Power spectral density

This concept is one that took me a long time to get my head around. Most sources just jump into the math without a good introduction.

A good introduction to the need for the concept of power spectral density is presented by Tustin [1]. I will paraphrase briefly. Suppose that Alice is trying to measure a random vibration. Assume that it is a stationary vibration (not changing in character with time) and that it is white (having the same amplitude at all different frequencies). We'll clarify these terms more later. Instead of a fancy spectrum analyzer, all she has is a voltmeter that reads volts RMS. For some reason, perhaps to eliminate 60 and 180 Hz electrical noise (it would be 50 and 150 in some countries), Alice puts an analog bandpass filter between her vibe pick and voltmeter. Suppose it is a passband from 95 Hz to 105 Hz. Alice reads 1V RMS on her volt meter. Suppose Bob attempts to measure the same source, but uses a different bandpass filter, say 90 Hz to 110 Hz. Then Bob reads 1.41 V RMS. What is going on here? They both measured the same vibration, but got different answers? To figure the situation, they ask Charlie to measure the vibration. He happens to have a filter from 80 Hz to 120 Hz, and he gets 2 V RMS. So which is correct? 1V, 1.41V, or 2V?

Conceptually, one should see that the three different answers are all ``correct'', as long as the experimenter states what bandwidth filter he/she used. A wider filter ``lets in'' more power, and so the meter reads a higher number. Rather that state the bandwidth all the time, engineers typically normalize their random vibration readings as follows: square the amplitude and divide by the bandwidth. In this case, Alice calculates 1²/10 = 0.1 (with units V² / Hz), Bob gets 1.41²/20 = 0.1 and Charlie gets 2²/40 = 0.1. This is called power spectral density, commonly abbreviated to PSD. When comparing PSD, all three measurements give the exact same answer.

Here is one subtle point: the choice of V²/Hz is not unique. We could have picked a different normalization if we wanted to. For example, some people use V/ $\sqrt{{Hz}}$ . This sometimes confuses people at first. V² is proportional to power, and power is a useful physical concept, so V²/Hz is straightforward enough to understand, but what does $\sqrt{{Hz}}$ really mean? Returning to our example of Alice, Bob, and Charlie, imagine that each person, instead of squaring the amplitude and dividing by the bandwidth, simply took the amplitude and then divided by the square root of the bandwidth. So Alice calculates 1/ $\sqrt{{10}}$ = 0.316 (with units V/ $\sqrt{{Hz}}$ ), Bob gets 1.41/ $\sqrt{{20}}$ = 0.316, and Charlie gets 2/ $\sqrt{{40}}$ = 0.316. Observe that $\sqrt{{0.1}}$ = 0.316. So if we use V/ $\sqrt{{Hz}}$ it's the same thing as just taking the square root of the quantity expressed in V²/Hz. The choice of V²/Hz is probably more common, but as it is a somewhat arbitrary convention, some specializations will use one more often than the other.

Now, what happens if, instead of measuring a random signal, the crew were measuring a periodic input? For example, suppose we measure a pure sine wave of 1 V RMS at 100 Hz. Then, the choice of filter bandwidth doesn't matter. 95 - 105 Hz, or 90 - 110 Hz, or 90 - 120 Hz, all of these pass the sine wave equally with no changes. The meter reads 1 V RMS in all three cases, which is what we would expect. But if we naively try to take the power spectral density, normalizing by the filter bandwidth, we get different answers in all three cases ( 0.1 V² / Hz, 0.05 V²/ Hz, and 0.025 V^2/Hz).

So then, here is a key point: the measurements for periodic vibration and random vibration are different, and you must pick the correct one. It makes no sense to talk about the power spectral density of a sine wave, and it makes no sense to talk about the RMS voltage of a random vibration.

Now, what if our company upgrades and Alice, Bob, and Charlie now have fancy FFT spectrum analyzers instead of analog bandpass filters and voltmeters. Do we still need to worry about PSD? Yes. Let's say that all three are using digital spectrum analyzers with a rate of 1024 S/s. They all start measuring the exact same random, white-noise signal at the exact same time, but Alice records 16 seconds worth of data and takes a 16384 point FFT, Bob records 2 seconds and takes 2048 point FFT, and Charlie only records 0.25s worth of data for a 256 point FFT. Do all of the signals have the same power? Because we are talking about a stationary random signal, the answer is yes (we'll define that term more in a second). But if they have their analyzers set for Fourier spectrum magnitude, they will get different answers. Look at the left graph in figure 24.

Why does this happen? Remember that a DFT is like a set of bandpass filters (6.4) where the width of the filter depends on the length of the window. So Alice's FFT has a large number of narrow filters, while Charlie's FFT has a smaller number of wide filters. In the middle graph, the PSD has been computed by squaring the amplitude in each FFT line and dividing by the FFT bin width. This process of normalizing an FFT is called the periodogram. The three lines now fall almost on top of each other, although the plots are very jagged. Wasn't there supposed to be the same amplitude at each frequency (i.e. white noise)? This leads to the next section

**Figure 24:** a) time history of a simulated random signal b) FFT magnitude of the signal in (a). c) Power spectral density estimated by the periodogram (squaring the FFT and normalizing by bin width). Note that the estimate is very noisy, and does not get any better as the sample length gets longer. A 16s sample is just as noisy as a 0.25s sample. d) Power spectral density estimated by Barlett's method. The estimate is much cleaner, and now the estimate does improve with more samples. e) PSD estimated by Welch's method. A slight improvement on Bartlett's method.
$\includegraphics[bb=0bp 150bp 575bp 700bp,clip]{fig7A}$