Re: Improving Frequency Resolution in Time Constrained Signals

On Oct 30, 8:22 am, "tzoom84" <tszumow...@xxxxxxxxx> wrote:
For simplicity, suppose we are interested in detecting presence of either a
500Hz, 1kHz, and a 1.5kHz sinusoid in a signal that is being sampled at
100kHz. But lets say we only have 100 samples, or 1ms of data. This is
anywhere between 0.5 to 2 'cycles' of the sinusoids. So essentially, the
signal is well oversampled, but is not sampled long enough in time to
provide clean frequency resolution.

Question is: Are there particular techniques to improve frequency
resolution for oversampled, but time constrained, signals?

(One random thought considered was to decimate at different points in
time, say take every tenth sample starting at sample #1, then every tenth
starting at sample #2, to create 10 independent signals of same length.
Then concatenate to artificially lengthen the same signal. But I wasn't
sure if that would increase resolution since it doesn't add any new


With a sample rate = 100k and N = 100, you’ve got 100k/100 = 1k bin
spacing in the frequency domain (e.g.: DFT frequency index k = 0
corresponds to 0Hz, index k = 1 corresponds to 1kHz, index k = 2 is
2kHz, etc.). So perhaps you’d do 3 DFTs, one at k = .5 (corresponds
to 500 Hz), one at k = 1 (corresponds to 1 kHz), and one at k = 1.5
(corresponds to 1.5 kHz). Then compute magnitude squared of each and
find the largest. Alternately, you could zero pad an FFT to 200
points and use outputs k = 1, 2 and 3 to represent f = 500 Hz, 1 kHz
and 1.5 kHz. Keep in mind, of course, that zero padding doesn’t
really improve resolution - it provides an interpolation. If you
start with a sampling interval that’s too short, you get a lousy
result. Zero padding gives you more (interpolated) points of a lousy

Of course, I’m presuming that only one tone is present at a time,
their amplitudes are the same, and you don’t have a transition between
the tones in your 100 point sampling interval.

There is a technique that will give you very good frequency
resolution, but it requires that you have at least a few bins’ spacing
between your tonal inputs. So you’re operating well below the margins
at which it will work.

Kevin McGee