Alternative Wave Variables

We have thus far considered discrete-time simulation of transverse displacement $ y$ in the ideal string. It is equally valid to choose velocity $ v\isdeftext {\dot y}$, acceleration $ a\isdeftext {\ddot y}$, slope $ y'$, or perhaps some other derivative or integral of displacement with respect to time or position. Conversion between various time derivatives can be carried out by means integrators and differentiators, as depicted in Fig.C.10. Since integration and differentiation are linear operators, and since the traveling wave arguments are in units of time, the conversion formulas relating $ y$, $ v$, and $ a$ hold also for the traveling wave components $ y^\pm , v^\pm , a^\pm $.

Figure C.10: Conversions between various time derivatives of displacement: $ y = $ displacement, $ v = {\dot y}= $ velocity, $ a = {\ddot y}= $ acceleration, where $ {\dot y}$ denotes $ dy/dt$ and $ {\ddot y}$ denotes $ d^2y/dt^2$.
\includegraphics[scale=0.9]{eps/fwaveconversions}

Differentiation and integration have a simple form in the frequency domain. Denoting the Laplace Transform of $ y(t,x)$ by

$\displaystyle Y(s,x) \isdefs {\cal L}_s\{y(\cdot,x)\} \isdefs \int_0^\infty y(t,x) e^{-st} dt,$ (C.36)

where ``$ \cdot$'' in the time argument means ``for all time,'' we have, according to the differentiation theorem for Laplace transforms [284],

$\displaystyle {\cal L}_s\{{\dot y}(\cdot,x)\} \eqsp s Y(s,x) - y(0,x).$ (C.37)

Similarly, $ {\cal L}_s\{\dot y^{+}\} = s Y^{+}(s) - y^{+}(0)$, and so on. Thus, in the frequency domain, the conversions between displacement, velocity, and acceleration appear as shown in Fig.C.11.

Figure C.11: Conversions between various time derivatives of displacement in the frequency domain.
\includegraphics[scale=0.9]{eps/ffdwaveconversions}

In discrete time, integration and differentiation can be accomplished using digital filters [362]. Commonly used first-order approximations are shown in Fig.C.12.

Figure C.12: Simple approximate conversions between time derivatives in the discrete-time case: a) The first-order difference $ {\hat v}(n) = y(n) - y(n-1)$. b) The first-order ``leaky'' integrator $ {\hat y}(n) = v(n) + g {\hat y}(n-1)$ with loss factor $ g$ (slightly less than $ 1$) used to avoid infinite dc build-up.
\includegraphics[width=\twidth]{eps/fdigitaldiffint}

If discrete-time acceleration $ a_d(n)$ is defined as the sampled version of continuous-time acceleration, i.e., $ a_d(n) \isdeftext
a(nT,x)={\ddot y}(nT,x)$, (for some fixed continuous position $ x$ which we suppress for simplicity of notation), then the frequency-domain form is given by the $ z$ transform [485]:

$\displaystyle A_d(z) \isdefs \sum_{n=0}^\infty a_d(n) z^{-n}$ (C.38)

In the frequency domain for discrete-time systems, the first-order approximate conversions appear as shown in Fig.C.13.

Figure C.13: Frequency-domain description of the approximate conversions between time derivatives in the discrete-time case. The subscript ``$ d$'' denotes the ``digital'' case. A ``hat'' over a variable indicates it is an approximation to the variable without the hat.
\includegraphics[scale=0.6]{eps/ffddigitaldiffint}

The $ z$ transform plays the role of the Laplace transform for discrete-time systems. Setting $ z=e^{sT}$, it can be seen as a sampled Laplace transform (divided by $ T$), where the sampling is carried out by halting the limit of the rectangle width at $ T$ in the definition of a Reimann integral for the Laplace transform. An important difference between the two is that the frequency axis in the Laplace transform is the imaginary axis (the ``$ j\omega $ axis''), while the frequency axis in the $ z$ plane is on the unit circle $ z = e^{j\omega T}$. As one would expect, the frequency axis for discrete-time systems has unique information only between frequencies $ -\pi/T$ and $ \pi/T$ while the continuous-time frequency axis extends to plus and minus infinity.

These first-order approximations are accurate (though scaled by $ T$) at low frequencies relative to half the sampling rate, but they are not ``best'' approximations in any sense other than being most like the definitions of integration and differentiation in continuous time. Much better approximations can be obtained by approaching the problem from a digital filter design viewpoint, as discussed in §8.6.

Spatial Derivatives

In addition to time derivatives, we may apply any number of spatial derivatives to obtain yet more wave variables to choose from. The first spatial derivative of string displacement yields slope waves

$\displaystyle y'(t,x)$ $\displaystyle \isdef$ $\displaystyle \frac{\partial}{\partial x}y(t,x)$  
  $\displaystyle =$ $\displaystyle y'_r(t-x/c) + y'_l(t+x/c)$  
  $\displaystyle =$ $\displaystyle -\frac{1}{c} {\dot y}_r(t-x/c) + \frac{1}{c}{\dot y}_l(t+x/c),
\protect$ (C.39)

or, in discrete time,
$\displaystyle y'(t_n,x_m)$ $\displaystyle \isdef$ $\displaystyle y'(nT,mX)$  
  $\displaystyle =$ $\displaystyle y'_r\left[(n-m)T\right]+ y'_l\left[(n+m)T\right]$  
  $\displaystyle \isdef$ $\displaystyle y'^{+}(n-m) + y'^{-}(n+m)$  
  $\displaystyle =$ $\displaystyle -\frac{1}{c} \dot y^{+}(n-m) + \frac{1}{c}\dot y^{-}(n+m)$  
  $\displaystyle \isdef$ $\displaystyle -\frac{1}{c} v^{+}(n-m) + \frac{1}{c}v^{-}(n+m)$  
  $\displaystyle =$ $\displaystyle \frac{1}{c} \left[v^{-}(n+m) - v^{+}(n-m) \right].
\protect$ (C.40)

From this we may conclude that $ v^{-}= cy'^{-}$ and $ v^{+}= -cy'^{+}$. That is, traveling slope waves can be computed from traveling velocity waves by dividing by $ c$ and negating in the right-going case. Physical string slope can thus be computed from a velocity-wave simulation in a digital waveguide by subtracting the upper rail from the lower rail and dividing by $ c$.

By the wave equation, curvature waves, $ y''= {\ddot y}/c^2$, are simply a scaling of acceleration waves, in the case of ideal strings.

In the field of acoustics, the state of a vibrating string at any instant of time $ t_0$ is often specified by the displacement $ y(t_0,x)$ and velocity $ {\dot y}(t_0,x)$ for all $ x$ [317]. Since velocity is the sum of the traveling velocity waves and displacement is determined by the difference of the traveling velocity waves, viz., from Eq.$ \,$(C.39),

$\displaystyle y(t,x) \eqsp \int_0^{x} y'(t,\xi)d\xi
\eqsp -\frac{1}{c}\int_0^{x} \left[v_r(t-\xi/c) - v_l(t+\xi/c)\right]d\xi,
$

one state description can be converted to the other.

In summary, all traveling-wave variables can be computed from any one, as long as both the left- and right-going component waves are available. Alternatively, any two linearly independent physical variables, such as displacement and velocity, can be used to compute all other wave variables. Wave variable conversions requiring differentiation or integration are relatively expensive since a large-order digital filter is necessary to do it right (§8.6.1). Slope and velocity waves can be computed from each other by simple scaling, and curvature waves are identical to acceleration waves to within a scale factor.

In the absence of factors dictating a specific choice, velocity waves are a good overall choice because (1) it is numerically easier to perform digital integration to get displacement than it is to differentiate displacement to get velocity, (2) slope waves are immediately computable from velocity waves. Slope waves are important because they are a simple scaling of force waves.


Force Waves

Figure C.14: Transverse force propagation in the ideal string.
\includegraphics[width=\twidth]{eps/fStringForce}

Referring to Fig.C.14, at an arbitrary point $ x$ along the string, the vertical force applied at time $ t$ to the portion of string to the left of position $ x$ by the portion of string to the right of position $ x$ is given by

$\displaystyle f_{rl}(t,x) = K\sin(\theta) \approx K\tan(\theta) = Ky'(t,x)$ (C.41)

assuming $ \left\vert y'(t,x)\right\vert \ll 1$, as is assumed in the derivation of the wave equation. Similarly, the force applied by the portion to the left of position $ x$ to the portion to the right is given by

$\displaystyle f_{lr}(t,x) = - K\sin(\theta) \approx - Ky'(t,x).$ (C.42)

These forces must cancel since a nonzero net force on a massless point would produce infinite acceleration. I.e., we must have $ f_{rl} +
f_{lr} \equiv 0$ at all times $ t$ and positions $ x$.

Vertical force waves propagate along the string like any other transverse wave variable (since they are just slope waves multiplied by tension $ K$). We may choose either $ f_{rl}$ or $ f_{lr}$ as the string force wave variable, one being the negative of the other. It turns out that to make the description for vibrating strings look the same as that for air columns, we have to pick $ f_{lr}$, the one that acts to the right. This makes sense intuitively when one considers longitudinal pressure waves in an acoustic tube: a compression wave traveling to the right in the tube pushes the air in front of it and thus acts to the right. We therefore define the force wave variable to be

$\displaystyle f(t,x) \isdefs f_{lr}(t,x) = -Ky'(t,x).$ (C.43)

Note that a negative slope pulls up on the segment to the right. At this point, we have not yet considered a traveling-wave decomposition.


Wave Impedance

Using the above identities, we have that the force distribution along the string is given in terms of velocity waves by

$\displaystyle f(t,x) = \frac{K}{c} \left[{\dot y}_r(t-x/c) - {\dot y}_l(t+x/c) \right], \protect$ (C.44)

where $ K/c \isdef K/\sqrt{K/\epsilon } = \sqrt{K\epsilon }$. This is a fundamental quantity known as the wave impedance of the string (also called the characteristic impedance), denoted as

$\displaystyle R\isdefs \sqrt{K\epsilon } \eqsp \frac{K}{c} \eqsp \epsilon c.$ (C.45)

The wave impedance can be seen as the geometric mean of the two resistances to displacement: tension (spring force) and mass (inertial force).

The digitized traveling force-wave components become

\begin{displaymath}\begin{array}{rcrl} f^{{+}}(n)&=&&R\,v^{+}(n) \\ f^{{-}}(n)&=&-&R\,v^{-}(n) \end{array} \protect\end{displaymath} (C.46)

which gives us that the right-going force wave equals the wave impedance times the right-going velocity wave, and the left-going force wave equals minus the wave impedance times the left-going velocity wave.C.4Thus, in a traveling wave, force is always in phase with velocity (considering the minus sign in the left-going case to be associated with the direction of travel rather than a $ 180$ degrees phase shift between force and velocity). Note also that if the left-going force wave were defined as the string force acting to the left, the minus sign would disappear. The fundamental relation $ f^{{+}}=
Rv^{+}$ is sometimes referred to as the mechanical counterpart of Ohm's law for traveling waves, and $ R$ in c.g.s. units can be called acoustical ohms [261].

In the case of the acoustic tube [317,297], we have the analogous relations

\begin{displaymath}\begin{array}{rcrl} p^+(n) &=& &R_{\hbox{\tiny T}}\, u^{+}(n)...
...p^-(n) &=& -&R_{\hbox{\tiny T}}\, u^{-}(n) \end{array} \protect\end{displaymath} (C.47)

where $ p^+(n)$ is the right-going traveling longitudinal pressure wave component, $ p^-(n)$ is the left-going pressure wave, and $ u^\pm (n)$ are the left and right-going volume velocity waves. In the acoustic tube context, the wave impedance is given by

$\displaystyle R_{\hbox{\tiny T}}= \frac{\rho c}{A}$   (Acoustic Tubes) (C.48)

where $ \rho$ is the mass per unit volume of air, $ c$ is sound speed in air, and $ A$ is the cross-sectional area of the tube. Note that if we had chosen particle velocity rather than volume velocity, the wave impedance would be $ R_0=\rho c$ instead, the wave impedance in open air. Particle velocity is appropriate in open air, while volume velocity is the conserved quantity in acoustic tubes or ``ducts'' of varying cross-sectional area.


State Conversions

In §C.3.6, an arbitrary string state was converted to traveling displacement-wave components to show that the traveling-wave representation is complete, i.e., that any physical string state can be expressed as a pair of traveling-wave components. In this section, we revisit this topic using force and velocity waves.

By definition of the traveling-wave decomposition, we have

\begin{eqnarray*}
f&=&f^{{+}}+f^{{-}}\\
v&=&v^{+}+v^{-}.
\end{eqnarray*}

Using Eq.$ \,$(C.46), we can eliminate $ v^{+}=f^{{+}}/R$ and $ v^{-}=-f^{{+}}/R$, giving, in matrix form,

$\displaystyle \left[\begin{array}{c} f \\ [2pt] v \end{array}\right] = \left[\b...
...ay}\right]
\left[\begin{array}{c} f^{{+}} \\ [2pt] f^{{-}} \end{array}\right].
$

Thus, the string state (in terms of force and velocity) is expressed as a linear transformation of the traveling force-wave components. Using the Ohm's law relations to eliminate instead $ f^{{+}}=
Rv^{+}$ and $ f^{{-}}=-Rv^{-}$, we obtain

$\displaystyle \left[\begin{array}{c} f \\ [2pt] v \end{array}\right] = \left[\b...
...d{array}\right]\left[\begin{array}{c} v^{+} \\ [2pt] v^{-} \end{array}\right].
$

To convert an arbitrary initial string state $ (f,v)$ to either a traveling force-wave or velocity-wave simulation, we simply must be able to invert the appropriate two-by-two matrix above. That is, the matrix must be nonsingular. Requiring both determinants to be nonzero yields the condition

$\displaystyle 0 < R < \infty.
$

That is, the wave impedance must be a positive, finite number. This restriction makes good physical sense because one cannot propagate a finite-energy wave in either a zero or infinite wave impedance.

Carrying out the inversion to obtain force waves $ (f^{{+}},f^{{-}})$ from $ (f,v)$ yields

$\displaystyle \left[\begin{array}{c} f^{{+}} \\ [2pt] f^{{-}} \end{array}\right...
...ft[\begin{array}{c} \frac{f+Rv}{2} \\ [2pt] \frac{f-Rv}{2} \end{array}\right].
$

Similarly, velocity waves $ (v^{+},v^{-})$ are prepared from $ (f,v)$ according to

$\displaystyle \left[\begin{array}{c} v^{+} \\ [2pt] v^{-} \end{array}\right] = ...
...[\begin{array}{c} \frac{v+f/R}{2} \\ [2pt] \frac{v-f/R}{2} \end{array}\right].
$


Power Waves

Basic courses in physics teach us that power is work per unit time, and work is a measure of energy which is typically defined as force times distance. Therefore, power is in physical units of force times distance per unit time, or force times velocity. It therefore should come as no surprise that traveling power waves are defined for strings as follows:

\begin{displaymath}\begin{array}{rcrl} {\cal P}^{+}(n) &\isdef &&f^{{+}}(n)v^{+}(n) \\ {\cal P}^{-}(n) &\isdef &-&f^{{-}}(n)v^{-}(n) \end{array}\end{displaymath}    

From the Ohm's-law relations $ f^{{+}}=
Rv^{+}$ and $ f^{{-}}=-Rv^{-}$, we also have

\begin{eqnarray*}
{\cal P}^{+}(n)&=&R[v^{+}(n)]^2 \eqsp [f^{{+}}(n)]^2/R,
\\
{\cal P}^{-}(n)&=&R[v^{-}(n)]^2 \eqsp [f^{{-}}(n)]^2/R.
\end{eqnarray*}

Thus, both the left- and right-going components are nonnegative. The sum of the traveling powers at a point gives the total power at that point in the waveguide:

$\displaystyle {\cal P}(t_n,x_m) \isdefs {\cal P}^{+}(n-m) + {\cal P}^{-}(n+m)$ (C.49)

If we had left out the minus sign in the definition of left-going power waves, the sum would instead be a net power flow.

Power waves are important because they correspond to the actual ability of the wave to do work on the outside world, such as on a violin bridge at the end of a string. Because energy is conserved in closed systems, power waves sometimes give a simpler, more fundamental view of wave phenomena, such as in conical acoustic tubes. Also, implementing nonlinear operations such as rounding and saturation in such a way that signal power is not increased, gives suppression of limit cycles and overflow oscillations [432], as discussed in the section on signal scattering.

For example, consider a waveguide having a wave impedance which increases smoothly to the right. A converging cone provides a practical example in the acoustic tube realm. Then since the energy in a traveling wave must be in the wave unless it has been transduced elsewhere, we expect $ {\cal P}^{+}$ to propagate unchanged along the waveguide. However, since the wave impedance is increasing, it must be true that force is increasing and velocity is decreasing according to $ {\cal P}^{+}= R(v^{+})^2 = (f^{{+}})^2/R$. Looking only at force or velocity might give us the mistaken impression that the wave is getting stronger (looking at force) or weaker (looking at velocity), when really it was simply sailing along as a fixed amount of energy. This is an example of a transformer action in which force is converted into velocity or vice versa. It is well known that a conical tube acts as if it's open on both ends even though we can plainly see that it is closed on one end. A tempting explanation is that the cone acts as a transformer which exchanges pressure and velocity between the endpoints of the tube, so that a closed end on one side is equivalent to an open end on the other. However, this view is oversimplified because, while spherical pressure waves travel nondispersively in cones, velocity propagation is dispersive [22,50].


Energy Density Waves

The vibrational energy per unit length along the string, or wave energy density [317] is given by the sum of potential and kinetic energy densities:

$\displaystyle W(t,x) \isdefs \underbrace{\frac{1}{2} Ky'^2(t,x)}_{\mbox{potential}} + \underbrace{\frac{1}{2} \epsilon {\dot y}^2(t,x)}_{\mbox{kinetic}}$ (C.50)

Sampling across time and space, and substituting traveling wave components, one can show in a few lines of algebra that the sampled wave energy density is given by

$\displaystyle W(t_n,x_m) \isdefs W^{+}(n-m) + W^{-}(n+m),$ (C.51)

where

\begin{eqnarray*}
W^{+}(n) &=& \frac{{\cal P}^{+}(n)}{c} \,\mathrel{\mathop=}\,\...
...ht]^2 \,\mathrel{\mathop=}\,\frac{\left[f^{{-}}(n)\right]^2}{K}.
\end{eqnarray*}

Thus, traveling power waves (energy per unit time) can be converted to energy density waves (energy per unit length) by simply dividing by $ c$, the speed of propagation. Quite naturally, the total wave energy in the string is given by the integral along the string of the energy density:

$\displaystyle {\cal E}(t) \,\mathrel{\mathop=}\,\int_{x=-\infty}^\infty W(t,x)dx \approx \sum_{m = -\infty}^\infty W(t,x_m)X$ (C.52)

In practice, of course, the string length is finite, and the limits of integration are from the $ x$ coordinate of the left endpoint to that of the right endpoint, e.g., 0 to $ L$.


Root-Power Waves

It is sometimes helpful to normalize the wave variables so that signal power is uniformly distributed numerically. This can be especially helpful in fixed-point implementations.

From (C.49), it is clear that power normalization is given by

\begin{displaymath}\begin{array}{rclrcl} \tilde{f}^{+}&\isdef & f^{{+}}/\sqrt{R}...
...rt{R}\qquad & \tilde{v}^{-}& \isdef & v^{-}\sqrt{R} \end{array}\end{displaymath} (C.53)

where we have dropped the common time argument `$ (n)$' for simplicity. As a result, we obtain

\begin{displaymath}\begin{array}{rcccl} {\cal P}^{+}& = & f^{{+}}v^{+}&=& \tilde...
...\ &=&(f^{{+}})^2 / R&=& (\tilde{f}^{+})^2 \nonumber \end{array}\end{displaymath}    

and

\begin{displaymath}\begin{array}{rcccl} {\cal P}^{-}& = & -f^{{-}}v^{-}&=& -\til...
... &=&(f^{{-}})^2 / R&=& (\tilde{f}^{-})^2. \nonumber \end{array}\end{displaymath}    

The normalized wave variables $ \tilde{f}^\pm $ and $ \tilde{v}^\pm $ behave physically like force and velocity waves, respectively, but they are scaled such that either can be squared to obtain instantaneous signal power. Waveguide networks built using normalized waves have many desirable properties [174,172,432]. One is the obvious numerical advantage of uniformly distributing signal power across available dynamic range in fixed-point implementations. Another is that only in the normalized case can the wave impedances be made time varying without modulating signal power. In other words, use of normalized waves eliminates ``parametric amplification'' effects; signal power is decoupled from parameter changes.


Total Energy in a Rigidly Terminated String

The total energy $ {\cal E}$ in a length $ L$, rigidly terminated, freely vibrating string can be computed as

$\displaystyle {\cal E}(t)$ $\displaystyle \isdef$ $\displaystyle \int_{0}^L W(t,x)dx = \int_{t_0}^{t_0+2L/c}{\cal P}(\tau,x) d\tau$ (C.54)
  $\displaystyle \approx$ $\displaystyle \sum_{m=0}^{N/2-1} W(t_n,x_m)X = \sum_{n=1}^{N}{\cal P}(t_0 + t_n,x_m) T$ (C.55)

for any $ x\in[0,L]$. Since the energy never decays, $ t$ and $ t_0$ are also arbitrary. Thus, because free vibrations of a doubly terminated string must be periodic in time, the total energy equals the integral of power over any period at any point along the string.
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Scattering at Impedance Changes
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The Dispersive 1D Wave Equation