Author Topic: Voltage at the end of an open transmission line Stub  (Read 2002 times)

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Offline GeckoTopic starter

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Voltage at the end of an open transmission line Stub
« on: August 27, 2015, 10:57:51 pm »
A question for the RF gurus:
I have (simplified) a transmission line circuit like in the attached image: Just a simple open ended stub on a transmission line. In my case the stub has a length of lambda/8. But I think for my question that doesn't really matter.

My question is: How can I calculate the Voltage (Amplitude) at the end of the stub (node B)?
My first Idea was to use the equation
V(z) = Vf * (1+r(z))
where Vf is the amplitude of the forward Voltage wave, which should be 0.5 Volt in the example if I'm not totally wrong. V(z) is the voltage at a point "z" on the transmission line. r(z) the reflection coefficient at point "z".
But with "z" chosen at the end of the stub, where r=1, I would get a result of 2V. But the simulation differs from that  :-- (0.633V with -26° phaseshift).

What am I doing wrong. How can I get the correct result?
« Last Edit: August 27, 2015, 11:00:24 pm by Gecko »
 

Offline rfeecs

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Re: Voltage at the end of an open transmission line Stub
« Reply #1 on: August 28, 2015, 12:17:17 am »
First, since you have an extra 50 ohm termination, you are driving the transmission line at point A with an equivalent 25 ohm source of amplitude .25V at point A.  (This would be the voltage if you terminated point A with a 25 ohm load instead of the open circuit line.)

So with the open circuit line, you are driving a 50 ohm line with a 25 ohm source, you have a standing wave along the line.  The current at the open circuit end equals zero.

There are lots of ways to solve it.  You have boundry conditions defined on each end of the 50 ohm line.  You can then take the transmission line equations and solve for the voltage and current at any point on the line.

Another way is:  calculate the impedance looking into the open stub: -jZ0cot(bl).  With that you can calculate the voltage at point A.  Now you have a 50 ohm open circuit line with a standing wave and you know the voltage at one end, so calculate the voltage at the open circuit end.

Of course the easy way you did already.  Throw it into a circuit simulator.
 

Offline GeckoTopic starter

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Re: Voltage at the end of an open transmission line Stub
« Reply #2 on: August 28, 2015, 08:38:54 pm »
Thanks rfeecs, you pushed me into the right direction and I finally got it!  :clap:

One of my problems was, that I neglected the phase of the reflection coefficient in my calculations, but (1+|r|) not= (1+r).
So with that error ruled out I could calculate the Voltage at Node A. Here is how I did it. Maybe it helps someone:
VA=VP*(1+r) = 0.447 exp(-j26°)   with VP=0.5V and r=0.447 exp(-j116°)  (r can either be calculated from the values with ((R1||ZX) - Z0)/((R1||ZX) + Z0) with ZX being the impedance when looking into the stub. Or simply read it from a Smith Chart)

With VA the voltage at Node B can be calculated with VA=V_B * cosh(gamma *L) + IB*ZL*sinh(gamma L). Because the stub is open IB=0.
So VB=VA/cosh(gamma L).

With a lossless line gamma=j*beta and hence
VB=VA/cos(beta L).

In my example L=lambda/8 and so: VB=VA*sqrt(2) = 0.633V exp(-j26°)

« Last Edit: August 28, 2015, 08:41:30 pm by Gecko »
 


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