Alas...our ultraparanoid server at work has excised your drawings....or I'd be able to help. Could you attach the files to an email to me?Nope, not getting it. This whole concept of transmission line is somehow so different from anything I've done before. Let's try baby steps. So it takes 2.5 microseconds for the pulse to reach point P. But how do I determine the voltage drop in R1? Can I use normal voltage division? So it would be \(U_{R1}= \frac{R_1}{R_1+Z_{01}+Z_{02}+R_L}E=2V\)
I wish I was there.I'm currently teaching a comprehensive section on the Smith Chart in my A.C. theory class.
Fall in Alaska, can't waitYou can enroll in the Fall semester.
eric
There are two points where reflections will occur. Where the two transmission lines meet (with unequal characteristic impedances) and at the load - due to load mismatch with the second line.I'm so lost I don't even know where to start. All I know is, that there's no reflection between R1 and Z01.
Any tips on how to start are appreciated!
Alaska? I moved to Georiga from Michigan 8 years ago 'cause 40 years of winter was enough! No thanks, I'll stay here and enjoy the warm weather and year around hiking. Besides, "Georgia HOPE" pays for me to go to school here.You can enroll in the Fall semester.
eric
Will the reflected signals turn to heat completely at R1? And will the signal that reaches the load turn to heat as well?There are two points where reflections will occur. Where the two transmission lines meet (with unequal characteristic impedances) and at the load - due to load mismatch with the second line.
Yes.Will the reflected signals turn to heat completely at R1? And will the signal that reaches the load turn to heat as well?
Not sure where the 4V came from.So what I get now, is: voltage drop in the first resistor and TL is 4V, so the voltage goin into the second TL is 8V. Reflection coefficient between first and second TL is .5, so the magnitude of signal that gets transmitted, is 1.5 times greater, 12V. And the pulse that gets "trapped" in the second TL gets half the magnitude and opposite sign every reflection.
There are two things bothering me. My logics are against my course material, which says that the transmission coefficient can be calculated using \(\tau=1+ \rho\), where rho is the reflection coefficient. Are we getting free energy? And the second thing: is the trapped pulse really going to be there for ever?
Unfortunately, no. This is an assignment for extra points in the next mid-term.Do you actually have access to a solution for the problem? Say from the source text ...
Doubt if that's the way to go ....I got the 4V from voltage division: \(U=\frac{R_1+Z_{01}}{R_1+Z_{01}+Z_{02}+R_L}E=4V\).
This sounds a bit more logical to me, but aren't we still getting free energy?Hence a pulse of E volts magnitude traversing the line intersection would have a reflected component of 0.5*E and a transmitted component of 0.75*E.
Does this really change something? The pulse goes through the whole line, so isn't it same, if we replace the line with resistor with the same resistance. If not, how do you get the 6V?Just consider the first transmission line as having 100Ω impedance as "viewed" from the source feed point.
by Duane Benson
by Jake Hertz
by Duane Benson