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Issue 292: EQ Answers

Problem 1—Let’s talk about noise! There are different types of noise that might be present in a system, and it’s important to understand how to deal with them.

For example, analog sensors and other types of active devices will often have AWGN, or Additive White Gaussian Noise, at their outputs. Any sort of analog-to-digital converter will add quantization noise to the data. What is the key difference between these two types of noise?

Answer 1—The key difference between AWGN and quantization noise is the PDF, or Probability Density Function, which is a description of how the values (voltage or current levels in analog systems, or data values in digital systems) are distributed.

The values from AWGN have a bell-shaped distribution, known variously as a Gaussian or Normal distribution. The formula for this distribution is:292-EQ-equation

µ represents the mean value, which we take to be zero in discussions about noise. σ is known as the “standard deviation” of the distribution, and is a way to characterize the “width” of the distribution.

It looks like this:


Source: Wikipedia (

While the curve is nonzero everywhere (from –∞ to +∞) it is important to note that the values will be within ±1 σ of the mean 68% of the time, within ±2 σ of the mean 95% of the time, and within ±3 σ of the mean 99.7% of the time. In other words, although the peak-to-peak value of this kind of noise is theoretically infinite, you can treat it as being less than 4σ 95% of the time.


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On the other hand, the values from quantization noise have a uniform distribution — the values are equally probable, but only over a fixed span that’s equal to the quantization step size of the converter. The peak-to-peak range of this noise is equal to the converter’s step size (resolution).

However, it’s important to note that both sources of noise are “white”, which is a shorthand way of saying that their effects are uniformly distributed across the frequency spectrum.

Problem 2—Signal-to-noise ratios are most usefully described as power ratios. How does one characterize the power levels for both AWGN and quantization noise?

Answer 2—The power of a noise signal is proportional to the square of its RMS value.

The RMS value of AWGN is numerically equal to its standard deviation.

The RMS value of quantization noise is simply the peak-to-peak value (the step size of the converter) divided by √12, or VRMS = 0.2887 VPP. This is easily derived if you characterize the quantization noise signal as a small sawtooth wave that gets added to the analog signal.

Question 3—When you have multiple sources of noise in a system, how can you characterize their combined effect on the overall system performance?

Answer 3—When combining noise sources, you can’t simply add their RMS voltage or current values together. From one sample to the next, one noise source might partially cancel the effects of the other noise source(s).

Instead, you add the individual noise power levels to come up with an overall noise power level. Since power is proportional to voltage (or current) squared, this means that you need to square the individual RMS measurements, add them together, and then take the square root of the result in order to get an equivalent overall RMS value.

VRMS(total) = √(VRMS(n1)2 + VRMS(n2)2 + …)

Problem 4—Broadband analog sensors and other active devices often specify their noise levels in units of “microvolts per root-Hertz” (µV/√Hz) or “nanoamps per root-Hertz” (nA/√Hz). Where does this strange unit come from, and how do you use it?


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Answer 4—As described in the previous answer, uncorrelated noise sources are added based on their power. With AWGN, the noise in one “segment” of the frequency spectrum is not correlated with another segment of the spectrum, so if you have a particular voltage level of noise in, say, a 1-Hz band of frequencies, you’ll have √2 times as much noise in a 2-Hz band of frequencies. In general, the RMS noise level for any particular bandwidth is going to be proportional to the square root of that bandwidth, which is why the devices are characterized that way.

So, if you have an opamp that’s characterized as having a noise level of 2 µV/√Hz, and you want to use this in an audio application with a bandwidth of 20 kHz, the overall noise at the output of the opamp will be 2 µV × √20000, or about 283 µVRMS. If your signal is a sinewave with a peak-to-peak value of 1V (353 mVRMS), you’ll have a signal-to-noise ratio of about 124 dB.

Contributed by David Tweed

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Issue 292: EQ Answers

by Circuit Cellar Staff time to read: 3 min