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I was seeing this video from khan academy about linear circuit systems(linear system with a single input), at 3:24 Willy McAllister says that if it is a linear circuit, then we can assume all components have the same 'omega'. I don't get why this is true and some searching on SE led me to this post. In the answer by jramsay, it is said that it is because linear system corresponds to a linear differential equation with the following relation between input x(t) and output y(t):

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

They state that if \$x(t)=\sin(\omega t)\$ then RHS can only contain sine and cosine term to first power, and then he says that this implies LHS must contain similar terms with the same frequency. The premise is intuitive for me but I don't get how the conclusion follows (The statements on LHS)

Tl;dr: I want to understand why linear differential equations governing a circuit imply each component has the same frequency.

Edit: I found this link helpful see pg-6 and 7

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    \$\begingroup\$ I don't know how to prove it, and this isn't what this site is all about. Clarification: If you have a linear system with a single frequency input, all components will have this same frequency. A linear system cannot produce new frequencies. Multiple frequency inputs can be analyzed with superposition. \$\endgroup\$
    – Mattman944
    Commented Mar 2, 2021 at 8:15
  • \$\begingroup\$ Thank you, I've edited that point into my question \$\endgroup\$ Commented Mar 2, 2021 at 8:21
  • \$\begingroup\$ Buraian, If you just spend a little time with sine, cosine, exp, and hyperbolic functions, it will all just fall in place. Just an hour or two. Then go look up Euler's. \$\endgroup\$
    – jonk
    Commented Mar 2, 2021 at 9:52
  • \$\begingroup\$ related: electronics.stackexchange.com/questions/368181/… \$\endgroup\$ Commented Mar 2, 2021 at 11:28
  • \$\begingroup\$ I do have familarity with those functions @jonk , the problem is I can't really transfer the ideas related to those functions to a 'somewhat' rigorous proof of the statement about the equation I've mentioned in the quesiton \$\endgroup\$ Commented Mar 2, 2021 at 11:51

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I want to understand why linear differential equations governing a circuit implies each component has same frequency.

If you differentiate a sine wave you get a cosine wave of exactly the same frequency. No matter how many times you do this, you get the same waveform shape albeit shifted in time and maybe amplitude (but not in frequency). Differentiating does not produce new harmonics when the original signal is sine shaped. To "generate" distortion or non-linearity requires harmonics to be present. It can't happen with linear differential equations.

If you take the series definition of a sinewave for instance: -

$$x - \dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+....$$

And then, if you differentiated it you'd get this: -

$$1 - 3\dfrac{x^2}{3!}+5\dfrac{x^4}{5!}-7\dfrac{x^6}{7!}+....$$

Which equals: -

$$1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+....$$

And this is the series definition of a cosine wave. No harmonics are introduced: -

enter image description here

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  • \$\begingroup\$ Correct, but to make it clearer to the op it might be useful to explicit the fact that when you diiferentiate a sinusoidal functuon you get another sinusoidal function of the same frequency \$\endgroup\$ Commented Mar 2, 2021 at 10:28
  • \$\begingroup\$ @SredniVashtar didn't I say that in my answer: If you differentiate a sine wave you get a cosine wave - maybe you mean something else? Something more mathy? \$\endgroup\$
    – Andy aka
    Commented Mar 2, 2021 at 10:29
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    \$\begingroup\$ Nope, just making it explicit. When a student has a doubt like this it helps in expliciting the obvious. The cosine wave you get is obviously at the same frequency of the sine wave you differentiated, but that's what the OP did not seem to realize. \$\endgroup\$ Commented Mar 2, 2021 at 10:33
  • \$\begingroup\$ @SredniVashtar OK, maybe you are right - I've added a few more words. \$\endgroup\$
    – Andy aka
    Commented Mar 2, 2021 at 10:36
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    \$\begingroup\$ I understood the concept of the series definition of the sine and cosine wave and I also understand the distortion differentiating causes just a scale up of amplitude, but I'm still unable to follow the proof as said by the answerer in linked answer \$\endgroup\$ Commented Mar 2, 2021 at 12:03

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