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VinnyCee

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## Homework Statement

A multisection transmission line consists of three lossless transmission lines used to connect an ideal step source of 5V amplitude and 6[itex]\Omega[/itex] output impedance to two separate load resistances of 66[itex]\Omega[/itex] each. All three lines are characterized by the line parameters L = 364.5 nH/m and C = 125 pF/m and length of 40cm. Sketch voltages Vs and [itex]V_L[/itex] versus t for 0 < t < 20 ns.

http://img23.imageshack.us/img23/5678/problem218lq7.jpg

## Homework Equations

[tex]\Gamma_S\,=\,\frac{R_S\,-\,Z_0}{R_S\,+\,Z_0}[/tex]

[tex]\Gamma_L\,=\,\frac{R_L\,-\,Z_0}{R_L\,+\,Z_0}[/tex]

[tex]Z_0\,=\,\sqrt{\frac{L}{C}}[/tex]

[tex]v_p\,=\,\frac{1}{\sqrt{L\,C}}[/tex]

[tex]t_d\,=\,\frac{l}{v_p}[/tex]

## The Attempt at a Solution

http://img168.imageshack.us/img168/8978/problem218solnpu5.jpg

I found all the variables quite easily using the formulas above. I started making the bounce diagram and got a few [itex]t_d[/itex] into it. But when I get to 3[itex]t_d[/itex]=8.1 ns, I am confused by one of the bounced voltages. I have the answer, but I have no idea where the writers are getting a 0.2V bounce in the diagram.

On the right side, between 8.1ns < t < 10.8ns, there are three voltages. 0.8V, -0.1V and the mysterious 0.2V. I understand that the 0.8V line comes from the 1.2V on the left, and the -0.1V line comes from the reflection of the 0.3V line above. Actually there should be two of the 0.1V lines considering the reflection from both of the right hand lines. But where is that third 0.2V line coming from?

Please help!

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