Electrical waves supply excellent examples of all kinds of waves, and are often used for simulation of other kinds of waves. In this article, we shall consider harmonic waves on an ideal lossless transmission line made from linear circuit elements. Such an ideal line can be very closely approached by one constructed from actual components. Also, a line with distributed parameters can be closely approached by a line of sections made from discrete components. The lines we consider will be considered to be made up from such sections, where appropriate.
This article is intended to explain the backward wave, one in which the phase velocity is opposite to the energy velocity, a very counter-intuitive concept. However, backward waves are by no means nonexistent, and are simply a normal phenomenon that can occur with dispersive propagation. Dispersive propagation occurs when the phase velocity depends on frequency. Most of our acquaintance with waves is with nondispersive propagation, in which the phase velocity is independent of the frequency. In that case, the phase velocity, the energy velocity, and the group velocity are all equal to one another and are constants. This simple relation does not hold in general.
There is one common example of dispersive wave propagation that is easily observable, that of surface waves on water. If the depth h of the water is much less than the wavelength, then the phase velocity v = √(gh) is independent of wavelength and the propagation is nondispersive. On the other hand, if the depth of the water is much greater than the wavelength, the phase velocity is vp = √(g/k), where k = 2π/λ is the propagation constant, and λ is the wavelength. As we shall see, the phase velocity is the ratio ω/k, where ω = 2πf is the angular freqency. Therefore, for deep water waves, ω2 = gk is the relation between ω and k, called the dispersion relation. The group velocity vg is defined as the derivative dω/dk, and is about the speed at which a localized disturbance will travel, and is often the speed at which energy travels, as it is in the case of these waves. The group velocity of deep-water waves is, therefore, vg = vp/2.
If you drop a pebble in still water, a circular ring of disturbance moves out at the group velocity. If you look closely at the ring, it will be seen that waves are being created at the inner boundary of the disturbance that move outwards faster than the disturbance itself, and disappear at the forward edge. This is an excellent demonstration of the difference between phase and group velocities, and it is also clear that the energy moves at the group velocity. For this to succeed, the water must be considerably deeper than the wavelength of the waves in the disturbance, of course. In shallow water, the disturbance and the wavelets will move in unison.
We are going to consider voltages and currents that depend on time like x(t) = a cos(ωt + α), called harmonic or sinusoidal. There are two degrees of freedom in such a signal, the amplitude a and the phase α. We can express any sinusoidal signal in this form. Since we often have to consider linear combinations of signals of various amplitudes and phases, it is very convenient to express signals as the real part of a complex exponential ejωt = cos ωt + j sin ωt. A function like x(t) can be expressed as Re[Aejωt], where A = aejα is a complex amplitude called a phasor. Since all quantities include the time dependence ejωt, this factor can be omitted, and we can work with the phasors only. If y(t) = b cos(ωt + β), we can replace x(t) by A and y(t) by B, and write, for example, C = A + B, where C is the sum of the two complex numbers A and B (easy to compute), and the function x(t) + y(t) is the real part of Cejωt.
The phasors defined so far represent the peak values of the signals. It is often convenient to work with rms values instead, which are peak values divided by √2. In this article, we use peak value phasors only, though in most AC circuit work rms phasors are used. The one thing that cannot be done with phasors is multiplication and division. This certainly does not mean that we cannot multiply and divide them by complex numbers, but only that the product of two phasors has no meaning. One multiplication that we often do have to do is that invoved in calculating power, for example. If the instantaneous voltage is x(t) and the instantaneous current is y(t), then the instantaneous power is p(t) = x(t)y(t). What we are usually interested in is the time-average power P. In terms of the phasors, this is given by P = Re[AB*/2] = Re[A*B/2], where the * indicates complex conjugation (j → -j). It is easy to prove this by finding the explicit time average of the product of the sinusoids, using the formula for cos(A + B), and showing that it is equal to the phasor expression. Note that if we are using rms phasors, there is no division by 2.
The use of phasors makes the relation of current and voltage easy. We write V = IZ, where V and I are the phasor voltage and current, and Z is a complex number called the impedance. For R, L and C in series, Z = R + jωL + 1/jωC. Or, we write I = VY, where Y = 1/Z is a complex number called the admittance. For R, L and C in parallel, Y = 1/R + jωC + 1/jωL. These things should be familiar from AC circuit theory.
The energy stored in an inductance L is u(t) = Li(t)2/2. The average energy stored in the inductance is then U = L|I|2/4. Similarly, the energy stored in a capacitor is u(t) = Cv(t)2/2, so the average stored energy is U = C|V|2/4. The average power dissipated in a resistance R is P = R|I|2/4 = |V|2/4R.
A transmission line may be represented as in the diagram. A source, which we have represented as a sinusoidal voltage source, say Vejωt is connected across the input end of the line. The load is represented as a resistance R, which dissipates a power P = |V'|2/4R, where V' is the voltage at the load end of the line. This resistance is represented as having the value R = Zo, equal to the characteristic impedance of the line. The characteristic impedance is, in fact, normally resistive and has the special property that in a line so terminated the voltage wave from the source is completely absorbed, so that there is no reflected wave. The transmission line, in effect, transmits power from the source to the load. Note that the coordinate z identifies position along the line, starting from z = 0 at the source.
An infinitesimal length of the line dz is represented as at the left, characterized by a series impedance Z per unit length, and a shunt admittance Y per unit length. The voltages and currents at the input and output of the section are shown. The current I decreases by dI = -VYdz, and the voltage by dV = -IZdz in the length of the section, so we find the first-order differential equations dI/dz = -YV and dV/dz = -ZI, which give the dependence of V and I on z. If we eliminate I between the two equations, we get the second-order equation d2V/dz2 = ZYV in V alone. The product ZY is a constant, which we set equal to (jk)2. The reason for this choice is that we already know the outcome, and that ZY will be negative for the line we shall consider. This means that V is the solution of V" + k2V = 0 (the primes stand for differentiation with respect to z), which is very familiar to us. The general solution of this equation is V = Aejkz + Be-jkz. Then, I can be found as I = -(1/Z)dV/dz = (jk/Z)(-Aejkz + Be-jkz) = √(Y/Z)(Be-jkz - Aejkz). The characteristic impedance of the line is √(Z/Y), giving the ratio of voltage to current.
Now we need to interpret these solutions. If V = Aejkz, then the full expression for the resulting voltage is v(t) = Re[aej(kz + ωt + α). The expression (kz + ωt + α) is called the phase of the signal. To each value of the phase from 0 to 2π corresponds a value of the signal. If the phase is to be constant, so that we keep an eye on a particular value of the signal, then kz + ωt = const., or z = (const. - ωt)/k, which says that z is moving to the left, decreasing, with a velocity of ω/k. This is a phase velocity vp = -ω/k. Taking the other solution, V = Be-jkz, we find a phase of (-kz + ωt + β). Now, for constant phase, z = +(ω/k)t, so the wave is moving to the right. Phase velocity is illustrated in the diagram. Its meaning should be thoroughly understood.
In general, then, we have waves moving in both directions along the line, and their sum must satisfy the boundary conditions at the source and at the load. For a wave moving to the right, V = Be-jkz and I = (B/Zo)e-jkz. The ratio of these is V/I = Zo at any point, and in particular at the load, so the boundary condition is satisfied there. Conditions at the source give B = V. We now know the voltage and current at any point on the line.
Suppose that we have a signal composed of two sinusoids of neighboring frequencies, say ej(ωt - kz) + ej[(ω + dω)t - (k + dk)z]. Factoring, this is ej(ωt - kz)[1 + ej(tdω - zdk)]. This represents an amplitude-modulated sinusoid, with the modulation amplitude in square brackets. (If you work this out in detail, it turns out to be beats.) We only need to note that a particular point on the envelope moves in keeping tdω - zdk constant, which implies that z increases as (dω/dk)t. The group velocity vg is defined as dω/dk. Wave packets with a dominant frequency ω and propagation vector k move, approximately, with this velocity.
We will now consider a line where Z = jωL and Y = jωC. This line has series inductance and shunt capacitance, but no resistive losses, due to either a series resistance or a shunt conductance. This is the familiar case of parallel conductors, a pole line above the earth or a coaxial cable. For this sort of line, (jk)2 = (jωL)(jωC), or k2 = ω2LC. This equation, or k = ω√(LC), is called the dispersion relation for this line. When ω is plotted as a function of k, it is a straight line of constant slope ω/k = 1/√(LC) = vp. dω/dk = ω/k = 1/√(LC), so vg = vp. The propagation is nondispersive, and a wave packet travels at the phase velocity without change of shape. The characteristic impedance Zo = √(Z/Y) = √(L/C).
The wave moving to the right will be V = Be-jkz and I = (B/Zo)e-jkz, and the power dissipated in the matched load resistor will be P = |B|2/2Zo. Since the line is lossless, this will also be the power supplied by the source, as is easily checked by setting z = 0 in V and I. The energy per unit length of the line will be U = L|B|2/4Zo2 + C|B|2/4, summing the magnetic and electric energies. This gives U = (|B|2/4)[L(C/L) + C] = |B|2C/2. Note that the magnetic and electric energies are equal. All these energies are, of course, time averages.
The energy velocity vE will be the power P delivered to the load divided by the energy U stored per unit length in the line. This gives vE = (|B|2/2Zo)(2/|B|2C) = 1/√(LC) = vp = vg. The energy velocity is equal to the phase velocity and the group velocity. This is characteristic of nondispersive propagation. In the case of small dispersion, it is found that the energy velocity is equal to the group velocity, but is different from the phase velocity.
Now let us consider a line with Z = 1/jωC and Y = 1/&jωL. Here C is not capacitance per unit length, but 1/C is proportional to length, and similarly for L. That is, the units of C are F-m, not F/m, and L is H-m, not H/m. A discrete realization of such a line is shown in the figure. C and L are the values of the components, if we let one section correspond to one unit of distance. It is a high-pass line, unlike the forward-wave line we just considered, which is a low-pass line. The dispersion relation is k2 = 1/ω2LC, or vp = ω/k = 1/k2√(LC). This line is very strongly dispersive.
The group velocity, found in the usual way by differentiating the dispersion relation comes out to be vg = -1/k2√(LC) = -vp. It is the negative of the phase velocity! Of course, if the wave travels in the other direction, the group velocity reverses too. The wavelets in a wave packet would appear to travel in a direction opposite to that of the packet.
Now we must take a wave V = Aejkz, which travels from load to source with negative phase velocity, in order to transmit energy from source to load. The power delivered is again P = |A|2/2Zo. The energy density is U = (L/4)|VY|2 + (C/4)|IZ|2, which works out to U = |A|2k2C/2 or |A|2/2ω2L. Again, magnetic and electric energies are equal on average. The energy velocity vE = P/U = 1/k2√(LC) = vg, equal to the group velocity. This is not at all unusual.
Dispersion relations for the forward-wave line and the backward-wave line are plotted in the diagram. These are the curves for these particular simple lines. Other systems may have different curves, but the important features are shown here. The phase velocity is given by the slope of the line from the origin to a point on the dispersion curve, while the group velocity is given by the slope of a tangent to the dispersion curve. If the dispersion curve has only tangents in the same direction as the radius from the origin, then the phase and group velocities are in the same direction, and all the waves are forward. If the slope of the tangents are opposite, then there are backward waves.
Backward waves such as this happen at frequencies within bands of strong absorption, where the frequency decreases as the propagation vector increases (wavelength decreases). Phase velocity that decreases with increasing propagation constant is called anomalous dispersion, and backward waves are just one of the interesting phenomena that occur in such regions.
S. Ramo, J. R. Whinnery and T. Van Duzer, Fields and Waves in Communication Electronics (New York: John Wiley & Sons, 1965). pp. 51-53.
Composed by J. B. Calvert
Created 15 August 2003