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

$\displaystyle \left\Vert\,x\,\right\Vert _2^2
\isdef \int_{-\infty}^\infty \le...
...\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 L2 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

$\displaystyle \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 L2 norms are defined by

$\displaystyle \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$,

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

and we have the requirement

$\displaystyle \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 functionE.4.3). Thus, the allpass condition becomes

$\displaystyle \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

$\displaystyle \left\vert H(j\omega)\right\vert = 1, \quad \forall\, \omega\in(-\infty,\infty). \protect$ (E.15)

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

$\displaystyle 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.15) implies

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

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

$\displaystyle 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

$\displaystyle \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.,

    $\displaystyle 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

$\displaystyle 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

$\displaystyle 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.


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