are wrong or >0.5/1 = 2/1 = 1. > > >Nothing like interviewing someone, trying to screw them up, and >being completely wrong, eh ?? > >The zeros are not on the unit circle. > COMPLETELY wrong? Half wrong perhaps, (and technically not even that). Linear phase FIR filters have zeros either on the unit circle in complex conj. pairs, or in pairs with a reciprocal magnitude (and their complex conj's). I ignored the second class (because I forgot about them :( ). I *think* the remez exchange will generate filters with both kinds of zeros. Using your filter as an example, changing the second coefficient to anything less then two will generate complex zeros - which should be on the unit circle. The time response is still symmetric, so it meets the necessary and sufficent condition for linear phase, but it does have single zeros on the unit circle. Example: H(z) = 1 + (2/sqrt[2])z^-1 + z^-2 H(z) = [(z^-1) - (e^j*pi/4)]*[(z^-1) - (e^-j*pi/4)] (That's hard to read, should represent a zero an +- 45 deg on the unit circle, if I didn't make another mistake) The point I was trying to make is there are filters which satisfy the conditions for linear phase, but which place a zero on the unit circle. I don't know what the significance of this is because I have never used the linear phase property of FIR filters. I expect it would give unexpected results if a filtered signal was summed with one which was just delayed. As for trying to screw someone up, I think it is important to know why you know something. The reason we know symmetric filters have linear phase is the fourier transform of a symmetric signal is real. But negative numbers are real. So if you plot |gain| and phase, there will be 180 deg phase shifts. As an example consider a rectangular pulse, symetric about 0. It has a sinc function as a transform - which is real, but bi-phasic. It's real, so it has no phase shift. If it is the impulse response of a filter, does it just invert some frequencies, or does it have 180 deg phase shift? Does anyone use the linear phase property of FIR filters in practice? Is it necessary avoid zeros on the unit circle? jt -- ----------------------------------------------------------------------------- Jeff Taylor Physio Control Corp. -----------------------------------------------------------------------------