Andor wrote:> Jerry Avins schrieb: > >> Andor wrote: >>> Jerry Avins wrote: >>> ... >>>> My question was intended to provoke the realization that inversion is >>>> not a phase shift of any kind, even though it may sometimes be expedient >>>> to call it one. >>> Now you are really trolling, aren't you? If not, then it might be >>> interesting (shock therapy) to write down the transfer function of >>> >>> a) a 180� phase shifter >>> >>> and >>> >>> b) an inverter. >>> >>> Now compare the two ... >> Now compare their group delays. Transfer functions with different group >> delays are not identical. > > But equal transfer functions have equal group delay ... did you read > that post I linked regarding this exact same discussion 15 years ago in > this forum?Yes, and not for the first time. It's one of the few instances where I think Max was off base. (Max: Where are you?) By its nature, phase shift implies time shift. Inversion doesn't. If math tells you that cascaded Hilbert transformers produce the same output at the same time as an inverter, you need to fix the math. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
How does an inverter affect phase?
Started by ●June 15, 2006
Reply by ●June 17, 20062006-06-17
Reply by ●June 17, 20062006-06-17
Jerry Avins wrote:> By its nature, phase shift implies time shift.Let's look at a p radian phase shifter. Do you agree that this is a linear, time-invariant system? If you agree with me on that, then it follows that the frequency response of the p radian phase shifter (phase "advancer" would be a better term) is H_p(w) =3D exp(j p), 0 <=3D w < pi =3D exp(-j p), pi <=3D w < 2 pi. Now if you set p =3D pi, you get the frequency response of the 180=B0 phase shifter H_pi(w) =3D -1, for all w. There is no time shift, or delay, because the phase response is constant, and therefore the group delay is zero for all frequencies.> Inversion doesn't.That is correct.> If math tells you that cascaded > Hilbert transformers produce the same output at the same time as an > inverter, you need to fix the math.There really is nothing broken with the math.
Reply by ●June 17, 20062006-06-17
Andor wrote:> Jerry Avins wrote: > >> By its nature, phase shift implies time shift. > > Let's look at a p radian phase shifter. Do you agree that this is a > linear, time-invariant system? If you agree with me on that, then it > follows that the frequency response of the p radian phase shifter > (phase "advancer" would be a better term) is > > H_p(w) = exp(j p), 0 <= w < pi > = exp(-j p), pi <= w < 2 pi. > > Now if you set p = pi, you get the frequency response of the 180� > phase shifter > > H_pi(w) = -1, for all w. > > There is no time shift, or delay, because the phase response is > constant, and therefore the group delay is zero for all frequencies.Great! Now build one that doesn't incorporate an inverter.>> Inversion doesn't. > > That is correct. > >> If math tells you that cascaded >> Hilbert transformers produce the same output at the same time as an >> inverter, you need to fix the math. > > There really is nothing broken with the math.I know. That worries me. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●June 17, 20062006-06-17
JA [Sat, 17 Jun 2006 15:57:33 -0400]: >Calling ketchup a vegetable doesn't make it one. Because it's fruit, after all. (R) -- 40th Floor - Software @ http://40th.com/ iPlay : the ultimate audio player for mobiles parametric eq, xfeed, reverb; all on a mobile
Reply by ●June 17, 20062006-06-17
By its nature, phase shift> implies time shift. Inversion doesn't. If math tells you that cascaded > Hilbert transformers produce the same output at the same time as an > inverter, you need to fix the math. >Well, consider a phase-lead (advance), by your argument it implies a time-advance and hence something uncausal. ie (1+sT1)/(1+sT2) where T1>T2. (say T1=10T2) M.P -- Posted via a free Usenet account from http://www.teranews.com
Reply by ●June 17, 20062006-06-17
Mad Prof wrote:> By its nature, phase shift >> implies time shift. Inversion doesn't. If math tells you that cascaded >> Hilbert transformers produce the same output at the same time as an >> inverter, you need to fix the math. >> > > Well, consider a phase-lead (advance), by your argument it implies a > time-advance and hence something uncausal. > > ie > > (1+sT1)/(1+sT2) where T1>T2. (say T1=10T2)Did I sat time advance? I thought I wrote "time shift". Andor's transfer function implies no time shift for the reasons he gives, but he needs an inverter -- multiplication by -1 -- to realize it. Although I can make a HT over a rather wide band with non-inverting amplifiers, Rs, and Cs, no number of those can create a delay-free inversion Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●June 18, 20062006-06-18
Jerry Avins wrote:> Andor wrote: >> Jerry Avins schrieb: >> >>> Andor wrote: >>>> Jerry Avins wrote: >>>> ... >>>>> My question was intended to provoke the realization that inversion is >>>>> not a phase shift of any kind, even though it may sometimes be >>>>> expedient to call it one. >>>> Now you are really trolling, aren't you? If not, then it might be >>>> interesting (shock therapy) to write down the transfer function of >>>> >>>> a) a 180� phase shifter >>>> >>>> and >>>> >>>> b) an inverter. >>>> >>>> Now compare the two ... >>> Now compare their group delays. Transfer functions with different group >>> delays are not identical. >> >> But equal transfer functions have equal group delay ... did you read >> that post I linked regarding this exact same discussion 15 years ago in >> this forum? > > Yes, and not for the first time. It's one of the few instances where I > think Max was off base. (Max: Where are you?) By its nature, phase shift > implies time shift. Inversion doesn't. If math tells you that cascaded > Hilbert transformers produce the same output at the same time as an > inverter, you need to fix the math. > > JerryIMO, you cannot have the same output with causal HTs. Inverter (i.e., multiplication by -1) has a group delay of 0 samples and HT with more than 1 taps has larger that 0 sample group delay. At least in practice, and in digital world. I know nothing about analog signal processing (<- a disclaimer ;)) Phase shift, in general, does not imply time shift. Only if the phase shift is _not_ a constant as a function of frequency, it implies time shift. Inverter is an example where phase shift does not imply time shift ;) -- Jani Huhtanen Tampere University of Technology, Pori
Reply by ●June 18, 20062006-06-18
"Jani Huhtanen" <jani.huhtanen@kolumbus.fi> wrote in message news:rj7lg.5499$qB5.4265@reader1.news.jippii.net...> Jerry Avins wrote: > > > Andor wrote: > >> Jerry Avins schrieb: > >> > >>> Andor wrote: > >>>> Jerry Avins wrote: > >>>> ... > >>>>> My question was intended to provoke the realization that inversionis> >>>>> not a phase shift of any kind, even though it may sometimes be > >>>>> expedient to call it one. > >>>> Now you are really trolling, aren't you? If not, then it might be > >>>> interesting (shock therapy) to write down the transfer function of > >>>> > >>>> a) a 180� phase shifter > >>>> > >>>> and > >>>> > >>>> b) an inverter. > >>>> > >>>> Now compare the two ... > >>> Now compare their group delays. Transfer functions with differentgroup> >>> delays are not identical. > >> > >> But equal transfer functions have equal group delay ... did you read > >> that post I linked regarding this exact same discussion 15 years ago in > >> this forum? > > > > Yes, and not for the first time. It's one of the few instances where I > > think Max was off base. (Max: Where are you?) By its nature, phase shift > > implies time shift. Inversion doesn't. If math tells you that cascaded > > Hilbert transformers produce the same output at the same time as an > > inverter, you need to fix the math. > > > > Jerry > > IMO, you cannot have the same output with causal HTs. Inverter (i.e., > multiplication by -1) has a group delay of 0 samples and HT with more than > 1 taps has larger that 0 sample group delay. At least in practice, and in > digital world. I know nothing about analog signal processing (<- a > disclaimer ;)) > >No offence but how can anybody know digital signal processing but not the analogue counterpart. Universities always teach analogue first surely. I have never heard of a DSP course that starts with digital - or am I wrong. M.P -- Posted via a free Usenet account from http://www.teranews.com
Reply by ●June 18, 20062006-06-18
Mad Prof wrote:> > "Jani Huhtanen" <jani.huhtanen@kolumbus.fi> wrote in message > news:rj7lg.5499$qB5.4265@reader1.news.jippii.net... >> Jerry Avins wrote: >> >> > Andor wrote: >> >> Jerry Avins schrieb: >> >> >> >>> Andor wrote: >> >>>> Jerry Avins wrote: >> >>>> ... >> >>>>> My question was intended to provoke the realization that inversion > is >> >>>>> not a phase shift of any kind, even though it may sometimes be >> >>>>> expedient to call it one. >> >>>> Now you are really trolling, aren't you? If not, then it might be >> >>>> interesting (shock therapy) to write down the transfer function of >> >>>> >> >>>> a) a 180� phase shifter >> >>>> >> >>>> and >> >>>> >> >>>> b) an inverter. >> >>>> >> >>>> Now compare the two ... >> >>> Now compare their group delays. Transfer functions with different > group >> >>> delays are not identical. >> >> >> >> But equal transfer functions have equal group delay ... did you read >> >> that post I linked regarding this exact same discussion 15 years ago >> >> in this forum? >> > >> > Yes, and not for the first time. It's one of the few instances where I >> > think Max was off base. (Max: Where are you?) By its nature, phase >> > shift implies time shift. Inversion doesn't. If math tells you that >> > cascaded Hilbert transformers produce the same output at the same time >> > as an inverter, you need to fix the math. >> > >> > Jerry >> >> IMO, you cannot have the same output with causal HTs. Inverter (i.e., >> multiplication by -1) has a group delay of 0 samples and HT with more >> than 1 taps has larger that 0 sample group delay. At least in practice, >> and in digital world. I know nothing about analog signal processing (<- a >> disclaimer ;)) >> >> > No offence but how can anybody know digital signal processing but not the > analogue counterpart. > Universities always teach analogue first surely. I have never heard of a > DSP course that starts with digital - or am I wrong.Non-taken. More or less I was referring to the practical side of analog signal processing. However, at least in TUT and especially on CS side, DSP starts with only a little analog signal processing. Very very little about how to implement - say - analog butterworth filter and much more about sampling theory, laplace transform and fourier transform. But those are handled on a single course. After that it's pretty much dsp (audio and image processing, filter design, dsp processors, compression, etc). In short, its all about algorithms here.. -- Jani Huhtanen Tampere University of Technology, Pori
Reply by ●June 18, 20062006-06-18
Jani Huhtanen wrote: ...> Phase shift, in general, does not imply time shift. Only if the phase shift > is _not_ a constant as a function of frequency, it implies time shift. > Inverter is an example where phase shift does not imply time shift ;)wt + phi = wt', where t' = t + phi/w. If phi is the phase shift, phi/w is the time shift. If an inverter is an example of phase shift, does the phase lead, or lag? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
