A ladder network
We would like now to consider an interesting circuit which can be analyzed in terms of series and parallel combinations. Suppose we start with the circuit of Fig. 22–18(a). We can see right away that the impedance from terminal $a$ to terminal $b$ is simply $z_1+z_2$. Now let’s take a little harder circuit, the one shown in Fig. 22–18(b). We could analyze this circuit using Kirchhoff’s rules, but it is also easy to handle with series and parallel combinations. We can replace the two impedances on the right-hand end by a single impedance $z_3=z_1+z_2$, as in part (c) of the figure. Then the two impedances $z_2$ and $z_3$ can be replaced by their equivalent parallel impedance $z_4$, as shown in part (d) of the figure. Finally, $z_1$ and $z_4$ are equivalent to a single impedance $z_5$, as shown in part (e). Now we may ask an amusing question: What would happen if in the network of Fig. 22–18(b) we kept on adding more sections forever—as we indicate by the dashed lines in Fig. 22–19(a)?  Can we solve such an infinite network? Well, that’s not so hard. First, we notice that such an infinite network is unchanged if we add one more section at the “front” end. Surely, if we add one more section to an infinite network it is still the same infinite network. Suppose we call the impedance between the two terminals $a$ and $b$ of the infinite network $z_0$; then the impedance of all the stuff to the right of the two terminals $c$ and $d$ is also $z_0$. Therefore, so far as the front end is concerned, we can represent the network as shown in Fig. 22–19(b). Forming the parallel combination of $z_2$ with $z_0$ and adding the result in series with $z_1$, we can immediately write down the impedance of this circuit: \begin{equation*} z=z_1\!+\!\frac{1}{(1/z_2)\!+\!(1/z_0)}\quad\text{or}\quad z=z_1\!+\!\frac{z_2z_0}{z_2\!+\!z_0}. \end{equation*}  But this impedance is also equal to $z_0$, so we have the equation \begin{equation} z_0=z_1+\frac{z_2z_0}{z_2+z_0}.\notag \end{equation}  We can solve for $z_0$ to get \begin{equation} \label{Eq:II:22:27} z_0=\frac{z_1}{2}+\sqrt{(z_1^2/4)+z_1z_2}. \end{equation}  So we have found the solution for the impedance of an infinite ladder of repeated series and parallel impedances. The impedance $z_0$ is called the characteristic impedance of such an infinite network. Let’s now consider a specific example in which the series element is an inductance $L$ and the shunt element is a capacitance $C$, as shown in Fig. 22–20(a). In this case we find the impedance of the infinite network by setting $z_1=i\omega L$ and $z_2=1/i\omega C$. Notice that the first term, $z_1/2$, in Eq. (22.27) is just one-half the impedance of the first element. It would therefore seem more natural, or at least somewhat simpler, if we were to draw our infinite network as shown in Fig. 22–20(b). Looking at the infinite network from the terminal $a'$ we would see the characteristic impedance \begin{equation} \label{Eq:II:22:28} z_0=\sqrt{(L/C)-(\omega^2L^2/4)}. \end{equation}  Now there are two interesting cases, depending on the frequency $\omega$. If $\omega^2$ is less than $4/LC$, the second term in the radical will be smaller than the first, and the impedance $z_0$ will be a real number. On the other hand, if $\omega^2$ is greater than $4/LC$ the impedance $z_0$ will be a pure imaginary number which we can write as \begin{equation*} z_0=i\sqrt{(\omega^2L^2/4)-(L/C)}. \end{equation*}  We have said earlier that a circuit which contains only imaginary impedances, such as inductances and capacitances, will have an impedance which is purely imaginary. How can it be then that for the circuit we are now studying—which has only $L$’s and $C$’s—the impedance is a pure resistance for frequencies below $\sqrt{4/LC}$? For higher frequencies the impedance is purely imaginary, in agreement with our earlier statement. For lower frequencies the impedance is a pure resistance and will therefore absorb energy. But how can the circuit continuously absorb energy, as a resistance does, if it is made only of inductances and capacitances? Answer: Because there is an infinite number of inductances and capacitances, so that when a source is connected to the circuit, it supplies energy to the first inductance and capacitance, then to the second, to the third, and so on. In a circuit of this kind, energy is continually absorbed from the generator at a constant rate and flows constantly out into the network, supplying energy which is stored in the inductances and capacitances down the line. This idea suggests an interesting point about what is happening in the circuit. We would expect that if we connect a source to the front end, the effects of this source will be propagated through the network toward the infinite end. The propagation of the waves down the line is much like the radiation from an antenna which absorbs energy from its driving source; that is, we expect such a propagation to occur when the impedance is real, which occurs if $\omega$ is less than $\sqrt{4/LC}$. But when the impedance is purely imaginary, which happens for $\omega$ greater than $\sqrt{4/LC}$, we would not expect to see any such propagation.