Transmission lines. Shouri Chatterjee. October 22, 2014

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1 Transmission lines Shouri Chatterjee October 22, 2014 The transmission line is a very commonly used distributed circuit: a pair of wires. Unfortunately, a pair of wires used to apply a time-varying voltage, is not ideal when the wavelength of the signal applied is comparable to the length of the wires. The following is a brief analysis of the transmission line. 1 The general distributed circuit A small section, of length x, of a transmission line, is modelled as a combination of a series resistance, a series inductance, a shunt capacitance, and a shunt conductance. A transmission line is made up of a cascade of several of these small sections. The values of the resistance, inductance, capacitance and conductance are proportional to the length of the transmission line section, and are typically denoted with the help of per-unit-length quantities. A transmission-line section is shown in Fig. 1. One method of analysis could be to compute the transmission (ABCD) matrix for the two-port network in Fig. 1, and cascade an infinite number of these. However, this analysis style will not be pursued in this discussion. i(x, t) R x L x i(x + x, t) + + v(x, t) G x v(x + x, t) - C x - x Figure 1: A section of a transmission line, of length x. The inductance, capacitance, resistance and conductance per unit length are L, C, R and G respectively. 1

2 One can construct other distributed circuits. The analyses of other distributed circuits is typically similar to the analysis we will present here for transmission lines. 2 Analysis technique Let us first evaluate the Kirchoff s equations for the transmission-line section in Fig. 1. Kirchoff s voltage and current laws give the following equations, (1), (2), respectively. v(x, t) v(x + x, t) = L di(x, t) x + i(x, t)r x (1) dt i(x, t) i(x + x, t) = C dv(x + x, t) x + v(x + x, t)g x (2) dt For this treatment of transmission lines, let us assume that they are loss-less, i.e, R and G are zero. We will extend our analysis of loss-less transmission lines to lossy transmission lines later. Further, let us analyze transmissionline sections of infinitesimally small length, at the limit of x tending to zero. (1) then reduces to: v(x, t) v(x + x, t) lim x 0 x or, v x di(x, t) = L dt i = L Note that we have started using partial derivatives because both i and v are functions of two variables, x and t. Reduction of (2) leads us to: (3) i(x, t) i(x + x, t) lim x 0 x v(x + x, t) = C (4) With x tending to zero, the term v(x+ x,t) (2) will therefore, further simplify to: is going to be equal to v(x,t). i x = C v (3), (5) together form a pair of coupled partial differential equations in two variables. They are known as the Telegrapher s equations. The solution of the Telegrapher s equations is the solution for the transmission-line problem. (5) 2

3 The Telegrapher s equations: v x i x i = L v = C The solution to the Telegrapher s equations is the solution to the transmission-line problem. (6) (7) 3 The wave equation The first step towards solving the Telegrapher s equations is to decouple the two equations. We can take the time derivative of (6) and compare with the spacial derivative of (7). The outcome of this is as follows: 2 v = L 2 i x 2 and 2 i = C 2 v x 2 x This gives: 2 i = L C 2 i x 2 2 (8) Likewise, we can take the time derivative of (7) and compare with the spacial derivative of (6). This results in a similar equation: 2 v x = 2 L C 2 v (9) 2 (8) and (9) are decoupled from each other. They happen to be the same equation, one for voltage and the other for time. In the mathematics literature, the form of this equation is known as the wave equation. The solution to the wave equation is the sum of a forward moving wave and a backward moving wave. If f(x) is any function of the variable x, then f(x ct) is a forward moving wave, f(x + ct) is a backward moving wave. c is the velocity of the wavefront. Plugging in f(x ct) as a solution to i(x, t) in (8), we get: f (x ct) = c 2 ( 1) 2 f (x ct) or in other words, the velocity of the wavefront, c, is given by 1/ L C. Plugging f(x + ct) as a solution gives an identical result. Working with (9) instead of (8) gives the same result, but for v(x, t). 3

4 For a second order differential equation, there should be two degrees of freedom. The two degrees of freedom are accounted for when we independently choose functions for the forward and backward moving waves. The general solution for both, v(x, t) and i(x, t), is therefore: f 1 (x ct) + f 2 (x + ct) f 1 and f 2 are arbitrary functions. c = 1/ L C. 4 Putting it all together So far, we have solved the wave equation for voltages and currents, but not the pair of Telegrapher s equations. In the solution for voltages and currents with the help of the wave equation, f 1 and f 2 are arbitrary functions, where f 1 is the forward moving wave, f 2 is the backward moving wave. Let us say that v(x, t) is made up of a forward moving function called V + (x ct) and a backward moving function called V (x + ct). Also, say i(x, t) is made up of a forward moving I + (x ct) and a backward moving I (x + ct). v(x, t) = V + (x ct) + V (x + ct) (10) i(x, t) = I + (x ct) + I (x + ct) (11) The above equations (10), (11), now can be plugged into the telegrapher s equations, (6) and (7). V +(x ct) + V (x + ct) = L ( ci +(x ct) + ci (x + ct)) (12) I +(x ct) + I (x + ct) = C ( cv +(x ct) + cv (x + ct)) (13) In (12) and (13), we can solve for I + and I in terms of V + and V. Further, plugging in the value of c as 1/ L C, we obtain the following. I +(x ct) = V +(x ct)/z 0 (14) I (x + ct) = V (x + ct)/z 0 (15) L where, Z 0 = (16) C The function I + is therefore V + /Z 0 with the possible addition of an arbitrary constant. The function I is therefore V /Z 0 with the possible addition of an arbitrary constant. Z 0 = L /C is defined to be the characteristic 4

5 impedance of the transmission line. Note: Z 0 is only a mathematical quantity. It is not a resistor or an impedance. 5 Interpretation of the solution What the analysis of the earlier sections tell us is that over a transmission line, voltages and currents move as waves. Kirchoff s laws are valid at all locations at all times; however the current going into a transmission line at the input port might not be the current coming out of the transmission line at the output port, because of intermediate paths through the distributed capacitors. The voltage along a transmission line is the sum of a forward moving voltage wave, and a backward moving voltage wave. The current travelling along a transmission line is also the sum of a forward moving current wave and a backward moving current wave. The voltage and current waves are related through Z 0, the characteristic impedance. The velocity of all the wavefronts is c. Z 0 is given by L /C, and c is given by 1/ L C. 1 The forward moving current wave is the forward moving voltage wave divided by Z 0. The backward moving current wave is the negative of the backward moving voltage wave divided by Z 0. The above can be summarized as the following. Over a transmission line, there will be a forward moving voltage wave, V +, and a backward moving voltage wave, V. At any point on the transmission line, and at any instant of time, the voltage is V + + V, and the current is (V + V )/Z 0. 6 Reflection coefficient Consider a transmission line of characteristic impedance Z 0, terminated by a load, Z L, as shown in Fig. 2. At the load end, if the forward and backward moving waves are V + and V respectively, then the voltage is V + + V, and the current into the load is (V + + V )/Z L. This also happens to be given by the sum of the forward and 1 A computation of L, the inductance per unit length and C, the capacitance per unit length, for simple geometries of transmission lines, will reveal that c is actually the speed of light in the given medium. Sometimes, this can be a quick way to estimate C if you know L, or vice-versa. 5

6 Z S V in + Z 0 Z L Figure 2: A source with source impedance Z S is being used to excite a load impedance, Z L, over a transmission line of characteristic impedance Z 0. backward moving current waves, (V + V )/Z 0. or or V + + V = V + V Z L Z 0 V + + V = Z L V + V Z 0 Γ L = V V + = Z L Z 0 Z L + Z 0 (17) In (17), we have defined Γ L to be the ratio of the backward moving wave to an impinging forward moving wave at the load. If a forward moving wave, V +, hits a load Z L, through a transmission line of characteristic impedance Z 0, then there will be a reflection from the load given by V = Γ L V +. Γ L is called the reflection coeffecient. (17) shows that the reflection coefficient is 0 if the load is also Z 0, the reflection coefficient is +1 if the load is an open circuit, and the reflection coefficient is 1 if the load is a short circuit. 7 The response to a pulse Let us now consider the situation where the voltage source in Fig. 2 is a step function, that is, V in = V 0 u(t). Before time t = 0, clearly, both the forward and backward moving waves should be equal to 0. Things should start changing only at time t = 0 +. At this instant, the input voltage is V 0. The voltage across the input terminals of the transmission lines is not known; let us assume this to be V x. The current going into the input terminals will then be (V 0 V x )/Z S. The voltage V x is made up of a forward moving wave that has just started, and a backward moving wave that is zero. This means, the forward moving wave is V x. The current, therefore, has to be V x /Z 0. (Remember, the current is (V + V )/Z 0. V + is V x, V is 0.) This gives: V 0 V x Z S = V x Z 0 or, V x = V 0 Z 0 Z S + Z 0 (18) 6

7 (18) shows that the voltage appearing at the input of the transmission line is the result of a resistive division of the applied voltage, as if the transmission line is a resistor of value Z 0. This is not really true, and just happens to be the case. The forward moving wave, V x, moves forward at a velocity c. After a duration l/c (where l is the length of the transmission line), it reaches the load. At the load end, the wave V x impinges on the load and creates a reflection, Γ L V x, at time t = (c/l) +. Now the voltage at the load is V x (1+Γ L ). The backward moving wave is now Γ L V x. The backward moving wave, Γ L V x, moves backwards and impinges on the source resistor at time t = 2l/c. The reflection coefficient at the source (now look at the transmission line backwards, the backward moving wave is your new forward direction, the source resistor is your new load resistor) is Γ S = Z S Z 0 Z S +Z 0. The impinging backward moving wave will cause a reflection, a new forward moving wave, Γ S Γ L V x, at time t = (2l/c) +. The voltage at the source is now going to be V x (1 + Γ L + Γ L Γ S ). The waves sloshing back and forth now happens forever - till the voltages settle to their final values. Graphically, this process is described in Fig. 3. At any point in space and at any time, the voltage is the sum of all the voltage waves that have passed till then. The voltage at the load as the system settles to steady state is going to be given by: lim V (l) = (V x Γ i SΓ i L + V x Γ i SΓ i+1 t L ) = V x Γ i SΓ i L(1 + Γ L ) i=0 = V x (1 + Γ L ) i=0 i=0 (Γ L Γ S ) i = V x 1 + Γ L 1 Γ L Γ S (19) We can plug values for Γ L, Γ S, and V x, into (19) as follows. Γ L = (Z L Z 0 )/(Z L + Z 0 ), Γ S = (Z S Z 0 )/(Z L + Z 0 ), and V x = V in Z 0 /(Z S + Z 0 ). This results in: lim V (l) = V in t Z 0 Z S + Z Z L Z 0 Z L +Z 0 1 Z L Z 0 Z L +Z 0 Z S Z 0 Z L = V in Z Z S +Z L + Z S 0 The result is indeed intuitive. After all, at DC the inductors and capacitors in the transmission line are all short-circuits and open-circuits respectively. This means that at DC, the transmission line is just a wire. Voltage division between Z L and Z S is therefore natural. Exercise 1. Is the above derivation valid when Γ L = 1 and Γ S = 1? 7

8 t = 0 Γ L V + t = 2l/c Γ 2 L Γ SV + t = 4l/c Γ 3 L Γ2 S V + t = 6l/c Γ 4 L Γ3 S V + t = 8l/c x = 0 V + t = l/c Γ L Γ S V + t = 3l/c Γ 2 L Γ2 S V + t = 5l/c Γ 3 L Γ3 S V + t = 7l/c Γ 4 L Γ4 S V + x = l Figure 3: Response to a pulse input. The x-axis is distance, the y-axis is time, going downwards. At any point, x, one can draw a section. The voltage at x at any point of time t, will be the sum of all the waves that have passed x till time t. Exercise 2. Show that the voltage at the source end of the transmission line also settles to V in Z L /(Z L + Z S ) as time progresses. (i.e., the same result as the load end of the transmission line.) Exercise 3. Plot the voltage at the load end of the transmission line as a function of time, when Z 0 is 50 Ω, Z L is 100 Ω, and Z S is 200 Ω. Exercise 4. A 220 V DC generator has a source impedance of close to 0 Ω. A transmission line of length 1 km, and characteristic impedance of 10 Ω is being used to deliver power to a load. The load suddenly switches OFF (i.e., it becomes an open circuit). Show that the switch being used to switch the load off, needs to be rated at least for 440 V. 8 Steady-state sinusoid stimulation Without any loss of generality, the forward and backward moving waves can be phasors. In other words, V + (x ct) and V (x + ct) can be given by V + e j(ωt kx) and V e j(ωt+kx) respectively. Here k is the wave number, and is given as k = ω/c = 2πf/c = 2π/λ, where λ is the wavelength corresponding to the frequency of excitation. The phasor corresponding to the forward moving wave is, then, V + e j2πx/λ. The backward moving wave is V e j2πx/λ. 8

9 V + and V are complex numbers, in general. The current at any point is given by the difference of the forward and backward moving voltage waves, divided by Z 0. Needless to say, we are working completely in the phasor domain. The important thing to note about analysis with phasors is the reflection coefficient. The reflection coefficient at the load is still given by: Γ L = Z L Z 0 Z L + Z 0 where Z L can be any complex impedance. This is because the same analysis as (17) holds. Now the forward moving wave at any point ahead of the load will be the same wave, advanced in time. The backward moving wave at any point ahead of the load will be the same wave, delayed in time. For example, at the source end, the forward moving wave will be e j2πl/λ times the forward moving wave at the load. The backward moving wave will be e j2πl/λ times the backward moving wave at the load. The reflection coefficient at the source is the ratio of the backward to the forward moving waves at the source. This tells us that Γ(0) = Γ L e j4πl/λ. In fact, now we can compute the reflection coefficient at any point in space, over the entire length of the transmission line. Γ(x) = Γ L e j4π(l x)/λ (20) With the knowledge of the reflection coefficient, if we know the voltage, we can evaluate the forward and backward moving waves. Once we know the forward and backward moving waves, we can evaluate the current. Exercise 5. A source of impedance 40 Ω is used to transmit power over a 50 Ω transmission line of length 10λ to a load of 60 Ω. VSWR (voltage standing wave ratio) is defined to be the ratio of the maximum voltage amplitude over the entire length of the transmission line, to the minimum voltage amplitude over the entire length of the transmission line. Evaluate the VSWR. Exercise 6. A 100 kv 50 Hz generator of source impedance 1 Ω is used to transmit power over a 5 Ω 10 km underground high-tension cable, to a load of impedance 1 + j2 Ω. Find the rating of the insulation required for the cable. 9 Quarter-wavelength lines From the preceding discussion, it follows that transmission lines of length λ/4 are special. For such transmission lines, from (20), Γ(0) is given by Γ L. 9

10 The input impedance, looking into the source end of the transmission line, is given by: V + + V 1 + Γ(0) Z in = V/I = Z 0 = Z 0 (21) V + V 1 Γ(0) Given that Γ(0) is nothing but Γ L, we obtain the input impedance as 1 Γ Z L 0 1+Γ L. With the value of Γ L from (17), we find that the input impedance is Z0/Z 2 L. For an open-circuit load, the input impedance looks like a short-circuit. For a short-circuit load, the input impedance looks like an open-circuit. For a capacitive load, the input impedance looks like an inductor, and vice-versa. A λ/4 long transmission line inverts the load impedance, normalized to Z 0. Note, this property is valid only at frequencies at which the length of the transmission line is exactly an odd multiple of λ/4. 10 Additional topics This treatment of transmission lines is quite incomplete. The following additional topics are not being touched upon. Lossy transmission lines have not been discussed at all; Z 0 for lossy transmission lines will be complex. One can solve the entire system of equations in the frequency domain, right from the beginning. This is the easiest way to deal with lossy transmission lines. Power transmission over lossy lines is a required study for power transmission and distribution. Microwave circuits frequently use transmission-line based filters. One can use high-z 0 transmission lines to realize inductors, low-z 0 transmission lines to realize capacitors. Richard s transformation warps the frequency axis for transmission-line based filters. There are several transmission-line based microwave components, for example, rat-race couplers, directional couplers. These can be analyzed easily given the basics. 10

ECE145A/218A Course Notes

ECE145A/218A Course Notes Last note set: Introduction to transmission lines 1. Transmission lines are a linear system - superposition can be used 2. Wave equation permits forward and reverse wave propagation