It is assumed that there is a low frequency signal , which contains three frequencies, , (23) where is the sampling rate. We perform FFT, apFFT, CZT and apCZT using MATLAB programs, respectively. Give the values , which are low frequencies. The values of and sampling points N are 200Hz and 3999, respectively. The resolution of traditional transforming is 0.05Hz, but the interval between 21.12Hz and 21.13Hz is 0.01Hz, which is less than 0.05Hz. Specific measurement setup processes that apCZT analyzes signal is as following block diagrams. In Fig 4, is starting frequency, is stopping frequency, is sampling rate. We set .
We perform apCZT as processes in Fig 3, and Fig 4 represents parameters of MATLAB czt function that is used in apCZT. The formula of Hanning window is written as…show more content… Fig 5 represents FFT, apFFT, CZT, apCZT and zoomFFT of (magnitudes shown, phase omitted). ig 5(a) and (b) indicate that it is hard to obtain the accurate frequencies, which is assumed. Fig 5(c) and (e) shows that there are two obvious peaks corresponding to 21.1328Hz and 21.3047Hz, respectively, while three frequencies are supposed in amplitude spectrum. However, three peaks, 21.0859Hz, 21.1563Hz and 21.3047Hz, appear in Fig 5(d), which represents amplitude spectrum using apCZT, and the frequency errors compared with theoretical values are 0.16%, 0.12% and 0.02%. ApCZT has better performance because of its three peaks.
We add random noise to x(n) and set Signal to Noise Ratio (SNR) to be -20dB. The result is as Fig
