Using the above identities, we have that the force distribution along
the string is given in terms of velocity waves by
![$\displaystyle f(t,x) = \frac{K}{c} \left[{\dot y}_r(t-x/c) - {\dot y}_l(t+x/c) \right], \protect$](http://www.dsprelated.com/josimages_new/pasp/img3483.png) |
(C.44) |
where
![$ K/c \isdef K/\sqrt{K/\epsilon } = \sqrt{K\epsilon }$](http://www.dsprelated.com/josimages_new/pasp/img3484.png)
. This is a fundamental
quantity known as the
wave impedance of the string (also called
the
characteristic impedance), denoted as
![$\displaystyle R\isdefs \sqrt{K\epsilon } \eqsp \frac{K}{c} \eqsp \epsilon c.$](http://www.dsprelated.com/josimages_new/pasp/img3485.png) |
(C.45) |
The
wave impedance can be seen as the geometric mean of the two
resistances to
displacement: tension (
spring force) and
mass (
inertial
force).
The digitized traveling force-wave components become
![\begin{displaymath}\begin{array}{rcrl} f^{{+}}(n)&=&&R\,v^{+}(n) \\ f^{{-}}(n)&=&-&R\,v^{-}(n) \end{array} \protect\end{displaymath}](http://www.dsprelated.com/josimages_new/pasp/img3486.png) |
(C.46) |
which gives us that the right-going force wave equals the wave
impedance times the right-going velocity wave, and the left-going
force wave equals
minus the wave impedance times the left-going velocity wave.
C.4Thus, in a
traveling wave, force is always
in phase with
velocity (considering the minus sign in the left-going case to be
associated with the direction of travel rather than a
![$ 180$](http://www.dsprelated.com/josimages_new/pasp/img1335.png)
degrees
phase shift between force and velocity). Note also that if the
left-going force wave were defined as the string force acting to the
left, the minus sign would disappear. The fundamental relation
![$ f^{{+}}=
Rv^{+}$](http://www.dsprelated.com/josimages_new/pasp/img3488.png)
is sometimes referred to as the mechanical counterpart of
Ohm's law for traveling waves, and
![$ R$](http://www.dsprelated.com/josimages_new/pasp/img9.png)
in c.g.s. units
can be called
acoustical ohms [
261].
In the case of the acoustic tube [317,297], we have the
analogous relations
![\begin{displaymath}\begin{array}{rcrl} p^+(n) &=& &R_{\hbox{\tiny T}}\, u^{+}(n)...
...p^-(n) &=& -&R_{\hbox{\tiny T}}\, u^{-}(n) \end{array} \protect\end{displaymath}](http://www.dsprelated.com/josimages_new/pasp/img1336.png) |
(C.47) |
where
![$ p^+(n)$](http://www.dsprelated.com/josimages_new/pasp/img1337.png)
is the right-going traveling
longitudinal pressure
wave component,
![$ p^-(n)$](http://www.dsprelated.com/josimages_new/pasp/img1338.png)
is the left-going
pressure wave, and
![$ u^\pm (n)$](http://www.dsprelated.com/josimages_new/pasp/img1339.png)
are the left and right-going
volume velocity waves. In the acoustic
tube context, the wave impedance is given by
(Acoustic Tubes) |
(C.48) |
where
![$ \rho$](http://www.dsprelated.com/josimages_new/pasp/img1197.png)
is the mass per unit volume of air,
![$ c$](http://www.dsprelated.com/josimages_new/pasp/img125.png)
is
sound speed in
air, and
![$ A$](http://www.dsprelated.com/josimages_new/pasp/img251.png)
is the
cross-sectional area of the tube.
Note that if we had chosen
particle velocity rather than volume
velocity, the wave impedance would be
![$ R_0=\rho c$](http://www.dsprelated.com/josimages_new/pasp/img1347.png)
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.
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