Passive String Terminations

When a traveling wave reflects from the bridge of a real stringed instrument, the bridge moves, transmitting sound energy into the instrument body. How far the bridge moves is determined by the driving-point impedance of the bridge, denoted $R_b(s)$ . The driving-point impedance is the ratio of the Laplace-transform of the force on the bridge, $F_b(s)$ , divided by the velocity-of-motion that results, $V_b(s)$ . That is, $R_b(s)\isdeftext F_b(s)/V_b(s)$ .

For passive systems (i.e., for all unamplified acoustic musical instruments), the driving-point impedance $R_b(s)$ is positive real (a property defined and discussed in §C.11.2). Being positive real has strong implications on the nature of $R_b(s)$ . In particular, the phase of $R_b(j\omega)$ cannot exceed plus or minus $90$ degrees at any frequency, and in the lossless case, all poles and zeros must interlace along the $j\omega$ axis. Another implication is that the reflectance of a passive bridge, as seen by traveling waves on the string, is a so-called Schur function (defined and discussed in §C.11); a Schur reflectance is a stable filter having gain not exceeding 1 at any frequency. In summary, a guitar bridge is passive if and only if its driving-point impedance is positive real and (equivalently) its reflectance is Schur. See §C.11 for a fuller discussion of this point.

At $x=0$ , the force on the bridge is given by (§C.7.2)

$$f_b(t) \eqsp Ky'(t,0) \eqsp - f(t,0)$$

where $K$ is the string tension as in Chapter 6, and $y'(t,0)$ is the slope of the string at $x=0$ . In the Laplace frequency domain, we have

$$F_b(s) \eqsp KY'(s,0) \eqsp - F(s,0),$$

due to linearity, and the velocity of the string endpoint is therefore

$$V(s,0) \equiv V_b(s) \isdefs \frac{F_b(s)}{R_b(s)} \eqsp -\frac{F(s,0)}{R_b(s)}.$$