>On Aug 30, 10:37=A0am, Jerry Avins <j...@ieee.org> wrote: > >> >> Not so fast! A resistor's temperature coefficient doesn't invalidate >> V=3DIR. At any temperature, give any two of those quantities it's easy >> compute the third. > >I agree, and apologize for not stating my point clearly. >If I apply a voltage to a resistor at time t =3D 0, there is >an initial current flow which decreases slightly as the >resistor warms up and reaches thermal equilibrium with >its surroundings. This effect is not captured in the >V =3D IR model according to which a steady voltage >applied at t =3D 0 results in a steady (time-invariant) >current I =3D V/R for all time. We use the DC model >V =3D IR all the time and even V(t) =3D I(t)R which works >for time-varying voltages and currents, but neither >model includes changes in the value of R as a result >of current flow. > >--Dilip Sarwate >useful results,You're just guilty of believing what it says on the resistor packet. Measure the resistance, and V=IR at all times for a constant V (DC) and with time integration sufficient to suppress thermal noise below your measurement threshold. Let me see. How would I measure the R at all times to verify this? I know. Lets measure the V and the I, and calculate V/I. :-) Steve
PSD of Integral of White Noise
Started by ●February 10, 2006
Reply by ●August 31, 20092009-08-31
Reply by ●August 31, 20092009-08-31
On Aug 30, 11:25�am, "Michael Plante" <michael.pla...@gmail.com> wrote:> > The integral of white noise yields a random walk, which is not WSS > anymore. �You can write the autocorrelation (of two arguments) for the > result down, but I don't know that a PSD makes sense.Any linear operation on a Gaussian distributed RV gives a Gaussian distributed RV. Integration, filtering, differentiation etc. are all linear operations and therefore, this property is widely used in communications analysis.
Reply by ●August 31, 20092009-08-31
On Aug 29, 5:54�am, "dvsarw...@yahoo.com" <dvsarw...@gmail.com> wrote:> wizard12342...@yahoo.com asked > > >Question: I recall learning that if a linear time-invariant system is > >driven by a WSS stochastic process, the PSD of the output equals the > >PSD of the input multiplied by the square of the magnitude of the > >transfer function. �So, if I have white noise with some PSD=C, and I > >pass this through an integrator, can I compute the PSD of the output by > >dividing C by w^2, to yield PSD OUT = C/w^2? > >Somehow, I recall this being an improper way of solving the problem. > >Can anyone help? > > The result that "the PSD of the output equals the PSD of the input > multiplied by the square of the magnitude of the transfer function" > does not hold when the impulse response of the linear time invariant > system is *not* square-integrable, that is, you cannot calculate the > output PSD the way the result claims to do when > > integral {from -infinity to + infinity} |h(t)|^2 dt = infinity. > > The impulse response of the integrator is the unit step function which > is not square-integrable, and so the conclusion that PSD OUT = C/w^2 > cannot be justified. > > Incidentally, 1/w is *not* the transfer function for the integrator; > you > are missing a term.... > > --Dilip SarwateThis result is infact used widely in communications analysis. Actually, I believe one reason it is used is because many times it is assumed that there is a predetection filter at the front of the receiver (at least in FM analysis). In FM analysis, this very property is used when dealing with the limiter-discriminator (LD) receiver which is esentially a differentiator (H=jw). For example, one common reference is the "Principles of Communications Systems" by Taub and Schilling (1971), section 9.2, pp. 299-301 where they compute the noise at the output of the LD. This same computation is used widely. If someone can further expand on the assumptions that allow this computation, that would help everyone.
Reply by ●September 1, 20092009-09-01
On 30 Aug, 15:54, "dvsarw...@yahoo.com" <dvsarw...@gmail.com> wrote:> On Aug 29, 11:06 pm, robert bristow-johnson > > <r...@audioimagination.com> wrote: > > > > and so the conclusion that PSD OUT = C/w^2 > > > cannot be justified. > > > not without hand-waving. but whenever we Neaderthal engineers dabble > > with dirac impulses, white noise, nasty little singularities and > > infinities somewhere, we manage to justify all sorts of evil. > > The output PSD > C/w^2 that you are proudly defending means that the output > process has infinite variance: every random variable in the > process has infinite variance. So, if the input noise is a > white Gaussian noise process, then the output process is > also Gaussian process, and all the Gaussian random variables > comprising the process have infinite variance. How useful is > this result?This result is meaningless as stated. The signal at a single point, while conceivable mathematically, can never be measured. Actually, the signal at a single point need never be conceived of mathematically either, and probably should not be. It is better to conceive of white noise as a sequence of probability distributions at different scales of averaging related by an (in this case very simple) equation. Rather, what is measured is the time or space average of the signal (or a function of it), and these averages have finite variance. Once measured at one scale, they determine all other expectation values for the white noise system because of the equation relating different scales. This is all that matters for the definition of the measure. Conceiving of the signal at a point and its infinite variance commits a sin that Gauss (among others) warned us against: of taking infinity too seriously, as a number. Infinite quantities are limits, that is all. They represent what is larger (or smaller) than the largest (or smallest) quantity in which we are interested, and the existence of a limit shows that what we are calculating does not depend too much on these uninteresting (and usually unknown) things. In general, as Gauss recommended, limits should not be taken before the end of a calculation, for otherwise we risk creating problems. Sometimes it works, but often it does not.> Similarly, white noise is a useful model and it gives very > useful results when we think of second-order systems > (that is, systems whose impulse responses are > square-integrable) but stretching the results to where > they make no sense (whether accompanied by vigorous > hand-waving or not) is not a useful exercise.What does not make sense is to conceive of a system that is not square integrable, for two reasons. First, if we are really talking about power (as opposed to the square of some abstract amplitude), then clearly a system with infinite power is impossible. Second, there is always a frequency cut-off, meaning that the signal at a single point is not meaningful. This is not surprising: it is not even clear that spacetime is a continuous entity. A frequency cut-off is effectively assumed in the computations in the other posts, at least at intermediate stages, and this renders them correct. illywhacker;
Reply by ●September 1, 20092009-09-01
On 30 Aug, 15:54, "dvsarw...@yahoo.com" <dvsarw...@gmail.com> wrote:> On Aug 29, 11:06 pm, robert bristow-johnson > > <r...@audioimagination.com> wrote: > > > > and so the conclusion that PSD OUT = C/w^2 > > > cannot be justified. > > > not without hand-waving. but whenever we Neaderthal engineers dabble > > with dirac impulses, white noise, nasty little singularities and > > infinities somewhere, we manage to justify all sorts of evil. > > The output PSD > C/w^2 that you are proudly defending means that the output > process has infinite variance: every random variable in the > process has infinite variance. So, if the input noise is a > white Gaussian noise process, then the output process is > also Gaussian process, and all the Gaussian random variables > comprising the process have infinite variance. How useful is > this result?You seem to be ruling out Brownian motion as a useful model, which is a bit drastic! The only reason the variance is infinite is that the zero frequency is not controlled. This is because, for example, the initial value of the signal is not given. As soon as it is, the variance becomes finite for finite times. (Alternatively, one can view the process as a probability distribution on signals modulo constants.)> Similarly, white noise is a useful model and it gives very > useful results when we think of second-order systems > (that is, systems whose impulse responses are > square-integrable) but stretching the results to where > they make no sense (whether accompanied by vigorous > hand-waving or not) is not a useful exercise.What does not make sense is systems of infinite power, of infinite extent, or that extend down to infinitely small distances (e.g. the value at a point), except as limits that are useful in some circumstances, obstructive in others. illywhacker;
Reply by ●September 1, 20092009-09-01
On Sep 1, 4:35�am, illywhacker <illywac...@gmail.com> wrote:> You seem to be ruling out Brownian motion as a useful > model, which is a bit drastic!I did not rule out Brownian motion, which is a different issue entirely. As you well know, Brownian motion is not a wide-sense-stationary process, and the usual notion of power spectral density as it is used in the second-order theory of random processes in linear systems is not applicable. The question posed by the OP was about the power spectral density of the output of a linear time-invariant system driven by white noise. The linear system is stated to be an integrator, that is, the impulse response is the unit step function, and so the output of the linear system at time t is the integral of the input from -infinity to t (not from 0 to t as is the case for Brownian motion). The usual second-order theory is not applicable because the linear system is not BIBO-stable, but if one does go through the motions anyway (as engineers are apt to do), the resulting process turns out to have infinite variance. Even this would not faze some people, and so I reminded people that if the input were Gaussian white noise, then the output process would be a Gaussian process, and would consist of Gaussian random variables all of which have infinite variance, thus making it a bit hard to write down the probability density functions.>What does not make sense is systems of infinite power, of >infinite extent, or that extend down to infinitely small >distances (e.g. the value at a point), except as limits >that are useful in some circumstances, obstructive in >others.True, and one might say that a linear time-invariant system whose impulse response is a step function does not make sense because it assumes that the it has been busily integrating its input from -infinity up till now, which is not true in any practical sense. But many commonly used mathematical models do give us systems whose response extends to infinity. The mathematical model for a capacitor discharging through a resistor says that there is residual charge at all finite times; it is only the limiting value of the charge (as t approaches infinity) that is zero . Most engineers would say that after some time has elapsed (say 5 time constants), the capacitor is for all practical purposes fully discharged. Let me remind people: The OP asked>Question: I recall learning that if a linear time-invariant system is >driven by a WSS stochastic process, the PSD of the output equals the >PSD of the input multiplied by the square of the magnitude of the >transfer function. So, if I have white noise with some PSD=C, and I >pass this through an integrator, can I compute the PSD of the output by>dividing C by w^2, to yield PSD OUT = C/w^2?>Somehow, I recall this being an improper way of solving the problem.to which my response is that the conclusion that PSD OUT = C/w^2 cannot be justified in the context of the theory "PSD of the output equals the PSD of the input multiplied by the square of the magnitude of the transfer function." The conclusion may well be true, and there may be other ways of proving its validity, but what the OP recalls -- that his way of reaching the conclusion is an improper way of solving the problem -- is perfectly correct. --Dilip Sarwate
Reply by ●September 1, 20092009-09-01
illywhacker wrote:> On 30 Aug, 15:54, "dvsarw...@yahoo.com" <dvsarw...@gmail.com> wrote: >> On Aug 29, 11:06 pm, robert bristow-johnson >> >> <r...@audioimagination.com> wrote: >> >>>> and so the conclusion that PSD OUT = C/w^2 >>>> cannot be justified. >>> not without hand-waving. but whenever we Neaderthal engineers dabble >>> with dirac impulses, white noise, nasty little singularities and >>> infinities somewhere, we manage to justify all sorts of evil. >> The output PSD >> C/w^2 that you are proudly defending means that the output >> process has infinite variance: every random variable in the >> process has infinite variance. So, if the input noise is a >> white Gaussian noise process, then the output process is >> also Gaussian process, and all the Gaussian random variables >> comprising the process have infinite variance. How useful is >> this result? > > This result is meaningless as stated. The signal at a > single point, while conceivable mathematically, can never > be measured.Does the same apply to slope? Some clarification would have been useful. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●September 2, 20092009-09-02
On 1 Sep, 15:16, "dvsarw...@yahoo.com" <dvsarw...@gmail.com> wrote:> On Sep 1, 4:35 am, illywhacker <illywac...@gmail.com> wrote: > > > You seem to be ruling out Brownian motion as a useful > > model, which is a bit drastic! > > I did not rule out Brownian motion, which is a different > issue entirely. As you well know, Brownian motion is > not a wide-sense-stationary process, and the usual > notion of power spectral density as it is used in the > second-order theory of random processes in linear > systems is not applicable.Brownian motion strictly defined indeed has a fixed starting point. However, without a fixed starting point, it can also be viewed as a measure on functions on the real line. Without the fixed point at zero, the variance is of course infinite. (This is no more mysterious than the fact that a definite integral as a function of its upper limit has an unknown value if the lower limit is not fixed.) One solution to this is to fix a point. Another is to regard it as a measure on functions modulo constants. In the particular case at hand, however, all this is moot, because there will be a fixed starting point for the integration.> The question posed by the > OP was about the power spectral density of the output > of a linear time-invariant system driven by white noise. > The linear system is stated to be an integrator, that is, > the impulse response is the unit step function, and so > the output of the linear system at time t is the integral > of the input from -infinity to t (not from 0 to t as is the > case for Brownian motion). The usual second-order > theory is not applicable because the linear system is > not BIBO-stable, but if one does go through the motions > anyway (as engineers are apt to do), the resulting > process turns out to have infinite variance. Even this > would not faze some people, and so I reminded people > that if the input were Gaussian white noise, then the > output process would be a Gaussian process, and > would consist of Gaussian random variables all of > which have infinite variance, thus making it a bit hard > to write down the probability density functions.A minor point: you cannot write down probability density functions (in a mathematically strict sense, which is what you seem to favour) for white noise either: you can only write down correlation functions. A major point: you seem to be wrong anyway. The Wiener-Khinchin theorem guarantees exactly what the OP wants, as a glance at the 'Applications' section of the relevant Wikipedia page would tell you. This result can be found in a far easier way, however, by heeding the advice of Gauss, as follows. While it is good for our sense of our own cleverness and power to be able to manipulate infinite mathematical objects directly, one must remember that infinity is not a number, and does not exist except as a limit. Thus it is a cause of errors to take a limit too early: limits should always and only be taken at the end of a calculation. The existence of a limit then just shows that the results at finite values of the parameter that is going to the limit are not too sensitive to the precise value provided it is large enough. Of course, once we are familiar enough with certain types of limit, we can take them earlier (e.g. we manipulate derivatives directly), but we must never forget that there is always the possibility of errors creeping in; if this happens we must render everything finite and only take the limits at the end. This is what engineers and physicist effectively do when hand-waving. In the present case, by imposing low- and high-frequency cut-offs, everything is rendered finite. The above result can then be derived and the cut-offs removed. This is a better way to do it than using more complex theory: simpler, more transparent, and more illustrative of the meaning of the limit.> >What does not make sense is systems of infinite power, of > >infinite extent, or that extend down to infinitely small > >distances (e.g. the value at a point), except as limits > >that are useful in some circumstances, obstructive in > >others. > > True, and one might say that a linear time-invariant > system whose impulse response is a step function > does not make sense because it assumes that the > it has been busily integrating its input from -infinity > up till now, which is not true in any practical sense. > But many commonly used mathematical models > do give us systems whose response extends to infinity. > The mathematical model for a capacitor discharging > through a resistor says that there is residual charge at > all finite times; it is only the limiting value of the charge > (as t approaches infinity) that is zero . Most engineers > would say that after some time has elapsed (say 5 time > constants), the capacitor is for all practical purposes > fully discharged.The advice of Gauss applies. There is nothing wrong with infinities, provided they are always regarded as limits, and if necessary, regularized at intermediate stages of a calculation. illywhacker;
Reply by ●September 2, 20092009-09-02
On 2 Sep, 10:50, illywhacker <illywac...@gmail.com> wrote: Illywhacker, A few monts ago you posted a link to a paper on directional statistics, in the context of unscented Kalman filters. The search engine on google groups is broken so I can't find the original thread, and I can't find anything that looks relevant in IEEExplore. Could you please re-post the link to the paper? Rune
Reply by ●September 2, 20092009-09-02
On 2 Sep, 11:33, Rune Allnor <all...@tele.ntnu.no> wrote:> On 2 Sep, 10:50, illywhacker <illywac...@gmail.com> wrote: > > Illywhacker, > > A few monts ago you posted a link to a paper on directional > statistics, in the context of unscented Kalman filters. The > search engine on google groups is broken so I can't find the > original thread, and I can't find anything that looks relevant > in IEEExplore. > > Could you please re-post the link to the paper? > > RuneHi Rune, Do you recall whether it was a paper or a Wikipedia link? illywhacker;
