Andor wrote:> Hello Clay, > > thanks for your public domain work (including the other papers on your > site - I especially liked the oscillator paper).You are welcome and thanks. I hope people find them useful.> > I have a query with the Nyquist noise paper. You develop your formular > in differential form, for example > > P(f) = h f / ( exp(h f / k T) - 1) df, > > keeping the integral operator df. Further down you integrate and get > > P_bar = integral_0^b P(f) df. > > which, when substituting the term above becomes > > P_bar = integral_0^b h f / ( exp(h f / k T) - 1) df^2, > > which is probably not what you want (note the df^2 term). Can't you > just delete the differential operator in your derivation and set for > example > > P(f) = h f / ( exp(h f / k T) - 1)? > > Regards, > AndorHello Andor, Thanks for pointing out my sloppiness. The thing about the df term is wanting to show during the density of states transformation that we are going from the summation to an integration, so I really don't want to drop the df. So I fixed the last set of equations so as to not have a df squared term. I think this is okay since we modified something inside of a summation to become something inside of an integration. This pretty much follows the standard development in statistical physics texts for handling the summation. Thanks, Clay

# Johnson Nyquist Noise

Started by ●December 30, 2006

Reply by ●January 10, 20072007-01-10

Reply by ●January 10, 20072007-01-10

Clay wrote:> Andor wrote: > > Hello Clay, > > > thanks for your public domain work (including the other papers on your > > site - I especially liked the oscillator paper).You are welcome and thanks. > I hope people find them useful.I certainly did.> Thanks for pointing out my sloppiness. The thing about the df term is > wanting to show during the density of states transformation that we are > going from the summation to an integration, so I really don't want to > drop the df. So I fixed the last set of equations so as to not have a > df squared term. I think this is okay since we modified something > inside of a summation to become something inside of an integration.Except that the notation of summation does not require a "df". I would prefer that P(f) be a proper power density function, which one can integrate over f and use the standard notation P_bar = integral_0^b P(f) df. Also, when you plot P(f) the "df" operator magically disappears, and P(f) becomes a standard function. Strictly speaking, you are plotting P(f) / df = h f / ( exp(h f / k T) - 1) and not P(f) = h f / ( exp(h f / k T) - 1)T df, except that P(f) / df has a different meaning altogether (never mind the factor k T).> This pretty much follows the standard development in statistical > physics texts for handling the summation.You must admit that something like integral_0^b P(f) looks awful. I never did like statistical physics (except quantum information theory of course :-). Regards, Andor

Reply by ●January 10, 20072007-01-10

Andor wrote:> > You must admit that something like > > integral_0^b P(f) > > looks awful. I never did like statistical physics (except quantum > information theory of course :-). > > Regards, > AndorHello Andor, Yes I admit, it looks odd to say the least. I'll see if I can work out a "cleaner" way of doing this. Stat. Phys. is a slippery subject to say the least. I have a book[1] that takes a nonstandard approach, and the author wants to refute much of the standard paradigm about how Stat. Phys. is approached. When I get time I'm going to revisit his ideas and see if I can sort them out. Thanks, Clay [1] Lavenda, Bernard, "Statistical Physics, a Probabilistic Approach" John Wiley, 1991