Translations:Phase rotation/31/en

Let us look at an example. The fast Fourier transform (FFT) uses the same number of sample points in both the time and frequency domains. Let the number of sample points be $N={64}$ (note that N is a power of 2). Suppose that the sampling interval is dt = 0.004 ms, in which case the Nyquist frequency is $f_{n}={1/}\left(2dt\right)={125}$. The Nyquist range is $2f_{n}={250Hz}$. The frequency sampling interval is $df={250/64=3.90625}$ Hz. Let the envelope spectrum be a rectangular spectrum of amplitude 1 centered at the origin, with a total bandwidth equal to one-eighth of the Nyquist range. The Nyquist range is ${250=64}df$, so one-eighth of the Nyquist range is 8 df, and the bandwidth is 8 $df={31.25}$ Hz. One-half of the bandwidth is 4 $df={15.625}$ Hz. The envelope spectrum is equal to 1 within the passband from the lower frequency –15.625 Hz to the upper frequency 15.625 Hz, and it is equal to 0 outside this passband (Figure 3b). The inverse Fourier transform of the envelope spectrum gives a symmetric envelope wavelet (Figure 3a). In fact, the envelope is called a positive envelope because its major peak is positive.