VSWR and transmission lines help

In summary: I don't know how to explain it. I think its because the voltage at the end is really disproportinate to the other voltages.low impedance- much lower than characteristic impedance acts as some sort of short, so there's a change of polarity aswell?maybe I am not on the right level, but wouldn't that just be 3 volts? or 3 volts each away..
  • #1
Livethefire
51
0
I had set up an inductive and capacitative ladder network to model a coaxial cable( or low pass filter)- 20 sections of L= 0.1 H and C=0.11 microF

That works out as a characteristic impedance of 953ohms.

The circuit starts at a signal generator, then 5ohms resistor before the low pass filter.

Using a signal generator set at 500Hz and +1V I sent a pulse down the line, terminated with:
A) 100ohms
B)953ohms
C)10000ohms

Measuring the output Voltage at equally spaced pins down the line allowed a plot of Voltage out over Voltage in.

B and C work out fine. B gave a straight line at one volt and C gave a line which the wave clearly destructively interfered with.. I can explain those with what i understand.

However what I cannot explain is when the line terminated with 100ohms. The plot I get with that seems really disproporatinate. It is a standing wave pattern that peaks at just over 6 volts. I can explain the shape and the reason for the shape (standing wave), but not WHY it peaks at 6 volts. I would have presumed 1+1= 2 asuming 100% reflection at the end (because I only put one volt in).

The blue one:
graph.gif


I tried the measurements again after doing a few differents things with the experiment and they seem to be stable at maxing out at 6.

I have looked into resonance but to my knowledge the resonant frequency of this circuit is around 1500Hz.

Any help or suggestions?
 
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  • #2
Hint -- you are mismatched with low impedance at *both* ends of the transmission line in that case...
 
  • #3
Your getting my cogs going... thank you.

So one volt in, mismatched gets reflected from the end, and from the start ... but I am still confused.

low immpedance- much lower than characteristic impedance acts as some sort of short, so there's a change of polarity aswell?

Maybe I am not on the right level, but wouldn't that just be 3 volts? or 3 volts each away.. I am getting confused. haha.

youve given me a bit more to think about though thanks, any other hints would be greatly apprechiated.
 
  • #4
Livethefire said:
Your getting my cogs going... thank you.

So one volt in, mismatched gets reflected from the end, and from the start ... but I am still confused.

low immpedance- much lower than characteristic impedance acts as some sort of short, so there's a change of polarity aswell?

Maybe I am not on the right level, but wouldn't that just be 3 volts? or 3 volts each away.. I am getting confused. haha.

youve given me a bit more to think about though thanks, any other hints would be greatly apprechiated.

Have you learned about the Reflection Coefficient in transmission lines yet?

http://en.wikipedia.org/wiki/Reflection_coefficient

.
 
  • #5
yeah, 1 for full reflection, eg infinity impedance, -1 for full reflection opposite polarity, eg short 0 impedance. Then 0 for matched line.

And then the equation in a few forms. One for (1+A)/(1-A)
Etc.
 
  • #6
Looks like a resonance effect.

Did you try sweeping across a range of frequencies to check this?
For example, monitor the voltage at the center of the line and tune the signal generator to see if you get a peak there at any frequency.
 
  • #7
I think that 1/sqrt(LC) = 1 x 10-4 seconds, so 500 Hz would be a full wavelength in 20 sections. So at 500 Hz, the input impedance at the signal generator input (using a sine wave) would be about the same as the termination impedance. It might be a good idea to back terminate in the calculated characteristic impedance, 953 ohms, to damp all resonances. 1500 Hz, the observed resonance, is the third harmonic resonance (3 wavelengths).
Bob S
 
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  • #8
Ive used a few different formulas and i get different values for the reflection coeff...

some near 0.9 and 0.8, and in one particular case 0.03. Obvviously I am doing something wrong and need a bit more help. With 0.9 and 0.8 I tried an infinite power series, ie, reflects off one end then the other and then the other until the signal dies out.

1+0.9^2 +0.9^3...

But I get an answer well over 6 and well under 6 for the various cases when I do around 20-30 terms.
And as well each end is like a short, so there is a polarity change on the reflection?
I don't really know what to do about that. It would seem it would just cancel itself out.
 
  • #9
Are you back terminated in 953 ohms? Your signal generator is a voltage source. Can you use a sine wave signal rather than pulses? You should get minimal reflections if you are both forward and back terminated in characteristic impedance.
Bob S
 
  • #10
Oh sorry,
Its an acoustic wave generator so tthey are sinusodial...pulse was a bad word. My fault.
Well

Theres 5 ohms between the generator and the line, and then the the line in this blue case was terminated with 100ohmns

Its not the reflections that's bothering me, its just that I can't predict 6 volts using the equations I know of. You would think there would be a way of predicting that sort of thing is all.

ie using reflection coffecients- I was thinking a power series like my last post or something like it with two different reflection cofficients, one for each side.
 
  • #11
There are reflections when the wave hits the source/input. If your input impedance is poorly matched than the reflections from the load will be sent back out to the load again when it hits the input port. This can cause the reflections to continue building up. When you assumed 1+1=2 as being a max, you assumed that the reflected wave from the load is fully absorbed at the input. How familiar are you with transmission lines and 2-port networks? The input impedance mis-match is a common occurance and is handled in any treatment of the two.

A quick reference to look at:

http://www.amanogawa.com/archive/docs/D-tutorial.pdf <- Look at slide #89, the exact equations that you want can be found in this pdf.
http://www.amanogawa.com/archive/docs/E-tutorial.pdf <- This set of slides gives a nice conceptual explanation of what is happening in a transmission line.

EDIT: Oh yeah, don't forget that the source may have its own impedance in addition to what ever resistance you place in line with it to feed the circuit. When we used function generators to estimate the RC time constant of unknown circuits, many of my students would forget about the function generator's impedance when they did the calculation.
 
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What is VSWR?

VSWR stands for Voltage Standing Wave Ratio, which is a measure of how well a transmission line is matched to the input signal. It is a ratio of the maximum voltage to the minimum voltage on the line and is used to determine the efficiency and quality of a transmission line.

What are some common causes of high VSWR?

High VSWR can be caused by a variety of factors, including impedance mismatches, poor connections, and damage to the transmission line. Environmental factors such as temperature and humidity can also affect VSWR.

How does VSWR affect signal quality?

A high VSWR can result in signal loss, distortion, and reflections, which can degrade the quality of the signal. This can lead to communication errors and reduced performance in systems that rely on high-quality signals, such as radio and telecommunications.

What is the purpose of using transmission lines?

Transmission lines are used to transfer signals or energy from one point to another with minimal loss or distortion. They are commonly used in electronic devices, communication systems, and power distribution networks.

What are some common types of transmission lines?

Some common types of transmission lines include coaxial cables, twisted pair cables, and waveguides. Different types of transmission lines have varying characteristics and are used for different applications depending on factors such as frequency, power level, and distance of transmission.

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