Impulse function applied to RLC circuit

AI Thread Summary
The discussion centers on the application of the impulse function to a series RLC circuit, specifically analyzing the current i(t) at time t=0+. The author references a voltage source Vo*delta(t) and questions how the current can be Io = Vo/L given the instantaneous voltage change. A key point raised is the misunderstanding of the inductor and capacitor behavior at this moment; the inductor acts as a short circuit while the capacitor behaves as an open circuit. To resolve the confusion, it's suggested to derive the current by writing the differential equation and integrating it between the intervals of 0- and 0+. Understanding these principles is crucial for accurately analyzing RLC circuits under impulse conditions.
marc.orr
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Hello, I've been reviewing some material from my undergraduate studies. In the book "Signals and Systems" by Haykin in the first chapter he gives an example of a voltage source with the value of Vo*delta(t) applied to a series RLC circuit.

Vo is some arbitrary constant and delta(t) is the impulse function. The question is what is the value of the current i(t) at time t=0+ where 0+ is at 0 looking from the more positive side of time.

He says the answer is (without any sort of derivation) Io = Vo/L. How is this possible? I figure that as there is an instantaneous change in voltage, the inductor will look like a short circuit and the capacitor will look like an open circuit. So how can there be any current flow?

Thankyou in advance.

Marc
 
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marc.orr said:
I figure that as there is an instantaneous change in voltage, the inductor will look like a short circuit and the capacitor will look like an open circuit.
You have this reversed.
 
If you write the differential equation and integrate it between 0- and 0+ you can find Io.
 
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