Scattering Solution

Define the junction pressure $p_j$ and junction velocity $v_j$ by

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

Then we can write

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

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

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

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

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:

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