Timeline for Does power spectral density change with sampling rate?
Current License: CC BY-SA 4.0
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| Jan 13, 2023 at 1:53 | history | edited | Dan Boschen | CC BY-SA 4.0 |
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| Jan 13, 2023 at 1:28 | comment | added | Dan Boschen | @Nick I think I see what your confusion was; see update at start of my answer | |
| Jan 13, 2023 at 1:28 | history | edited | Dan Boschen | CC BY-SA 4.0 |
added 837 characters in body
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| Jan 13, 2023 at 1:03 | comment | added | Dan Boschen | (Or a spectrum analyzer; with your RBW comment): either way that is a measurement of what the noise density is but doesn’t set the noise density; correct measurement does not change what’s there. If we don’t amplify that resistor noise sufficiently before we sample, then the noise we measure is the quantization noise (and affected by the sampling rate)- if we do amplify enough (and filter properly) then the noise we measure is the amplified thermal noise plus noise figure and is not affected by the sampling rate. Hope that helps clear it up. | |
| Jan 13, 2023 at 0:50 | comment | added | Dan Boschen | This is with reference to the formula I gave: a converter with an ENOB of 10 bits at a given sampling rate would have a total noise power (when integrated from -fs/2 to +fs/2) of -61.8 dB below the power level of a full scale sine wave right at clipping. This noise if white (and usually is except when we sample a sine wave at an integer multiple of its rate) will be evenly distributed across that bandwidth, so the noise density would be that total power level divided by that sampling rate | |
| Jan 13, 2023 at 0:45 | comment | added | Dan Boschen | White noise from a resistor is not at all affected by the sampling rate unless you sample in a way that causes aliasing (you must low pass filter so that any noise above half the sampling rate, for real signals, does not alias). Regarding your first comment; the noise density frequency is quantization noise IS the total noise divided by the sampling rate. You might be thinking of what you get from an FFT since you mention resolution BW (that is not what I am talking about): if the total quantization noise is 10 mW and that noise is spread across 10 Hz then the noise density is 1mW/Hz. | |
| Jan 12, 2023 at 16:03 | comment | added | Nick | Thanks. However, I don't fully understand. Also, the noise density is not the noise divided by the sampling rate fs, it is the noise divided by the frequency resolution bandwidth df. Also, I'm talking about a white noise signal, like thermal noise from a resistor--not quantization noise due to sampling. Also, it seems like your second paragraph contradicts your first: are you saying that if the noise is from the signal itself, then no, the sampling rate will not change the noise density? Can you please explain that further? Thanks a lot! | |
| Jan 12, 2023 at 0:43 | history | answered | Dan Boschen | CC BY-SA 4.0 |