A Short-Time Fourier Transform

A.1 Parseval’s formula

Parseval’s formula for the Fourier transform can be transposed to the STFTs of two signals.

Theorem A.1 (Parseval's formula) Consider the STFT of two signals \(s1\) and \(s2\) taken over the windows \(w_1\) and \(w_2\) respectively. \[\begin{equation} \left\langle V_{w_1}s_1,V_{w_2}s_2\right\rangle_{L^2(\mathbb{R}^2)} = \left\langle s_1,s_2\right\rangle_{L^2(\mathbb{R})} \overline{\left\langle w_1,w_2\right\rangle}_{L^2(\mathbb{R})} \end{equation}\]

Proof. In order to prove Parseval’s formula, we use Fubini’s theorem to interchange the order of integration, we assume that \(\overline{w_1}w_2\in L^1(\mathbb{R})\). We also make use of the fact that \[\begin{equation} \delta(t)=\int_\mathbb{R}e^{-2\pi iwt}\mathrm{d}\omega,\quad\forall t\in\mathbb{R} \end{equation}\] where \(\delta\) is the Dirac distribution \[\begin{align} \left\langle V_{w_1}s_1,V_{w_2}s_2\right\rangle_{L^2(\mathbb{R}^2)} &= \int_{\mathbb{R}^2} \left\langle s_1,M_\omega T_\tau w_1\right\rangle \overline{\left\langle s_2,M_\omega T_\tau w_2\right\rangle} \mathrm{d}\tau\mathrm{d}\omega\\ &= \int_{\mathbb{R}^2} \int_\mathbb{R}s_1(t)\overline{M_\omega T_\tau w_1(t)}\mathrm{d}t \int_\mathbb{R}\overline{s_2(t')} M_\omega T_\tau w_2(t') \mathrm{d}t' \mathrm{d}\tau\mathrm{d}\omega\\ &= \int_{\mathbb{R}^4} s_1(t)\overline{T_\tau w_1(t)} \overline{s_2(t')} T_\tau w_2(t') e^{-2\pi i\omega(t-t')} \mathrm{d}t\mathrm{d}t'\mathrm{d}\tau\mathrm{d}\omega\\ &= \int_{\mathbb{R}^3} s_1(t)\overline{T_\tau w_1(t)} \overline{s_2(t')} T_\tau w_2(t') \overline{\int_\mathbb{R}e^{2\pi i\omega(t-t')}\mathrm{d}\omega} \mathrm{d}t\mathrm{d}t'\mathrm{d}\tau\\ &= \int_{\mathbb{R}^3} s_1(t)\overline{T_\tau w_1(t)} \overline{s_2(t')} T_\tau w_2(t') \overline{\delta(t-t')} \mathrm{d}t\mathrm{d}t'\mathrm{d}\tau\\ &= \int_{\mathbb{R}^2} s_1(t)\overline{T_\tau w_1(t)} \int_\mathbb{R}\overline{s_2(t')} T_\tau w_2(t') \overline{\delta(t-t')}\mathrm{d}t'\mathrm{d}t\mathrm{d}\tau\\ &= \int_{\mathbb{R}^2} s_1(t)\overline{T_\tau w_1(t)} \overline{s_2(t)} T_\tau w_2(t)\mathrm{d}t\mathrm{d}\tau\\ &= \int_\mathbb{R}s_1(t)\overline{s_2(t)} \int_\mathbb{R}\overline{T_\tau w_1(t)} T_\tau w_2(t)\mathrm{d}\tau\mathrm{d}t\\ &= \int_\mathbb{R}s_1(t)\overline{s_2(t)} \int_\mathbb{R}\overline{w_1(t-\tau)} w_2(t-\tau)\mathrm{d}\tau\mathrm{d}t\\ &= \left\langle s_1,s_2\right\rangle_{L^2(\mathbb{R})} \overline{\left\langle w_1,w_2\right\rangle}_{L^2(\mathbb{R})} \end{align}\]

which implies that \(\left\lVert V_w s\right\rVert_2 = \left\lVert s\right\rVert_2 \left\lVert w\right\rVert_2\), and for \(\left\lVert w\right\rVert_2=1\) we have \[\begin{equation}\label{eq:isometry} \left\lVert V_w s\right\rVert_2 = \left\lVert s\right\rVert_2,\quad s\in L^2(\mathbb{R}) \end{equation}\]

This means that the STFT operator \(V_w\) is an isometry from \(L^2(\mathbb{R})\) to \(L^2(\mathbb{R}^2)\) if \(\left\lVert w\right\rVert_2=1\).

A.2 Inverse Short-Time Fourier Transform

Theorem A.2 (STFT inversion formula) Let \(w,h\in L^2(\mathbb{R})\) with \(\left\langle w,h\right\rangle\neq0\). Then for all \(s\in L^2(\mathbb{R})\). \[\begin{equation} s(t) = \frac{1}{\left\langle w,h\right\rangle} \iint_{\mathbb{R}^2}V_w s(\tau,\omega)M_\omega T_\tau h(t) \mathrm{d}\omega\mathrm{d}\tau \end{equation}\]

Proof. Since \(V_w s\in L^2(\mathbb{R}^2)\), let \(\tilde s\in L^2(\mathbb{R})\) be defined as \[\begin{equation} \tilde s(t) = \frac{1}{\left\langle w,h\right\rangle} \iint_{\mathbb{R}^2}V_w s(\tau,\omega)M_\omega T_\tau h(t) \mathrm{d}\omega\mathrm{d}\tau \end{equation}\] We therefore have, for all \(\varphi\in L^2(\mathbb{R})\), using Parseval’s formula \[\begin{align} \left\langle\tilde s, \varphi\right\rangle_{L^2(\mathbb{R})} &= \frac{1}{\left\langle w,h\right\rangle} \iint_{\mathbb{R}^2} V_w s(\tau,\omega) \left\langle M_\omega T_\tau h, \varphi\right\rangle_{L^2(\mathbb{R})} \mathrm{d}\omega\mathrm{d}\tau\\ &= \frac{1}{\left\langle w,h\right\rangle} \iint_{\mathbb{R}^2} V_w s(\tau,\omega) \overline{\left\langle\varphi,M_\omega T_\tau h\right\rangle}_{L^2(\mathbb{R})} \mathrm{d}\omega\mathrm{d}\tau\\ &= \frac{1}{\left\langle w,h\right\rangle} \iint_{\mathbb{R}^2} V_w s(\tau,\omega) \overline{V_h \varphi(\tau,\omega)} \mathrm{d}\omega\mathrm{d}\tau\\ &= \frac{1}{\left\langle w,h\right\rangle} \left\langle V_w s, V_h \varphi\right\rangle_{L^2(\mathbb{R}^2)}\\ &= \left\langle s, \varphi\right\rangle_{L^2(\mathbb{R})} \end{align}\]

Since \(\left\langle\tilde s,\varphi\right\rangle_{L^2(\mathbb{R})}=\left\langle s,\varphi\right\rangle_{L^2(\mathbb{R})},\forall\varphi\in L^2(\mathbb{R})\) then in the weak sense \(\tilde s = s\).

It is worth noting that this proof is better expressed with concepts of operator algebra. Let’s consider \(F\) a function on \(E\) such that \(F(x)\in B\) forall \(x\in E\), where \(B\) is a Banach space, we therefore have \[\begin{equation} f = \int_E F(x)\mathrm{d}x\implies \left\langle f,\varphi\right\rangle = \int_E \left\langle F(x),\varphi\right\rangle \mathrm{d}x \end{equation}\] for all \(\varphi\in B^*\) where \(B^*\) is the dual space to \(B\). Moreover, if the mapping \(\ell(\varphi)\mapsto\int_E \left\langle F(x),\varphi\right\rangle\mathrm{d}x\) is a bounded linear operator on \(B^*\), then \(\ell\) defines a unique element \(f\in B^{**}\). If \(B\) is a reflexive Banach space \(\left(B^{**}=B\right)\), we have a \(f\in B\). In our case, \(E=\mathbb{R}^2\) and \(F\in L^2(\mathbb{R}^2)\). Let \(f\) be defined as \[\begin{equation} f = \iint_{\mathbb{R}^2} F(\tau,\omega)M_\omega T_\tau \gamma \mathrm{d}\omega\mathrm{d}\tau \end{equation}\] The operator \(\ell\) is then defined as \[\begin{equation} \ell(\varphi) = \iint_{\mathbb{R}^2} F(\tau,\omega)\left\langle M_\omega T_\tau \gamma,\varphi\right\rangle_{L^2(\mathbb{R})} \mathrm{d}\omega\mathrm{d}\tau = \iint_{\mathbb{R}^2} F(\tau,\omega) \overline{\left\langle V_\gamma \varphi\right\rangle}_{L^2(\mathbb{R})} \mathrm{d}\omega\mathrm{d}\tau \end{equation}\] We prove \(\ell\) is a bounded operator on \(B^*=L^2(\mathbb{R})\) with the help of the Cauchy-Schwartz inequality and Parseval’s formula.

\[\begin{equation} \forall\varphi\in L^2(\mathbb{R}),\left\lvert\ell(\varphi)\right\rvert \leq \left\lVert F\right\rVert_2 \left\lVert V_\gamma \varphi\right\rVert_2 = \left\lVert F\right\rVert_2 \left\lVert\gamma\right\rVert_2 \left\lVert\varphi\right\rVert_2 \end{equation}\] Which implies that \(\ell\) defines a unique function \(f\in L^2(\mathbb{R})\) such that \(\left\langle f,\varphi\right\rangle_{L^2(\mathbb{R})} = \ell(\varphi),\forall\varphi\in L^2(\mathbb{R})\).