Why do we get ripples in short time fourier transform?
Answers
The short-time Fourier transform (STFT), is a Fourier-related transform used to determine the sinusoidal frequency and phase content of local sections of a signal as it changes over time.
In practice, the procedure for computing STFTs is to divide a longer time signal into shorter segments of equal length and then compute the Fourier transform separately on each shorter segment.
This reveals the Fourier spectrum on each shorter segment. One then usually plots the changing spectra as a function of time.
It can also be explained with reference to the sampling and Nyquist frequency.
Take a window of N samples from an arbitrary real-valued signal at sampling rate fs . Taking the Fourier transform produces N complex coefficients. Of these coefficients only half are useful (the last N/2 being the complex conjugate of the first N/2 in reverse order, as this is a real valued signal).
These N/2 coefficients represent the frequencies 0 to fs/2 (Nyquist) and two consecutive coefficients are spaced apart by fs/N Hz.
To increase the frequency resolution of the window the frequency spacing of the coefficients needs to be reduced. There are only two variables, but decreasing fs (and keeping Nconstant) will cause the window size to increase — since there are now fewer samples per unit time. The other alternative is to increase N, but this again causes the window size to increase. So any attempt to increase the frequency resolution causes a larger window size and therefore a reduction in time resolution—and vice versa.
TY PLZ MARK MY BRAINLIST