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 reactance (§7.1). 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)

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

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:

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

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

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

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.