# How to interpret spectrogram correctly?

I have a pressure transducer signal from a reacting mixture available here, which has a sampling frequency of 50kHZ. I have also obtained the spectrogram of the signal with NFFT=256, noverlap=4 as shown below: I expected to see one high power frequency after pressure rise, instead of multiple frequencies (after t=0.03s). Why are multiple frequencies detected? Am I using the correct NFFT and noverlap parameters? How are NFFT and noverlap selected for this signal?

• Is the power spectral density legend inverted ? Because it says darker red = lower PSD while it seems to be the contrary on the spectrogram Feb 29, 2016 at 10:02
• No, they are not inverted. Feb 29, 2016 at 10:37
• The cylinder pressure line graph shows a lot of peaks too. The thing is ringing like a church bell. Mar 1, 2016 at 16:17

The problem is not the spectrogram parameters, these are correct since they only depend on what resolution you want in time and frequency domain. Also, the spectrogram interpretation is correct, there are multiple frequency peaks. The problem may be:

I expected to see one high power frequency after pressure rise, instead of multiple frequencies

Why? If you are not in free-field conditions (and you mention a cylinder) there will be multiple wave reflections, resonances, standing waves and 3D pressure patterns. A sudden pressure step will distribute its power over the spectrum, but frequency peaks will appear because you are essentially "tuning" your cylinder.

Recall that pressure waves travel at sound speed. Considering for example the habitual 340 m/s; if your cylinder measures 1m, the wave front will travel back and forth each 0.0029s, adding multiple frequency components to the measurement.

As for standing wave patterns, we can consider the case of a closed cylinder, where we can compute the frequency of the fundamental frequency as $f_1 = \frac{a}{4L}$ with subsequent harmonics at $nf_1$ where $n$ are odd integers (if your cylinder is indeed closed, does it have a length of ~20cm?). Regarding the problem of measuring higher frequency waves in ducts, consider that according to Eriksson, it may be assumed that waves in a circular duct propagate in a planar fashion (i.e., pressure is constant in every cross-sectional area of the duct) only below a cut-off frequency of: $$f = 1.84 \frac{a}{\pi D}\sqrt{1-M^2}$$

Where a is the sound speed, D the duct diameter and M the Mach number of the mean flow. Above this frequency, the pressure distribution in a given section of the duct will not be uniform, it will have nodes. See the following animation for an idea on how pressure varies in a given planar section, noting that only the Z(0,0) is a typical planar wave. Beyond the planar wave frequency it is difficult to correctly measure the pressure spectrum with only one transducer. In any case, I believe that the signal processing that you are performing is correct; the only problem is that you are witnessing the acoustic effects of your cylinder.

• Great explanation! In this case, the fundamental frequency would be 3000 Hz, right? or would it be the 3rd harmonics? The cylinder is closed and has a shape given in this video youtube.com/watch?v=rksoa_JG7lo Mar 1, 2016 at 18:00
• @nxkryptor It appears so! But note that these expressions are for a straight pipe, in a complex geometry like yours there could be multiple reflections/standing waves. As I see you're dealing with CFD in diesel combustion chambers, maybe this paper from some colleagues is useful regarding their acoustic properties: dx.doi.org/10.1108/02644400710718583 Best of luck! :) Mar 2, 2016 at 14:26

The answer from @orgeGT is quite detailed. It really looks like, to me, a pressure signal from a knocking gazoline engine, where you observe a wide band effect (the impulse part of the knock) and resonant frequencies and their harmonics (the wiggling part of the knock), and a (slight) decay in their amplitude, related to the variations of the volume of the cylinder volume, and the change in temperature. While the knock phenomenon may be unrelated to your, you can check the literature, as signal analysis sounds close

For plotting spectrogram, size of the window, and percentage overlap depends on the signal you are using, usually window length is taken as the time till which you believe that the signal will be stationary.

if window length is small(number of samples) than you will get better resolution in time domain. and if window length is large you will get more resolution in frequency domain.

I have plotted the spectrogram of your signal for various window lengths(with 50% overlap), I am always getting multiple frequency components. 