Which noise parameters?

I'm working on a project in which I need to calculate some noise parameters of a signal in order to give an idea how the system noise changes in the time: in short, the user should measure this parameters every week, in order to see if the noise has changed.

Well, I have the signal in Volt, during the noise measurement the output would be a constant value (i.e. a read for all the time of the measure a constant value i.e. 100 mV +/- the noise in the range of 0.1 mV), then this signal is sampled by a 24 bit A/D, then I get the signal in my code.

My problem is that I do not know exactly which parameter I have to calculate in order to give to the user an idea of the increasing/decreasing noise on the systems.

As first I thought a simple peak-peak difference in the measuring interval and the value of signal average. But then, which other parameter could I calculate? S/N would make no sense, because I have no signal, just a constant value

Any suggestion?

Thanks for your reply,.

Reply to
Giobibo
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Your signal is a constant DC value? Presumably then the noise is an AC signal on top of it. Maybe you want to consider performing a Fourier analysis of the sampled signal to find out how much power is in the noise frequency components, and where the power lies.

Reply to
Greg Neill

The signal is the output of a UV-Sensor lighted (during the noise measuring) with a costant UV-light: thus the output signal could be seen as a DC signal

Well, this could be an idea, but how could I use the information to see if the system noise increase/decrease? I should offer some simple parameters to the users, in order they could simply understand if the sensor is getting worse with the time...

Thank you for your reply.

Reply to
Giobibo

If you have a sensor signal, and it has a certain expected range, then you actually do have a numerator for S/N. The denominator, the noise, can be measured as E[n(t)^2], which is the average power of the noise. You can measure the signal over a time period where the sensor signal should be substantially constant. Get a set of some number of readings of the signal, maybe 100 or 1000. Subtract the mean value of the samples from each sample. That gets n(t). Square the result, which gives n(t)^2. Sum those squares and divide the result by the number of samples. That will be the average noise power in that interval. If you want to make an S/N out of it, you can use the range of valid sensor signals squared (to get power), divided by the average noise power. So if your sensor has a valid signal range of say, 0 to 2 V, and you get a series of readings that go like this...

1.0, 1.1, 1.2, 0.9, 1.0, the mean would be 1.04. Subtract 1.04 from each value to get-

-.04, +0.06, +.16, -0.14, -.04. Then square those and sum them to get 0.052. Divide that by 5, the number of samples to get 0.0104. This is the average noise power E[n(t)^2]. The full signal range is 0 to 2 V. If you reckon the signal power as a sinusoid of full scale amplitude (to choose an arbitrary figure), it would have a one-sided amplitude of 1, and a power of (1^2)/2, or 1/2. The S/N is then (1/2)/.0104 = 48, or 16.8 db.

-- John

Reply to
John O'Flaherty

Thank you very much for your clear explanation.

The value E[n(t)^2] is really a good parameter.

What is still not clear to me is the sense of S/N in this case. Ok, I understand that I can image as my signal the sinusoid of full scale amplitude and in this way I can calculate my S/N, but the numerator is always the same value (0.5) in all my calculation (because instead of my true signal I have to use always that sinusoid).

At this point, would not be better to use just the E[n(t)^2], instead of the S/N?

Thanks again for your reply. Gio

Reply to
Giobibo

Sure, you could just use noise power. Using the maximum signal value gives the "peak signal-to-noise ratio". That gives a reference point to compare the noise level to, but you could just use some arbitrary level, like the original noise power, as zero db, and subsequent changes could be referenced to that. The change in dbs would be the same.

-- John

Reply to
John O'Flaherty

Thanks again for your help.

Gio

Reply to
Giobibo

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