Frequency resolution and time resolution

UltraVox XT applies the Fast Fourier Transform algorithm applied to SFT, for this reason the frame length N is always a power of 2 (64, 128,..., 2048). The higher N, the smaller the frequency spacing df.

If the SFT frame is 2048 samples long, the SFT analysis gives you 2048 equally-spaced frequency bins from 0 Hz up to the sampling frequency divided by two. Increasing SFT length means to reduce the spacing of frequency bins according to the formula df = Sampling frequency/SFT length. This increases the waveform frequency resolution.

However, any timing resolution that occurs within a SFT frame is lost in the analysis, since all temporal changes are lumped together in a single frame. If SFT length is 2048, the frame duration is, at a sampling frequency of 384 kHz, 2048/384000 = 5.33 ms. This means that the first spectrum is created at t= 0, and the next spectrum is created at t=5.33 ms. Two subsequent events occurring within this time would not be distinguishable.

The following table shows the inverse relation between frequency resolution df (=sampling frequency / SFT length) and time resolution dt (=SFT length/sampling frequency), when sampling frequency is 384 kHz.

SFT length (samples)

df (=384000/SFT length) (Hz)

dt (=SFT length/384000 (ms)

64

6000

0.17

128

3000

0.33

256

1500

0.67

512

750

1.33

1024

375

2.67

2048

187.5

5.33

The time resolution dt is the width of the spectrogram pixel, when Overlap = 0.

If you increase the SFT length, there will be more “pixels” along the frequency axis of the spectrogram, however they are now “wider”, since they represent a longer time interval; the higher the SFT, the longer this time. This relationship represents the trade-off between frequency resolution and time resolution.

See also

The following videos:

Short-time Fourier Transform and the Spectrogram

http://www.youtube.com/watch?v=NA0TwPsECUQ

FFT basic concepts

http://www.youtube.com/watch?v=z7X6jgFnB6Y