What Happens When a Transmission Line Source is Switched Off?

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Discussion Overview

The discussion revolves around the behavior of a lossless transmission line when the source is switched off, particularly focusing on the implications of the source resistance and the resulting circuit configuration. Participants explore the theoretical aspects of voltage behavior in response to switching actions, considering both ideal and practical scenarios.

Discussion Character

  • Debate/contested
  • Technical explanation

Main Points Raised

  • One participant proposes that switching off the source results in an open circuit, equating it to a source resistance of infinity.
  • Another participant questions this interpretation, suggesting that turning off the source voltage is equivalent to creating a short circuit, as it sets the voltage across the source to zero.
  • A different participant emphasizes the importance of defining the action taken when the source is turned off, noting that a real power supply would ramp down the output voltage due to output capacitance, while an in-line switch would create an open circuit.
  • There is uncertainty regarding the correct interpretation of the circuit behavior after the source is switched off, with conflicting views on whether it behaves as an open circuit or a short circuit.

Areas of Agreement / Disagreement

Participants express disagreement on the consequences of switching off the source, with multiple competing views on whether it results in an open circuit or a short circuit. The discussion remains unresolved.

Contextual Notes

Participants highlight the need for clarity on the definitions and assumptions regarding the source's behavior when switched off, as well as the implications of practical versus ideal conditions in the analysis.

seang
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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))

<br /> V^+ = \frac{V_o * R_0}{R_s + R_0}<br />

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

Could I maybe?

<br /> V^+ = \frac{V_o * R_0}{R_L + R_0}<br />
 
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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?
 
Yes sir. I think anyway. I'll attach some pictures later today
 
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.
 
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.
 
seang said:
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.

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.
 

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