# Transmission Line Discharging:

1. Jan 24, 2007

### seang

I'm looking at a lossless transmission line where the load is a resistor, and the input is a step. The source resistance is zero, and the system has reached steady state.

So in this case, what happens when the source is switched off? You'd have the resistor on the right, and to the left of the t-line, an open circuit. Isn't this the same as a source resistance of infinity?

So to obtain the initial, positive direction voltage, I'd say:

V = (V(steady) * R (characteristic) / (R (source) + R (characteristic))

$$V^+ = \frac{V_o * R_0}{R_s + R_0}$$

which equals 0. I doubt this is correct, but I don't know what I'm doing wrong.

Could I maybe?

$$V^+ = \frac{V_o * R_0}{R_L + R_0}$$

Last edited: Jan 24, 2007
2. Jan 25, 2007

### Staff: Mentor

This is a hard question, at least as stated. Are you saying that the IC is that there is a Vo source with Rs=0, the system is settled, and then you open circuit the source?

3. Jan 25, 2007

### seang

Yes sir. I think anyway. I'll attach some pictures later today

4. Jan 25, 2007

### Manchot

If you switch off the source at t=0, the result is not an open circuit: it is a short circuit! Turning off the source voltage is equivalent to setting the voltage across it to zero, which is the same as a short circuit. If you'd like, you can think about it as adding another voltage signal opposite to the original one.

5. Jan 26, 2007

### seang

I think you're incorrect. The source is 'turned off' via a switch, which forms the open circuit right? The source doesn't go to zero; I think that's what you had in mind.

6. Jan 26, 2007

### Staff: Mentor

Well, as Manchot says, turning off a real power supply and leaving it connected does ramp the output voltage down to zero (slowly usually because of all the output capacitance in the power supply). On the other hand, you can open circuit the connection between the power supply and the transmission line with an in-line switch. It's important to define exactly what you are doing in this problem.