Passive String Terminations

In Chapter 4 we analyzed the effect of rigid string terminations on traveling waves. We found that waves derived by time-derivatives of displacement (displacement, velocity, acceleration, and so on) reflect with a sign inversion, while waves defined in terms of the first spatial derivative of displacement (force, slope) reflect with no sign inversion. In this appendix, we will look at the more realistic case of yielding terminations for strings. This analysis can be considered a special case of the loaded string junction analyzed in §G.10.

Yielding string terminations (at the bridge) have a large effect on the sound produced by acoustic stringed instruments. Rigid terminations can be considered a reasonable model for the solid-body electric guitar in which maximum sustain is desired for played notes. Acoustic guitars, on the other hand, must transduce sound energy from the strings into the body of the instrument, and from there to the surrounding air. All audible sound energy comes from the string vibrational energy, thereby reducing the sustain (decay time) of each played note. Furthermore, because the bridge vibrates more easily in one direction than another, a kind of ``chorus effect'' is created from the detuning of the horizontal and vertical planes of string vibration (as discussed further in §4.12.1). A perfectly rigid bridge, in contrast, cannot transmit any sound into the body of the instrument, thereby requiring some other transducer, such as the magnetic pickups used in electric guitars, to extract sound for output.^L.1

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 Laplace transform of the force on the bridge $F_b(s)$ to the velocity of motion that results $V_b(s)$.

For passive systems (i.e., for all unamplified acoustic musical instruments), the driving-point impedance is positive real (see §L.4) [432,528], which means (1) $R_b(s)$ is real when $s$ is real, and (2) the real part of $R_b(s)$ is nonnegative when the real part of $s$ is nonnegative, i.e., re$\left\{s\right\}\geq0\implies$   re$\left\{R_b(s)\right\}\geq0$.
This seemingly simple property has deep 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.

At $x=0$, the force on the bridge is given by

$$f_b(t) = Ky'(t,0) = - f(t,0)$$

where $K$ is the string tension as in Chapter 4, and $y'$ is the slope of the string at $x=0$. In the frequency domain, we have

$$F_b(s) = KY'(s,0) = - F(s,0)$$

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

$$V(s,0) \equiv V_b(s) \isdef \frac{F_b(s)}{R_b(s)} = -\frac{F(s,0)}{R_b(s)}$$

How do we take a continuous-time driving-point impedance $R_b(s)$ into the digital domain? This is analogous to the problem of converting an analog electrical filter into a corresponding digital filter--a problem which has been well studied [349].

The bilinear transform has the advantages of being (1) free of aliasing, and (2) order invariant. The entire $j\omega$ axis maps exactly once from the $s$ plane onto the unit circle in the $z$ plane (rather than summing around it infinitely many times, or ``aliasing'' as it does in ordinary sampling). The right-half $s$ plane maps to the exterior of the unit circle in the $z$ plane, and re$\left\{s\right\}<0$ maps to $\vert z\vert<1$; this means stability is preserved as it must be. As a result of the one-to-one mapping, the bilinear transform preserves the positive-real property of passive impedances, where it is appropriate to replace re$\left\{s\right\}\geq 0$ with $\vert z\vert\geq1$ in the definition of positive real (see §L.4).

``Order invariant'' means an $N$th-order $s$-plane transfer function carries over to an $N$th-order $z$-plane transfer function. Order in both cases equals the degree of the rational transfer function (the maximum of the degrees of the numerator and denominator polynomials). In continuous time, the order is incremented once for each independently moving mass or spring. In discrete time, the order is increased by one when a sample of delay is added to the system state, and the number of multiplies needed to implement a digital simulation is bounded by twice the order plus one.

If the bridge couples the string to a simple mass-spring system, depicted schematically in Fig. L.1, then the driving-point impedance is second order, and we have

$$R_b(s) = ms + \mu + {k/s}$$

where $m$ is the mass, $k$ is the spring constant, and $\mu$ is the dashpot resistance.

Figure L.1: A second-order driving-point impedance terminating the ideal string.

The general second-order impedance above can be viewed as the sum of the mass, spring, and dashpot impedances. That means they are formally ``in series'' with each other, following an analogy with circuit theory. Thus, the driving-point impedance of the mass $m$ is $ms$, that of the spring is $k/s$, and the dashpot $\mu$ is a real impedance. In general, no matter how complicated the interconnection of masses, springs, and damping elements, simple ``resistor network'' analysis yields the driving-point impedance $R_b(s)$ ``seen'' at the bridge. To find $R_b(s)$, it is only necessary to know that impedances add in series (as in the above example) and that admittances add in parallel, where an admittance is the reciprocal of an impedance. Of course, musical instrument bodies are not simple mass-spring systems. However, they are often modeled as such by assigning a second-order section to each important resonance observed in the body.