Transmission Line Analysis

Contents

Introduction

A transmission line is a cable for carrying an electrical signal from one place to another. We usually don't think of transmission lines as cables, but they do provide some filter-like functions, and their behavior is interesting. Examples of transmission lines are coaxial cables and twisted pair cables. The conductors have resistance and inductance. The gap between them creates capacitance. In the early days of cable-making, there would be current leaking through the insulation as well, but in modern cables, such leakage is negligible. The inductance and capacitance dominate the behavior of all good transmission lines. After a certain length of cable, the resistance will cause significant attenuition of the signal.

Wave Equation

If we imagine a perfect transmission line, it will take carry the voltage from one end to the other in a fixed time interval, regardless of the rate at which the voltage changes. If we put a signal V(t), where t is time, in at one end, the signal at the other end will be V(t−τ), where τ is the time it takes for a signal to propagate from one end of the cable to the other. We accept that τ cannot be zero because no signal can travel faster than the speed of light.

Let us determine the signal-carrying properties of a transmission line by modelling it with as a distributed inductor, resistor, and capacitor, as shown below.

Our transmission line element is a short length, δx, of cable. Distance along the cable is x. Although the capacitance, inductance, and resistance of a transmission line are distributed and mingled with one another, we lump them into three separate components in our model of a transmission line element. As δx→0, our lumped model becomes a distributed model.

Let us consider the voltage across the inductor and resistor, which are in series, and experience the current I(x,t).

The rate of change of voltage with x at a particular time is a function of the rate of change of current with time and the current itself.

The rate of change of current with x at a particular time is proportional to the rate of change of voltage with time. Let us differentiate (1) with respect to x and (2) with respect to t. We use the resulting equations to eliminate terms in I.

We arrive at a partial differential equation in V. If we assume R is zero, we are left with the second derivative in x being proportional to the second derivative in t. These are the conditions under which a sinusoidal wave will propagate without distortion or attenuition. Consider a sinusoide of frequency f = ω/2π, as shown below.

The sinusoidal wave has the unique property that its derivatives have the same shape as the original. There is some scaling of the amplitude of the waveform as we differentiate, and it is this scaling that constrains the solution to our transmission line equation. If we set t = √(LC)x, we see the movement of the positive zero-crossing of the sinusoid (the value of sine when its angle is zero). We have dx/dt = 1/√(LC). The velocity of the sine wave is 1/√(LC). Provided that L and C remain constant with ω, the velocity of all sine waves will be the same. If we represent our input V(0,t) as a sum of sinusoids using a Fourier transform, all these sinusoids will propagate along the transmission line at the same speed, so that their sum will remain undistorted as it propagates, and we will have our ideal transmission line: V(x,t) = V(t−τ), with τ = x√(LC).

Common Lines

The following table gives the values of L and C for various mediums, including free space. The capacitance of free space is the same as the permittivity of free space. The inductance of free space is the same as the permeability of free space.

The high-frequency resistance of wires is proportional to √ω, due to the skin effect, which we will discuss later. For now, the table gives R for f = ω/2π ≤ 1 kHz. We obtained R for the cables by adding the conductor and shield DC resistance per meter. In a vacuum, there is no heat-generating resistance to the movement of charges so R = 0. We are not confident that we can defend our assignation of R = 0 in water, but that's our best guess. Also in the table is the characteristic impedance of the medium.

Characteristic Impedance

Consider the relationship between voltage and current at the input of our transmission line.

When we let R = 0, we see that V(x)/I(x) is not a function of t, nor even of x. At any x, V/I = √(L/C). So far as the source of V(0,t) is concerned, the transmission line behaves in exactly the same way as a resistor of value √(L/C). We call this resistance the characteristic impedance of the transmission line. The characteristic impedance of free space, for waves propagating through a vacuum, is 377 Ω. The characteristic impedance of water is 42 Ω. Waves propagate through a vacuum at 299 m/μs, but they propagate through water at only 34 m/μs.

General Solution

Our results so far have been subject to the assumption that R = 0. Let us now consider the case when R > 0. We have the second-order differential equation in V given above. Instead of a simple sinusoidal solution, we propose a solution in which the sinusoid amplitude decreases exponentially with distance as it propagates along the cable at a fixed velocity. We have good reason to suggest a solution of this form. No circuit made of capacitors, inductors, and resistors can distort the shape of a sinusoid, so we are confident that a sinusoid propagating down a transmission line will remain a sinusoid. For small values of R, the cable beyond an element of transmission line behaves like a pure resistance, Z. The element's resistance will divide the sinusoidal amplitude by 1/(1+Rδx/Z), which implies exponential attenuition with distance.

The parameter γ, is the wavenumber of the signal propagating down the transmission line. We have γ = 2π/λ, where λ is the wavelength. The parameter b is the attenuition constant, which defines how the amplitude of our sine wave decreases with distance along the line. In the case where R = 0, we would have b = 0 and γ = ω√(LC). With the complex exponential representation, our derivatives are simpler, as you can see below.

Now we substitute the Derivatives of the General Solution (7) into the Transmission Line Equation (3). We compare the real and imaginary parts of the resulting identity and so obtain two equations in b and γ, which we solve for γ to obtain Equation 8.

The wave velocity, v = ω/γ,is the speed with which a peak in the wave propagates along the transmission line. The wavelength, lambda = 2π/γ, is the distance between peaks in the wave at a particular point in time. The following table gives γ, v, and λ under various conditions.

As we increase ω from zero, the wave velocity increases from zero and approaches 1 / √(LC). Thus high-frequency waves travel more quickly than low-frequency waves. Consider RG58/U cable at 1 kHz. We have ωL = 1.7 mΩ/m << R = 53 mΩ/m. So v = 25 m/μs at 1 kHz. As ω → ∞ we have v → 200 m/μs.

Before we start to imagine higher-frequency waves getting ahead of low-frequency waves, let us solve for the attenuition constant, b. If the high-frequency waves are attenuated more quickly than low frequency waves, it won't matter how much the high-frequency waves get ahead of the low-frequency waves, because the high-frequency waves will become vanishingly small.

Here we see two familiar terms in the solution for b. We have √(C/L) ≡ 1/Z and also R/ωL. The wave amplitude after one meter is aexp(−b). The following table gives b and exp(−b) under various conditions.

Consider RG58/U cable at 1 kHz. We have ωL = 1.7 mΩ/m << R = 53 mΩ/m. So b = 0.00012/m and the attenuition will be 0.001 dB/m. As ω → ∞, b → 0.0093 R. If R did not increase with frequency, but remained constant at 53 mΩ/m, we would have b → 0.00049/m and the attenuition would be 0.0043 dB. Meanwhile, our 1-kHz waves would move at 25 m/μs (see above) and our high-frequency waves would move at 200 m/μs (see above). If we apply a step change in V, the rising edge of the step will travel along 100 m of cable in 0.5 μs, while the 1 kHz component of the step will arrive in 4 μs.

Skin Effect

But the resistance of conductors in cables does increase with frequency, as a result of the skin effect. High-frequency currents induce magnetic fields that in turn apply a force to the electrons in the high-frequency current itself. These fields drive the moving electrons away from the center of the current flow. In a wire, the current is pushed to the surface. The skin depth at a particular frequency, ω, is the depth that includes 63% (1/e) of the total current flow.

δ = √(2 ρ / ω μ)

Where δ = skin depth, ρ = resistivity of the material, and μ is the permeability of the material. For copper, we have ρ = 1.7 ×10⁻⁸ Ωm and μ = 4π ×10⁻⁷ N/A². Thus δ = 160/√ω mm = 64/√f mm. At 1 kHz, δ = 2 mm. At 1 GHz, δ = 2.1 μm.

The central, solid copper conductor of an RG58/U cable has radius 400 μm. At 1 kHz, the skin depth is five times the radius, and so has negligible effect upon the current flow. That's why we say R remains constant below 1 kHz. But at 1 GHz, the skin depth is only 2.1 μm, and only a small fraction of the copper in the wire carries current. The current density, J, at depth x decreases from a maximum, J_o, at the surface in the following way.

J = J_oexp(−x/δ)

If we intergrate J with respect to x, we find that the total current around the surface of a wire of radius r is 2πrJ_oδ. We want to know is the effective series resistance of a one-meter section of wire of radius r at frequency ω. We do this by an argument of conservation of energy. The inductance and capacitance of a transmission line do not dissipate heat, while R dissipates I²R. To determine R, we calculate the heat dissipated in a one-meter wire and divide by I². We set up the problem like this, integrate like this, and obtain the following equation.

R = ρ / (4 π r δ) = √(ρωμ) / 4√2 π r

The above relationship applies when ω >> 2π kHz. For copper wires we have:

R = 21 √f / r nΩ/m

For our 400-μm radius copper wire at 1 GHz we have R = 1.6 Ω/m, so b = 0.029/m and the attenuition of 100 m of the central conductor is 13 dB. But we have not taken account of the resistance of the return conductor, which is the shield. The shield diameter is 3.5 mm, for which R = 0.18 Ω/m at 1 GHz. With the shield resistance added into R, we have b = 0.034/m, and the attenuition of a 100 m RG58/U cable is 15 dB.

When we look up the actual attenuition of RG58/U cable, we find it to be anywhere from 0.3 dB/m to 0.5 dB/m, depending upon the manufacturer, even though all RG58/U has the same diameter copper conductor and shield. Our calculation is only 0.15 dB/m. We measured the attenuition of several cables at 900 MHz, as we show here, and our measurements agreed roughly with the manufacturer's data sheets.

The discrepancy between our prediction and the measured attenuition of coaxial cables is the result of surface roughness and impurities in the copper, primarily the former. In order to obtain the 0.15 dB/m performance of our prediction, the surface of the copper conductor must be polished to better than 1 μm, or else the skin through which the current flows will be an undulating series of valleys and mountains that greatly increase the effective length of the wire. Furthermore, any surface corrosion of the copper will dramatically reduce the resistivity of the surface material. Thus our 0.15 dB/m is the theoretical minimum resistance of a 400-μm radius copper conductor with surface roughness ±100 nm and no corrosion.

The following table shows our perfect-surface attenuition for a 400-μ radius copper conductor, and compares it to the observed attenuition from various manufacturers at a range of frequencies.

The typical surface roughness of drawn copper wire is around 2 μm (0.0001"). At 100 MHz, the skin depth in copper is 6.6 μm, so surface roughness will start to affect the resistance of the wire at this frequency. High-performance coaxial cables have polished silver conductors in the center, like these.

Distortion

According to our analysis, a 100-m length of RG58/U cable will attenuate the amplitude of a 1-GHz signal by a minimum 15 dB, when the copper is pure on its surface, and perhaps as much as 54 dB. A 1-kHz signal, meanwhile, will be attenuated by only 0.10 dB. The 1-GHz signal will get to the end of the cable in 500 ns, but it will be a 33 times smaller. The 1-kHz signal will get to the end in 4 μs, and it will be no smaller. The separation and attenuition of different frequencies by a transmission line causes distortion of the signals it carries. The longer the transmission line, the greater the distortion. The higher the frequency of the input, the greater the distortion. When we apply a step input, we will find that the sharp edge of the step proceeds sharply at first, but becomes rounded off as it moves farther along the cable.

Sometimes the rounding off of a sharp step in a cable is called dispersion, but dispersion is distortion caused by different frequencies moving at different speeds, usually higher frequencies lagging behind lower frequencies. Most cable distortion, for practical cables like RG58/U, is caused by attenuition of higher frequencies, not by dispersion, and what dispertion there is works to slow down the low-frequency signals, not the high-frequency signals.

Conclusion

Our discusion answers the most important questions about the behavior of transmission lines. We show that it is the attenuition of higher frequencies that causes the rounding of logic voltage transitions as they propagate along a transmission line. Higher frequencies travel faster than lower frequencies, so dispersion cannot be the cause of the rounding. Very low frequencies travel several times slower than high frequencies, and the result is that the low-frequency content of a voltage transition might arrive several microseconds later than the initial sharp edge. This late-arrival explains the settling of transitions that we observe at the far end of long transmission lines. We show that the attenuition of higher frequencies is dominated by the skin effect, which serves to increase the effective resistance of the transmission line in proportion to the square root of the frequency. Once we get 100 MHz, the surface polish of the conductor starts to dominate the performance of the cable, and we come to understand why the highest-performance coaxial cables use polished silver conductors.