# Reflections in Transmission lines

#### likephysics

Why do reflections occur in transmission lines. Why does the wave travel back from the load?

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#### lalbatros

Any discontinuity along the transmission line will produce a reflexion in addition to a transmission through the discontinuity.

This is just the same as in optics: a discontinuity of the refractive index implies a reflexion. In optics it eventually produces also refraction-transmission because of the two(three)-dimensional propagation.

You know from optics that the reflexion can be reduced. For example a thin coting with a properly chosen refractive index and a precise thickness can reduce partly the reflexion in a range of wavelengths. This also applies in transmission lines.

Of course a transmission lines has an end. This end may be open or totally closed. In both case this will be a discontinuity and produice reflexion. It is also possible to close the transmission line by a properly chosen impedance (or load). When the impedance is properly matched, there will be no reflexion any more and the power will be fully absorbed in the load. This is called impedance matching. You have maybe noticed that 50 Ohm terminations are extensively used in labs and that every physicist using electronic devices has a few 50 Ohm load in a drawer. These load are used for impedance matching to avoid reflexions in an experimental setup (in the MHz range here). Similar load are used also for waveguide setup for the same reason. Reflexion are often pertubing measurements and must handled properly by impedance matching.

Why is it that reflexion occurs?
On the surface of a glass in otpics, this is because of the change in the refractive index. The dense medium (glass) has a high polarisability, which implies effective charges building at a very frequency on its surface and sending a signal back.

The same think happens in transmission lines.
A special case is an open line.
In a transmission, charges build up on all along the line in a wave pattern.
If the line is infinite, this results in a forward propagation without reflexion.
If you cancel all the charges building up in the second half of the infinite line, you should not be surprised to signal propagating in the opposite direction. This would be just like exciting the exact oppoiste charges on the second half of the line and will imply a backward signal. It is easy to cancel all these charges: just cut the line and remove on half of it.
Another way to see what happens is to realise that the charge discontinuity at the end of a transmission line is an excitation that will propagate bacward. Impedance matching is doing exactly what is needed to avoid this excitation.

The maths of all that can be found in any textbook on the subject. It all goes back to the Maxwell's equations.

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#### likephysics

lalbatros, thanks for the explanation. Optics analogy is great. But how to explain it to ppl who come from circuit background(non-Electromagnetics).

#### Bob S

In a transmission line, like RG-8 for example with Z0 = 50 ohms, the forward TEM signal has a voltage to current ratio of 50:1. The Poynting vector for a forward traveling wave is PF = E x H. If there is a short at the end of the transmission line, there is current but no voltage at the short. The only solution that can support this boundary condition is another TEM wave with PR = -E x H, which is the Poynting vector for a backward reflected wave.

Bob S

#### Born2bwire

Gold Member
lalbatros, thanks for the explanation. Optics analogy is great. But how to explain it to ppl who come from circuit background(non-Electromagnetics).
I do not think there is any better way to do it. Circuits, disregarding RF and microwave circuits, largely ignore wave theory. So there isn't a direct analog along those lines. But there are a large number of typical daily experiences that we can look towards. How about the reflection and transmission of sound? When you yell at a barrier, some of the sound is reflected (echo) and some is transmitted (some bloke on the other side hears your muffled screams). This happens for the same reasons, the barrier represents a change in the impedance of the medium. Why does a change in impedance cause reflections? Well, one way to think about it is conservation of momentum. When light goes from one medium to the next its velocity changes (if there is a change in the index of refraction). This means that the momentum of the wave changes and for the momentum to be conserved there must be some reflection. This is probably the most physical analogue that we can use though it requires us to think more quantum mechanically than the classical theory that we use to derive the behavior.

#### lalbatros

lalbatros, thanks for the explanation. Optics analogy is great. But how to explain it to ppl who come from circuit background(non-Electromagnetics).
This is straigthforward, using a lumped elements model of the transmission line.
The picture below shows three elements of a lossy line model.

http://www.ece.uci.edu/docs/hspice/hspice_2001_2-269.html

Using small lumped elements in very large numbers leads to the wave equation for transmission lines.
The same can be derived from the Maxwell's equations, of course.
It is even possible to study discontinuities or smooth variations of the line parameters.
The maths and the principles will be the same for the 1-D transmission lines equations, or the circuit model, or the Maxwell's equation (applied to 1-D).

The main effect that can come out of such a model is the wave propagation, including reflexions.
These effects are obviously related to the 1-D aspect, which is represented by a large number of elements connected in series. In this sense, there is no analogy possible with "usual circuit", since "usual circuit" are supposed to be made from a few lumped elements, not a long chain of elements.

The maths can be made very simple using the two-port network point of view.
You can easily derive the impedance matrix from the structure of one of the three lumped elements shown on the picture above. The input voltage and current will be related to the output voltage and current by a relation like: (impedance matrix)

The coefficients of the impedance matrix are related to the values of the distributed inductances, capacities and resistances.
From the impedance matrix, you will easily derive what I usually call the "scattering matrix", but this is just my pet-name for this matrix: It would be better to call this the "propagation matrix" since from this input-output relation it is straight-forward to calculate the whole transmission line or any part of it.
If the parameters are constant along the line, then you simply need to multiply the same matrix a large number of time to get the "propagation matrix" for the full line. This leads you to the exponential of this matrix.
And this is where the waves will pop out: when losses are small enough, then the exponential of the matrix will exhibit the wave propagation in the transmission line. If the losses are high you will see the wave damping in the line.
With this simple matrix formalism, you can analyse absolutely every aspect of transmission line.
You can also drill down the results to understand the physics, of course.
But does it matter to understand the physics further than the picture shown above?

Michel

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