# Homework Help: Transmission lines bounce diagram

1. Sep 22, 2010

### likephysics

1. The problem statement, all variables and given/known data
This is not a HW problem. Just refreshing Tx line theory(reflections).
I have a pulse of 3.3v, period 36nsecs. (Ton=18ns, Toff=18ns) with a source impedance of 50 ohms, connected to a tx line of impedance 50 ohms. The load is high impedance or infinity.
So reflection at load is 1 and reflection at source is 0.
I drew the pulse waveform, but the pulse ends up at -1.65v. This does not agree with my Signal integrity simulation. So I wanted to double check if my work is right. Pls see attachment.

2. Relevant equations

To draw bounce digram for a pulse, I made 2 sources. One with +3.3v and the other with -3.3v.
Also, when should I stop drawing the bounce diagram. In my case $$\Gamma$$g is 0. so, after 3ns : ($$\Gamma$$L * $$\Gamma$$g* V1+) will be zero.

L is load and g is for generator.

3. The attempt at a solution

#### Attached Files:

• ###### Hithesh.pdf
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2. Sep 22, 2010

### Staff: Mentor

It depends on whether your square wave driven is 0V to +3.3V, or -3.3V to +3.3V as you have drawn.

If it is +/-3.3V as you have drawn, then your bounce diagram is correct. Just keep on extending the timeline of the v(t) waveform, and you'll see it is correct.

Since the Zo of the TL is matched to the Zout of the source, there is a voltage division, and the initial wave headed down the TL is 1.65V. When it reflects off the far end, you get a full positive reflected wave, and as it propagates back up the TL, it raises the voltage along the TL to the full 3.3V output of the source. When that reflected wave hits the source, there are no further reflections, so the TL stays charged up to the DC 3.3V.

When the source transitions down to -3.3V, you again get the voltage division against Zo of the TL, so you get -1.65V, until the full (same direction) reflection from the far end of the TL can push the voltage down all the way to -3.3V.