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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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    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. – Mattman944 Mar 02 '21 at 08:15
  • Thank you, I've edited that point into my question – Reine Abstraktion Mar 02 '21 at 08:21
  • 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. – jonk Mar 02 '21 at 09:52
  • related: https://electronics.stackexchange.com/questions/368181/why-in-a-passive-circuit-with-a-sinusoidal-input-do-all-voltages-and-currents – Sredni Vashtar Mar 02 '21 at 11:28
  • 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 – Reine Abstraktion Mar 02 '21 at 11:51
  • Thank you, @SredniVashtar that was a great post to read – Reine Abstraktion Mar 02 '21 at 11:52
  • @Buraian Okay. It's late at night for me. I'll see about helping in the day when I can access a PC. – jonk Mar 02 '21 at 11:53
  • Thank you ^^ @jonk – Reine Abstraktion Mar 02 '21 at 11:54
  • @SredniVashtar I'm able to answer this question from the post you've linked but I'm unable to understand the idea discussed in the linked post – Reine Abstraktion Mar 02 '21 at 11:56
  • @Buraian Is Andy's response of any help yet? I can see how it may not be -- it's a bit over-general. But it does approach an answer. So I'm just curious how you see it at this time. – jonk Mar 02 '21 at 11:58
  • I've left a comment on Andy's answer @jonk – Reine Abstraktion Mar 02 '21 at 12:03
  • @Buraian Got it. Thanks. That tells me more of what you need. I'll get some sleep first. – jonk Mar 02 '21 at 12:06

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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

Andy aka
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  • 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** – Sredni Vashtar Mar 02 '21 at 10:28
  • @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? – Andy aka Mar 02 '21 at 10:29
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    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. – Sredni Vashtar Mar 02 '21 at 10:33
  • @SredniVashtar OK, maybe you are right - I've added a few more words. – Andy aka Mar 02 '21 at 10:36
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    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 – Reine Abstraktion Mar 02 '21 at 12:03
  • @Buraian what proof are you unable to follow? Can you be more explicit in this? What linked answer do you refer to? – Andy aka Mar 02 '21 at 12:05
  • In my question body, I had linked an [answer](https://electronics.stackexchange.com/a/430284/236654) in another post. In it,OP says that if LHS must contain only sine and cosine terms with same frequency as LHS, I'm not sure how it is implied from x(t) = sin(wt). The part I understand is that the right hand side is some combination of sin(wt) and cos(wt), not sure how that leads back to a statement about 'y' though – Reine Abstraktion Mar 02 '21 at 12:26
  • @Buraian it's not very clear in his answer. I think he appears to be saying the same as my answer (above) in that if you differentiate a sinusoid, you cannot introduce harmonics. Neither has that answer been "accepted" by the OP (original proposer) and not the guy answering. – Andy aka Mar 02 '21 at 12:43
  • @Andyaka I think this answer needs the addition of hyperbolic equivalents: \$\cos\left(\omega t+\phi\right)=\frac{e^{j\left(\omega t+\phi\right)}+e^{-j\left(\omega t+\phi\right)}}2\$ and \$\sin\left(\omega t+\phi\right)=\frac{e^{j\left(\omega t+\phi\right)}-e^{-j\left(\omega t+\phi\right)}}{2j}\$. (Based upon \$\cos\left(x\right)=\cosh\left(ix\right)\$ and \$\sin\left(x\right)=-i\sinh\left(ix\right)\$.) This is where everything suddenly simplifies. – jonk Mar 04 '21 at 05:38
  • Though the frequency btohers me another thing which bothers me is that in the linked post he has seemingly time varying coefficients in the series – Reine Abstraktion Mar 04 '21 at 21:07
  • Could you elaborate how that simplifies? @jonk – Reine Abstraktion Mar 04 '21 at 21:08
  • @Buraian It separates out the time varying parts from the constant parts, makes the derivatives easy, and makes the collection of terms easy. I thought I'd have time before now, but I loaded down for a little while yet. – jonk Mar 05 '21 at 00:27