In article Enr1b.4302$Jk5.4200214@feed2.centurytel.net, Fred Marshall at fmarshallx@remove_the_x.acm.org wrote on 08/22/2003 12:29:> "robert bristow-johnson" <rbj@surfglobal.net> wrote in message > news:BB6AE62D.31A6%rbj@surfglobal.net... >> In article 3F453609.571AA6E5@ieee.org, Jerry Avins at jya@ieee.org wrote on >> >> There are a lot of LTV systems that change the frequency of an input >> (imagine DC going into an amp and someone wiggling the volume control at 1 >> Hz, lot'sa multiples on 1 Hz coming out and DC going in). it may not be >> time-invariant, but it's linear. > > Well .... as we discussed a few months ago, the wiggles have to be > *perfectly* periodic for it to be linear and time-varying with respect to > the dc input.perhaps in my unrealistic example (in which i was probably being two specific, but i was just trying to make a point), but not for the case in general. if you have a good linear amplifier that is good all the way down to DC and your input is some non-zero DC level and someone is fiddling with the gain control, even if it's not a periodic fiddling, frequency components are showing up in the output that are not in the input (which only contains a non-zero component at 0 Hz). the system in linear and there are new frequency components generated in the output that weren't in the input. that's because it's LTV. it cannot happen in an LTI system where: LTI{ exp(j*w*t) } = H(j*w) * exp(j*w*t)> My hand is steady enough but not that steady!if you drink as much caffeine as me, that's bound to happen.> But since you said "dc", that makes a linear time-invariant system with > respect to the input called "wiggles".this i don't understand at all. the gain, g(t), is a parameter not the input. i understand the symmetry of the equation y(t) = g(t) * x(t) but then, if you insist on that interpretation, then let the knob be some kinda tone control or delay control of a linear system. that'll map non-linearly to most parameters of the output. In article bi5grm$hec$1@newslocal.mitre.org, Stan Pawlukiewicz at stanp@nospam_mitre.org wrote on 08/22/2003 12:37:> robert bristow-johnson wrote:...>> >> okay, the frequency shift is not linear with speed. fine. what i was >> thinking about when i would say that some box or "effect" or whatever is not >> linear is that there are cases when the superposition does not hold. that >> is does >> >> Doppler{x(t) + y(t)} = Doppler{x(t)} + Doppler{y(t)} >> >> ? >> >> There are a lot of LTV systems that change the frequency of an input >> (imagine DC going into an amp and someone wiggling the volume control at 1 >> Hz, lot'sa multiples on 1 Hz coming out and DC going in). it may not be >> time-invariant, but it's linear. >> >> i think i understand what you were saying, but we have a different semantic. >>> If you use only f as a state variable, the system is linear. > > If you use f and v as state variables, the state equations are nonlinear.i guess it's semantics, but i don't view either f, f', or v as states but as respective parameters the input signal, output signal, and the Doppler Effect system. since the Doppler Effect system is an Linear Time-Variant (LTV) system (according to my semantic), it should be surprising that frequency components can appear in the output that weren't there in the input. the mapping of the parameters f and v (and c) to f' or f'-f is some mathematical thingie that depends on the physics governing the system. but how is that qualitatively different from other LTV systems? the effect of a tone control out the output amplitude is probably some messy non-linear equation but that would not compel me to conclude that the system itself is non-linear. at least not with the definitions of things that i am working with.> This article might be overkill, but look at the state equations in: > > Sonar tracking of multiple targets using joint probabilistic data association > Fortmann, T.; Bar-Shalom, Y.; Scheffe, M.; > Oceanic Engineering, IEEE Journal of , Volume: 8 Issue: 3 , Jul 1983 > Page(s): 173 - 184hmmm. kinda makes me wish i had joined IEEE back in 1983 (... NOT!). :-) i'll have to see if they have it at the UVM engineering library. anyway, i understand constructing an analysis or tracking system where frequencies of Doppler shifted signals are somehow detected or extracted from the sonar signal and those parameter estimates (f') are inputted into a linear or non-linear system (and become states thereof) that finagles a position and/or velocity of some hypothetical or actual object that is moving and generating or reflecting the sonar signal. that's perhaps a non-linear system. fine. but i still think the correct semantic for the Doppler Effect itself is a Linear Time-Variant system. that is because Doppler{x1(t) + x2(t)} = Doppler{x1(t)} + Doppler{x2(t)} . if the superposition property holds, it's a linear system. it's similar to a moving-tapped delay line, from how i understand it. r b-j
Is the Doppler effect really linear?
Started by ●August 19, 2003
Reply by ●August 22, 20032003-08-22
Reply by ●August 22, 20032003-08-22
"robert bristow-johnson" <rbj@surfglobal.net> wrote in message news:BB6BF63D.320F%rbj@surfglobal.net...> In article Enr1b.4302$Jk5.4200214@feed2.centurytel.net, Fred Marshall at > fmarshallx@remove_the_x.acm.org wrote on 08/22/2003 12:29: > > > "robert bristow-johnson" <rbj@surfglobal.net> wrote in message > > news:BB6AE62D.31A6%rbj@surfglobal.net... > >> In article 3F453609.571AA6E5@ieee.org, Jerry Avins at jya@ieee.orgwrote on> >> > >> There are a lot of LTV systems that change the frequency of an input > >> (imagine DC going into an amp and someone wiggling the volume controlat 1> >> Hz, lot'sa multiples on 1 Hz coming out and DC going in). it may notbe> >> time-invariant, but it's linear. > > > > Well .... as we discussed a few months ago, the wiggles have to be > > *perfectly* periodic for it to be linear and time-varying with respectto> > the dc input. > > perhaps in my unrealistic example (in which i was probably being two > specific, but i was just trying to make a point), but not for the case in > general. if you have a good linear amplifier that is good all the waydown> to DC and your input is some non-zero DC level and someone is fiddlingwith> the gain control, even if it's not a periodic fiddling, frequencycomponents> are showing up in the output that are not in the input (which onlycontains> a non-zero component at 0 Hz). the system in linear and there are new > frequency components generated in the output that weren't in the input. > that's because it's LTV.Robert, Just to be picky, you can't say "the system is linear" without stating with respect to which "input". Here you seem to be saying that the dc is the input. So, with that caveat, so far so good. Nope, this one isn't LTV if the wiggles are general. It's only LTV if the wiggles are periodic. Superposition only applies in the latter case. Put in f1(t) at one time. Put in f2(t) at another time. Put in f1(t) + f2(t) at a third time. The frequencies at the output won't be the same unless the control is periodic. The frequencies have to be the same in order for superposition to apply - and here they don't. Fred
Reply by ●August 22, 20032003-08-22
Stan Pawlukiewicz <stanp@nospam_mitre.org> wrote in message news:<bi5e9v$dqm$1@newslocal.mitre.org>...> Rune Allnor wrote: > > Stan Pawlukiewicz <stanp@nospam_mitre.org> wrote in message news:<bi30ll$ldt$1@newslocal.mitre.org>... > > > >>Eric Jacobsen wrote: > >> > >>>On Wed, 20 Aug 2003 22:06:40 -0400, Jerry Avins <jya@ieee.org> wrote: > >>> > >>> > >>> > >>>>Raymond Toy wrote: > >>>> > >>>>... > >>>> > >>>> > >>>>>So, if I were blindfolded, I could tell if I were in a train moving > >>>>>towards a whistle instead of the whistle moving towards me? > >>>>> > >>>> > >>>>Only if you knew the original pitch and the speed very accurately. Let's > >>>>say that the train is moving at Mach .2 (fast, but possible) and its > >>>>whistle has a frequency of f. when the train is approaching, its > >>>>whistle's frequency sounds like f/(1 - .2), or 1.25f. If the train is in > >>>>the station and you are on another approaching it at Mach .2, the pitch > >>>>you would hear would be f(1 + .2) or 1.2f. The nonlinearity that Clay > >>>>wrote of -- I hadn't thought of it explicitly before, but its tickling > >>>>was the reason I started this thread -- is that the magnitude of the > >>>>pitch shift from a moving source is different for the same speeds to and > >>>>away. > >>>> > >>>>Jerry > >>> > >>> > >>>I'm not sure if that's really a nonlinearity, since the cases are > >>>different. Writing the expression as you did with the terms for the > >>>emitter and receiver separated clarifies the issue, I think. > >>> > >>>There's actually a lot of information about the situation that can be > >>>obtained merely by observing the Doppler history of things moving > >>>around you (this was the topic of my MS thesis). Back in the day I > >>>set out to show that one could even determine how far away the emitter > >>>was by the Doppler history observed at a single sensor (i.e., > >>>microphone). After some initial failed attempts I told my advisor > >>>that I was going to drop that parameter from my analysis, and he > >>>emphatically replied in his Korean accent, "No! Must do range!" I > >>>then set out to prove that range could not be determined in this > >>>manner and after a weekend hunched over a pad of engineering paper I > >>>had an algorithm for determining range from the Doppler observation. > >>>It worked pretty well. > >>> > >>>So my effort to prove that it couldn't be done failed, too. > >>> > >>> > >>>Eric Jacobsen > >>>Minister of Algorithms, Intel Corp. > >>>My opinions may not be Intel's opinions. > >>>http://www.ericjacobsen.org > >> > >>There's a derivation in Quinn and Hannan's "The Estimation and Tracking > >> of Frequency" > > > > > > Eh... what's the trick? Assuming a stationary reciever, the Doppler > > history would show some atan type of behaviour. If you observe for long > > enough (and conditions otherwise allow), you will get the asymptotic > > boundaries of the frequency, which means you can determine the time of > > the Closest Point of Approach and thus the emitted frequency. So far so > > good. > > > > If you want to estimate CPA range, you need to know the target velocity > > as well. You don't know that, do you? Or maybe the velocity is given by > > the asymptotics of the atan? > > > > Rune > > Hi Rune, > > I recall you said you had a bunch of Jasa on CD. Look at > > Quinn, B.G. "Doppler speed and range estimation using frequency and > amplitude estimates", JASA 98:5 part 1, pp 2560-2566 (1995).Yes I have. The only problem is that they are in my office. 700 km away from here. I'll take a look the next time go there. Rune
Reply by ●August 22, 20032003-08-22
In article f56893ae.0308221459.28d50b7a@posting.google.com, Rune Allnor at allnor@tele.ntnu.no wrote on 08/22/2003 18:59:> Stan Pawlukiewicz <stanp@nospam_mitre.org> wrote in message > news:<bi5e9v$dqm$1@newslocal.mitre.org>......>> >> Hi Rune, >> >> I recall you said you had a bunch of Jasa on CD. Look at >> >> Quinn, B.G. "Doppler speed and range estimation using frequency and >> amplitude estimates", JASA 98:5 part 1, pp 2560-2566 (1995). > > Yes I have. The only problem is that they are in my office. 700 km away > from here. I'll take a look the next time go there.boy, that's a helluva commute. r b-j
Reply by ●August 23, 20032003-08-23
robert bristow-johnson wrote:>...> Doppler Effect itself is a Linear Time-Variant system. that is because > > Doppler{x1(t) + x2(t)} = Doppler{x1(t)} + Doppler{x2(t)} . >That doesn't hold when the source moves. m_1/(1 - m_1) + m_2/(1 - m_2) != (m_1 + m_2)/(1 - m_1 -m_2) It does hold when the observer moves. It nearly holds when m_x << 1. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●August 23, 20032003-08-23
In article 3F46F216.FE3E888D@ieee.org, Jerry Avins at jya@ieee.org wrote on 08/23/2003 00:48:> robert bristow-johnson wrote: >> > ... > >> Doppler Effect itself is a Linear Time-Variant system. that is because >> >> Doppler{x1(t) + x2(t)} = Doppler{x1(t)} + Doppler{x2(t)} . >> > That doesn't hold when the source moves. > > m_1/(1 - m_1) + m_2/(1 - m_2) != (m_1 + m_2)/(1 - m_1 -m_2) > > It does hold when the observer moves. It nearly holds when m_x << 1.i think it's semantic. i am not sure what your m_1 and m_2 quantities are. i am talking about the actually sound source generated and that is moving with respect to the observer. x(t) is the sound source, Doppler{x(t)} is whatever the observer is hearing. that operator is not linear? r b-j
Reply by ●August 23, 20032003-08-23
robert bristow-johnson wrote:> > In article 3F46F216.FE3E888D@ieee.org, Jerry Avins at jya@ieee.org wrote on > 08/23/2003 00:48: > > > robert bristow-johnson wrote: > >> > > ... > > > >> Doppler Effect itself is a Linear Time-Variant system. that is because > >> > >> Doppler{x1(t) + x2(t)} = Doppler{x1(t)} + Doppler{x2(t)} . > >> > > That doesn't hold when the source moves. > > > > m_1/(1 - m_1) + m_2/(1 - m_2) != (m_1 + m_2)/(1 - m_1 -m_2) > > > > It does hold when the observer moves. It nearly holds when m_x << 1. > > i think it's semantic. i am not sure what your m_1 and m_2 quantities are. > i am talking about the actually sound source generated and that is moving > with respect to the observer. x(t) is the sound source, Doppler{x(t)} is > whatever the observer is hearing. that operator is not linear? > > r b-jNo; t'aint. I call the Doppler shift D. That's f', the observed frequency less f, the emitted frequency. I call the relative Doppler shift d; D/f. I state speeds relative to c, the speed of sound in which the source moved. The speed of sound is unity and the speed of the emitter is m, for Mach. When there are several speeds, they are m_1, m_2, etc. Then, when the source approaches the stationary observer, d = m/(1 - m) d = 1/9 for m = .1 Linearity with respect to m requires that d = 2/9 for m = .2. In fact, d = 2/8 for m = .2. What semantics allow that to be called linear? (d --> infinity as m --> 1. Shock waves are interesting.) Jerry Side puzzler. If my ice cream vendor's Pachelbel drops an octave as he goes by, how fast is he going? :-) -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●August 23, 20032003-08-23
"Jerry Avins" <jya@ieee.org> wrote in message news:3F47CB2A.7BE430FF@ieee.org...>> > Side puzzler. If my ice cream vendor's Pachelbel drops an octave as he > goes by, how fast is he going? :-)Jerry, What's the air temp?? Clay> -- > Engineering is the art of making what you want from things you can get. > �����������������������������������������������������������������������
Reply by ●August 24, 20032003-08-24
Jerry Avins <jya@ieee.org> wrote in message news:<3F465150.E2667543@ieee.org>...> robert bristow-johnson wrote:...> > > > Doppler{x(t) + y(t)} = Doppler{x(t)} + Doppler{y(t)} > >...> > > > i think i understand what you were saying, but we have a different semantic. > >...> > I'll simplify by expressing velocities as ratios of the speed of sound. > Numerically, the speed of sound becomes 1 (one, not ell) and the speed > of motion becomes m, the Mach number. Also, I define D_s to be the > Doppler shift due to moving source and D_o the shift due to moving > observer. (D_x = f' - f for the appropriate case.) > > For an approaching observer, f' = f(1 + m), D_o = fm; clearly linear. > > For a approaching source, f' = f/(1-m), D_s = fm/(1 - m) = D_o/(1 - m). > That's linear in f, but not in m.this stuff i agree with (as a statement of physical phact).> How does semantics enter into this?probably in what we mean by "Doppler Effect". what you mean, if i read this right, is an operator that operates on a speed, m, as an input and barfs out f' - f as an output. my semantic regarding the definition of the Doppler Effect is a system that when given an input signal of x(t), something like y(t) = x(t/(1-m)) comes out. if given x1(t) + x2(t) as input then y(t) = y1(t) + y2(t) comes out where y1(t) = x1(t/(1-m)) and y2(t) = x2(t/(1-m)) i look at m and f and f' (or f'-f) as parameters or properties of the system (or "effect"), input, and output respectively. given my semantics of what the "Doppler Effect" is and what its input and output is, it's a Linear Time-Variant system. In article 3F47CB2A.7BE430FF@ieee.org, Jerry Avins at jya@ieee.org wrote on 08/23/2003 16:14:> Side puzzler. If my ice cream vendor's Pachelbel drops an octave as he > goes by, how fast is he going? :-)lessee: assuming you survive getting runned over by the truck, f' = f/(1-m) when he's moving toward you, and f' = f/(1+m) when he's moving away. so somehow, i guess, f/(1-m) = 2 * f/(1+m) or 1+m = 2*(1-m) m = 1 - 2*m 3*m = 1 m = 1/3 so v = c/3, over 100 m/s. so maybe you don't survive the passing of the truck. maybe some formula-1 Indie 500 car sells ice cream and has loudspeakers playing Pachebel's Canon on it, but it would have to be scooting fully flat out to drop it an octave. r b-j
Reply by ●August 24, 20032003-08-24
robert bristow-johnson wrote:> > Jerry Avins <jya@ieee.org> wrote in message > news:<3F465150.E2667543@ieee.org>... > > robert bristow-johnson wrote: > ... > > > > > > Doppler{x(t) + y(t)} = Doppler{x(t)} + Doppler{y(t)} > > > > ... > > > > > > i think i understand what you were saying, but we have a different semantic. > > > > ... > > > > I'll simplify by expressing velocities as ratios of the speed of sound. > > Numerically, the speed of sound becomes 1 (one, not ell) and the speed > > of motion becomes m, the Mach number. Also, I define D_s to be the > > Doppler shift due to moving source and D_o the shift due to moving > > observer. (D_x = f' - f for the appropriate case.) > > > > For an approaching observer, f' = f(1 + m), D_o = fm; clearly linear. > > > > For a approaching source, f' = f/(1-m), D_s = fm/(1 - m) = D_o/(1 - m). > > That's linear in f, but not in m. > > this stuff i agree with (as a statement of physical phact). > > > How does semantics enter into this? > > probably in what we mean by "Doppler Effect". what you mean, if i read this > right, is an operator that operates on a speed, m, as an input and barfs out > f' - f as an output. > > my semantic regarding the definition of the Doppler Effect is a system that > when given an input signal of x(t), something like > > y(t) = x(t/(1-m))I assumed that x(t) was some quantity x that is a function of time. Was I wrong? For simplicity in talking about pitch, we seemed to be talking about sine waves. What does x(t/(1-m)) represent? What other function of time are you representing here?> > comes out. > > if given x1(t) + x2(t) as input then > > y(t) = y1(t) + y2(t) > > comes out where > > y1(t) = x1(t/(1-m)) and y2(t) = x2(t/(1-m)) > > i look at m and f and f' (or f'-f) as parameters or properties of the system > (or "effect"), input, and output respectively. given my semantics of what > the "Doppler Effect" is and what its input and output is, it's a Linear > Time-Variant system. > > In article 3F47CB2A.7BE430FF@ieee.org, Jerry Avins at jya@ieee.org wrote on > 08/23/2003 16:14: > > > Side puzzler. If my ice cream vendor's Pachelbel drops an octave as he > > goes by, how fast is he going? :-) > > lessee: assuming you survive getting runned over by the truck, f' = f/(1-m) > when he's moving toward you, and f' = f/(1+m) when he's moving away. so > somehow, i guess, > > f/(1-m) = 2 * f/(1+m) > or > 1+m = 2*(1-m) > > m = 1 - 2*m > > 3*m = 1 > > m = 1/3 > > so > > v = c/3, over 100 m/s. > > so maybe you don't survive the passing of the truck. > > maybe some formula-1 Indie 500 car sells ice cream and has loudspeakers > playing Pachebel's Canon on it, but it would have to be scooting fully flat > out to drop it an octave. > > r b-jI wondered how they got the exhaust to sound sl lo with the engines screaming! I don't get what you mean because I'm interpreting your symbols wrong. What I'm getting at is that a plot of m/(1-m) vs. m from m = -1 to +1 is hardly a straight line. What else does "linear with m" mean? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
