Understanding Windowing And Overlapping Analysis

[latexpage]The following article was inspired by a question asked by a reader regarding my previous article – Data Windows : What, why and when?. Specifically, the reader asked “Could you please put an example about overlapping technique? Maybe with the same 10 or 9.5 Hz sinusoidal wave example so that we can see how overlapped window could make a difference in processed frequency spectrum.” 

Overlap analysis, just as the name suggests, means consecutive analysis blocks of the time data are overlapped by the designated percentage of the time record.  In the case of the examples presented here, the time block is selected as 1 second corresponding to a ${\Delta}f$ of 1 Hz.

Figure 1: Initial sine wave

As stated in the previous article, analysis of a stationary signal, Figure 1, with a selected overlap of 0%  would consist of analysis of time blocks with an applied Hanning windowing function (Figure 2).

Figure 2: Hanning Window

Once we apply our window then the signal would look like that shown in Figure 3.  The spectrum would be calculated from this signal.

Figure 3: Original signal with window applied

Now, what happens if the signal is not  stationary, such as displayed in Figure 4.  You can see an event that occurs at periodic intervals (1 second intervals) that might well be important to the analysis. (NOTE: This signal was created by adding an impulsive component containing a 100 Hz sinusoid)

Figure 4: Sine wave with periodic 100Hz sinusoid added

Overlaid in red in Figure 4 is the Hanning window function. Applying this window to the signal with 0% overlap would result in the analysis signal being almost exactly the same as in Figure 3 because the Hanning window function zeros out the beginning and end of each time record. And, in our example case, this coincides with the periodic impulse.

Overlap processing (50% used for this example) would capture the events of interest (100 Hz sinusoid pulsed at 1 second intervals) because with the overlapping of the time records used to calculate the spectrum,  the Hanning window function would align with the events thus including them in the calculated spectrum. This is shown in figure 5.

Figure 5: Sine wave plus impulse with 50% overlap

The difference in the resulting spectrum is shown in Figure 6.  The blue curve is with no overlap and the red curve is with the 50% overlap as shown in the above Figures 4 & 5 respectively.

Figure 6: Spectra using 0% & 50% overlap

It is plain that with no overlap all of the 100Hz information is missing. This shows the potential for missing events by not using adequate overlap when applying a windowing function.

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John Mathey

Senior Technical Specialist (Retired) at Prosig
John Mathey graduated with a MS degree from the University of Toledo in 1972. John has over 35 years of experience with instrumentation, measurement, and analysis. Twenty-five of those years were spent at Ford Motor Company solving and providing training for vehicle noise, vibration, and harshness (NVH) issues. He was a technical specialist at Prosig USA, Inc. where he provided technical support to Prosig customers in the U.S.A.

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SeanL
SeanL
9 years ago

Thank you for the example.
With 50% overlapped window, it could pick up 100Hz information which is missed with 0% overlapped window.
Therefore, is the true information an addition of two results shown in Figure6, or I guess overlapping?
If there is 50Hz content occurring at 0.5 second with 1 second interval,
the 50% overlap window will miss this 50Hz content, but 0% window will catch.
So, how the true information will be estimated? – 0% overlap will catch10 & 50 Hz, and 50% will do 10 & 100Hz. Neither window will get true contents of 10, 50 & 100 Hz.
Thank you again,

Madhavi
Madhavi
1 year ago

Hi John,
Thank you very much for your clear explanation. I have some questions regarding the postprocessing:
1. When we consider overlapping windows, we obtain a matrix from which we can plot the spectrogram. Can you explain how you have obtained the frequency spectrum shown in figure 6?
2. Should the length of the window be equal to the length of the FFT? For example if my FFT length is 512, can my window length be 400 or so?
3. If yes, how to maintain a particular frequency resolution of the windowed signal? For example if I calculate the window length with 2 Hz frequency resolution (in my case it is important to obtain window length from frequency resolution and sampling frequency) and then my FFT length is the next power of 2, then the frequency resolution will be lesser than 2 Hz.

Thanks a lot again.

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