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Why is a combination of R, L, C linear time-invariant system? But not diode, or transistor?

I looked over V-I equation for R, L, C, but still don't know how to prove it.

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  • $\begingroup$ Votes and best answer validation are required $\endgroup$ – Laurent Duval Jul 28 at 12:01
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actually, the diode and transistor are also time-invariant components. the transistor is even reasonably linear in a range of operation between cutoff and saturation. even, for very small displacements, a diode can be sorta linear, but a diode in normal use is clearly not linear.

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Linear

The voltage-current equation of a diode and a transistor is related by an exponential function, while for RLC components they are linear operators (derivatives, integrals and multiplication by constants)

Time-invariant

This will depend on the pre-defined voltages and currents, i.e., initial conditions. For example, take an RLC with a capacitor fully charged. As the circuit starts, the capacitor and the inductor will keep throwing energy back and forth (oscillation) and the resistor will eat some of that energy away (decay). Depending on when you apply an input, the response will be different. Hence, by definition, the system is time-variant.

A circuit with a diode/transistor CAN be time-invariant, but it can't be linear (unless you look at it with the small-signal model, but that's a whole other story).

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Probably an RLC circuit, with a resistor (R), an inductor (L), and a capacitor (C), with parameters that do not vary over time. In the series circuit for instance, with constant voltage, you are led to a linear differential equation.

So the circuit, with inputs and outputs, is a system. With constant parameters, it is time-invariant. It is linear, governed by a linear equation.

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In response to the first question by the OP: RLC circuits are indeed LTI systems because the relationships are linear and the response DUE TO THE INPUT does not vary if the input is delayed, other than the delay itself. If some capacitor was to be charged prior to the signal being introduced, that charging step would also be delayed along with delaying the introduction of the signal because that charging step is also an input. The responses would be identical.

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