- #1
seang
- 184
- 0
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))
[tex]
V^+ = \frac{V_o * R_0}{R_s + R_0}
[/tex]
which equals 0. I doubt this is correct, but I don't know what I'm doing wrong.
Could I maybe?
[tex]
V^+ = \frac{V_o * R_0}{R_L + R_0}
[/tex]
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))
[tex]
V^+ = \frac{V_o * R_0}{R_s + R_0}
[/tex]
which equals 0. I doubt this is correct, but I don't know what I'm doing wrong.
Could I maybe?
[tex]
V^+ = \frac{V_o * R_0}{R_L + R_0}
[/tex]
Last edited: