A Physical Derivation of Wave Digital Elements

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This section provides a ``physical'' derivation of Wave Digital Filters (WDF), which contrasts somewhat with the more formal derivation common in the literature. The derivation is presented as a numbered series of steps (some with rather long discussions):

To each element, such as a capacitor or inductor, attach a length of waveguide (electrical transmission line) having wave impedance

, and make it infinitesimally long. (Take the limit as its length goes to zero.) A schematic depiction of this is shown in Fig.F.1a. For consistency, all signals are Laplace transforms of their respective time-domain signals. The length must approach zero in order not to introduce propagation delays into the signal path.

**Figure F.1:** a) Physical schematic for the derivation of a wave digital model of driving-point impedance . The inserted waveguide impedance is real and positive, but otherwise arbitrary. b) Expanded view of the interior of the infinitesimal waveguide section, also representing the termination impedance as an impedance-step within the waveguide.
$\includegraphics{eps/wdelt}$

Points to note:

The infinitesimal waveguide is terminated by the element. The element reflects waves as if it were a new waveguide section at impedance , as depicted in Fig.F.1b.
The interface to the element is recast as traveling-wave components $F^{+}(s)$ and $F^{-}(s)$ at impedance . In terms of these components, the physical force on the element is obtained by adding them together: $F(s) = F^{+}(s)+F^{-}(s)$ .
The waveguide impedance is arbitrary because it has been physically introduced. We will need to know it when we connect this element to other elements. The element's interface to other elements is now a waveguide (transmission line) at real impedance .
The junction is ``parallel'' (cf. §7.2):
- Force (voltage) must be continuous across the junction, since otherwise there would be a finite force across a zero mass, producing infinite acceleration.
- The sum of velocities (currents) into the junction must be zero by conservation of mass (charge).

Reflectance of a General Lumped Waveguide Termination

Calculate the reflectance of the terminated waveguide. That is, find the Laplace transform of the return wave divided by the Laplace transform of the input wave going into the waveguide. In general, the reflectance of an impedance step for force waves (voltage waves in the electrical case) is

$\displaystyle \fbox{$\displaystyle \hat{\rho}_R(s) \isdef \frac{F^{-}(s)}{F^{+}(s)} = \frac{R(s)-R_0}{R(s)+R_0}$} \protect$

(F.1)

This is easily derived from continuity constraints across the junction. Specifically, referring to Fig.F.1b, let $F_R(s) = F^{+}_R(s) + F^{-}_R(s)$ denote the physical force and its traveling-wave components within the ``pseudo-infinitesimal-generalized-waveguide'' defined by the element impedance

, with the `

' superscript denoting a right-going wave.^F.1 Similarly, let $V(s) = V^{+}(s)+V^{-}(s)$ denote the velocity and its component wave variables on the side of the junction at impedance

, and let $V_R(s) = V^{+}_R(s)+V^{-}_R(s)$ denote the corresponding quantities on the element-side of the junction at impedance

. Again, the `

' superscript denotes travel to the right. Then the physical continuity constraints imply

$\begin{eqnarray*} F(s) &=& F_R(s)\\ 0 &=& V(s) + V_R(s) \end{eqnarray*}$

By the definition of wave impedance in a waveguide, we have

$\begin{eqnarray*} F^{+}(s) &=& \quad\! R_0 V^{+}(s)\\ F^{-}(s) &=& - R_0 V^{-}(... ... &=& \quad\! R(s) V^{+}_R(s)\\ F^{-}_R(s) &=& - R(s) V^{-}_R(s) \end{eqnarray*}$

Thus,

$\begin{eqnarray*} 0 &=& V(s) + V_R(s)\\ &=& \left[V^{+}(s)+V^{-}(s)\right] + ... ...s)}\right] &=& \frac{2}{R_0}F^{+}(s) + \frac{2}{R(s)}F^{+}_R(s) \end{eqnarray*}$

Defining $\Gamma_0 \isdef 1/R_0$ and $\Gamma(s) \isdef 1/R(s)$ , we have

$\displaystyle F(s) = \frac{2\Gamma_0}{\Gamma_0+\Gamma(s)} F^{+}(s) + \frac{2\Ga... ...a(s)} F^{+}_R(s) \isdef {\cal A}(s) F^{+}(s) + {\cal A}_R(s)F^{+}_R(s) \protect$

(F.2)

Now that we've solved for the junction force

, the outgoing waves are simply obtained from the force continuity constraint, $F(s) = F^{+}(s)+F^{-}(s) = F^{+}_R(s)+F^{-}_R(s)$ :

$\displaystyle F^{-}(s)$	$\displaystyle =$	$\displaystyle F(s) - F^{+}(s) \protect$	(F.3)
$\displaystyle F^{-}_R(s)$	$\displaystyle =$	$\displaystyle F(s) - F^{+}_R(s) \protect$	(F.4)

Finally, the force-wave reflectance of an impedance step from

can be found by solving Eq. $\,$ (F.3) and (F.2) for $F^{-}(s)/F^{+}(s)$ with $F^{+}_R(s)$ set to zero:

$\begin{eqnarray*} \hat{\rho}(s) &\isdef & \frac{F^{-}(s)}{F^{+}(s)} = \frac{F(s)... ...mma_0-\Gamma(s)}{\Gamma_0+\Gamma(s)} = \frac{R(s)-R_0}{R(s)+R_0} \end{eqnarray*}$

as claimed.

Reflectances of Elementary Impedances

We now derive the reflectances of the elements used in LTI analog electric circuits, viz., the capacitor, inductor, and resistor.

Capacitor Reflectance

For a capacitor of Farads, the driving-point impedance is (see §7.1.3)

$\displaystyle R_C(s)=\frac{1}{Cs}$

(or

for a spring with constant

). Substituting into Eq. $\,$ (F.1) gives the reflectance

$\displaystyle \hat{\rho}_C(s) = \frac{R_C(s)-R_0}{R_C(s)+R_0} = \frac{1 - R_0 C s}{1 + R_0 C s} \protect$

(F.5)

Inductor Reflectance

For an inductor of Henrys, we have

$\displaystyle R_L(s)$	$\displaystyle =$	$\displaystyle Ls$
$\displaystyle \,\,\Rightarrow\,\,\hat{\rho}_L(s)$	$\displaystyle =$	$\displaystyle \frac{Ls-R_0}{Ls+R_0} = \frac{ s - R_0/L }{ s + R_0/L} \protect$	(F.6)

Resistor Reflectance

Finally, for a resistor of Ohms, we get

$\displaystyle \hat{\rho}_R(s) = \frac{R-R_0}{R+R_0} = \frac{1 - R_0/R }{ 1 + R_0/R } \protect$

(F.7)

Note that both the capacitor and inductor reflectances are stable allpass filters, as they must be. Also, the resistor reflectance is always less than 1, no matter what waveguide impedance we choose.

Choosing Impedance to Simplify Element Reflectance

Observe that there is a natural choice for each waveguide impedance which will give us a normalized, ``universal reflectance'' for each element:

For the capacitor, setting gives

$\displaystyle \fbox{$\displaystyle \hat{\rho}_C(s) = \frac{1 - s}{1 + s}$} \protect$ (F.8)
For the inductor, setting gives

$\displaystyle \fbox{$\displaystyle \hat{\rho}_L(s) = - \frac{1 - s}{1 + s}$} \protect$ (F.9)
And for the resistor, we set to obtain

$\displaystyle \fbox{$\displaystyle \hat{\rho}_R(s) = 0$} \protect$ (F.10)

Digitizing Elementary Reflectances by Bilinear Transform

Going to discrete time via the bilinear transform means making the substitution

$\displaystyle s = c \frac{1-z^{-1}}{1+z^{-1}}$

(F.11)

where

is an arbitrary real constant, usually taken to be

Solving for $z^{-1}$ gives us the inverse bilinear transform:

$\displaystyle z^{-1}= \frac{1-s/c}{1+s/c} \protect$

(F.12)

In this case, we see that setting further simplifies our universal reflectances in the digital domain:

For the ``wave digital capacitor'' (or spring), Eq. $\,$ (F.8) becomes

$\displaystyle \fbox{$\displaystyle \hat{\rho}_C\left(\frac{z-1}{z+1}\right) = z^{-1}$}$
For the ``wave digital inductor'' (or mass), Eq. $\,$ (F.9) becomes

$\displaystyle \fbox{$\displaystyle \hat{\rho}_L\left(\frac{z-1}{z+1}\right) = - z^{-1}$}$
And for the ``wave digital resistor'' (or dashpot), Eq. $\,$ (F.10) becomes

$\displaystyle \fbox{$\displaystyle \hat{\rho}_R\left(\frac{z-1}{z+1}\right) = 0$}$
as before in the continuous-time case.

Note that this choice of is also the only one that eliminates delay-free paths in the fundamental elements. This allows them to be used as building blocks for explicit finite difference schemes.

We may still obtain the above results using the more typical value (instead of ) in the bilinear transform. From Eq. $\,$ (F.12), it is clear that changing amounts to a linear frequency scaling of $s=j\omega$ . Such a scaling may be compensated by choosing the waveguide (port) impedances to be (instead of ) for the inductor, and (instead of ) for the capacitor.