Total energy of a causal impulse response (tricky integral) The total energy of a causal impulse response is given by:
$$
\begin{align}
E
&= {1 \over 2 \pi} \int_{-\pi}^{\pi} \left|{H(e^{i\omega})}\right|^2d\omega
\\
&= {1 \over 2 \pi i} \oint H(z)H(z^{-1})z^{-1}dz
\end{align}
$$
Where $z = e^{-i\omega}$ and $H(z)$ is given by:
$$
H(z) = {b_0 + b_1 z^{-1} + b_2 z^{-2} \over 1 + a_1 z^{-1} + a_2 z^{-2}}
$$
Apparently the determination of this integral in the z-domain is known, does anyone know what it is?

Use the residue theorem. Note that the poles of $H(z)$ and of $H(z^{−1})$ are reciprocals of each other, so unless they're on the unit circle (in which case the integral would be undefined), two are inside and two are outside (and $z^{−1}$ contributes a third pole at $z=0$, with residue $b_0b_2/a_2$). I don't see any way to make use of the special structure of the problem beyond that.
