Reflection Transfer Function

Given the bridge driving-point impedance $R_b(s)$, how do we incorporate it in a digital waveguide model? Let $\hat R_b(z)$ denote the digital impedance transfer function. If the bilinear transform was used, we have

$$\hat R_b(z) \isdef R_b\left(\frac{2}{T} \frac{1-z^{-1}}{1+z^{-1}}\right)= \frac{F_b(z)}{V_b(z)}$$

where $F_b(z)$ is the $z$ transform of the force applied to the bridge, in discrete time, and $V_b(z)$ is the $z$ transform of the bridge vertical velocity. Since the bridge and string move together, $V_b(z)=V(z)$. Also, the force applied to the bridge by the string is the one which acts to the left, or $f_l(t,0)$. Since we chose the right-acting force $f_r$ for our force wave variable $f$, we have $F_b(z) = -F(z,0)$. We have thus related the impedance of the termination to quantities wholly within the string wave state:

$$\hat R_b(z) = -\frac{F(z)}{V(z)} = -\frac{F^{+}(z) + F^{-}(z)}{V^{+}(z)+V^{-}(z)} = -R\frac{F^{+}(z) + F^{-}(z)}{F^{+}(z)-F^{-}(z)}$$

In a simulation context, we know the bridge reflectance $\hat R_b(z)$ and string wave impedance $R$, and we know the wave variable impinging on the bridge, in this case $F^{-}(z)$ since we took the bridge to be on the far left. We therefore need to solve for $F^{+}(z)$ in the above expression. This yields the reflection transfer function at the bridge,

$$S_b(z) \isdef \frac{F^{+}(z)}{F^{-}(z)} = \frac{\hat R_b(z) - R}{\hat R_b(z) + R}$$

Note that this same result can be obtained from the general formula for scattering at a loaded waveguide junction for the case of a single waveguide ($N=1$) terminated by a lumped load.

Because $\hat R_b(z)$ is positive real, $S_b(z)$ is a Schur function, i.e., $\vert S_b(z)\vert\leq1$ for $\vert z\vert\leq1$. Schur functions become allpass filters as damping goes to zero, and they cannot provide gain at any frequency, i.e., the gain is less than or equal to $1$, as needed for string loop stability. Note that reflection filters always have an equal number of poles and zeros, as can be seen from the above expression.

The reflection transfer function is defined for force waves. Note that as the bridge impedance goes to infinity (becomes rigid), $S_b(z)$ approaches $1$, a result which agrees with an analysis of rigid terminations. Since typical bridges are quite rigid, $S_b(z)\approx 1$ in all practical cases. Similarly, if the bridge impedance goes to zero, $S_b(z)$ goes to $-1$ which also agrees with the physics of a string with a free end. In all cases, we have $\vert S_b(e^{j\omega T})\vert\leq1$ for all $\omega$.

If the bridge driving-point impedance were $R$, i.e., matched to the wave impedance, the reflection transfer function would vanish; in that case, the bridge is indistinguishable from a continuation of the string, and the incident wave propagates into it with no reflection. The ``matched impedance'' case is the one which delivers maximum power from the string to the instrument body through the bridge, but it is clearly a useless case in practice. Plucking a string with such a bridge would produce the sound of the pluck, but no sustained oscillation; there would be no ``pitch'' to the sound. In natural instruments, there is always a problematic tradeoff between radiated energy and energy which is reflected back to sustain oscillation.

The velocity-wave version of the bridge reflection transfer function follows immediately from the force-wave reflectance, since

$$\frac{V^{+}(z)}{V^{-}(z)} = \frac{F^{+}(z)/R}{-F^{-}(z)/R} = - S_b(z) .$$

It is simply the negative of the force wave version.

The velocity reflectance is unchanged if we switch to displacement waves or acceleration waves, since

$$S_b(z) = -\frac{V^{-}(z)}{V^{+}(z)} = -\frac{H(z)V^{-}(z)}{H(z)V^{+}(z)}$$

whether $H(z)$ is an integrator, differentiator, or any other transfer function free of zeros on the unit circle.

Since the reflection transfer function $S_b(z)$ specifies what is reflected from the bridge at each frequency, the transmission transfer function from the string into the body must be in some sense its complement. This is true, and it is quick to show that the bridge transmittance is $V_b(z)/V^{+}(z) = 1-S_b(z)$ for velocity waves and $F_b(z)/F^{+}(z) = 1+S_b(z)$ for force waves. Note that the velocity-wave bridge reflectance and transmittance are allpass complementary [509]. The same is true also for force waves if the reflected wave is converted from a left-going wave to a right-going wave so that it can be compared with the transmitted wave. (The reflected and transmitted velocity waves are comparable without a sign flip because both left and right-going velocity waves are positive in the ``up'' direction.)

It is also quick to show that the reflectance and transmittance are also power complementary [509] for both force and velocity waves. The average power incident at the bridge at frequency $\omega$ can be expressed in the frequency domain as $F^{+}(e^{j\omega T})\overline{V^{+}(e^{j\omega T})}$. The reflected power is then $F^{-}\overline{V^{-}} = -\left\vert S_b\right\vert^2F^{+}\overline{V^{+}}$. Flipping the sign so as to make the reflected power ``right going,'' (or, equivalently, to obtain reflected power without an added sign which indicates its direction of travel), we obtain the power reflection transfer function $\left\vert S_b\right\vert^2$. The power transmittance is given by

$$F_b\overline{V_b} = [(1+S_b)F^{+}][\overline{1-S_b}\overline{V^{+}}] = (1-\left\vert S_b\right\vert^2)F^{+}\overline{V^{+}} .$$

Adding the reflected and transmitted power gives $1$ for all $\omega$, as it must to conserve energy.