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.(adsbygoogle = window.adsbygoogle || []).push({});

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]

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# Transmission Line Discharging:

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