mistake, should read "at audio frequency bandwidths" (example: from 16Hz to
16kHz we usually struggle mainly with 1/f noise. In wideband radio systems,
for example LTE with 18 MHz bandwidth, shot noise is usually negligible
against thermal noise).
Reply by mnentwig●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 dbd●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 robert bristow-johnson●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 R
what'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 mnentwig●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 brent●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 robert bristow-johnson●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 glen herrmannsfeldt●September 9, 20112011-09-09
> 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 mnentwig●September 9, 20112011-09-09
>> sqrt(Watts) per Hertz
sqrt(Watts) per _SQRT_(Hertz) :-)
Reply by mnentwig●September 9, 20112011-09-09
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)".