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let's consider this situation in which the two ports of a transmission line are connected to a short circuit and to an inductor:

enter image description here

The transmission line is supposed to be lossless. My book makes a mathematical analysis in order to find the resonance frequency (precisely, resonance frequencies) of this network, which may start to oscillate.

But my question is: how can it start to oscillate on its own (since it is a passive device)? I have understood that there are no losses, but I think it is necessarily a form of excitement or accumulated energy in the circuit (for example, to make an LC circuit oscillate, the capacity must be pre-charged at the time of closing the switch).

Oscillation means voltage + current and so energy. And it cannot create energy from 0.

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  • \$\begingroup\$ It will not oscillate unless some sort of energy is added somehow, as you have mentioned, because that is impossible. Perhaps some stray magnetic fields from something else will excite the inductor. \$\endgroup\$ – user253751 Nov 26 '19 at 11:12
  • \$\begingroup\$ May some noise be sufficient to start oscillations? \$\endgroup\$ – Kinka-Byo Nov 26 '19 at 13:24
  • \$\begingroup\$ Well noise is adding energy so yes. Thermal noise by itself will be very small. \$\endgroup\$ – user253751 Nov 26 '19 at 13:35
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how can it start to oscillate on its own

It will not oscillate in steady state (assuming either the lines or the inductor are real devices with some loss). If the inductor is initially charged (by some other circuit, not shown in your diagram), it may oscillate for some time until the stored energy is dispersed through the loss mechanisms.

to make an LC circuit oscillate, the capacity must be pre-charged at the time of closing the switch

Not true. You could leave the capacitor uncharged, and pre-charge the inductor (with an initial current) instead.

Oscillation means voltage + current and so energy. And it cannot create energy from 0.

Correct. This circuit will not start oscillating on its own. It needs some other way to obtain an initial stored energy before it can oscillate.

In the real world (no lossless components) it will (like any other oscillator) need continuous energy input to maintain a steady state oscillation.

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  • \$\begingroup\$ Perfect, thank you very much \$\endgroup\$ – Kinka-Byo Nov 26 '19 at 20:19

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