Summary of STFT Computation Using the FFT

1. Read $M$ samples of the input signal $x$ into a local buffer of length $N\geq M$ which is initially zeroed
   $$\tilde{x}_m(n) \isdef x(n+mR), \qquad n=-M_h, \ldots\,,-1,0,1,\ldots\,, M_h$$
   where $x_m$ is called the $m$th frame of the input signal, and $M\isdef 2M_h+1$ is the frame length (which we assume is odd for reasons to be discussed later). The time advance $R$ (in samples) from one frame to the next is called the hop size or step size.

2. Multiply the data frame pointwise by a length $M$ spectrum analysis window $w(n), n=-M_h,\ldots\,,M_h$ to obtain the $m$th windowed data frame:
   $$\tilde{x}_m^w(n) \isdef \tilde{x}_m(n) w(n), \qquad n=-M_h,\ldots\,,M_h$$

3. Extend $\tilde{x}_m^w$ with zeros on both sides to obtain a zero-padded windowed data frame:
   $$\tilde{x}_m^{w,z}(n) \isdef \left\{\begin{array}{ll} \tilde{x}_m^... ...frac{N}{2}}-1 \\ [5pt] 0, & -{\frac{N}{2}}\leq n < -M_h \\ \end{array}\right.$$
   where $N$ is chosen to be a power of two larger than $M$. The number $N/M$ is the zero-padding factor; as discussed in §4.3, the zero-padding factor is the interpolation factor for the spectrum, i.e., each FFT bin is replaced by $N/M$ bins, interpolating the spectrum using ideal bandlimited interpolation [243], where the ``band'' in this case is the $M$-sample nonzero duration of $\tilde{x}_m^w$ in the time domain.

4. Take a length $N$ FFT of $\tilde{x}_m^{w,z}$ to obtain the STFT at time $m$:
   $$\tilde{X}_m^{w,z}(e^{j\omega_k })=\sum _{n=-N/2}^{N/2-1} \tilde{x}_m^{w,z}(n) e^{-j\omega_k n T}$$
   where $\omega_k = 2\pi k f_s / N$, and $f_s=1/T$ is the sampling rate in Hz. The STFT bin number is $k$.