# Proof of Paley-Wiener criterion for causality

The Paley-Wiener criterion for causality is that $\displaystyle\int_{\mathbb{R}}\frac{A(\omega)}{1 + \omega^2}\mathrm{d}\omega$ exists and is finite, where $A(\omega) = \left|\mathcal{F}[f]\right|$ is the modulus of the Fourier transform of a function $f: \mathbb{R}\to\mathbb{R}$

How is this derived from the usual statement of the Paley-Wiener theorem?

I have also seen that this proves that a causal filter can't provide total rejection for a band of frequencies, but this appears to me to follow directly from the usual statement that the Fourier transform of a square-integrable distribution vanishing for negative reals is holomorphic and the identity theorem for holomorphic functions.

This is nice question which I have recently encountered while studying ideal filters. Suppose transfer function $H(p)$ which is complex rational and holomorphic (or meromorphic?) in $p = \sigma + j\omega$ and satisfy Cauchy-Riemann conditions.

We can rewrite transfer function as $H(p) = e^{-\Gamma(p)}=10^{-\frac{1}{20}\Gamma(p)}$ with respect to nepers and decibels. Since exponetital function is holomorphic so the $\Gamma$ is. If we restrict ourselves to imaginary axis $p = j\omega$ we have

$\Gamma(j\omega) = A(j\omega) + jB(j\omega)$

where $A(j\omega)$ is an attentuation charateristics and $B(j\omega)$ is a phase characteristics. In general we have identity $\ln z = \ln |z| + j\arg z$ for any $z \in \mathbb{C}$. So we can write

$A(j\omega) = \ln|H(j\omega)|$ and $B(j\omega)=\arg H(j\omega)=\arctan\frac{\Im\lbrace H(j\omega)\rbrace}{\Re\lbrace H(j\omega)\rbrace}$.

Then consider pair of Hilbert transforms given by integrals (Cauchy principal value are taken in each of the integrals)

\begin{equation} A(\omega) = \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{B(j\xi)}{\omega - \xi}\,\mathrm{d}\xi \quad\quad B(\omega) = -\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{A(j\xi)}{\omega - \xi}\,\mathrm{d}\xi \end{equation}

Evaluating these integrals for module characteristic of ideal low pass filter - say let $A(\omega)$ is constant at $a$ from $-\omega_1$ to $\omega_1$ (module characteristics are symetric) gives \begin{equation} B(\omega) = - \frac{a}{\pi}\ln\frac{|\omega + \omega_1|}{|\omega - \omega_1|} \end{equation} Thus $B(\omega)$ has points in infinity at $\pm\omega_1$.

See picture bellow: Similarly for linear phase characteristics $B(\omega) = a\omega$.

This is saying to us that we can't reach filter transfer function $H(p)$ with both ideally rectangular module characteristics and ideally linear phase characteristics at the same time.

We see that $B(\omega)$ with points at $\pm \infty$ can't realise any phase characteristics $\arg H(j\omega)$ of a given feasible filter. That's the reason why \begin{equation} -\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{A(j\xi)}{\omega - \xi}\,\mathrm{d}\xi < \infty \end{equation} must converge and so the \begin{equation} -\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{|A(j\xi)|}{\omega^2 + \xi^2}\,\mathrm{d}\xi < \infty \end{equation} must.