# On a transmission line, why is there a reflected wave when a voltage/current wave encounters a different impedance?

Why, physically, does this happen? I can analogously to other phenomenon understand why this occurs if the impedance is greater than the characteristic impedance, but why is there still a reflected way if the impedance is less, such as when the impedance is 0?

I do not intuitively understand why this happens. The wave should find the medium easier to go through.

• This is the same phenomenon that happens with light: Light is reflected from glass to air for example. Commented Jan 28, 2017 at 9:15
• Write the wave equation on each side of the interface where the impedance changes under the form left traveling wave + right traveling wave (4 waves in all) and adjust the coefficients so that 1) it includes your incident wave coming from infinity (2 equations) and 2) the sum is continuous as well as its derivative at the interface (2 equations). You will see that in any case, you get a non zero reflected wave (unless of course there is no change of impedance).
– user130529
Commented Jan 28, 2017 at 15:43
• "The wave should find the medium easier to go through." - you are, perhaps, inferring an "emotional" desire on the part of the wave to make forward progress, but it is equally "happy" to bounce backwards given any excuse ... Commented Mar 24, 2019 at 16:22

I think it's common to try to think of transmission lines like hoses carrying water, and the impedance corresponds to something like the radius of the hose, so the bigger the hose the easier it is for the water to get through. Unfortunately, that type of thinking doesn't really translate to discontinuities in transmission lines.

The short answer is that whenever there is any kind of a material discontinuity, a wave is going to at least partially bounce off of it. And it doesn't matter whether it's from a higher impedance to lower impedance material, or vice versa. Consider that you can see a reflection of yourself at the surface of a pool of water, both from above and from below. The water has higher impedance than air, but some light from below is still reflected back down.

There are going to be many, many ways to think about this from a physical standpoint. Some might resonate with you, some might not. Here's how I like to think of it.

Consider a coaxial waveguide transmission line, with an inner conductor, an outer conductor, and a dielectric material within. There are basically two types of impedance discontinuities: (1) same material, with a change in radius of inner and/or outer conductor, and (2) same inner and outer conductors, but with a different dielectric material. Let's think about these separately in the case of an incident electric field approaching the discontinuity.

(1) If there is a geometric discontinuity, that means there will be a sharp metal corner. Sharp metal corners diffract the electric field, radiating it in all directions locally. This translates to some, but not all, of the diffracted energy continuing forward, and some energy being bounced back the way it came from. That's true even if it's a smaller cross-section opening into a larger cross-section.

(2) The more interesting case is if the cross-section is the same, but the dielectric material changes.

(2a) If the coax goes from vacuum to a dielectric, once the electric field hits the dielectric, it will cause the electrons in the molecules of the dielectric to vibrate, which will radiate a secondary electric field. Some of that field will propagate forward, and some will go back the way it came.

(2b) If the material in the coax goes from dielectric to vacuum, what happens?

It's important to understand what's really happening when an electric field is propagating in a dielectric. From a macroscopic view, it looks like the wave is propagating more slowly, and with a shorter wavelength than a wave propagating in a vacuum. What's really happening inside is that there is a wave propagating just like in a vacuum, but it's vibrating the electrons within each molecule (a dipole) which are then radiating their own electric fields (electric polarization vector). When you combine the original field with the contributions from all the radiation of the vibrating electrons, and look at it from a macroscopic level, it looks like one wave with a different wavelength and propagation constant. Note that each dipole is radiating in all directions, and it's the cumulative sum of all of them that make the whole thing look like another wave propagating in only one direction.

At the interface where the dielectric ends and vacuum begins, the vibrating dipoles radiate forward into the vacuum, but they also radiate backward into the dielectric section. Since there are no more dipoles on the other side, the fields don't combine the way they do in a homogeneous dielectric medium, and you end up with another backward propagating wave in the dielectric.