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? Thanks, -JJ

# PSD of Integral of White Noise

Started by ●February 10, 2006

Reply by ●February 11, 20062006-02-11

<wizard12342002@yahoo.com> wrote in message news:1139620284.889061.14200@o13g2000cwo.googlegroups.com...> 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? > > > Thanks, > > > -JJ >It's right. Tom

Reply by ●August 29, 20092009-08-29

>dividing C by w^2, to yield PSD OUT = C/w^2?Y is ur observed noisy signal, X signal of interset Syy = Sxx + w^2 thus Sxx = Syy - w^2 w^2 = sum(y)/N; Sxx(dB) = Syy(dB) / 10log(w^2) [Check the last line] The problem is calculating the PSD for your observed signal, that's why we go to smoothed or averaged periodograms ( most famous Welch )

Reply by ●August 29, 20092009-08-29

wizard12342002@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 Sarwate

Reply by ●August 30, 20092009-08-30

On Aug 29, 8:54�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.it represents a signal that has infinite power. but so also is theoretical white noise. it ain't square-integrable either. but somehow we wave our hands and speak of the inverse Fourier transform of its PSD.> The impulse response of the integrator is the unit step function which > is not square-integrable,yet, somehow, we hand-wave a little and come up with U(f) = (1/2)*delta(f) + 1/(j*2*pi*f) when u(t) = (1/2)*( 1 + sgn(t) )> 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.> Incidentally, 1/w is *not* the transfer function for the integrator; > you are missing a term....Dilip, if it's not a leaky integrator, what term do you mean? the delta(f)? funny that delta(f) is a worry when 1/f is going all to hell when f=0 anyway. the other funny thing is, that all sorts of textbooks identify integrators as just 1/s or 1/(jw) anyway. how would we represent it accurately in the s-plane? H(s) = (1/2)*delta(s/(j*2*pi) + 1/s ? r b-j

Reply by ●August 30, 20092009-08-30

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.Robert: Neanderthal engineers (amongst whom I count myself) do all kinds of things but ultiimately, if the results do not make sense as they apply to reality, they discard them. 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? Basic circuit theory that is taught (or should be taught) to all electrical engineers says that the voltage V across a resistor and the current I through it are related as V = IR where R is the resistance. Wonderful model, very useful in lots of applications. But you know what? I applied a gazillion volts to a 1-ohm resistor and I didn't observe a gazillions amperes flowing through it: I saw a small flash of light and a puff of smoke. More realistically, the model V = IR is flawed in the sense that it doesn't account for the resistor heating up and thus causing changes in R, it doesn't account for the skin effect if the applied signal is at high frequency or the fact that I cannot transmit light through a copper wire, and so on and so forth. But, nonetheless V = IR is a useful model and it gives very useful and applicable results in a wide variety of circumstances but by no means in all circumstances. 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. Here endeth the homily for the day. --Dilip Sarwate

Reply by ●August 30, 20092009-08-30

dvsarwate@yahoo.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. > > Robert: > > Neanderthal engineers (amongst whom I count myself) do all > kinds of things but ultiimately, if the results do not make sense > as they apply to reality, they discard them. 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? > > Basic circuit theory that is taught (or should be taught) to > all electrical engineers says that the voltage V across a > resistor and the current I through it are related as V = IR > where R is the resistance. Wonderful model, very useful > in lots of applications. But you know what? I applied a > gazillion volts to a 1-ohm resistor and I didn't observe a > gazillions amperes flowing through it: I saw a small flash > of light and a puff of smoke. More realistically, the model > V = IR is flawed in the sense that it doesn't account for > the resistor heating up and thus causing changes in R, > it doesn't account for the skin effect if the applied signal > is at high frequency or the fact that I cannot transmit light > through a copper wire, and so on and so forth. But, > nonetheless V = IR is a useful model and it gives very > useful and applicable results in a wide variety of > circumstances but by no means in all circumstances. > 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. > > Here endeth the homily for the day.Not so fast! A resistor's temperature coefficient doesn't invalidate V=IR. At any temperature, give any two of those quantities it's easy compute the third. That V=IR is a mere approximation is shown (among other ways) by considering the Johnson noise across an open-circuited resistor. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������

Reply by ●August 30, 20092009-08-30

On Aug 30, 10:37�am, Jerry Avins <j...@ieee.org> wrote:> > Not so fast! A resistor's temperature coefficient doesn't invalidate > V=IR. 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 = 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 = IR model according to which a steady voltage applied at t = 0 results in a steady (time-invariant) current I = V/R for all time. We use the DC model V = IR all the time and even V(t) = 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,

Reply by ●August 30, 20092009-08-30

>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 byThe 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. One book that discusses differential equations fed by random processes restricts discussion to BIBO-stable systems. I'm not sure if this is overly-restrictive, but an integrator does not fall in that category (pole at the origin, outside the open left half plane).

Reply by ●August 30, 20092009-08-30

dvsarwate@yahoo.com wrote:> On Aug 30, 10:37 am, Jerry Avins <j...@ieee.org> wrote: > >> Not so fast! A resistor's temperature coefficient doesn't invalidate >> V=IR. 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 = 0, there is > an initial current flow which decreases slightly as the > resistor warms up and reaches thermal equilibrium with > its surroundings.Not necessarily so. The temperature coefficient may be negative. Once got a *ATTABOY* for making use of that -- back when (ATTABOY + $0.25) = (1 cup coffee) This effect is not captured in the> V = IR model according to which a steady voltage > applied at t = 0 results in a steady (time-invariant) > current I = V/R for all time. We use the DC model > V = IR all the time and even V(t) = 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, >