Thermal White Noise - Johnson–Nyquist noise

In summary, Dave is trying to solve the equation for the frequency of thermal noise, but is stuck because there is an exponential expression in it. He suggests that you use the wrong formula for your situation, and I remind him that Johnson-Nyquist noise is classical thermodynamics.
  • #1
Mechatron
38
0
I'm trying to measure the thermal white noise generated by chemical batteries.
So far I've measured the current noise, the voltage noise (V noise) and the bandwidth (delta v).
From the equation below, I'm trying to solve the equation for the frequency.
The problem is that there's an exponential expression in it. But it gets worse, the frequency f is in it.

http://s24.postimg.org/jbunj5z1h/Vnoise.jpg

All I could come up with is to put hf/kt on one side of the equation and I'm stuck with ln (4hfr delta v / V^2). Do you have a solution?
 
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  • #2
Looks like you are using the wrong formula for you situation.

The formulas for thermal noise that we typically use in EE are derived from classical thermodynamics which leads to power spectral density being independent of frequency. Of course this implies that there is an infinite amount of thermal noise power in every resistor when we include all frequencies up to infinity (look up "ultraviolet catastrophe").

The resolution to this paradox (thanks to Planck) is that when frequencies get very high, classical thermodynamics breaks down because the quantization of energy (E=hv) becomes significant.

Your formula is a general formula that includes quantum effects, and thus shows a dependency on frequency.
 
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  • #3
the_emi_guy said:
Looks like you are using the wrong formula for you situation.

The formulas for thermal noise that we typically use in EE are derived from classical thermodynamics which leads to power spectral density being independent of frequency. Of course this implies that there is an infinite amount of thermal noise power in every resistor when we include all frequencies up to infinity (look up "ultraviolet catastrophe").

The resolution to this paradox (thanks to Planck) is that when frequencies get very high, classical thermodynamics breaks down because the quantization of energy (E=hv) becomes significant.

Your formula is a general formula that includes quantum effects, and thus shows a dependency on frequency.

I think you are wrong. The equation calculates the frequency with respect to (1) the current running through an internal resistance, and (2) the bandwidth of the white noise. What you end up with is simply a complex frequency, which is solved using the Lambert W equation. Might as well throw in the towel and call it a simple frequency.
 
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  • #4
the_emi_guy said:
Looks like you are using the wrong formula for you situation.

The formulas for thermal noise that we typically use in EE are derived from classical thermodynamics which leads to power spectral density being independent of frequency. Of course this implies that there is an infinite amount of thermal noise power in every resistor when we include all frequencies up to infinity (look up "ultraviolet catastrophe").

The resolution to this paradox (thanks to Planck) is that when frequencies get very high, classical thermodynamics breaks down because the quantization of energy (E=hv) becomes significant.

Your formula is a general formula that includes quantum effects, and thus shows a dependency on frequency.

And by the way, for the record, Johnson–Nyquist noise is classical thermodynamics.
 
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  • #5
the_emi_guy said:
Looks like you are using the wrong formula for you situation.

The formulas for thermal noise that we typically use in EE are derived from classical thermodynamics which leads to power spectral density being independent of frequency. Of course this implies that there is an infinite amount of thermal noise power in every resistor when we include all frequencies up to infinity (look up "ultraviolet catastrophe").

The resolution to this paradox (thanks to Planck) is that when frequencies get very high, classical thermodynamics breaks down because the quantization of energy (E=hv) becomes significant.

Your formula is a general formula that includes quantum effects, and thus shows a dependency on frequency.

You can not, I repeat, you can not exclude the frequency when calculating white noise. You don't know noise is measured in Hertz?
 
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  • #7
Mechatron said:
You can not, I repeat, YOU CAN NOT exclude the frequency when calculating white noise. Have you lost your mind? You don't know noise is measured in Hertz?

noise ( thermal) is measured in K ( Kelvin) frequency is measured in Hz ( Hertz )

Dave
 
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  • #8
Color me confused. How is noise measured in Hertz?

I thought noise was expressed in power or voltage or current. The amount of power may depend on the bandwidth ( so noise voltage is sometime expressed as volts/root-hertz, etc)

At very high frequencies you can determine the power spectral density by using hf. Do a unit analysis of hf and you get volts, or eV.
 
  • #9
uh huh

when I am dealing with thermal noise in low noise RF preamplifiers etc its always measured in KD
 
  • #10
Mechatron said:
And by the way, for the record, Johnson–Nyquist noise is classical thermodynamics.

Yes, in his paper Nyquist used classical thermodynamic arguments including the classical equipartition law (along with transmission line theory) to derive
[tex]
v_n^2 = 4 R k_B T \delta \nu
[/tex]
in order to explain the experimental results published by Johnson. This is the classical result. Notice that it is not the equation you posted. Yours has Planck's constant in it, which does not show up in classical physics. Nyquist does state the equation you use at the end of his paper to indicate what happens at very high frequencies and/or low temperatures, when the classical equipartition law no longer holds and quantum effects must be included. Of course the classical result is all that was needed to explain Johnson's data. If you take the limit as [itex]h \rightarrow 0[/itex] (or equivalently assume [itex]h f << k_B T[/itex]) in your expression then you recover the classical result.

May I ask how you are doing this measurement? Are you really in a regime where it is not true that [itex]h f << k_B T[/itex]? What is the bandwidth of your measurement? etc. If you really want help (as opposed to simply looking to "in your face" people) then we need more information. For example, usually we know [itex]f[/itex] and [itex]\delta \nu[/itex] from the measurement setup, so it isn't clear why you want to solve for [itex]f[/itex] at all ...jason
 
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  • #11
davenn said:
uh huh

when I am dealing with thermal noise in low noise RF preamplifiers etc its always measured in K


D
I don't doubt it, but technically speaking I think you are referring to Noise Temperature, which of course has units of temperature. We do similar things at my workplace, where we might say, "that amp has 2 dB of noise" when we are actually referring to the noise figure. In any case, the noise performance of amps is one step beyond the noise from a resistor, which I think is all the OP is concerned with.

jason
 
  • #12
meBigGuy said:
Color me confused. How is noise measured in Hertz?

I thought noise was expressed in power or voltage or current. The amount of power may depend on the bandwidth ( so noise voltage is sometime expressed as volts/root-hertz, etc)

At very high frequencies you can determine the power spectral density by using hf. Do a unit analysis of hf and you get volts, or eV.

I am speaking about the frequency disturbances caused by thermal radiation. As far as I understand, the frequency generated from current passing the internal resistor in a battery can be found in the unit for frequency in the equation in the link below:

http://s24.postimg.org/4cqk2wbet/Vnoise.jpg
 
  • #13
jasonRF said:
Yes, in his paper Nyquist used classical thermodynamic arguments including the classical equipartition law (along with transmission line theory) to derive
[tex]
v_n^2 = 4 R k_B T \delta \nu
[/tex]
in order to explain the experimental results published by Johnson. This is the classical result. Notice that it is not the equation you posted. Yours has Planck's constant in it, which does not show up in classical physics. Nyquist does state the equation you use at the end of his paper to indicate what happens at very high frequencies and/or low temperatures, when the classical equipartition law no longer holds and quantum effects must be included. Of course the classical result is all that was needed to explain Johnson's data. If you take the limit as [itex]h \rightarrow 0[/itex] (or equivalently assume [itex]h f << k_B T[/itex]) in your expression then you recover the classical result.

May I ask how you are doing this measurement? Are you really in a regime where it is not true that [itex]h f << k_B T[/itex]? What is the bandwidth of your measurement? etc. If you really want help (as opposed to simply looking to "in your face" people) then we need more information. For example, usually we know [itex]f[/itex] and [itex]\delta \nu[/itex] from the measurement setup, so it isn't clear why you want to solve for [itex]f[/itex] at all ...jason

Thank you for your reply. I will be as clear as I possible can.
Please go to the following website regarding 1995 IEEE international frequency control symposium:

http://tf.nist.gov/general/pdf/1133.pdf

In general, it is accepted that a common DC battery only supplies a DC voltage.
However, a small frequency somewhere arround 20 - 1000 Hz is generated, due to heat generated in the battery. I call it thermal noise, forget I ever said white noise. You can calculate the current noise with the instrument shown on page 370 (page 4 in the pdf). You can then use the current noise to calculate the voltage noise. Subsequently you use the voltage noise and the measured temperature of the battery, which would be between 300 - 323 Kelvin, to find the bandwidth of the frequency/noise. Finally, you can add this data into the equation in the link below, as shown on page 1:

http://s24.postimg.org/4cqk2wbet/Vnoise.jpg

But you see, there's a problem, you've got exp / e in the equation. By simplifying the equation, you can get y = ln (x*y -1) / w. You can solve this equation with respect to y, which represents the frequency. You end up with an expression including a product log:

http://www.wolframalpha.com/input/?i=y/w+=+ln((x*y)-1)+with+respect+to+y

But the problem doesn't stop here. Now you're stuck with a complex number.
But really, this is a representation of a complex frequency.

Concept of complex FrequencyDefinition: A type of frequency that depends on two parameters ; one is the ” σ” which controls the magnitude of the signal and the other is “w”, which controls the rotation of the signal ; is known as “complex frequency”.

Makes sense, right? A complex number is just a two dimensional number:
http://m.eet.com/media/1068017/lyons_pt2_3.gif

So I think I just need to know, is the frequency in the equation, the frequency which you'd see on an oscillator, generated from the heat from the battery?

Michael
 
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  • #14
You are using the wrong equation.

You are using equation (1) of the NIST document.
You should be using equation (2) "At low frequencies Johnson noise can be approximated by...", which is the same equation that jasonRF has posted.

You are dealing with kHz right, not GHz or THz.
 
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  • #15
Mechatron said:
Thank you for your reply. I will be as clear as I possible can.
Please go to the following website regarding 1995 IEEE international frequency control symposium:

http://tf.nist.gov/general/pdf/1133.pdf

In general, it is accepted that a common DC battery only supplies a DC voltage.
However, a small frequency somewhere arround 20 - 1000 Hz is generated, due to heat generated in the battery. I call it thermal noise, forget I ever said white noise. You can calculate the current noise with the instrument shown on page 370 (page 4 in the pdf). You can then use the current noise to calculate the voltage noise. Subsequently you use the voltage noise and the measured temperature of the battery, which would be between 300 - 323 Kelvin, to find the bandwidth of the frequency/noise.
I read the paper. The noise is essentially white except for the lowest frequencies, as indicated by their data and by their statements. They do not calculate the bandwidth from the voltage noise and the temp; instead, the resolution of their spectrum analyzer defines the bandwidth. In fact, "measurement bandwidth" is always what the bandwidth number in Nyquists formula means. This is even stated in the beginning of the paper you link. Note that their voltage noise measurements are presented in spectral density = Volts / sqrt(Hz). To place their measurements in this form they had to know the bandwidth already (after all, they set up the spectrum analyzer, right?).
Mechatron said:
Finally, you can add this data into the equation in the link below, as shown on page 1:

http://s24.postimg.org/4cqk2wbet/Vnoise.jpg
You are still mistaken - that equation describes the frequency dependence of thermal noise. Noise is generated at all frequencies! In any case, for parameters relevant to your case: T = 300K and f = 1000 Hz,
[tex]
\frac{h f}{k_B T} \approx 10^{-10}
[/tex]
so surely the classical result is all that matters. Just look at the data in the paper. It is flat with frequency except at the lowest frequencies where other processes are dominating. It is not frequency dependent (edit: it is not frequency depended in the portion of frequency space you seem to care about).

jason
 
  • #16
jasonRF said:
I read the paper. The noise is essentially white except for the lowest frequencies, as indicated by their data and by their statements. They do not calculate the bandwidth from the voltage noise and the temp; instead, the resolution of their spectrum analyzer defines the bandwidth. In fact, "measurement bandwidth" is always what the bandwidth number in Nyquists formula means. This is even stated in the beginning of the paper you link. Note that their voltage noise measurements are presented in spectral density = Volts / sqrt(Hz). To place their measurements in this form they had to know the bandwidth already (after all, they set up the spectrum analyzer, right?).

You are still mistaken - that equation describes the frequency dependence of thermal noise. Noise is generated at all frequencies! In any case, for parameters relevant to your case: T = 300K and f = 1000 Hz,
[tex]
\frac{h f}{k_B T} \approx 10^{-10}
[/tex]
so surely the classical result is all that matters. Just look at the data in the paper. It is flat with frequency except at the lowest frequencies where other processes are dominating. It is not frequency dependent (edit: it is not frequency depended in the portion of frequency space you seem to care about).

jason

Exactly, the equation describes the frequency dependence of thermal noise.
The noise you put in the equation is a specific one. The frequency I want to calculate using the equation, is the same frequency I measure using an oscilloscope. Please stop saying it's not frequency dependent when the frequency is in the equation, and the frequency is what I want to calculate. In other words, it's a DC voltage signal I'm looking for, with oscillations.

I've added a graph:

http://s22.postimg.org/hp31y8dnl/Graph.png

Here's why it's so important to me:

http://www.acs.psu.edu/drussell/Demos/superposition/pulses.gif

The small wave represents the wave/frequency generated from heat from the battery, and the larger wave is the desired wave which I generate in a circuit. The DC voltage with oscillations is introduced into the circuit, and I believe with a feedback/loop, thing's just get worse.
 
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  • #17
Mechatron,

For the third time...you are using the wrong forumla. Use equation (2) in NIST document, not equation (1).

We don't need to invoke Planck's constant when working with signals in the kHz range.
 
  • #18
jasonRF said:
I read the paper. The noise is essentially white except for the lowest frequencies, as indicated by their data and by their statements. They do not calculate the bandwidth from the voltage noise and the temp; instead, the resolution of their spectrum analyzer defines the bandwidth. In fact, "measurement bandwidth" is always what the bandwidth number in Nyquists formula means. This is even stated in the beginning of the paper you link. Note that their voltage noise measurements are presented in spectral density = Volts / sqrt(Hz). To place their measurements in this form they had to know the bandwidth already (after all, they set up the spectrum analyzer, right?).




You are still mistaken - that equation describes the frequency dependence of thermal noise. Noise is generated at all frequencies! In any case, for parameters relevant to your case: T = 300K and f = 1000 Hz,
[tex]
\frac{h f}{k_B T} \approx 10^{-10}
[/tex]
so surely the classical result is all that matters. Just look at the data in the paper. It is flat with frequency except at the lowest frequencies where other processes are dominating. It is not frequency dependent (edit: it is not frequency depended in the portion of frequency space you seem to care about).

jason

"They do not calculate the bandwidth from the voltage noise and the temp".
I think you need to read the paper again. They calculate the Voltage noise with an equation which consists of the bandwidth and the temperature. You can rearrange this equation and solve it with respect to frequency. Please look at the following image in the link below!

http://s30.postimg.org/c86jz3fkh/frequency.png

This frequency F, is this the frequency I would see on an oscillator? Is this the frequency generated from heat from the generator?
 
  • #19
the_emi_guy said:
Mechatron,

For the third time...you are using the wrong forumla. Use equation (2) in NIST document, not equation (1).

We don't need to invoke Planck's constant when working with signals in the kHz range.

I think I understand what you're trying to say, but you don't want me to use the equation 1, just because an approximation would be good enough, which would exclude a very important data.
 
  • #20
the_emi_guy said:
Mechatron,

For the third time...you are using the wrong forumla. Use equation (2) in NIST document, not equation (1).

We don't need to invoke Planck's constant when working with signals in the kHz range.

Which Boltzmann's constant is being used? Is it 8.617 3324×10−5 eV/K?
 
  • #21
I am encouraging you to employ equation (2) because you will avoid the problems you struggling with dealing with the exponential and the frequency term. At kHz frequencies you are not losing any significant accuracy by avoiding the use of quantum mechanics. Kind of like ignoring general relativity and sticking to Newton's laws to calculate the flight of a cannon ball.

Stick with the SI units for Boltzmann constant (JK^-1) since all other electrical quantities volts, amps etc. are SI.
 
  • #22
jasonRF said:
I read the paper. The noise is essentially white except for the lowest frequencies, as indicated by their data and by their statements. They do not calculate the bandwidth from the voltage noise and the temp; instead, the resolution of their spectrum analyzer defines the bandwidth. In fact, "measurement bandwidth" is always what the bandwidth number in Nyquists formula means. This is even stated in the beginning of the paper you link. Note that their voltage noise measurements are presented in spectral density = Volts / sqrt(Hz). To place their measurements in this form they had to know the bandwidth already (after all, they set up the spectrum analyzer, right?).




You are still mistaken - that equation describes the frequency dependence of thermal noise. Noise is generated at all frequencies! In any case, for parameters relevant to your case: T = 300K and f = 1000 Hz,
[tex]
\frac{h f}{k_B T} \approx 10^{-10}
[/tex]
so surely the classical result is all that matters. Just look at the data in the paper. It is flat with frequency except at the lowest frequencies where other processes are dominating. It is not frequency dependent (edit: it is not frequency depended in the portion of frequency space you seem to care about).

jason

You said it's not frequency dependant, but in equation 2, bandwidth is included.
Is there any way of calculating what the frequency is based on the bandwidth of the signal?
 
  • #23
Mechatron said:
Exactly, the equation describes the frequency dependence of thermal noise.
The noise you put in the equation is a specific one. The frequency I want to calculate using the equation, is the same frequency I measure using an oscilloscope. Please stop saying it's not frequency dependent when the frequency is in the equation, and the frequency is what I want to calculate. In other words, it's a DC voltage signal I'm looking for, with oscillations.

I've added a graph:

http://s22.postimg.org/hp31y8dnl/Graph.png

Is this what you are measuring on your oscilloscope? If so, what exactly is the circuit you are measuring? Is there a very narrowband filter in there somewhere (either in the circuit or in the way you setup your measurement device)? If not, then the oscillation you see isn't from Nyquist thermal noise (it could be from something else, just not Nyquist thermal noise).

You should try plotting the function you are using from 10 Hz up to 10 MHz for T=300K. It should look quite flat. In fact, it will look flat until your frequency gets close to [itex] k_B T / h [/itex] which is the order of [itex] 10^{13} Hz[/itex] if I did the arithmetic correctly.

EDIT: Listen the_emi_guy and use consistent units. Actually, listen to him on everything - he is steering you the right way every time as far as I can tell.

jason
 
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  • #24
Equation (2) only requires bandwidth, not frequency. The idea is that the spectral noise density is flat. There is the same amount of noise power in the 1KHz-2KHz bandwidth as there is in the 10KHz-11KHz bandwidth.

Of course we are only talking about thermal noise. At low frequencies other noise mechanisms become very significant (1/f noise in particular).
 
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  • #25
Thread closed for Moderation and cleanup...
 
  • #26
Thread re-opened.
 
  • #27
Thermal Noise in batteries - Signal to Noise

I am building a transmitter which will generate a 20 kHz signal from a DC voltage supply (battery).
The battery is generating thermal noise which will interfere with the signal. The thermal noise is generating an electrical signal (alternating current) which can be measured in Hz, as far as I'm concerned. I expect this AC signal to be anywhere between 20 - 1000 Hz. Below I am giving you a more technical definition of the terms that concern me.

Thermal noise is approximately white, meaning that its power spectral density is nearly equal throughout the frequency spectrum. The amplitude of the signal has very nearly a Gaussian probability density function.

A noise signal is typically considered as a linear addition to a useful information signal. Typical signal quality measures involving noise are signal-to-noise ratio.

Signal-to-noise ratio is defined as the power ratio between a signal (meaningful information) and the background noise (unwanted signal).


I want to know what the linear addition to the 20 kHz signal would be.
If I were able to calculate the signal-to-noise ratio, how can I use it to calculate this linear addition? Can Voltage Noise (dBV/Hz) be used to find the linear addition? How can I calculate the linear addition?

I've added a link to a pdf file regarding 1995 IEEE International Frequency Control Symposium.
http://tf.nist.gov/general/pdf/1133.pdf

Using Planck's law does not seem useful to me, because I end up with a very high value.
If the frequency is low enough:
f = KbT/h = =1.3806488*10^-23 J/K * 300 K / =6.62606957*10^-34 Js = 6.2 THz (Tera Hz).
So I think this frequency is not the same as the "linear addition".

Reference to Planck's law:
http://en.wikipedia.org/wiki/Johnson–Nyquist_noise

Final words:
I am interpreting thermal noise as the linear addition to the desired signal in this manner;
Frequency disturbance from thermal noise + signal = 20Hz + 20 000 Hz = 20 020Hz.
The purpose is to find a suitable filter capacitor which can remove the thermal noise / linear addition. I know it might seem unnecessary, but it's for respect to science research purpose, but also because I want to build more sensitive systems. I just want things to be perfect.
 
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  • #28
Mechatron said:
I am interpreting thermal noise as the linear addition to the desired signal in this manner;
Frequency disturbance from thermal noise + signal = 20Hz + 20 000 Hz = 20 020Hz.
This is where you are getting off track, noise does not add in the manner you are describing.

Mechatron said:
I am building a transmitter which will generate a 20 kHz signal from a DC voltage supply (battery).
We need to know the *bandwidth* of the 20KHz signal, or more specifically, the bandwidth of the receiver that will be trying to recover this signal. I'll explain why shortly.
Mechatron said:
The battery is generating thermal noise which will interfere with the signal. The thermal noise is generating an electrical signal (alternating current) which can be measured in Hz, as far as I'm concerned. I expect this AC signal to be anywhere between 20 - 1000 Hz. Below I am giving you a more technical definition of the terms that concern me.

Thermal noise is approximately white, meaning that its power spectral density is nearly equal throughout the frequency spectrum. The amplitude of the signal has very nearly a Gaussian probability density function.

Here you are saying that the thermal noise is both white and limited to the very small 20-1000Hz band. These are mutually exclusive statements. White means its power spectral density at 1MHz is the same as at 20Hz.

So let's say that the power spectral density is 1nW per Hz of bandwidth. This means that there will be 1nW of noise power in every 1Hz of bandwidth. If I build a perfect (doesn't create any noise of its own) receiver that tunes to 20KHz with a bandwidth of 1Hz (receives everything between 20,000Hz and 20,001Hz, rejects everything else) then I will have 1nW of noise power coming out of this receiver.

On the other hand, let's say your receiver has 1KHz of bandwidth (20KHz - 21KHz received) then we will have 1000*1nW = 1uW of noise power received.

If your 20KHz signal is 1W, then our SNR is 1W/1uW.

Hope this clarifies things, we need to know the bandwidth of the 20KHz receiver, and the noise power in whatever units (dBm/Hz, nW/Hz).
 
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  • #29
Is this as simple as:

1. At moderate frequencies the noise is not frequency dependent.

2. At all frequencies the measured noise power is bandwidth dependent.

Even though both the frequency and bandwidth are expressed in Hz., they are completely independent.
 
  • #30
the_emi_guy said:
This is where you are getting off track, noise does not add in the manner you are describing.


We need to know the *bandwidth* of the 20KHz signal, or more specifically, the bandwidth of the receiver that will be trying to recover this signal. I'll explain why shortly.



Here you are saying that the thermal noise is both white and limited to the very small 20-1000Hz band. These are mutually exclusive statements. White means its power spectral density at 1MHz is the same as at 20Hz.

So let's say that the power spectral density is 1nW per Hz of bandwidth. This means that there will be 1nW of noise power in every 1Hz of bandwidth. If I build a perfect (doesn't create any noise of its own) receiver that tunes to 20KHz with a bandwidth of 1Hz (receives everything between 20,000Hz and 20,001Hz, rejects everything else) then I will have 1nW of noise power coming out of this receiver.

On the other hand, let's say your receiver has 1KHz of bandwidth (20KHz - 21KHz received) then we will have 1000*1nW = 1uW of noise power received.

If your 20KHz signal is 1W, then our SNR is 1W/1uW.

Hope this clarifies things, we need to know the bandwidth of the 20KHz receiver, and the noise power in whatever units (dBm/Hz, nW/Hz).

Please be aware that I am not taking the receiver into consideration. We need to focus on the transmitter only. It's the signal transmission that matters, not the reception.

So you're saying that heat generated from a battery does not output an alternating current in a circuit? The transmitter generates a square wave, so it has infinite bandwidth. However, the oscillator indicated that the transmitter wasn't generating a perfect 20 kHz signal, you could see some oscillations on top of the square wave, and the oscillator indicated it was swinging between 20 000 - 20400 Hz. I understand now the relationship between noise power and bandwidth.

But I need to know how the heat generated from the battery, will change the 20 kHz signal, or better yet, how it affects a DC signal alone. Just take a good look in the pdf file again. I could use equation 2 to find the bandwidth of the thermal noise from the battery. Just forget about the bandwidth of the receiver, it's not needed. Because no battery can produce a perfect DC voltage due to heat, which is but not limited to its internal resistance.

From the first sentence in the link below:
"In electronics, noise is a random fluctuation in an electrical signal"
http://en.wikipedia.org/wiki/Noise_(electronics)

This electric signal is produced by thermal noise, and is added to the 20 kHz. I'm not measuring this thermal noise in volts, but in Hz in the oscillator. They are added just like illustrated in the link below (from a mathematical perspective, physically they travel the same way):

http://www.acs.psu.edu/drussell/Demos/superposition/pulses.gif
 
  • #31
the_emi_guy said:
This is where you are getting off track, noise does not add in the manner you are describing.


We need to know the *bandwidth* of the 20KHz signal, or more specifically, the bandwidth of the receiver that will be trying to recover this signal. I'll explain why shortly.



Here you are saying that the thermal noise is both white and limited to the very small 20-1000Hz band. These are mutually exclusive statements. White means its power spectral density at 1MHz is the same as at 20Hz.

So let's say that the power spectral density is 1nW per Hz of bandwidth. This means that there will be 1nW of noise power in every 1Hz of bandwidth. If I build a perfect (doesn't create any noise of its own) receiver that tunes to 20KHz with a bandwidth of 1Hz (receives everything between 20,000Hz and 20,001Hz, rejects everything else) then I will have 1nW of noise power coming out of this receiver.

On the other hand, let's say your receiver has 1KHz of bandwidth (20KHz - 21KHz received) then we will have 1000*1nW = 1uW of noise power received.

If your 20KHz signal is 1W, then our SNR is 1W/1uW.

Hope this clarifies things, we need to know the bandwidth of the 20KHz receiver, and the noise power in whatever units (dBm/Hz, nW/Hz).

I'm sorry if things seem to be off track, but I think we're close to solving our problem.
I wanted to add that I calculated the bandwidth using the second equation in the pdf with value; 300 K, (r+R) 1000,2 Ohm, -204dBV/Hz and the Boltzmann's constant of 1,38*10^-3 which gave me a bandwidth of 2,51 *10^21 (Zetta Hz), which in the electromagnetic specter would be gamma radiation (if it was a frequency), which doesn't seem very reasonable to me. And as previously mentioned, the f=KbK/h of 6,2 THz didn't make much sense either. So I'm really, really confused.
 
  • #32
So you're saying that heat generated from a battery does not output an alternating current in a circuit? The transmitter generates a square wave, so it has infinite bandwidth. However, the oscillator indicated that the transmitter wasn't generating a perfect 20 kHz signal, you could see some oscillations on top of the square wave, and the oscillator indicated it was swinging between 20 000 - 20400 Hz. I understand now the relationship between noise power and bandwidth.
Have you even considered that the variations in the osc freq is due to phase noise being generated by the oscillator and has nothing to do with the battery at all ?

cheers
Dave
 
  • #33
Closed pending moderation.
 
  • #34
Thanks Evo! Thread re-opened after merging two threads on the same subject.
 
  • #35
@Mechatron -- it sounds like the ringing or noise you are seeing is caused by something else. Batteries are not noisy. You can always follow the battery with a low-dropout linear voltage regulator to be extra sure that your power supply is quiet.
 

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