Short-time Fourier transform

Remarks

The integral converges absolutely, since $f \cdot \overline{g(\cdot - x)}$ is integrable by the Cauchy–Schwarz inequality. So $V_g f(x, \cdot)$ is the Fourier transform of $f \cdot \overline{g(\cdot - x)}$. In inner product form, $V_g f(x, \omega) = \ip{f}{M_\omega T_x g}$, where $T_x g(t) = g(t - x)$ and $M_\omega g(t) = e^{2\pi i \omega t} g(t)$. Moyal's formula states $\norm{V_g f}_{L^2(\R^2)} = \norm{f}_2 \norm{g}_2$. The squared modulus $\abs{V_g f}^2$ is the spectrogram. In engineering texts the variable $\omega$ often denotes angular frequency, and the kernel is $e^{-i \omega t}$.

Example: let $f = g = 1_{[0, 1]}$. For $0 \le x \le 1$, the product $f \cdot g(\cdot - x)$ is the indicator of $[x, 1]$, so $V_g f(x, 0) = 1 - x$. For $\abs{x} > 1$ the supports are disjoint and $V_g f(x, \omega) = 0$.

Non-example: with the constant window $g = 1$, which is not in $L^2(\R)$, the formula gives $\hat f(\omega)$ for every $x$. It carries no time information, and this choice is excluded by the definition.