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Why is the frequency of the response same as that of the forcing function in a linear circuit? What's the case when the circuit isn't linear?

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  • \$\begingroup\$ Hi, Adrash. You might have observed that your question attracted some downvotes. This is mainly because people feel that this question is a homework question, which is okay to post it here if you show some effort. \$\endgroup\$ – Hazem Apr 2 at 7:19
  • \$\begingroup\$ If this is actually a homework question, and depending on the way the course is taught, then they might be looking for an answer something like "because the complex exponentials are eigenfunctions of the derivative operator". The trick is relating that mathematical explanation to the real world. \$\endgroup\$ – The Photon Apr 2 at 16:04
  • \$\begingroup\$ @Hazem I get your point, but it's not a HW question, you're right I should have shown some effort. But I couldn't get a starting point. \$\endgroup\$ – Adarsh Kumar Apr 2 at 16:46
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If a circuit (or any system) is linear, then the output is governed by a linear differential equation. This means that the input signal and all its derivatives are not cubed, squared or anything like that.

So if out input is \$x(t)\$ and the output is \$y(t)\$ we can write an equation of the form

$$ a_0 + a_1(t)y(t) + a_2(t)\frac{dy(t)}{dt}+a_3(t)\frac{d^2y(t)}{dt^2} + ... = b_0 + b_1(t)x(t) + b_2(t)\frac{dx(t)}{dt}+b_3(t)\frac{d^2x(t)}{dt^2} + ... $$

So if x(t) is \$sin(wt)\$ the RHS can contain only terms of sin or cosine to the first power. Thus the LHS can only contain terms of sin or cosine to the first power. These all have the same frequency.

If there was a squared term, or some higher power, then there would be a sine squared term which oscillates at twice the original frequency, or another multiple.

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  • \$\begingroup\$ The equation you provided kinda confused me, I would appreciate if you could explain the LHS of it. What I understand of the equation is that for a linear circuit the KVL will be an integro-differential equation which when solved results in a term(if any) of order 2 at max. so the frequency remains the same, but even in this case the frequency of the term $$sin\omega t*cos\omega t$$ is twice as the original frequency? \$\endgroup\$ – Adarsh Kumar Apr 2 at 16:40
  • \$\begingroup\$ The equation I wrote was supposed to just be a general linear differential equation. Best to ignore it for now if you are familiar with the fact that KVL will give you an integro-differential equation. You are mistaken, however, that it can be at most of order 2. It can be of order N if there are N reactive elements (inductors and capacitors). The order of a linear differential equation refers to the highest derivative. No matter how many times you differentiate sine or cosine you can only get a sine or cosine of the same frequency. There will be no product terms like you wrote. \$\endgroup\$ – jramsay42 Apr 2 at 23:08
  • \$\begingroup\$ Yes you're right, i just misunderstood what you said. \$\endgroup\$ – Adarsh Kumar Apr 3 at 3:34
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'Forcing function' is the generic term for an input signal. If the forcing function is the unit impulse then it's response is the transfer function, since the Laplace transform of the unit impulse is unity. A transfer function must be a linear representation of a system for this to apply.

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A transfer function is defined as the output-to-input ratio as a function of frequency. It makes really no sense to find such a ratio for a sinusoidal input and an output that is non-sinusoidal (because the circuit is non-linear and introduces non-linear distortions). For this reason, transfer functions are given for linear circuits only, or better and more correct: ....for circuits which are assumed to be linear. In fact, there are no linear circuits - however, for sufficiently small amplitudes we treat the circuits as linear (and accept a small error).

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  • \$\begingroup\$ "(because the circuit is non-linear and introduces non-linear distortions)." So what? \$\endgroup\$ – Adarsh Kumar Apr 2 at 16:43
  • \$\begingroup\$ @AdarshKumar, ... so you cannot use any of the linear analysis tools, such as transfer function, hence can't use s -> jw to calculate a frequency response. Also cannot use superposition, etc... \$\endgroup\$ – Chu Apr 2 at 16:48
  • \$\begingroup\$ I get the point, but what's the relation between the transfer function and the frequency. Are you saying that we can't transform from time domain to the frequency domain, if yes why, is it related to the the transform function that you mentioned, how? \$\endgroup\$ – Adarsh Kumar Apr 2 at 18:06

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