A transmission line is a pair of electrical conductors carrying an electrical signal from one place to another. Coaxial cable and twisted pair cable are examples. The two conductors have inductance per unit length, which we can calculate from their size and shape. They have capacitance per unit length, which we can calculate from the dielectric constant of the insulation. In the early days of cable-making, there would be current leaking through the insulation, but in modern cables, such leakage is negligible. The electrical resistance of the conductors, however, is significant because it increases with frequency. The magnetic fields generated by high-frequency currents drive those currents to the outer edge of the conductor that carries them, so the higher the frequency, the thinner the layer of metal available to carry the current, and the higher the effective resistance of the cable. In this discussion, we derive and demonstrate the equations that govern the propagation of waves down a transmission line, and show how the frequency-dependent resistance of these cables gives rise to attenuation and distortion of high-frequency signals.
A perfect transmission line will carry an electrical signal from one place to another in a fixed time, regardless of the rate at which the voltage changes. If we apply a signal V(t) to one end of the transmission line, where t is time, the signal at the other end will be V(t−τ), where τ is a constant. We can model a real transmission line with a distributed inductance, capacitance, and resistance. We would like to calculate τ in terms of these distributed properties, and so determine the circumstances under which τ will be constant. The following drawing shows a small element of a transmission line: an element so small that we can assume its distributed properties are lumped into three components.
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. 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 attenuation. 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).
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.
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.
|Medium||C (pF/m)||L (nH/m)||v (m/μs)||Z (Ω)||R for f≤ 1kHz
|RG58/U Coaxial Cable||93.5||273||198||54||53|
|RG58C/U Coaxial Cable||101||252||198||50||50|
|RG59B/U Coaxial Cable||72.0||405||185||75||45|
|CAT-5 Twisted Pair (Solid)||49.2||495||203||100||180|
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.
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. Furthermore, if we consider the case when R is small, 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 attenuation 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 attenuation 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, λ = 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.
|R = 0||ω√(LC)||1 / √(LC)||2π / √(LC)|
|ωL >> R > 0||ω√(LC)||1 / √(LC)||2π / √(LC)|
|ω → ∞||ω√(LC)||1 / √(LC)||2π / √(LC)|
|ωL = R > 0||1.1 ω√(LC)||0.91 / √(LC)||1.8 π / √(LC)|
|ωL << R||0.71 √(ωRC)||0.71 √(ω / RC)||1.4 π / √(ωRC)|
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 attenuation 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 a exp(−b). The following table gives b and exp(−b) under various conditions.
|R = 0||0||1|
|ωL >> R > 0||0.50 R√(C / L) = 0.50 R / Z||exp(−0.50 R/Z)|
|ω → ∞||0.50 R√(C / L) = 0.50 R / Z||exp(−0.50 R/Z)|
|ωL = R > 0||0.45 R√(C / L) = 0.45 R / Z||exp(−0.45 R/Z)|
|ωL << R||0.71 √(ωRC)||exp(−0.71 √(ωRC))|
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 attenuation will be 0.001 dB/m. As ω → ∞, b → 0.0093R. If R did not increase with frequency, but remained constant at 53 mΩ/m, we would have b → 0.00049/m and the attenuation would be 0.0043 dB/m. 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 after 4 μs. Neither the high-frequency nor the 1-kHz components will be attenuated by more than 5%.
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, Jo, at the surface in the following way.
J = Joe−x/δ
If we integrate J with respect to x, we find that the total current around the surface of a wire of radius r is 2πrJoδ. 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 as shown below.
We integrate to solve for the resistance per unit length.
And so we obtain the following equation.
R = ρ / (4 π r δ) = √(ρωμ) / 4√2 π r = √(ρ f μ) / (4 r √π)
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 attenuation 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 attenuation of a 100 m RG58/U cable should be 15 dB. As we will see in the next section, however, the actual attenuation of such cable at 1 GHz is closer to 50 dB for 100 m.
When we look up the actual attenuation of RG58/U cable at 1 GHz, 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 attenuation 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 attenuation 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 conductivity 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 the perfect-surface attenuation for a 400-μm radius copper conductor, and compares it to the observed attenuation from various manufacturers at a range of frequencies.
|Frequency (MHz)||Perfect (dB/100m)||Measured (dB/100m)|
We see that the perfect and measured values are in good agreement at 1 MHz (0.4 dB difference), poor agreement at 100 MHz (9 dB difference), and gross disagreement at 1 GHz (39 dB difference). 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 1 GHz the skin depth is only 2 μm, so the path of the current is dominated by scratches and surface corrosion.
To carry a 1-GHz signal more than a few meters with metal conductors, we must polish the conductor surface. High-performance coaxial cables like these have polished silver conductors in the center.
According to our analysis, higher-frequency signals propagate more quickly down real transmission lines than low-frequency signals. But high-frequency signals are attenuated more quickly. A 100-m length of RG58/U cable will attenuate the amplitude of a 1-GHz signal by 55 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 almost six hundred times smaller in amplitude. The 1-kHz signal will get to the end in 4 μs and it will be almost as large as it started. The separation and attenuation 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.
The single biggest source of distortion in electrical transmission lines is the skin effect, which causes a dramatic increase in the effective resistance of the conductors as frequency increases. 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. Most cable distortion, for practical cables like RG58/U, is caused by attenuation of higher frequencies, not by dispersion, and what dispersion there is works to slow down the low-frequency signals, not the high-frequency signals.
Our discusion answers the most important questions about the behavior of transmission lines. We show that it is the attenuation of higher frequencies that causes the rounding off of a step-change in voltage, not the dispersion of higher frequencies. Indeed, we show that the dispersion of higher frequencies is such that higher frequencies will arrive sooner than lower frequencies. A 1-GHz signal will propagate down a 100-m RG58/U cable in 500 ns, while a 1-kHz signal will take 4 μs. But the amplitude of the 1-GHz signal will be six hundred times smaller when it gets to the other end, and the 1-kHz signal will be hardly attenuated at all. This late-arrival of the low-frequency component of a signal explains the settling of transitions that we observe at the far end of long transmission lines. We show that the attenuation 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 above 100 MHz, the surface polish of the conductor starts to dominate the performance of a cable, and we come to understand why the highest-performance coaxial cables use polished silver conductors.