I am having trouble understanding how noise affects the signals you can measure.

For example, imagine the real signal I want to measure of a sinewave with a peak-to-peak amplitude of 5uV.

This signal goes through a resistor before being amplified by an ideal amplifier. I don't understand the meaning or how to use the \$V_{rms}\$ that has \$V/\sqrt(Hz)\$. How do I calculate the bandwidth? What does \$V_{rms}\$ have to do with the 5uV peak-to-peak of the sine wave?

My question is whether, due to resistor noise, I will be able to measure the signal.

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2 Answers 2


You define the bandwidth, according to the wanted signal and your application.

Is that 5 uV signal a pure sinewave of known frequency? If so, you can look for the signal in a tiny bandwidth around the known frequency, perhaps +/- 3 Hz. Then you take the noise power in that 6 Hz bandwidth, and compare it to the signal power, and decide if it's detectable or not.

If that signal could be any where in an audio bandwidth, or a radio channel bandwidth, then you take the 20 kHz, or the 10 MHz or whatever, as your noise bandwidth, and repeat the power calculations.

Defining the p-p of a noise signal is tricky, as for true noise there is no peak, it just keeps on going, but getting rarer. Some people take 3 sigma, or 5 or 6 sigma of the noise as a peak. However, it's much more repeatable to work with everything in power.

  • \$\begingroup\$ Yes, the requirement that frequency is stable is key - I'd add that phase shouldn't vary as well, but that goes along with stable frequency. And that 5uV signal should persist for a long time, since digging signals out of noise takes time, Neil, you might add how long the signal must persist for that +/- 3Hz bandwidth example. \$\endgroup\$
    – glen_geek
    Commented Feb 1, 2023 at 14:21

Noise can be modeled as a random voltage or current. Since it's not deterministic, the best we can do is model it as a distribution. The normal gaussian distribution models it pretty well.

Now, from Fourier theory, you might remember that the autocorrelation (R(0)) is equivalent to the power spectral density when you take the Fourier transform. At the same time, autocorrelation is equivalent to the variance of your distribution.

Well, Nyquist found that the noise power spectral density in a resistor is given by \$4kTR\$ volts-squared/hertz.

However, this is the power concentrated only over a 1Hz. We'd like to know the total power over the band. This can be easily obtained by multiplying the expression above by the bandwidth of interest. If you want to find the voltage spectral density, simply take the square root of the value you get.

Now, if we have a noise value in \$V^2/Hz\$ or the squared-root equivalent (not necessarily the one from the resistor), we'd like to compare that to a fix number too, not a waveform. Turns out the RMS is equivalent to the sq. root of the variance (the standard deviation), and we can get the RMS of a sine wave by simply dividing it by \$\sqrt{2}\$. To get the equivalent of the std in noise, you take the square root of your noise power in \$V^2\$ unit (after you multiply by the bandwidth).

Now we have 2 values we can compare and take the ratios to compute the SNR. They both have the same dimensions.

For your particular situation, it turns out that it's the resistor current noise that is important, assuming you have an input voltage source with its associated source impedance. The bigger your resistor is, the smaller the current noise that will develop a voltage noise due to multiplication with your source impedance.

So, in short, you should be able to measure your signal provided that you have a high SNR, i.e. that your amplifier and system have an associated low noise.


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