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resolution bandwidth

Is my understanding correct in selecting ADC resolution? (Noise analysis)Delta Sigma Converters: ResolutionNoise Bandwidth of multichannel systemHow to calculate opamp input noise for DC input?Quantifying white noise with a spectrum analyzerAmplitude of PSD on spectrum analyzer varies strongly with RBWMeasuring PSR of a current reference chipMaximum Allowable Noise of ADC Input SignalWill noise floor dominate the ADC resolution?External signal noise versus ADC resolution

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6

$begingroup$

I have the following trivial question:

I have acquired a noise spectrum with a resolution bandwidth of $124mathrmHz$.
I would like to convert it to $mathrmV/sqrtHz$. Shall I simply multiply or divide the signal by $sqrt124$??

Regards

noise spectrum-analyzer resolution

share|improve this question

edited May 20 at 18:15

Daniele Tampieri

1,2742717

asked May 20 at 11:26

user222401user222401

311

New contributor

user222401 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The spectral shape determines everything, for which you have not defined
  $endgroup$
  – Sunnyskyguy EE75
  May 21 at 0:16
  

  
  
  
  

add a comment | 

6

$begingroup$

I have the following trivial question:

I have acquired a noise spectrum with a resolution bandwidth of $124mathrmHz$.
I would like to convert it to $mathrmV/sqrtHz$. Shall I simply multiply or divide the signal by $sqrt124$??

Regards

noise spectrum-analyzer resolution

share|improve this question

edited May 20 at 18:15

Daniele Tampieri

1,2742717

asked May 20 at 11:26

user222401user222401

311

New contributor

user222401 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The spectral shape determines everything, for which you have not defined
  $endgroup$
  – Sunnyskyguy EE75
  May 21 at 0:16
  

  
  
  
  

add a comment | 

$begingroup$

I have the following trivial question:

I have acquired a noise spectrum with a resolution bandwidth of $124mathrmHz$.
I would like to convert it to $mathrmV/sqrtHz$. Shall I simply multiply or divide the signal by $sqrt124$??

Regards

noise spectrum-analyzer resolution

share|improve this question

edited May 20 at 18:15

Daniele Tampieri

1,2742717

asked May 20 at 11:26

user222401user222401

311

New contributor

user222401 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this question

edited May 20 at 18:15

Daniele Tampieri

1,2742717

asked May 20 at 11:26

user222401user222401

311

New contributor

user222401 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited May 20 at 18:15

Daniele Tampieri

1,2742717

asked May 20 at 11:26

user222401user222401

311

New contributor

user222401 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited May 20 at 18:15

Daniele Tampieri

1,2742717

edited May 20 at 18:15

Daniele Tampieri

1,2742717

asked May 20 at 11:26

user222401user222401

311

New contributor

user222401 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked May 20 at 11:26

user222401user222401

311

3 Answers
3

active

oldest

votes

5

$begingroup$

Yes: however this requires a few words of explanation. Noise voltage has not the usual meaning since its exact value is obviously not exactly known. What can be measured and has a well defined and predictable behavior is the noise power adsorbed by a well defined load $Z_L$, in a frequency bandwidth $B$ centered around a frequency $f_o$, which has the following form
$$
P(B,f_o)=fraclangle V_n^2(f_o)rangleZ_L=frac1Z_Lintlimits_f_o-B/2^f_o+B/2S(f)mathrmdf
$$
i.e. it is the power spectral density $S(f)$ which gives the expected value. Since you have acquired a noise spectrum with a resolution bandwidth of $B=124mathrmHz$ the measured value you have acquired is precisely
$$
langle V_n^2(f_o)rangle=;intlimits_f_o-B/2^f_o+B/2S(f)mathrmdfimplies S(f)simeq fraclangle V_n^2(f)rangleB
$$
However, instead of $S(f)$ it is customary preferred to give the noise voltage density $v_n(f)$
$$
v_n(f)=sqrtS(f)=fraclangle V_n^2(f)rangle^frac12sqrtB
$$
In sum, if your spectrum analyzer gives you already a power spectrum in which each spectral line is expressed as a voltage, dividing it by the square root of its resolution bandwidth gives you the sought for noise voltage density.

share|improve this answer

edited May 21 at 12:10

answered May 20 at 13:12

Daniele TampieriDaniele Tampieri

1,2742717

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Shouldn’t the lower end of the integral be $$f_0-B/2$$?
  $endgroup$
  – Jonas Schäfer
  May 20 at 16:28
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @JonasSchäfer Yes, thank you! I'll immediately correct the typo.
  $endgroup$
  – Daniele Tampieri
  May 20 at 16:29
  

  
  
  
  

add a comment | 

3

$begingroup$

If you had a noise value of N volts per $sqrtHz$ then that would be a noise voltage of $Ncdot sqrt124$ volts in your 124 Hz bandwidth. Reversing the operation means dividing by $sqrt124$.

share|improve this answer

edited May 20 at 12:13

answered May 20 at 12:06

Andy akaAndy aka

246k11189430

$endgroup$

add a comment | 

-2

$begingroup$

Some great background to noise measurement in different bandwidths can be found in:

Agilent application note 1303 Agilent Spectrum Analyzer Measurements and Noise:

Measuring Noise and Noise-like Digital Communications Signals with a Spectrum Analyzer

Found all over the web, here is one copy.

share|improve this answer

answered May 20 at 12:49

tomnexustomnexus

5,46511029

$endgroup$

add a comment | 

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5

$begingroup$

Yes: however this requires a few words of explanation. Noise voltage has not the usual meaning since its exact value is obviously not exactly known. What can be measured and has a well defined and predictable behavior is the noise power adsorbed by a well defined load $Z_L$, in a frequency bandwidth $B$ centered around a frequency $f_o$, which has the following form
$$
P(B,f_o)=fraclangle V_n^2(f_o)rangleZ_L=frac1Z_Lintlimits_f_o-B/2^f_o+B/2S(f)mathrmdf
$$
i.e. it is the power spectral density $S(f)$ which gives the expected value. Since you have acquired a noise spectrum with a resolution bandwidth of $B=124mathrmHz$ the measured value you have acquired is precisely
$$
langle V_n^2(f_o)rangle=;intlimits_f_o-B/2^f_o+B/2S(f)mathrmdfimplies S(f)simeq fraclangle V_n^2(f)rangleB
$$
However, instead of $S(f)$ it is customary preferred to give the noise voltage density $v_n(f)$
$$
v_n(f)=sqrtS(f)=fraclangle V_n^2(f)rangle^frac12sqrtB
$$
In sum, if your spectrum analyzer gives you already a power spectrum in which each spectral line is expressed as a voltage, dividing it by the square root of its resolution bandwidth gives you the sought for noise voltage density.

share|improve this answer

edited May 21 at 12:10

answered May 20 at 13:12

Daniele TampieriDaniele Tampieri

1,2742717

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Shouldn’t the lower end of the integral be $$f_0-B/2$$?
  $endgroup$
  – Jonas Schäfer
  May 20 at 16:28
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @JonasSchäfer Yes, thank you! I'll immediately correct the typo.
  $endgroup$
  – Daniele Tampieri
  May 20 at 16:29
  

  
  
  
  

add a comment | 

5

$begingroup$

Yes: however this requires a few words of explanation. Noise voltage has not the usual meaning since its exact value is obviously not exactly known. What can be measured and has a well defined and predictable behavior is the noise power adsorbed by a well defined load $Z_L$, in a frequency bandwidth $B$ centered around a frequency $f_o$, which has the following form
$$
P(B,f_o)=fraclangle V_n^2(f_o)rangleZ_L=frac1Z_Lintlimits_f_o-B/2^f_o+B/2S(f)mathrmdf
$$
i.e. it is the power spectral density $S(f)$ which gives the expected value. Since you have acquired a noise spectrum with a resolution bandwidth of $B=124mathrmHz$ the measured value you have acquired is precisely
$$
langle V_n^2(f_o)rangle=;intlimits_f_o-B/2^f_o+B/2S(f)mathrmdfimplies S(f)simeq fraclangle V_n^2(f)rangleB
$$
However, instead of $S(f)$ it is customary preferred to give the noise voltage density $v_n(f)$
$$
v_n(f)=sqrtS(f)=fraclangle V_n^2(f)rangle^frac12sqrtB
$$
In sum, if your spectrum analyzer gives you already a power spectrum in which each spectral line is expressed as a voltage, dividing it by the square root of its resolution bandwidth gives you the sought for noise voltage density.

share|improve this answer

edited May 21 at 12:10

answered May 20 at 13:12

Daniele TampieriDaniele Tampieri

1,2742717

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Shouldn’t the lower end of the integral be $$f_0-B/2$$?
  $endgroup$
  – Jonas Schäfer
  May 20 at 16:28
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @JonasSchäfer Yes, thank you! I'll immediately correct the typo.
  $endgroup$
  – Daniele Tampieri
  May 20 at 16:29
  

  
  
  
  

add a comment | 

$begingroup$

Yes: however this requires a few words of explanation. Noise voltage has not the usual meaning since its exact value is obviously not exactly known. What can be measured and has a well defined and predictable behavior is the noise power adsorbed by a well defined load $Z_L$, in a frequency bandwidth $B$ centered around a frequency $f_o$, which has the following form
$$
P(B,f_o)=fraclangle V_n^2(f_o)rangleZ_L=frac1Z_Lintlimits_f_o-B/2^f_o+B/2S(f)mathrmdf
$$
i.e. it is the power spectral density $S(f)$ which gives the expected value. Since you have acquired a noise spectrum with a resolution bandwidth of $B=124mathrmHz$ the measured value you have acquired is precisely
$$
langle V_n^2(f_o)rangle=;intlimits_f_o-B/2^f_o+B/2S(f)mathrmdfimplies S(f)simeq fraclangle V_n^2(f)rangleB
$$
However, instead of $S(f)$ it is customary preferred to give the noise voltage density $v_n(f)$
$$
v_n(f)=sqrtS(f)=fraclangle V_n^2(f)rangle^frac12sqrtB
$$
In sum, if your spectrum analyzer gives you already a power spectrum in which each spectral line is expressed as a voltage, dividing it by the square root of its resolution bandwidth gives you the sought for noise voltage density.

share|improve this answer

edited May 21 at 12:10

answered May 20 at 13:12

Daniele TampieriDaniele Tampieri

1,2742717

$endgroup$

share|improve this answer

edited May 21 at 12:10

answered May 20 at 13:12

Daniele TampieriDaniele Tampieri

1,2742717

answered May 20 at 13:12

Daniele TampieriDaniele Tampieri

1,2742717

answered May 20 at 13:12

Daniele TampieriDaniele Tampieri

1,2742717

3

$begingroup$

If you had a noise value of N volts per $sqrtHz$ then that would be a noise voltage of $Ncdot sqrt124$ volts in your 124 Hz bandwidth. Reversing the operation means dividing by $sqrt124$.

share|improve this answer

edited May 20 at 12:13

answered May 20 at 12:06

Andy akaAndy aka

246k11189430

$endgroup$

add a comment | 

3

$begingroup$

If you had a noise value of N volts per $sqrtHz$ then that would be a noise voltage of $Ncdot sqrt124$ volts in your 124 Hz bandwidth. Reversing the operation means dividing by $sqrt124$.

share|improve this answer

edited May 20 at 12:13

answered May 20 at 12:06

Andy akaAndy aka

246k11189430

$endgroup$

add a comment | 

$begingroup$

If you had a noise value of N volts per $sqrtHz$ then that would be a noise voltage of $Ncdot sqrt124$ volts in your 124 Hz bandwidth. Reversing the operation means dividing by $sqrt124$.

share|improve this answer

edited May 20 at 12:13

answered May 20 at 12:06

Andy akaAndy aka

246k11189430

$endgroup$

share|improve this answer

edited May 20 at 12:13

answered May 20 at 12:06

Andy akaAndy aka

246k11189430

answered May 20 at 12:06

Andy akaAndy aka

246k11189430

answered May 20 at 12:06

Andy akaAndy aka

246k11189430

-2

$begingroup$

Some great background to noise measurement in different bandwidths can be found in:

Agilent application note 1303 Agilent Spectrum Analyzer Measurements and Noise:

Measuring Noise and Noise-like Digital Communications Signals with a Spectrum Analyzer

Found all over the web, here is one copy.

share|improve this answer

answered May 20 at 12:49

tomnexustomnexus

5,46511029

$endgroup$

add a comment | 

-2

$begingroup$

Some great background to noise measurement in different bandwidths can be found in:

Agilent application note 1303 Agilent Spectrum Analyzer Measurements and Noise:

Measuring Noise and Noise-like Digital Communications Signals with a Spectrum Analyzer

Found all over the web, here is one copy.

share|improve this answer

answered May 20 at 12:49

tomnexustomnexus

5,46511029

$endgroup$

add a comment | 

$begingroup$

Some great background to noise measurement in different bandwidths can be found in:

Agilent application note 1303 Agilent Spectrum Analyzer Measurements and Noise:

Measuring Noise and Noise-like Digital Communications Signals with a Spectrum Analyzer

Found all over the web, here is one copy.

share|improve this answer

answered May 20 at 12:49

tomnexustomnexus

5,46511029

$endgroup$

share|improve this answer

answered May 20 at 12:49

tomnexustomnexus

5,46511029

answered May 20 at 12:49

tomnexustomnexus

5,46511029

answered May 20 at 12:49

tomnexustomnexus

5,46511029