What is the voltage at a point of impedance mismatch? - Page 1
Forum section: electronics / beginners

author: Whales
author_rank: Super Contributor
timestamp: June 03, 2016, 12:04:40 pm
content: I have two transmission lines with different impedances connected directly together. If I send a voltage step over one line, then at the junction between the lines some of it is reflected and some of it is transmitted. Using reflection and transmission coefficients I can then work out how large these reflected and transmitted voltages are (& make a bounce diagram).EDIT: Don't worry about reading this second paragraph, just jump to the pictures in my post below.But what is the voltageatthe point of intersection throughout this process? I have been taught to sum the incoming and reflected voltage fronts to calculate the voltage at a source or load connected to a transmission line, but it does not seem to make sense to sum in the same way at an impedance mismatch point between two transmission lines (it makes more sense to only sum 'one side' of the reflection point).I've been researching this for several hours and I have not been able to find anything but examples for finding the voltage at points other than a point with partial transmission as well as reflection. Intuitively I'm having difficulty understanding why the junction point has a voltage that is not the average of both sides anyway.

author: danadak
author_rank: Super Contributor
timestamp: June 03, 2016, 01:43:16 pm
content: Discussion here -http://home.sandiego.edu/~ekim/e194rfs01/tlsmthek.pdfRegards, Dana.

author: Whales
author_rank: Super Contributor
timestamp: June 04, 2016, 12:20:21 am
content: Thanks Dana.Quoted from the document you linked:I have a feeling that "summing the voltages coming and going toward a point" still only applies at the 'ends' of bounce diagrams, not at points of discontinuity at their centre.Take first this example, where a mismatched resistive load is at the end of a transmission line:Here everything makes sense. The voltage just to the left of the resistor is the same as the voltage across the resistor.Now for two mismatching transmission lines: (EDIT: the 200 ohm line extends on forever to the right)If you do the calculations you will find voltage A + B = C. In other words the voltage slightly to the left of the junction is the same as slightly to the right after the reflection/transmission event.Here it makes no sense to me to add all of the incoming/outgoing waves to find the voltage at this junction, as you would get double the voltage that is present on either side. ie a discontinuity.Perhaps the rule should be 'add the voltageson one sideof the point', and my first diagram (with the resistor) should have been drawn like so:Am I understanding this correctly?

author: Rerouter
author_rank: Super Contributor
timestamp: June 04, 2016, 01:54:26 am
content: The short form in my head relies on dividers.Lets say its a 100 ohm 5V step,While traversing the 100 ohm line its a voltage divider at 2.5V. The wire acts as a 100 ohm load.once it reaches the termination resistor or impedance change, the divider is altered,So lets say at the end of your line you had a 200 ohm termination, and your measuring right at the signal source,Your signal begins down the transmission line, loading the generator with an ideal 100 ohms, resulting in 2.5V, On the oscilloscope this will be what appears as the first bump,Once the signal reaches the 200 ohm termination, the divider changes, and you get 3.33V, but you dont see this on the scope yet, it still has to go back along the transmission line to your measuring point,Then once it reaches back, you see the second bump to 3.33V and is steady until the next change,the same holds true for a 50 ohm termination, you see the 2.5V until it reaches and returns from the termination, at which point it shows a negative jump to 1.25V,Now in your example with the 100, 200, 400, etc impedance line, it gets a little weirder, lets assume all these are the same lengthYour signal like the previous ones would start off at 2.5V along the 100 ohm transmission line, This is your first bump.Once it encounters the 200 ohm impedance the divider changes, and 3.33V begins making its way back to the measuring pointThe 3.33V reaches the measuring point at the same time as for the 400 ohm line, so at the measuring point it bumps to 3.33V and at the beginning of the 400 ohm it bumps to 4V.This continues along, and you will see smaller and smaller bumps being added further and further apart, (longer time for round trip)

author: Whales
author_rank: Super Contributor
timestamp: June 04, 2016, 02:05:47 am
content: Quote from: Rerouter on June 04, 2016, 01:54:26 amApologies, I meant to indicate that the 200 line extends off for inifnity. I'm only interested in the voltage at the junction point immediately during and after the first 'bounce' event at that point.Quote from: Rerouter on June 04, 2016, 01:54:26 amI'm measuring the voltage at the junction between the 100 ohm and 200 ohm lines, not at the generator.

author: Rerouter
author_rank: Super Contributor
timestamp: June 04, 2016, 02:41:12 am
content: Assuming no parasitics, you would immediately see a single rise to 3.3V in my scenario,In reality it also matters what your measuring node is like, and the impedance discontinuity where the wires are joined,

author: Rerouter
author_rank: Super Contributor
timestamp: June 04, 2016, 02:46:15 am
content: I do actually have to thank you for your question, cleared up some misconceptions i myself had about the whole process.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 04, 2016, 03:29:09 am
content: The voltage at the node is something other than half the source (i.e., open circuit) voltage.Indeed, you can calculate the reflection coefficient easily, by coming up with a ratio Gamma which is 0 when the voltage ratio A is 1/2, Gamma = 1 when A = 1, and Gamma = -1 when A = 0.The outcome, in terms of impedances, is:\( \Gamma = \frac{Z_1 - Z_2}{Z_1 + Z_2} \)or something like that.As for measuring it in circuit, this formula comes in handy when evaluating a SPICE simulation (the formula can be constructed from SPICE primitives, or entered directly in the postprocessor), or it can be implemented (IRL) with a pair of transformers: a reflectance bridge.Tim

author: dmills
author_rank: Super Contributor
timestamp: June 04, 2016, 02:00:49 pm
content: Note that if the 200 ohm line extends to infinity, you can conceptually replace it with a 200 ohm resistor, the behaviour will be the same.Regards, Dan.