DWM, Cont'd

We just derived

$$F_m(s) - F^{+}_1(s) + F^{-}_1(s) + F^{+}_2(s) = 0$$

(no incoming wave from string 2)

Substitute

$$\begin{eqnarray*} F^{+}_i(s)&=&RV^{+}_i(s)\\ F^{-}_i(s)&=&-RV^{-}_i(s)\\ F_m(s)&=&ms V(s) \end{eqnarray*}$$

to obtain

$$msV - RV^{+}_1 + RV^{-}_1 + RV^{+}_2 = 0$$

We always have $V=V^{+}_1+V^{-}_1=V^{+}_2+V^{-}_2$
(series combination)
Since $V^{-}_2=0$, we have $V^{+}_2=V$, and

$$\begin{eqnarray*} && (R+ms)V - RV^{+}_1 + RV^{-}_1 \;=\; 0\\ \Rightarrow \; && (R+ms)(V^{+}_1+V^{-}_1) - RV^{+}_1 + RV^{-}_1 \;=\; 0. \end{eqnarray*}$$

Solving for the velocity reflection transfer function
(or velocity reflectance) of the mass from string 1 gives

$$\zbox{\rho^v_1 \isdef \frac{V^{-}_1}{V^{+}_1} = -\frac{ms}{ms+2R}}$$

By physical symmetry, the mass looks the same from string 2:

$$\zbox{\rho^v_2 \isdef \frac{V^{+}_2}{V^{-}_2} = \rho^v_1 = -\frac{ms}{ms+2R}}$$

(reflectance of a mass $m$ at the end of a string of wave impedance $R$)

Limiting Behavior:

• When $m=0$, reflectance is zero $\;$ (no reflected wave)
• When $m\to\infty$, $\rho^v_1\to-1$ $\;$ (rigid termination)