>> sqrt(Watts) per Hertzsqrt(Watts) per _SQRT_(Hertz) :-)
Question about noise voltage
Started by ●September 7, 2011
Reply by ●September 9, 20112011-09-09
Reply by ●September 9, 20112011-09-09
mnentwig <markus.nentwig@n_o_s_p_a_m.renesasmobile.com> wrote:> if you'd write "volts per Hertz", it would imply that volts are integrated > over frequency. That's not correct. > I guess one could try some substitution of variables in the integration, > getting via "watts per Hertz" and "sqrt(Watts) per Hertz" to "volts per > sqrt(Hertz)".So sqrt(watt ohms)/sqrt(Hz) would be better? Reminds me of the volt-amp unit used where the power factor isn't 1. -- glen
Reply by ●September 9, 20112011-09-09
On 9/7/11 9:43 PM, brent wrote:> I have been pondering the notation for noise voltage. The noise > voltage across a 50 ohm resistor is expressed as 1nV/sqrt(1Hz) >this is the same as (1 nV)^2 per Hz.> As I ponder this it seems to me that one is concerned about the noise > voltage period. For instance, if you have a 1Hz BW signal you would > say the noise voltage on a 50 ohm resistor is 1nV.you would say that the *power* dissipated in the 50 ohm resistor due to noise is (1 nV)^2 / (50 ohms) per Hz. that's what you know from this spec. -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."
Reply by ●September 9, 20112011-09-09
On Sep 9, 6:35�pm, robert bristow-johnson <r...@audioimagination.com> wrote:> On 9/7/11 9:43 PM, brent wrote: > > > I have been pondering the notation for noise voltage. �The noise > > voltage across a 50 ohm resistor is expressed as 1nV/sqrt(1Hz) > > this is the same as (1 nV)^2 per Hz. > > > As I ponder this it seems to me that one is concerned about the noise > > voltage period. �For instance, if you have a 1Hz BW signal you would > > say the noise voltage on a 50 ohm resistor is 1nV. > > you would say that the *power* dissipated in the 50 ohm resistor due to > noise is > > � � (1 nV)^2 / (50 ohms) �per Hz. > > that's what you know from this spec. > > -- > > r b-j � � � � � � � � �r...@audioimagination.com > > "Imagination is more important than knowledge."Even though what you wrote is obvious, it is helpful.
Reply by ●September 10, 20112011-09-10
Hello, what I meant is: Assume - a resistance R - a constant voltage density Vdens across R - an integration bandwidth f1..f2 The dissipated power is P = integrate(Vdens^2 / R, f, f1, f2) which simplifies to P = Vdens^2 / R * integrate(1, f, f1, f2); and finally (1) P = Vdens^2 / R * (f2-f1) Now I introduce an equivalent voltage "Vequiv" that dissipates the same power P over the same resistance: (2) P = Vequiv ^ 2 / R Combining (1) and (2): Vequiv^2 = Vdens^2 * (f2-f1) square root: Vequiv = Vdens * sqrt(f2-f1) This shows, for consistency of units: *** a voltage density must be volts per square root of Hertz. *** I hope this sheds some light on the formalism behind "V/sqrt(Hz)". PS: above uses maxima syntax for integrals and the "usual textbook assumptions for noise-like signals", whatever that means (now let's throw in some negative resistance...)
Reply by ●September 10, 20112011-09-10
On 9/10/11 5:30 AM, mnentwig wrote:> Hello, > > what I meant is: Assume > - a resistance R > - a constant voltage density Vdens across Rwhat's "voltage density"? i know what "current density" is, but i have never heard of the term "voltage density" until now. what units is "voltage density" expressed in? if it's "volts per meter", then the correct name for this quantity is "electrostatic field".> - an integration bandwidth f1..f2 > > The dissipated power is P = integrate(Vdens^2 / R, f, f1, f2)can you use something more conventional to define what your integration is? mnentwig, we might be able to address your question, but you have to "speak english". by that, i mean, use the language, including mathematical expression, that we can understand. don't make up your own lexicon for terms or for functions. -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."
Reply by ●September 10, 20112011-09-10
On Sep 7, 6:43�pm, brent <buleg...@columbus.rr.com> wrote:> I have been pondering the notation for noise voltage. �The noise > voltage across a 50 ohm resistor is expressed as 1nV/sqrt(1Hz) > ...Take a look at how instrument makers deal with it starting at page 29 of: Signals and Units http://www.bksv.com/pdf/Bv0031.pdf Dale B. Dalrymple
Reply by ●September 10, 20112011-09-10
>what's "voltage density"? i know what "current density" is, but i have >never heard of the term "voltage density" until now.Hi, the terms I'm using "should" be consistent with everyday use (mistakes may happen), but I believe the main challenge is that the question relates to electrical engineering in a DSP forum. There is absolutely nothing digital about it, but trying to cover the EE topic in full detail might call for a book chapter or two. It won't work here - and I remember being thoroughly confused after trying to make sense of said book chapters for the first time... The "densities" I'm referring to are spectral densities. That is, something needs to be integrated over frequency, for example power spectral density. Consider the following thought experiment: I couple a resistor "R" at temperature "T" via an ideal bandpass with bandwidth "B" to another resistor "R" at zero Kelvin temperature (more generally: conjugate matched impedance). Thermodynamics dictate that a noise power of N = kBT is transfered from the "hot" resistor to the cold one, where k is Boltzmann's constant. This holds true up to a couple of Terahertz - otherwise, there would be an obvious problem for infinite bandwidth. From the noise power delivered to the "cold" resistor, I can derive an equivalent voltage in the bandwidth of my filter. Or, in more general terms, a frequency-dependent voltage (spectral) density. The kBT "thermal" noise is rather small, but for example 1/f (shot) noise in semiconductor devices is hard to overlook even at room temperature. Comparing spectral power density (which is kT Watts/Hertz at temperature T) and spectral voltage density, the advantage of the former is that I don't need to know the resistor in the definition, since it's valid for a matched load (same resistance) and the R drops out. What I was trying to explain - maybe not too successfully - is the units of spectral voltage density, which is volts per square root of Hertz, and why "square root of Hz".
Reply by ●September 10, 20112011-09-10
