# Analytical proof that L and C are linear devices?

#### xopek

I mean I know they are linear since they obey the ohms law. But I don't quite understand the reasoning that since, say, V=Ldi/dt and taking a derivative is a linear operation therefore it is a linear device?? I can verify that sin'(x) = cos(x) or sin(x+90) so the signal is time shifted but its form remains the same. That sounds logical to me but that has nothing to do with the fact that differentiation is a linear operation (which I believe is related to limits and differentials dy=m*dx etc). But what if we feed some sort of a high order polynomial instead of sin into L? Then taking derivative would distort the signal and would make the transformation nonlinear. So is it only linear when the signal is sin or cos?

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#### tnich

Homework Helper
They are linear in the sense that if $f(.)$ is the transfer function of the inductor or capacitor, then given two signals $x_1(t)$ and $x_2(t)$, $$f(a_1x_1(t) +a_2x_2(t)) = a_1f(x_1(t)+a_2f(x_2(t))$$ That is $f(.)$ is a linear operator on functions of time. That means that you can use Fourier analysis to decompose a signal into a sum of $\sin(.)$and $\cos(.)$ signals, apply the transfer function to each one, and then sum them up to get the result of applying the transfer function to the original signal. So inductors and capacitors do not preserve phase relationships among frequency components.

• anorlunda

#### xopek

Thanks. So just to clarify f(.) is a derivative? What is a transfer function in this context? I/V, i.e. impedance?
If so, then for a capacitor, I(t)=CdV/dt, and V(t)=cos(wt), so the transfer function, if we can call it that, is Z(t) or simply Xc?

#### tnich

Homework Helper
The transfer function converts an input signal to an output signal. So in this case, if the input signal is $V(t)$ and the output signal is $I(t) = C\frac{dV}{dt}$, the transfer function is $f(.) = C\frac{d}{dt}$. You can easily show that it is linear in the sense that I defined it in post #2.

#### sophiecentaur

Gold Member
So is it only linear when the signal is sin or cos?
The term 'linear' implies that the same transfer function applies for any circuit with just RLC in it, irrespective of the amplitude of the signal. I think you are over thinking this. Differentiation still scales with the value of a constant multiplier.

• tnich

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