Moving Rigid Termination

It is instructive to study the ``waveguide equivalent circuit'' of the simple case of a rigidly terminated ideal string with its left endpoint being moved by an external force, as shown in Fig.6.4. This case is relevant to bowed strings9.6) since, during time intervals in which the bow and string are stuck together, the bow provides a termination that divides the string into two largely isolated segments. The bow can therefore be regarded as a moving termination during ``sticking''.

Figure 6.4: Moving rigid termination for an ideal string at time $ t_0$.
\includegraphics[width=\twidth]{eps/fMovingTermPhysical}

Referring to Fig.6.4, the left termination of the rigidly terminated ideal string is set in motion at time $ t=0$ with a constant velocity $ v_0$. From Eq.$ \,$(6.5), the wave impedance of the ideal string is $ R=\sqrt{K\epsilon }$, where $ K$ is tension and $ \epsilon $ is mass density. Therefore, the upward force applied by the moving termination is initially $ f_0=Rv_0$. At time $ t_0<L/c$, the traveling disturbance reaches a distance $ c t_0$ from $ x=0$ along the string. Note that the string slope at the moving termination is given by $ -v_0 t_0/(c t_0) = -v_0/c = -(f_0/R)/c = -f_0/K$, which derives the fact that force waves are minus tension times slope waves. (See §C.7.2 for a fuller discussion.)

Digital Waveguide Equivalent Circuits

Figure 6.5: Digital waveguide ``equivalent circuits'' for the rigidly terminated ideal string with a moving termination. a) Velocity-wave simulation. b) Force-wave simulation.
\includegraphics[width=\twidth]{eps/fMovingTerm}

Two digital waveguide ``equivalent circuits'' are shown in Fig.6.5. In the velocity-wave case of Fig.6.5a, the termination motion appears as an additive injection of a constant velocity $ v_0$ at the far left of the digital waveguide. At time 0, this initiates a velocity step from 0 to $ v_0$ traveling to the right. When the traveling step-wave reaches the right termination, it reflects with a sign inversion, thus sending back a ``canceling wave'' to the left. Behind the canceling wave, the velocity is zero, and the string is not moving. When the canceling step-wave reaches the left termination, it is inverted again and added to the externally injected dc signal, thereby sending an amplitude $ 2v_0$ positive step-wave to the right, overwriting the amplitude $ v_0$ signal in the upper rail. This can be added to the amplitude $ -v_0$ signal in the lower rail to produce a net traveling velocity step of amplitude $ v_0$ traveling to the right. This process repeats forever, resulting in traveling wave components which grow without bound, but whose sum is always either 0 or $ v_0$. Thus, at all times the string can be divided into two segments, where the segment to the left is moving upward with speed $ v_0$, and the segment to the right is motionless.

At this point, it is a good exercise to try to mentally picture the string shape during this process: Initially, since both the left end support and the right-going velocity step are moving with constant velocity $ c$, it is clear that the string shape is piece-wise linear, with a negative-slope segment on the left adjoined to a zero-slope segment on the right. When the velocity step reaches the right termination and reflects to produce a canceling wave, everything to the left of this wave remains a straight line which continues to move upward at speed $ v_0$, while all points to the right of the canceling wave's leading edge are not moving. What is the shape of this part of the string? (The answer is given in the next paragraph, but try to ``see'' it first.)


Animation of Moving String Termination and Digital Waveguide Models

In the force wave simulation of Fig.6.5b,7.4 the termination motion appears as an additive injection of a constant force $ f_0=Rv_0$ at the far left. At time 0, this initiates a force step from 0 to $ f_0$ traveling to the right. Since force waves are negated slope waves multiplied by tension, i.e., $ f^{{+}}=-Ky'^{+}$, the slope of the string behind the traveling force step is $ y'=-f_0/K$. When the traveling step-wave reaches the right termination, it reflects with no sign inversion, thus sending back a doubling-wave to the left which elevates the string force from $ f_0$ to $ 2f_0$. Behind this wave, the slope is then $ y'=-2f_0/K$. This answers the question of the previous paragraph: The string is in fact piecewise linear during the first return reflection, consisting of two line segments with slope $ -f_0/K$ on the left, and twice that on the right. When the return step-wave reaches the left termination, it is reflected again and added to the externally injected dc force signal, sending an amplitude $ 2f_0$ positive step-wave to the right (overwriting the amplitude $ f_0$ signal in the upper rail). This can be added to the amplitude $ f_0$ samples in the lower rail to produce a net traveling force step in the string of amplitude $ 3f_0$ traveling to the right. The slope of the string behind this wave is $ y'= -3f_0/K$, and the slope in front of this wave is still $ -2f_0/K$. The force applied to the string by the termination has risen to $ 3f_0$ in order to keep the velocity steady at $ v_0$. (We may interpret the $ f_0$ input as the additional force needed each period to keep the termination moving at speed $ v_0$--see the next paragraph below.) This process repeats forever, resulting in traveling wave components which grow without bound, and whose sum (which is proportional to minus the physical string slope) also grows without bound.7.5The string is always piecewise linear, consisting of at most two linear segments having negative slopes which differ by $ -f_0/K$. A sequence of string displacement snapshots is shown in Fig.6.6.

Figure 6.6: Successive snapshots of the rigidly terminated ideal string with a moving termination. For clarity, the string is plotted higher on each successive snapshot. (One can consider both endpoints to be moving at the same speed up to time 0, after which the left termination begins moving faster by a constant velocity offset.)
\includegraphics[width=\twidth]{eps/moveterm}


Terminated String Impedance

Note that the impedance of the terminated string, seen from one of its endpoints, is not the same thing as the wave impedance $ R=\sqrt{K\epsilon }$ of the string itself. If the string is infinitely long, they are the same. However, when there are reflections, they must be included in the impedance calculation, giving it an imaginary part. We may say that the impedance has a ``reactive'' component. The driving-point impedance of a rigidly terminated string is ``purely reactive,'' and may be called a reactance7.1). If $ f(t)$ denotes the force at the driving-point of the string and $ v(t)$ denotes its velocity, then the driving-point impedance is given by (§7.1)

$\displaystyle R(j\omega) \isdefs \left.\frac{F(s)}{V(s)}\right\vert _{s=j\omega},
$

where $ F(s)$ and $ V(s)$ denote the Laplace transforms of $ f(t)$ and $ v(t)$. In the case of a rigidly terminated string above, as well as in any system in which all energy delivered to the system is ultimately reflected back to the input, the impedance is purely imaginary at every frequency (a ``pure reactance''), as is easy to show:

$\displaystyle R(s) \isdefs \frac{F(s)}{V(s)}
\eqsp \frac{F^{+}+F^{-}}{V^{+}+V^...
...{-s2L/c}F^{+}}{V^{+}-e^{-s2L/c}V^{-}}
\eqsp R\frac{1+e^{-s2L/c}}{1-e^{-s2L/c}}
$

where $ L$ denotes the string length. Let $ P=2L/c$ denote the period of string vibration. Then on the frequency axis $ s=j\omega$ we have

$\displaystyle R(j\omega)
\eqsp R\frac{1+e^{-j\omega P}}{1-e^{-j\omega P}}
\eqsp R\frac{2\cos(\omega P/2)}{2j\sin(\omega P/2)}
\eqsp -jR\,\cot(\omega P/2).
$

Thus, the driving-point impedance of a rigidly terminated string is purely reactive (imaginary), with alternating poles and zeros along the $ j\omega $ axis. Impedance will be discussed further in §7.1 below.


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The Ideal Plucked String
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Rigid Terminations