# frequencies in STFT spectrogram

Question

The help file info for STFT Spectrogram says, in part:
STFT Spectrogram {X} is a 2D array that describes the time waveform energy
distribution in the joint time-frequency domain. The number of rows (time
axis) in STFT Spectrogram {X} is given by the following equation.
# rows = # elements in X / time increment,  then rounded down
The number of columns (frequency axis) in STFT Spectrogram {X} is given by
the following equation.
# columns = (# elements in X / 2) + 1,  then rounded down
By experimenting I have confirmed that the help file is correct, but I am
surprised that the number of columns (i.e. frequencies in the STFT) is what
it is.  If the window size is N (I assume that means N time points are
used to compute each spectrum estimate), then the number of frequencies in the
spectrum should be N/2 + 1, even if the number of elements is much
larger.  Can someone clarify? 
Thank you.
Maybe all the points in X outside the N point window ARE used, but set to zero. 
In essence this would amount to zero-padding an N point long signal to make
it as long as the original signal.  Why would this be a good idea?  It increases
the computational effort with no increase in actual information about the signal. 
It also can make one think the spectrum has a much
higher frequency resolution than it really does have.  The spectrum of a zero-padded
time series has no more "real" frequency resolution than the unpadded series -
it is just a resampling, with interpolation, of the spectrum of the unpadded signal.

Answer 1

That's an interesting question. If you open up the STFT Spectrogram Vi to look at the algorithm being employed, you can look at exactly what process is going on. It does appear as though a type of zero-padding is being used.
Zero-padding before the FFT to a power of 2 would actually decrease processing time due to the FFT algorithm used, but that doesn't seem to be the rationale here.
Jarrod S.
National Instruments