# How to demonstrate mathematically the conditions of reflection in transmission line?

Electrical Engineering Asked on October 29, 2020

We are all taught about transmission lines and its associates — characteristic impedance, reflection, standing wave, etc. It is well known what happens when a transmission line is terminated with different types of impedances:

1) Open – reflection occurs

2) Shorted – reflection occurs but inverted

3) Equal to Zo – no reflection

4) Between Zo and open/short – partial reflection

So far, so good. Every textbook and other reference will tell you this. But I cannot find a straightforward explanation why it is like that. Some demonstrations show the charging of capacitors through inductors and how the end capacitor doubles voltage due to collapsed magnetic field in the inductor, etc., but I find it very confusing. So maybe a mathematical demonstration would be convincing.

How can we show mathematically, for instance, that when the load impedance Zl is equal to Zo then reflection is zero, and so on?

## 4 Answers

Surprised there aren't any proper answers yet. The question is asking about how to demonstrate this property mathematically, not intuitively. This is a great question, so let's derive some stuff.

SHORT ANSWER

Solve the Telegrapher's equations, to get the general form of the voltage and current on the line. For the lossless case, this is: $$v(z) = V_1 e^{-jbeta z} + V_2 e^{j beta z}$$ For unknown complex "amplitude constants" V1 and V2, and constant beta, which is $$beta = omega sqrt{LC}$$ And the current is $$i(z) = frac{V1}{Z_0} e^{-jbeta z} - frac{V2}{Z_0} e^{jbeta z}$$ Where we introduce the constant Z0, $$Z_0 = sqrt{frac{L}{C}}$$ Which comes out of the math.

Constants V1 and V2 are determined by the boundary conditions. The condition at the load gives: $$V_2 = Gamma V_1$$ That is, the amplitude of the reflection is Gamma * V1. Gamma is called the reflection coefficient: $$Gamma = frac{R_L - Z_0}{R_L + Z_0}$$ When RL = Z0, Gamma = 0, and there is no reflected wave. When RL=0, Gamma is -1, implying full reflection with phase inversion (reflected wave out of phase with incident). When RL=infinity, Gamma is +1, implying full reflection with no phase inversion (reflected wave in phase with incident). The other results follow from plugging in values for Gamma.

MORE RIGOROUS ANSWER

Fundamentally, a distributed element or transmission line is a circuit element with a spatial dimension (call it z) such that, for real constants R, L, G, and C, and time- and space-dependent voltage and current V(t,z) and I(t,z), the following partial differential equations hold: $$frac{partial V}{partial z} = -L frac{partial I}{partial t} - RI$$ $$frac{partial I}{partial z} = -C frac{partial V}{partial t} - GV$$ These are called the telegrapher's equations, well-known PDEs describing transmission lines. Note that they are only PDEs because we have both time and space dependence.

If we limit ourselves to phasors, then V(t,z) and I(t,z) become v(z) and i(z), complex valued functions of only space. The Telegrapher's equations then become: $$frac{d v}{d z} = -i(R+jomega L)$$ $$frac{d i}{d z} = -v(G+jomega C)$$ For angular velocity omega, $$omega = 2 pi f$$ For assumed frequency f.

Great, ordinary differential equations. For simplicity, assume R = G = 0. The general solution to these equations is well-known; we find for v(z), $$v(z) = V_1 e^{-jbeta z} + V_2 e^{j beta z}$$ For unknown complex "amplitude constants" V1 and V2, and complex constant beta, which is $$beta = omega sqrt{LC}$$ (When R and G are not zero, beta becomes complex, with a real part the same as beta here, and an imaginary part representing attentuation with increasing z).

Using this expression, we can now solve for i, $$i(z) = frac{V1}{Z_0} e^{-jbeta z} - frac{V2}{Z_0} e^{jbeta z}$$ Where we introduce the constant Z0, $$Z_0 = sqrt{frac{L}{C}}$$

It now remains to determine the constants V1 and V2. These amplitude constants are determined by the particular problem at hand; specifically, we can provide two boundary conditions (e.g. V(0) = 1 and V(5) = 0), and then plug them into our expression for v(z) to find V1 and V2. Simple enough. To find boundary conditions, we use KVL and KCL, and Ohm's Law. KVL and KCL imply that at the boundary of the distributed element (where the transmission line connects to a lumped circuit) the current must be continuous and the voltage must be continuous. This is hugely helpful.

Let's solve the basic transmission line problem. This image shows the situation.

Note that the spatial coordinates only apply to the distributed element, and more particularly, to this distributed element. If there was another transmission line, it would have its own coordinate system.

For conventional reasons, make z=0 the end of the transmission line, right before the load RL, with +z in the direction shown, so that the beginning of the transmission line is at z = -length. We have then two boundary conditions. First, $$frac{v(0)}{i(0)} = R_L$$ Due to continuity at the boundary and Ohm's law. Second, using KVL, $$frac{V_s - v(-l)}{i(-l)} = R_s$$ Where l is the length of the line. We have our boundary conditions, and we can solve for the amplitude constants V1 and V2 (note that we haven't clarified what V1 and V2 represent intuitively, only that they are mathematical constants inherent in the problem as we constructed it).

From the first condition, we can substitute in v(z) and i(z) setting z=0 (recall that we have these, from above) to get: $$frac{v(0)}{i(0)} = frac{V_1 e^0 + V_2 e^0}{V_1/Z_0 e^0 - V_2/Z_0 e^0} = R_L$$ So: $$Z_0 frac{V_1 + V_2}{V_1 - V_2} = R_L$$ Which gives, by some algebra: $$V_2 = frac{R_L - Z_0}{R_L + Z_0} V_1$$ We will denote that constant fraction by Gamma, as shown: $$Gamma = frac{R_L - Z_0}{R_L + Z_0}$$ So that $$V_2 = Gamma V_1$$ And our expression for v(z) becomes: $$v(z) = V_1(e^{-jbeta z} + Gamma e^{jbeta z})$$ Now we're all done.

Let's interpret our result now (which, mind you, we have refrained from doing to avoid projecting our own intuition on something mathematical). v(z) is a phasor, meaning its "amplitude" is actually the amplitude of a sinusoidal signal, in time, varying with frequency w. The phase is the phase angle of the sinusoid. Essentially, v(z) represents the envelope of the actual voltage on the line, which has a frequency w.

The complex exponential in the expression for v(z) can be expanded as $$e^{-jbeta z} = cos beta z - j sin beta z$$ Which makes the 'oscillating' or 'wave' nature of the complex exponential terms clear. What we're seeing is, as we look at different positions z, the amplitude and phase of the phasor change, and they change sinusoidally, giving an envelope which looks like a sine wave, and as time goes in, the whole envelope moves to the +z direction. This is the "forward traveling wave."

By similar reasoning, when we have $$e^{jbeta z}$$
(with +j instead of -j), the same thing occurs, but now the envelope waveform slides backwards, a "backward traveling wave." And so, we can rewrite v(z) as $$v(z) = V^+(z) + V^-(z)$$ Where $$V^+(z) = V_1 e^{-jbeta z}$$ Is the forward-traveling wave with amplitude V1, and $$V^-(z) = Gamma V_1 e^{j beta z}$$ Is the backward-traveling wave, often called the reflection.

We see that the amplitude of the reflection is Gamma * V1. And, in turn, Gamma is: $$Gamma = frac{R_L - Z_0}{R_L + Z_0}$$ And Z0 is: $$Z_0 = sqrt{frac{L}{C}}$$ Which is characteristic of the line (the characteristic impedance, called an impedance due to its similarity to the fact that it relates v(z) and i(z), but note that it does not have units of ohms). When RL = Z0, Gamma = 0, and there is no reflected wave. When RL=0, Gamma is -1, implying full reflection with phase inversion (reflected wave out of phase with incident). When RL=infinity, Gamma is +1, implying full reflection with no phase inversion (reflected wave in phase with incident). The other results follow from plugging in values for Gamma.

Correct answer by Sam Gallagher on October 29, 2020

How can we show mathematically, for instance, that when the load impedance Zl is equal to Zo then reflection is zero

• Assume a transmission line of characteristic impedance $$Z_0$$
• Assume an applied voltage ($$V_F$$) at one end of the line
• From the above, the current that flows ($$I_F$$) is $$V_F$$ divided by $$Z_0$$

When the voltage and accompanying current reach the end of the line and meet $$Z_L$$ there could be a violation of ohm's law if $$Z_0$$ does not equal $$Z_L$$.

If $$Z_L$$ > $$Z_0$$ we have to consider ways in which we can avoid violating ohm's law.

This method could solve the problem (as a thought experiment): -

• Somehow make the voltage arriving at $$Z_L$$ a bit bigger and, at the same time
• Somehow make the current arriving at $$Z_L$$ a bit smaller
• The modified voltage and current would be "adjusted" to have a ratio suitable for $$Z_L$$

Or, algebraically we could say: -

$$dfrac{V_F + delta V_F}{I_F - delta I_F} = Z_L$$

$$therefore dfrac{V_F}{I_F}cdot dfrac{1 + delta}{1 - delta} = Z_Llongrightarrow Z_0cdot dfrac{1 + delta}{1 - delta} = Z_L$$

$$text{Hence,}hspace{1cm}delta Z_0 +delta Z_L = Z_L - Z_0$$

$$text{And,}hspace{1cm}delta = dfrac{Z_L-Z_0}{Z_L+Z_0}$$

But, of course, we call $$delta$$ by it's usual name (reflection coefficient) $$Gamma$$. $$delta$$ is just a device I invented to get through the thought experiment.

However, the important subtlety that prevents an ohm's law violation is the "bit" we add to voltage and the "bit" we subtract from current ($$delta V_F$$ and $$delta I_F$$). If we took their ratio it is $$Z_0$$. This means that they can naturally flow (together) back into the transmission line because they have the perfectly correct ratio to do so.

That is called a reflection and travels from load to source.

Clearly, if $$V_F$$ and $$I_F$$ were originally of a ratio that matched the load ($$Z_0$$) impedance (right from the start), we wouldn't need to set up the algebra that figured out how to deal with the unwanted "bits" and, there would be no thoughts of violating ohm's law nor talk of reflections.

Answered by Andy aka on October 29, 2020

There are lots of mathematical descriptions to be found out there, so maybe a something in between mathematical and intuitive would be useful. The key point is to remember that the voltage or current at any point on the transmission line is a sum of the forward-propagating ('incident') and backward-propagating ('reflected') waves. This is true at the termination point as well.

1. Open termination: The current has to be zero, the voltage does not. Therefore, to make the current at the termination point zero, the reflected wave has have equal magnitude, opposite sign current at the termination point. Therefore the reflected wave has opposite sign current, same sign voltage, as the incident wave.

2. Shorted termination: The voltage has to be zero, the current does not. Therefore, to make the voltage at the termination point zero, the reflected wave has to have equal magnitude, opposite sign voltage at the termination point. Therefore, the reflected wave has opposite sign voltage, same sign current, as the incident wave.

3. Equal to Zo: If you did not have the termination point, but instead the line continued, the impedance of the rest of the line measured at that same point (assuming the line went on to infinity) would be Zo. So if you terminate it with Zo instead, the voltage and current should behave in the same way as if the line continued off to infinity.

Answered by rpm2718 on October 29, 2020

Would have made a comment, but don't have the rep.

Interesting to see a question for a more mathematical explanation considering it's usually the intuition for transmission lines which isn't there for most learners.

"Fundamentals of Electronics 101" should have equipped you with the basic tools for understanding current, energy conservation laws and simple relationships.

"Electromagnetism 101" should have equipped you to work with transmission lines with/without loss. Expressing a signal as two sinusoids travelling in opposing directions, through negating the sign applied to the propagation constant (the 180° phase shift a product of the boundary conditions present at the load). A kinetic example of a boundary condition being a point held stationary on a string.

I'll be the first to admit that transmission line theory (and for that matter, all the physics which holds up electromagnetics) can be intimidating, but it's (genuinely) quite simple.

The reflection properties of transmission lines are a trivial result of some "EM 101" theory so it's probably best to hold off the maths for the moment. I recommend Ulaby and Ravaioli's Fundamentals of Applied Electromagnetics as a good primer.

Answered by Benjamin Crawford Ctrl-Alt-Tut on October 29, 2020

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