Lossless Analog Filters

As discussed in §B.2, the an allpass filter can be defined as any filter that preserves signal energy for every input signal $x(t)$ . In the continuous-time case, this means

$$\left\Vert\,x\,\right\Vert _2^2 \isdef \int_{-\infty}^\infty \left\vert x(t)\right\vert^2 dt = \int_{-\infty}^\infty \left\vert y(t)\right\vert^2 dt \isdef \left\Vert\,y\,\right\Vert _2^2$$

where $y(t)$ denotes the output signal, and $\left\Vert\,y\,\right\Vert$ denotes the $L_2$ norm of $y$ . Using the Rayleigh energy theorem (Parseval's theorem) for Fourier transforms [87], energy preservation can be expressed in the frequency domain by

$$\left\Vert\,X\,\right\Vert _2 = \left\Vert\,Y\,\right\Vert _2$$

where $X$ and $Y$ denote the Fourier transforms of $x$ and $y$ , respectively, and frequency-domain $L_2$ norms are defined by

$$\left\Vert\,X\,\right\Vert _2 \isdef \sqrt{\frac{1}{2\pi}\int_{-\infty}^\infty \left\vert X(j\omega)\right\vert^2 d\omega}.$$

If $h(t)$ denotes the impulse response of the allpass filter, then its transfer function $H(s)$ is given by the Laplace transform of $h$ ,

$$H(s) = \int_0^{\infty} h(t)e^{-st}dt,$$

and we have the requirement

$$\left\Vert\,X\,\right\Vert _2 = \left\Vert\,Y\,\right\Vert _2 = \left\Vert\,H\cdot X\,\right\Vert _2.$$

Since this equality must hold for every input signal $x$ , it must be true in particular for complex sinusoidal inputs of the form $x(t) = \exp(j2\pi f_xt)$ , in which case [87]

$$\begin{eqnarray*} X(f) &=& \delta(f-f_x)\\ Y(f) &=& H(j2\pi f_x)\delta(f-f_x), \end{eqnarray*}$$

where $\delta(f)$ denotes the Dirac ``delta function'' or continuous impulse function (§E.4.3). Thus, the allpass condition becomes

$$\left\Vert\,X\,\right\Vert _2 = \left\Vert\,Y\,\right\Vert _2 = \left\vert H(j2\pi f_x)\right\vert\cdot\left\Vert\,X\,\right\Vert _2$$

which implies

$$\left\vert H(j\omega)\right\vert = 1, \quad \forall\, \omega\in(-\infty,\infty). \protect$$ (E.13)

Suppose $H$ is a rational analog filter, so that

$$H(s) = \frac{B(s)}{A(s)}$$

where $B(s)$ and $A(s)$ are polynomials in $s$ :

$$\begin{eqnarray*} B(s) &=& b_M s^M + b_{M-1}s^{M-1} + \cdots + b_1 s + b_0\\ A(s) &=& s^N + a_{N-1}s^{N-1} + \cdots + a_1 s + a_0 \end{eqnarray*}$$

(We have normalized $B(s)$ so that $A(s)$ is monic ($a_N=1$ ) without loss of generality.) Equation (E.13) implies

$$\left\vert A(j\omega)\right\vert = \left\vert B(j\omega)\right\vert, \quad \forall\, \omega\in(-\infty,\infty). \protect$$

If $M=N=0$ , then the allpass condition reduces to $\vert b_0\vert=\vert a_0\vert=1$ , which implies

$$b_0 = e^{j\phi} a_0 = e^{j\phi}$$

where $\phi\in[-\pi,\pi)$ is any real phase constant. In other words, $b_0$ can be any unit-modulus complex number. If $M = N = 1$ , then the filter is allpass provided

$$\left\vert b_1j\omega + b_0\right\vert = \left\vert j\omega + a_0\right\vert, \quad \forall\, \omega\in(-\infty,\infty).$$

Since this must hold for all $\omega$ , there are only two solutions:

1. $b_0=a_0$ and $b_1=1$ , in which case $H(s)=B(s)/A(s)=1$ for all $s$ .
2. $b_0=\overline{a_0}$ and $b_1=1$ , i.e.,
   $$B(j\omega)=e^{j\phi}\overline{A(j\omega)}.$$

Case (1) is trivially allpass, while case (2) is the one discussed above in the introduction to this section.

By analytic continuation, we have

$$1 = \left\vert H(j\omega)\right\vert = \left\vert H(j\omega)\right\vert^2 = \left. H(s)\overline{H(s)}\right\vert _{s=j\omega}$$

If $h(t)$ is real, then $\overline{H(j\omega)} = H(-j\omega)$ , and we can write

$$1 = \left. H(s)H(-s)\right\vert _{s=j\omega}.$$

To have $H(s)H(-s)=1$ , every pole at $s=p$ in $H(s)$ must be canceled by a zero at $s=p$ in $H(-s)$ , which is a zero at $s=-p$ in $H(s)$ . Thus, we have derived the simplified ``allpass rule'' for real analog filters.