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Alright folks here it is, another transmission line question that has been bothering me. I understand the case where there is an abrupt change in impedance along a transmission line that leads to reflection of portion (or even all) of the signal.

Now, what is bothering me for a while is the case where we have a transmission line who's impedance varies in a predictable manner over its length. Lets suppose that we have a PCB trace who's characteristic impedance depends on it's width as per physics. Now suppose that this width is increasing linearly as the signal travels on it which leads to a continuosly linear change in it's impedance. I expect that signal would be reflected in this case too but continuously! But what I cannot imagine is how would the reflection look like in this case on the transmitting end and what the signal would look like on the receiving end. Besides this, how can one mitigate this type of impedance mismatch, I suppose that getting the correct receiver termination would be tricky in this case. hmmmmmmm...

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Continuous varying impedances are used all the time for impedance matching. If you have a very capacitive part of a trace (for example, where a large component pad might be), you can have a relatively inductive transition before or after it to "balance" it out.

What will end up happening is that the reflections will "stack up" but, instead of being at one point (a VSWR peak), it will be moderately spread out. You can still imagine it discretely, but in small steps.

And also remember, if you have a small reflection point, any backward reflection after THAT will be reflected slightly FORWARD, and so on.

Anyway, the good gents at Microwaves 101 always have a nice, in depth explanation.

edit: I didn't completely answer your question. "How it would look" is dependent a bit on how you are describing it. In the frequency domain, what you'll probably get is a VSWR that is "de-Q'd". You'll go from a nice sharp peak at midband to a more gradual, broader band response.

In the time domain....well, I don't work with the time domain as much but I would imagine you would have a lower amplitude, longer pulsewidth "ringing" or reflection.

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    \$\begingroup\$ Horn antennas are also a device that does the same - it converts one impedance (namely the waveguide feeding it) to the impedance of free space (377 ohms). \$\endgroup\$
    – Andy aka
    Oct 2, 2013 at 14:52
  • \$\begingroup\$ hmm stacking up eh ... that is what I was expecting. I did not understand the reflection being slightly forward point though. As far as I know the energy of a wave is absorbed by the termination impedance. If termination impedance is same as the source impedance than all energy is absorbed. This is explained as making the signal feel like there is infinite length transmission line. OK, but is the energy not absorbed as a signal travels on a PCB track with the same impedance throughout as well? With a linearly varying impedance there would be energy absorbed AND reflection as well, correct? \$\endgroup\$
    – quantum231
    Oct 2, 2013 at 15:10
  • \$\begingroup\$ @Andyaka I had put in an entire edit about horn antennas relating them to acoustic horns but decided to go with the frequency domain description. Good call! \$\endgroup\$
    – scld
    Oct 2, 2013 at 15:19
  • \$\begingroup\$ @quantum231 You are correct. There is dielectric loss and the signal will eventually dissipate. So, when you have a reflection backwards, even if it's a perfect reflection, you will have attenuation due to the board, components, etc. However, there will also be ANOTHER reflection because your source impedance is not perfect. And THAT reflection will be smaller in small part to board/component loss. At each little junction of mismatch, you can imagine this small ringing back and forth that eventually disappears because of the series attenuation that applies on each forward/backward reflection. \$\endgroup\$
    – scld
    Oct 2, 2013 at 15:23
  • \$\begingroup\$ .....the reflections also will be smaller because there's no such thing as a perfect reflection. But, you'll do better to concentrate on one non-ideal concept at a time. \$\endgroup\$
    – scld
    Oct 2, 2013 at 15:25
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What you're asking about is called a transmission line taper.

In general, there's no analytical solution to describe the reflections. The link in Chris L's answer (if you follow through to Klopfenstein's paper) gives some examples of specific taper shapes where something close to an analytical answer has been found.

The basic way to study it is to imagine breaking up the continuous taper into several segments, each with a slightly different Z0 value. You calculate the reflections at each discontinuity and how they add up to give the overall reflection and transmission characteristics.

Then you split up the taper into finer and finer steps (with smaller and smaller discontinuities in Z0) until you have a good enough approximation to the continuous taper. You could try to calculate the results by hand but it's much easier to just get a computer program to do do it. Luckily, this kind of program is pretty easy to find --- it's called a finite element simulation program.

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Please note, tapering is very effective and dramatically reduces the total magnitude of the reflection. As shown in scld's citation, the total magnitude of the reflection from a taper is much much less than the total magnitude of the reflection from an abrupt discontinuity.

enter image description here

In this example, the reflection coefficient can be easily designed to be <1% at the frequency of interest.

For a common-sense explanation, it is helpful to think of anti-reflective coatings that are used in optics. In optics, reflections are caused by an abrupt "impedance mismatch" between two materials with mismatched indexes of refraction. An anti-reflective coating significantly reduces the magnitude of the reflection, and the way it works is it consists of several layers of gradually increasing index of refraction, that together form a stair-step approximation of a continuous taper.

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