resolution bandwidth
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}
$begingroup$
I have the following trivial question:
I have acquired a noise spectrum with a resolution bandwidth of $124mathrm{Hz}$.
I would like to convert it to $mathrm{V/sqrt{Hz}}$. Shall I simply multiply or divide the signal by $sqrt{124}$??
Regards
noise spectrum-analyzer resolution
edited May 20 at 18:15
Daniele Tampieri
1,7482 gold badges7 silver badges18 bronze badges
asked May 20 at 11:26
user222401user222401
361 bronze badge
$endgroup$
add a comment |
$begingroup$
I have the following trivial question:
I have acquired a noise spectrum with a resolution bandwidth of $124mathrm{Hz}$.
I would like to convert it to $mathrm{V/sqrt{Hz}}$. Shall I simply multiply or divide the signal by $sqrt{124}$??
Regards
noise spectrum-analyzer resolution
edited May 20 at 18:15
Daniele Tampieri
1,7482 gold badges7 silver badges18 bronze badges
asked May 20 at 11:26
user222401user222401
361 bronze badge
$endgroup$
$begingroup$
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 $124mathrm{Hz}$.
I would like to convert it to $mathrm{V/sqrt{Hz}}$. Shall I simply multiply or divide the signal by $sqrt{124}$??
Regards
noise spectrum-analyzer resolution
edited May 20 at 18:15
Daniele Tampieri
1,7482 gold badges7 silver badges18 bronze badges
asked May 20 at 11:26
user222401user222401
361 bronze badge
$endgroup$
I have the following trivial question:
I have acquired a noise spectrum with a resolution bandwidth of $124mathrm{Hz}$.
I would like to convert it to $mathrm{V/sqrt{Hz}}$. Shall I simply multiply or divide the signal by $sqrt{124}$??
Regards
noise spectrum-analyzer resolution
noise spectrum-analyzer resolution
edited May 20 at 18:15
Daniele Tampieri
1,7482 gold badges7 silver badges18 bronze badges
asked May 20 at 11:26
user222401user222401
361 bronze badge
edited May 20 at 18:15
Daniele Tampieri
1,7482 gold badges7 silver badges18 bronze badges
asked May 20 at 11:26
user222401user222401
361 bronze badge
edited May 20 at 18:15
Daniele Tampieri
1,7482 gold badges7 silver badges18 bronze badges
edited May 20 at 18:15
Daniele Tampieri
1,7482 gold badges7 silver badges18 bronze badges
edited May 20 at 18:15
Daniele Tampieri
1,7482 gold badges7 silver badges18 bronze badges
1,7482 gold badges7 silver badges18 bronze badges
asked May 20 at 11:26
user222401user222401
361 bronze badge
asked May 20 at 11:26
user222401user222401
361 bronze badge
asked May 20 at 11:26
user222401user222401
361 bronze badge
361 bronze badge
$begingroup$
The spectral shape determines everything, for which you have not defined
$endgroup$
– Sunnyskyguy EE75
May 21 at 0:16
add a comment |
$begingroup$
The spectral shape determines everything, for which you have not defined
$endgroup$
– Sunnyskyguy EE75
May 21 at 0:16
$begingroup$
The spectral shape determines everything, for which you have not defined
$endgroup$
– Sunnyskyguy EE75
May 21 at 0:16
$begingroup$
The spectral shape determines everything, for which you have not defined
$endgroup$
– Sunnyskyguy EE75
May 21 at 0:16
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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
$$
P(B,f_o)=frac{langle V_n^2(f_o)rangle}{Z_L}=frac{1}{Z_L}intlimits_{f_o-B/2}^{f_o+B/2}S(f)mathrm{d}f
$$
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 $B=124mathrm{Hz}$ the measured value you have acquired is precisely
$$
langle V_n^2(f_o)rangle=intlimits_{f_o-B/2}^{f_o+B/2}S(f)mathrm{d}fimplies S(f)simeq frac{langle V_n^2(f)rangle}{B}
$$
However, instead of $S(f)$ it is customary preferred to give the noise voltage density $v_n(f)$
$$
v_n(f)=sqrt{S(f)}=frac{langle V_n^2(f)rangle^frac{1}{2}}{sqrt{B}}
$$
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.
answered May 20 at 13:12
Daniele TampieriDaniele Tampieri
1,7482 gold badges7 silver badges18 bronze badges
$endgroup$
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 |
$begingroup$
If you had a noise value of N volts per $sqrt{Hz}$ then that would be a noise voltage of $Ncdot sqrt{124}$ volts in your 124 Hz bandwidth. Reversing the operation means dividing by $sqrt{124}$.
answered May 20 at 12:06
Andy akaAndy aka
251k11 gold badges193 silver badges447 bronze badges
$endgroup$
add a comment |
$begingroup$
It's not a completely trivial question. Measurement of noise power is more difficult than measurement of tones.
Apparent noise power is affected by:
- Detector (if any) type and its response
- Non-linearities in the system (log amplifier in a spectrum analyser)
- Bandwidth of the filter used (noise equivalent bandwidth is not the same as -3 dB bandwidth)
For example - here's a diagram of one channel, and an adjacent one (dashed). The scalloping loss may not be equivalent to the adjacent channel leakage.
So when converting between "noise seen in a certain 124 Hz channel", and "noise power density in $text{dBm}/{text{Hz}}$ or $V/{sqrt{Hz}}$ you might need to correct for more than just the bandwidth. This is the case whether this is an FFT channel or a spectrum analyser setting.
If you're using a spectrum analyser, it will have a mode called "marker noise" which compensates as best as possible for these effects, to give you the effective noise power per Hz in a signal-free channel. If you're doing it yourself, I recommend reading Agilent application note 1303 Spectrum Analyzer Measurements and Noise: Measuring Noise and Noise-like Digital Communications Signals with a Spectrum Analyzer
This application note has some great background to noise measurement in different bandwidths. It's aimed at spectrum analysers but many of the principles are useful to other measurement systems.
Found all over the web, but not on the official site as far as I can see. Here is one copy.
answered May 20 at 12:49
tomnexustomnexus
5,5141 gold badge10 silver badges29 bronze badges
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
$$
P(B,f_o)=frac{langle V_n^2(f_o)rangle}{Z_L}=frac{1}{Z_L}intlimits_{f_o-B/2}^{f_o+B/2}S(f)mathrm{d}f
$$
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 $B=124mathrm{Hz}$ the measured value you have acquired is precisely
$$
langle V_n^2(f_o)rangle=intlimits_{f_o-B/2}^{f_o+B/2}S(f)mathrm{d}fimplies S(f)simeq frac{langle V_n^2(f)rangle}{B}
$$
However, instead of $S(f)$ it is customary preferred to give the noise voltage density $v_n(f)$
$$
v_n(f)=sqrt{S(f)}=frac{langle V_n^2(f)rangle^frac{1}{2}}{sqrt{B}}
$$
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.
answered May 20 at 13:12
Daniele TampieriDaniele Tampieri
1,7482 gold badges7 silver badges18 bronze badges
$endgroup$
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
$$
P(B,f_o)=frac{langle V_n^2(f_o)rangle}{Z_L}=frac{1}{Z_L}intlimits_{f_o-B/2}^{f_o+B/2}S(f)mathrm{d}f
$$
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 $B=124mathrm{Hz}$ the measured value you have acquired is precisely
$$
langle V_n^2(f_o)rangle=intlimits_{f_o-B/2}^{f_o+B/2}S(f)mathrm{d}fimplies S(f)simeq frac{langle V_n^2(f)rangle}{B}
$$
However, instead of $S(f)$ it is customary preferred to give the noise voltage density $v_n(f)$
$$
v_n(f)=sqrt{S(f)}=frac{langle V_n^2(f)rangle^frac{1}{2}}{sqrt{B}}
$$
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.
answered May 20 at 13:12
Daniele TampieriDaniele Tampieri
1,7482 gold badges7 silver badges18 bronze badges
$endgroup$
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
$$
P(B,f_o)=frac{langle V_n^2(f_o)rangle}{Z_L}=frac{1}{Z_L}intlimits_{f_o-B/2}^{f_o+B/2}S(f)mathrm{d}f
$$
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 $B=124mathrm{Hz}$ the measured value you have acquired is precisely
$$
langle V_n^2(f_o)rangle=intlimits_{f_o-B/2}^{f_o+B/2}S(f)mathrm{d}fimplies S(f)simeq frac{langle V_n^2(f)rangle}{B}
$$
However, instead of $S(f)$ it is customary preferred to give the noise voltage density $v_n(f)$
$$
v_n(f)=sqrt{S(f)}=frac{langle V_n^2(f)rangle^frac{1}{2}}{sqrt{B}}
$$
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.
answered May 20 at 13:12
Daniele TampieriDaniele Tampieri
1,7482 gold badges7 silver badges18 bronze badges
$endgroup$
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
$$
P(B,f_o)=frac{langle V_n^2(f_o)rangle}{Z_L}=frac{1}{Z_L}intlimits_{f_o-B/2}^{f_o+B/2}S(f)mathrm{d}f
$$
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 $B=124mathrm{Hz}$ the measured value you have acquired is precisely
$$
langle V_n^2(f_o)rangle=intlimits_{f_o-B/2}^{f_o+B/2}S(f)mathrm{d}fimplies S(f)simeq frac{langle V_n^2(f)rangle}{B}
$$
However, instead of $S(f)$ it is customary preferred to give the noise voltage density $v_n(f)$
$$
v_n(f)=sqrt{S(f)}=frac{langle V_n^2(f)rangle^frac{1}{2}}{sqrt{B}}
$$
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.
answered May 20 at 13:12
Daniele TampieriDaniele Tampieri
1,7482 gold badges7 silver badges18 bronze badges
edited May 26 at 18:10
answered May 20 at 13:12
Daniele TampieriDaniele Tampieri
1,7482 gold badges7 silver badges18 bronze badges
answered May 20 at 13:12
Daniele TampieriDaniele Tampieri
1,7482 gold badges7 silver badges18 bronze badges
answered May 20 at 13:12
Daniele TampieriDaniele Tampieri
1,7482 gold badges7 silver badges18 bronze badges
1,7482 gold badges7 silver badges18 bronze badges
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 |
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
2
2
$begingroup$
Shouldn’t the lower end of the integral be $$f_0-B/2$$?
$endgroup$
– Jonas Schäfer
May 20 at 16:28
$begingroup$
Shouldn’t the lower end of the integral be $$f_0-B/2$$?
$endgroup$
– Jonas Schäfer
May 20 at 16:28
2
2
$begingroup$
@JonasSchäfer Yes, thank you! I'll immediately correct the typo.
$endgroup$
– Daniele Tampieri
May 20 at 16:29
$begingroup$
@JonasSchäfer Yes, thank you! I'll immediately correct the typo.
$endgroup$
– Daniele Tampieri
May 20 at 16:29
add a comment |
$begingroup$
If you had a noise value of N volts per $sqrt{Hz}$ then that would be a noise voltage of $Ncdot sqrt{124}$ volts in your 124 Hz bandwidth. Reversing the operation means dividing by $sqrt{124}$.
answered May 20 at 12:06
Andy akaAndy aka
251k11 gold badges193 silver badges447 bronze badges
$endgroup$
add a comment |
$begingroup$
If you had a noise value of N volts per $sqrt{Hz}$ then that would be a noise voltage of $Ncdot sqrt{124}$ volts in your 124 Hz bandwidth. Reversing the operation means dividing by $sqrt{124}$.
answered May 20 at 12:06
Andy akaAndy aka
251k11 gold badges193 silver badges447 bronze badges
$endgroup$
add a comment |
$begingroup$
If you had a noise value of N volts per $sqrt{Hz}$ then that would be a noise voltage of $Ncdot sqrt{124}$ volts in your 124 Hz bandwidth. Reversing the operation means dividing by $sqrt{124}$.
answered May 20 at 12:06
Andy akaAndy aka
251k11 gold badges193 silver badges447 bronze badges
$endgroup$
If you had a noise value of N volts per $sqrt{Hz}$ then that would be a noise voltage of $Ncdot sqrt{124}$ volts in your 124 Hz bandwidth. Reversing the operation means dividing by $sqrt{124}$.
answered May 20 at 12:06
Andy akaAndy aka
251k11 gold badges193 silver badges447 bronze badges
edited May 20 at 12:13
answered May 20 at 12:06
Andy akaAndy aka
251k11 gold badges193 silver badges447 bronze badges
answered May 20 at 12:06
Andy akaAndy aka
251k11 gold badges193 silver badges447 bronze badges
answered May 20 at 12:06
Andy akaAndy aka
251k11 gold badges193 silver badges447 bronze badges
251k11 gold badges193 silver badges447 bronze badges
add a comment |
add a comment |
$begingroup$
It's not a completely trivial question. Measurement of noise power is more difficult than measurement of tones.
Apparent noise power is affected by:
- Detector (if any) type and its response
- Non-linearities in the system (log amplifier in a spectrum analyser)
- Bandwidth of the filter used (noise equivalent bandwidth is not the same as -3 dB bandwidth)
For example - here's a diagram of one channel, and an adjacent one (dashed). The scalloping loss may not be equivalent to the adjacent channel leakage.
So when converting between "noise seen in a certain 124 Hz channel", and "noise power density in $text{dBm}/{text{Hz}}$ or $V/{sqrt{Hz}}$ you might need to correct for more than just the bandwidth. This is the case whether this is an FFT channel or a spectrum analyser setting.
If you're using a spectrum analyser, it will have a mode called "marker noise" which compensates as best as possible for these effects, to give you the effective noise power per Hz in a signal-free channel. If you're doing it yourself, I recommend reading Agilent application note 1303 Spectrum Analyzer Measurements and Noise: Measuring Noise and Noise-like Digital Communications Signals with a Spectrum Analyzer
This application note has some great background to noise measurement in different bandwidths. It's aimed at spectrum analysers but many of the principles are useful to other measurement systems.
Found all over the web, but not on the official site as far as I can see. Here is one copy.
answered May 20 at 12:49
tomnexustomnexus
5,5141 gold badge10 silver badges29 bronze badges
$endgroup$
add a comment |
$begingroup$
It's not a completely trivial question. Measurement of noise power is more difficult than measurement of tones.
Apparent noise power is affected by:
- Detector (if any) type and its response
- Non-linearities in the system (log amplifier in a spectrum analyser)
- Bandwidth of the filter used (noise equivalent bandwidth is not the same as -3 dB bandwidth)
For example - here's a diagram of one channel, and an adjacent one (dashed). The scalloping loss may not be equivalent to the adjacent channel leakage.
So when converting between "noise seen in a certain 124 Hz channel", and "noise power density in $text{dBm}/{text{Hz}}$ or $V/{sqrt{Hz}}$ you might need to correct for more than just the bandwidth. This is the case whether this is an FFT channel or a spectrum analyser setting.
If you're using a spectrum analyser, it will have a mode called "marker noise" which compensates as best as possible for these effects, to give you the effective noise power per Hz in a signal-free channel. If you're doing it yourself, I recommend reading Agilent application note 1303 Spectrum Analyzer Measurements and Noise: Measuring Noise and Noise-like Digital Communications Signals with a Spectrum Analyzer
This application note has some great background to noise measurement in different bandwidths. It's aimed at spectrum analysers but many of the principles are useful to other measurement systems.
Found all over the web, but not on the official site as far as I can see. Here is one copy.
answered May 20 at 12:49
tomnexustomnexus
5,5141 gold badge10 silver badges29 bronze badges
$endgroup$
add a comment |
$begingroup$
It's not a completely trivial question. Measurement of noise power is more difficult than measurement of tones.
Apparent noise power is affected by:
- Detector (if any) type and its response
- Non-linearities in the system (log amplifier in a spectrum analyser)
- Bandwidth of the filter used (noise equivalent bandwidth is not the same as -3 dB bandwidth)
For example - here's a diagram of one channel, and an adjacent one (dashed). The scalloping loss may not be equivalent to the adjacent channel leakage.
So when converting between "noise seen in a certain 124 Hz channel", and "noise power density in $text{dBm}/{text{Hz}}$ or $V/{sqrt{Hz}}$ you might need to correct for more than just the bandwidth. This is the case whether this is an FFT channel or a spectrum analyser setting.
If you're using a spectrum analyser, it will have a mode called "marker noise" which compensates as best as possible for these effects, to give you the effective noise power per Hz in a signal-free channel. If you're doing it yourself, I recommend reading Agilent application note 1303 Spectrum Analyzer Measurements and Noise: Measuring Noise and Noise-like Digital Communications Signals with a Spectrum Analyzer
This application note has some great background to noise measurement in different bandwidths. It's aimed at spectrum analysers but many of the principles are useful to other measurement systems.
Found all over the web, but not on the official site as far as I can see. Here is one copy.
answered May 20 at 12:49
tomnexustomnexus
5,5141 gold badge10 silver badges29 bronze badges
$endgroup$
It's not a completely trivial question. Measurement of noise power is more difficult than measurement of tones.
Apparent noise power is affected by:
- Detector (if any) type and its response
- Non-linearities in the system (log amplifier in a spectrum analyser)
- Bandwidth of the filter used (noise equivalent bandwidth is not the same as -3 dB bandwidth)
For example - here's a diagram of one channel, and an adjacent one (dashed). The scalloping loss may not be equivalent to the adjacent channel leakage.
So when converting between "noise seen in a certain 124 Hz channel", and "noise power density in $text{dBm}/{text{Hz}}$ or $V/{sqrt{Hz}}$ you might need to correct for more than just the bandwidth. This is the case whether this is an FFT channel or a spectrum analyser setting.
If you're using a spectrum analyser, it will have a mode called "marker noise" which compensates as best as possible for these effects, to give you the effective noise power per Hz in a signal-free channel. If you're doing it yourself, I recommend reading Agilent application note 1303 Spectrum Analyzer Measurements and Noise: Measuring Noise and Noise-like Digital Communications Signals with a Spectrum Analyzer
This application note has some great background to noise measurement in different bandwidths. It's aimed at spectrum analysers but many of the principles are useful to other measurement systems.
Found all over the web, but not on the official site as far as I can see. Here is one copy.
answered May 20 at 12:49
tomnexustomnexus
5,5141 gold badge10 silver badges29 bronze badges
edited May 31 at 21:24
answered May 20 at 12:49
tomnexustomnexus
5,5141 gold badge10 silver badges29 bronze badges
answered May 20 at 12:49
tomnexustomnexus
5,5141 gold badge10 silver badges29 bronze badges
answered May 20 at 12:49
tomnexustomnexus
5,5141 gold badge10 silver badges29 bronze badges
5,5141 gold badge10 silver badges29 bronze badges
add a comment |
add a comment |
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$begingroup$
The spectral shape determines everything, for which you have not defined
$endgroup$
– Sunnyskyguy EE75
May 21 at 0:16