Lesson 3 derived the telegrapher’s equations from Kirchhoff’s laws on a short section of line. They hold for any signal, but they are coupled, and each one contains derivatives in both \(z\) and \(t\). Lesson 2 gave us the tool that removes the time derivatives. When the source is a sinusoid at frequency \(\omega\), every voltage and current on the line is a sinusoid at the same frequency, so each one is the real part of a phasor times \(e^{j\omega t}\). On a line the phasor depends on position, so we write it \(\widetilde{V}(z)\): its magnitude is the amplitude of the sinusoid at \(z\), and its angle is the phase there.

We take the derivative of \(v(z,t)\) with respect to time. Taking the real part and differentiating with respect to \(t\) commute, because \(t\) is real, so we can move the derivative inside. The phasor \(\widetilde{V}(z)\) does not depend on time, so the derivative acts only on \(e^{j\omega t}\), which brings down a factor of \(j\omega\). The result is again the real part of a phasor times \(e^{j\omega t}\), and that phasor is \(j\omega\widetilde{V}(z)\). Differentiating in time is the same as multiplying the phasor by \(j\omega\).

The derivative with respect to \(z\) works the other way. Now \(e^{j\omega t}\) does not depend on \(z\), so the derivative acts only on \(\widetilde{V}(z)\). Because the phasor depends on \(z\) alone, its derivative is an ordinary derivative, \(d\widetilde{V}/dz\), not a partial one.

We replace each quantity in the telegrapher’s equations with its phasor: \(v\) with \(\widetilde{V}\), \(i\) with \(\widetilde{I}\), each \(\partial/\partial t\) with \(j\omega\), and each \(\partial/\partial z\) with \(d/dz\). In the first equation the two terms on the right both multiply \(\widetilde{I}(z)\), so we factor it out; in the second, both multiply \(\widetilde{V}(z)\). The factor \(R' + j\omega L'\) is the series impedance of the line per meter, and \(G' + j\omega C'\) is its shunt admittance per meter. The equations are still coupled, but they are now ordinary differential equations in one variable, \(z\), with constant coefficients.

To get an equation in \(\widetilde{V}\) alone, we differentiate the first phasor equation with respect to \(z\). The left side becomes the second derivative of \(\widetilde{V}\), and on the right the constant \(R' + j\omega L'\) stays outside the derivative, leaving \(d\widetilde{I}/dz\). The second phasor equation says exactly what \(d\widetilde{I}/dz\) is: \(-\left(G' + j\omega C'\right)\widetilde{V}(z)\). Substituting it removes the current entirely. Multiplying both sides by \(-1\) and moving everything to the left gives the wave equation for the voltage. We name the product of the series impedance and the shunt admittance \(\gamma^2\), because the solution depends on its square root.

The square root \(\gamma\) is the propagation constant. It is complex, because the series impedance and the shunt admittance are complex, so we write it as a real part \(\alpha\) and an imaginary part \(\beta\). A square root has two values that differ only in sign; we choose the root with \(\alpha \ge 0\) and \(\beta > 0\), and the solution below carries both signs anyway. The units of \(\gamma\) are inverse meters, since \(\gamma z\) must be dimensionless. We mark the real part in nepers per meter and the imaginary part in radians per meter to keep track of what each one does: the solution will show that \(\alpha\) sets how fast the wave decays and \(\beta\) sets how fast its phase changes. Lesson 5 computes both for a real cable.

The same three steps uncouple the current. We differentiate the second phasor equation with respect to \(z\), substitute the first equation for \(d\widetilde{V}/dz\), and rearrange. The product of the two factors is the same \(\gamma^2\), because multiplication does not depend on order, so the current obeys exactly the same wave equation as the voltage. The two equations are now uncoupled: each one contains a single unknown function.

The wave equation is a second-order linear differential equation with constant coefficients, the same kind of equation that describes an RLC circuit in time. We solve it the same way. We assume a solution of the form \(A e^{sz}\), because the derivative of an exponential is the same exponential times a constant. Each derivative brings down a factor of \(s\), so the second derivative is \(s^2 A e^{sz}\). Substituting into the wave equation, the factor \(A e^{sz}\) appears in both terms and is never zero, so we can divide it out. What remains is the characteristic equation, \(s^2 - \gamma^2 = 0\), and its two roots are \(s = +\gamma\) and \(s = -\gamma\).

Each root gives one solution, and because the wave equation is linear, any sum of solutions is also a solution. A second-order equation has exactly two independent solutions, so the general solution is the sum of the two, each with its own constant. We name the constants \(V_0^+\) and \(V_0^-\) for reasons the next two frames make clear. The current has the same form with its own constants, \(I_0^+\) and \(I_0^-\). The four constants are not independent: the current and voltage must still satisfy the first-order phasor equations, and Lesson 5 uses that to find \(I_0^\pm\) from \(V_0^\pm\). The source and load set the rest, which is the subject of Lesson 6.

To see what the solution looks like on the line, we return to the time domain: multiply the phasor by \(e^{j\omega t}\) and take the real part. We write each complex constant as a magnitude and a phase, and split each exponential \(e^{\mp\gamma z}\) into a real part \(e^{\mp\alpha z}\) and an imaginary part \(e^{\mp j\beta z}\). The real exponentials and the magnitudes are real numbers, so they come outside the real part. The remaining imaginary exponentials combine into a single \(e^{j(\cdot)}\), and the real part of \(e^{j\theta}\) is \(\cos\theta\). The result is two sinusoids, each multiplied by an exponential in \(z\).

A crest of the first term sits wherever its cosine argument \(\omega t - \beta z + \phi^+\) is zero. Solving for its position gives \(z = (\omega t + \phi^+)/\beta\), which grows as \(t\) grows, so the crest moves in the \(+z\) direction. Every point of constant phase does the same, so the whole first term is a wave traveling toward the load, and its amplitude \(\left\vert V_0^+\right\vert e^{-\alpha z}\) shrinks as it goes. That is why we named its constant \(V_0^+\). In the second term the crest sits at \(z = -(\omega t + \phi^-)/\beta\), which moves in the \(-z\) direction, toward the source; its amplitude \(\left\vert V_0^-\right\vert e^{\alpha z}\) shrinks as \(z\) decreases, so this wave also decays in its own direction of travel. The voltage on any line is the sum of an incident wave and a reflected wave.

On a lossless line the product under the square root is \(j^2\omega^2 L'C'\), which is a negative real number. Its square root is purely imaginary, so the attenuation constant is zero and the phase constant is \(\omega\sqrt{L'C'}\). Both waves travel without decaying. For RG-58, with \(L' = 259\) nH/m and \(C' = 96.8\) pF/m from Lesson 3, at 100 MHz,

so the phase changes by one full cycle, \(2\pi\) radians, every \(2\pi/3.15 = 2.0\) m. That is the 2 m wavelength Lesson 3 found for this cable at 100 MHz, now derived from \(L'\) and \(C'\).

Today we solved the wave equation and found that the voltage on any line is the sum of two traveling waves. Two questions remain before we can use the solution. The first is how the current in each wave relates to its voltage; the answer is the characteristic impedance \(Z_0\), which Lesson 5 derives from the phasor equations. The second is what the propagation constant means in numbers: how fast the wave travels, how long a wavelength is, and how much a real cable attenuates it. Lesson 5 answers both, and Lesson 6 then terminates the line in a load and finds the reflected wave.

Before Lesson 5, read sections 3.7 through 3.9 of Ellingson, Electromagnetics, Vol. 1.