Plane-Wave Scattering

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**Figure C.15:** Plane wave propagation in a medium changing from wave impedance to .
$\includegraphics{eps/planewavescat}$

Consider a plane wave with peak pressure amplitude propagating from wave impedance into a new wave impedance , as shown in Fig.C.15. (Assume and are real and positive.) The physical constraints on the wave are that

pressure must be continuous everywhere, and
velocity in must equal velocity out (the junction has no state).

Since power is pressure times velocity, these constraints imply that signal power is conserved at the junction.^C.5Expressed mathematically, the physical constraints at the junction can be written as follows:

$\begin{eqnarray*} p^+_1+p^-_1 &=& p^+_2\quad\mbox{(pressure continuous across ju... ...+}_1+v^{-}_1 &=& v^{+}_2\quad\mbox{(velocity in = velocity out)} \end{eqnarray*}$

As derived in §C.7.3, we also have the Ohm's law relations:

$\begin{displaymath} \begin{array}{rcrl} p^+_i&=&&R_iv^{+}_i\\ p^-_i&=&-&R_iv^{-}_i \end{array}\end{displaymath}$

These equations determine what happens at the junction.

To obey the physical constraints at the impedance discontinuity, the incident plane-wave must split into a reflected plane wave and a transmitted plane-wave such that pressure is continuous and signal power is conserved. The physical pressure on the left of the junction is , and the physical pressure on the right of the junction is , since according to our set-up.

Scattering Solution

Define the junction pressure and junction velocity by

$\begin{eqnarray*} p_j &\isdef & p^+_1+p^-_1 = p^+_2\quad\mbox{(pressure at junct... ...f & v^{+}_1+v^{-}_1 = v^{+}_2\quad\mbox{(velocity at junction).} \end{eqnarray*}$

Then we can write

$\begin{eqnarray*} p^+_1+p^-_1 &=& p^+_2\;=\;p_j\\ [10pt] \,\,\Rightarrow\,\,R_1v... ...\\ [10pt] \,\,\Rightarrow\,\,2\,R_1v^{+}_1 - R_1 v_j &=& R_2 v_j \end{eqnarray*}$

$\displaystyle \,\,\Rightarrow\,\,\zbox {v_j = \frac{2\,R_1}{R_1 + R_2}v^{+}_1.}$

Note that $v_j=v^{+}_2$ , so we have found the velocity of the transmitted wave. Since $v_j = v^{+}_1+v^{-}_1$ , the velocity of the reflected wave is simply

$\displaystyle v^{-}_1 = v_j - v^{+}_1 = \left[\frac{2\,R_1}{R_1+R_2} - 1\right]v^{+}_1 = \frac{R_1-R_2}{R_1+R_2} v^{+}_1.$

We have solved for the transmitted and reflected velocity waves given the incident wave and the two impedances.

Using the Ohm's law relations, the pressure waves follow easily:

$\begin{eqnarray*} p^+_2 &=& R_2v^{+}_2 = R_2 v_j = \frac{2\,R_2}{R_1+R_2}p^+_1\\ [10pt] p^-_1 &=& -R_1v^{-}_1 = \frac{R_2-R_1}{R_1+R_2} p^+_1 \end{eqnarray*}$

Reflection Coefficient

Define the reflection coefficient of the scattering junction as

$\displaystyle \zbox {\rho = \frac{R_2-R_1}{R_1+R_2} = \frac{\mbox{Impedance Step}}{\mbox{Impedance Sum}}.}$

Then we get the following scattering relations in terms of $\rho$ for pressure waves:

$\begin{eqnarray*} p^+_2 &=& (1+\rho)p^+_1\\ [3pt] p^-_1 &=& \rho\,p^+_1 \end{eqnarray*}$

Signal flow graphs for pressure and velocity are given in Fig.C.16.

**Figure C.16:** Signal flow graph for the pressure and velocity components of a plane wave scattering at an impedance discontinuity :.
$\includegraphics{eps/planewavescatdm2}$

It is a simple exercise to verify that signal power is conserved by checking that $p^+_1v^{+}_1 = p^+_2v^{+}_2 + ( - p^-_1v^{-}_1)$ . (Left-going power is negated to account for its opposite direction-of-travel.)

So far we have only considered a plane wave incident on the left of the junction. Consider now a plane wave incident from the right. For that wave, the impedance steps from to , so the reflection coefficient it ``sees'' is $-\rho$ . By superposition, the signal flow graph for plane waves incident from either side is given by Fig.C.17. Note that the transmission coefficient is one plus the reflection coefficient in either direction. This signal flow graph is often called the ``Kelly-Lochbaum'' scattering junction [297].