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I am familiar that in any circuit composed of linear passive elements and a sinusoidal input, all voltages and currents through and across any element will exhibit the same sinusoidal behavior and frequency as the input; that's how passive filters work in fact. But I can't figure out or find a concrete/straightforward proof for why this happens, if not plain observation.

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  • \$\begingroup\$ You can prove for each component in question. Each component has a well defined behavior. \$\endgroup\$ – Eugene Sh. Apr 12 '18 at 15:38
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    \$\begingroup\$ Thr mather nature loves the sinewave. In the capacitor, for example, the current in the capacitor is directly proportional to the rate of change of voltage across its plates. I = C * dV/dt. So if the voltage is a sinewave surprise, surprise the derivative of a sine wave is a cosine wave (phase shift sine wave ). So, the mother nature must love the sine wave. The same is true for an inductor V = L* dI/dt. And if the voltage is a sine wave the current is a cosine wave. \$\endgroup\$ – G36 Apr 12 '18 at 15:45
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    \$\begingroup\$ I know We Don't Like Fun™ but your lecture about how mather nature loves the sine wave just made my day. \$\endgroup\$ – dlatikay Apr 12 '18 at 21:20
  • \$\begingroup\$ Using temperature coefficients of resistors, and the total thermal resistance (resistor, PCB trace, bolts to take heat off the PCB), you'll find the IP3 (3rd order distortion intercept point) of a Surface Mount resistor of value 100,000 ohms is about 1,000 volts. Of course that is a 10 watt dissipation in a SMT resistor. \$\endgroup\$ – analogsystemsrf Apr 13 '18 at 3:20
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    \$\begingroup\$ What kinds of passive elements are you talking about? Diodes are passive but I'll be damned if you can get sinusoids out of them... \$\endgroup\$ – Mehrdad Apr 13 '18 at 10:54
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I've been pouring my brains out and eventually I've found a nice mathematical approach to prove this and decided to answer my own question. In such a circuit, solving for any voltage/current across/through any component (I'll call that \$f\$) would always lead you to construct a differential equation that is always linear, with constant coefficients (due to linear properties of passive components) and non-homogeneous (due to the sinusoidal input). Such a differential equation will always take this form: $$a\frac{d^nf}{dt^n}+b\frac{d^{n-1}f}{dt^{n-1}}+...+j\frac{df}{dt}+kf=C\sin{(\omega t+\theta)}$$ where \$a...k\$ are constants (combinations of inductance, resistance, etc.), \$n\$ is the order of the differential equation (which reflects the number of energy storage elements in the circuit), and \$C\sin{(\omega t+\theta)}\$ is a generalized sinusoidal function that describes the input. A general solution to this differential equation will always take this form: $$f=\text{(general homogeneous solution)}+\text{(particular solution)}$$ where the particular solution \$=A\sin{(\omega t+\theta)}+B\cos{(\omega t+\theta)}\$ which is a sinusoidal function of the same frequency! Now, in AC circuit analysis, we are always looking at the circuit in steady state, when the homogeneous solution approaches zero (which inevitably happens because of resistances in the circuit).

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    \$\begingroup\$ We don't deserve people like you. A person who asks a good and well written question and then makes a good answer. \$\endgroup\$ – Harry Svensson Apr 12 '18 at 16:37
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    \$\begingroup\$ It's worth pointing out for future readers that the requirement for the circuit to be linear is not stated in the original question, but is required for this solution to apply (and for the result to be correct). Another way to say this is that sinusoids (and exponentials) are eigenfunctions of the derivative operator. \$\endgroup\$ – The Photon Apr 12 '18 at 16:43
  • \$\begingroup\$ Simply said: If the derivative of a sine has the same frequency, then ANY order derivative has the same freq. \$\endgroup\$ – Roland Apr 13 '18 at 15:23
  • \$\begingroup\$ How does your postulation address the condition of an ideal, resonant LC circuit where the transformation = 0? \$\endgroup\$ – Glenn W9IQ Apr 13 '18 at 20:49
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    \$\begingroup\$ A resonant LC circuit's output is just two sinusoids that exactly cancel out. Luckily, there is no such thing as an ideal LC circuit for the sinusoids to exactly cancel out, so the output is just a sinusoid with a very small amplitude. \$\endgroup\$ – mjtsquared Apr 14 '18 at 6:35
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This is only true for LTI (Linear Time-Invariant) circuits. If you have a non-ideal component (and they all are to one degree or another) you will see harmonics of the input frequency in the output. Inductors tend to be the worst of the lot, but all passive parts have such behavior. For example, capacitors can exhibit strong voltage coefficient and are not time invariant because of dielectric absorption.

For a straightforward (assuming roughly 2nd year University math knowledge) mathematical proof you can read these Berkeley course (EECS20N: Signals and Systems) notes. You can download the entire text here.

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  • \$\begingroup\$ Are inductors really the worst of the lot? Some core materials are certainly very nonlinear, but at least HF air toroidal inductors should be very linear indeed. \$\endgroup\$ – leftaroundabout Apr 13 '18 at 15:00
  • \$\begingroup\$ @leftaroundabout I suppose ceramic capacitors give them a run for their money. Inductors tend to be less ideal in a linear fashion because of the resistance of the wire. \$\endgroup\$ – Spehro Pefhany Apr 13 '18 at 17:56
  • \$\begingroup\$ If it is true in the case of LTI circuits, how do you address the condition of an ideal, resonant LC circuit where the transformation = 0? \$\endgroup\$ – Glenn W9IQ Apr 13 '18 at 23:41
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It happens because a sinewave is just one line in the frequency spectrum and no matter what you do with it using a linear filter or amplifier, all that happens is that the phase or amplitude shifts.

If it were a square wave (infinite harmonics) then applying a filter would attenuate or accenuate some frequencies more than others and the square wave would lose its recognizable square shape.

Square wave harmonics: -

enter image description here

Gif source

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  • \$\begingroup\$ if a square wave is like an apple, a sinusoidal input signal is like an orange \$\endgroup\$ – Roland Apr 13 '18 at 15:09
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The basic reason is that the constituent equations of ideal R, L and C components are linear, time invariant equations involving only derivatives and integrals (both linear operations) and that sine and cosine change into other sines and cosines when acted upon such linear operators.

The derivative and the integral of a sinusoidal function is another sinusoidal function of the same frequency (it can only change in amplitude and phase). KCL and KVL can only lead to algebraic sums of such sinusoidal functions, and that operation can only produce another sinusoidal function. So, in the end, when you connect R, L and C in a network, a sinusoidal input will always lead to a sinusoidal output.

See my other answer here.

All of this is a direct consequence of the self-similarity of the exponential function (related to sines and cosines by Euler's equation). You might want to read the first chapter in Giorgi, The Physics of Waves to get a complete explanation for that.

(Note that this property of transforming into a scaled and time-shifted copy of itself on an interval spanning from \$t=-\infty \$ to \$t=+\infty\$ its unique to generalized sinusoidal functions - all other functions will end up being 'deformed' by the linear time-invariant circuit. Solutions of a linear system that are scaled copies of themselves like in \$A \ x = \lambda \ x\$ (where \$\lambda\$ is a complex scalar carrying information on attenuation and phase shift) are called characteristic, or proper, or eigen- solutions of the systems. They can be used to build an orthogonal basis with the property that any other (well-behaved) function can be decomposed as a generalized sum of such elementary bricks - and this will lead you straight into Fourier series territory, but that's another story).

A concise explanation is given in the first answer to this question on Math SE: Why do we use trig functions in Fourier transforms, and not other periodic functions?

The Fourier basis functions \$e^{iωx}\$ are eigenfunctions of the shift operator \$S_h\$ that maps a function \$f(x)\$ to the function \$f(x−h)\$: \$e^{iω(x−h)}=e^{−iωh} e^{iωx}\$ for all \$x∈R\$.

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  • \$\begingroup\$ "So, in the end, when you connect R, L and C in a network, a sinusoidal input will always lead to a sinusoidal output." with a notable exception of a resonant LC circuit with an output of 0 - not a sine wave. \$\endgroup\$ – Glenn W9IQ Apr 14 '18 at 0:49
  • \$\begingroup\$ You mean A sin(w t + fi) for A=0 ? Still a sinusoid, just a bit too small to be appreciated. Same goes for placing two identical sinusoidal generators one against the other. \$\endgroup\$ – Sredni Vashtar Apr 14 '18 at 3:36
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This is true only when restricting passive elements to R,L,C, and maybe crystals that are properly driven - and even then, there are two exceptions, see below. Intentional and unintentional diodes, varistors, thermistors with a thermal mass, and other non-linear elements can quickly introduce distortions to a pure sinusoidal inputs. Overdriven crystals or ceramic filters might also behave rather nonlinear. If including two-terminal elements with negative resistance (gas discharge tubes, tunnel diodes) in the passive category, even more possibilities exist.

The exceptions:

Real-world parts tend to have imperfections that make them behave a bit like some nonlinear elements. Resistors can have "thermistor with a thermal mass" and even "varistor" behaviour. Capacitors can have voltage dependency in their value due to piezoelectric effects, electric fields yielding mechanical force, chemical effects (in electrolytics). Also, some electret-like effects seem to be documented for capacitors. Metal to metal joints can develop diode-like behaviour. Inductors can become nonlinear through core saturation, interaction of the magnetic field with nearby metal objects, etc...

All resistive components carrying a current exhibit some noise generating behaviours, the lower limits of which are defined by hard physics.

Mind that all real-life seemingly non-sinusoidal, repetitive signals can be perfectly described as a sum of sine waves of varying frequencies and phases.

Looking for the connection to nature will have you going in circles: Sine waves are the principal ingredient in making circles and ovals and round things, according to maths geeks (if you want to draw a circle on a computer, you will usually either use sine/cosine functions or use pythagoras' theorem directly in some way...) . Nature makes a lot of round things (hair, plant stalks, cherries, cherry stains, tornadoes, etc) and keeps an ample supply of sine waves around for that purpose.

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  • \$\begingroup\$ Your answer was from before adding 'linear' to the question. Yes, in practice, most things do not behave perfectly linear. But also, perfect sinus signals are hard to find in the real world. The tomatoe is not a perfect circle, neither is planet Earth or its orbit. Practical signals are indeed nice to model using multiple sines. \$\endgroup\$ – Roland Apr 13 '18 at 15:15
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    \$\begingroup\$ Actually a perfect sine is impossible to find in the real world. You need it to run from time \$-\infty\$ to \$+\infty\$ and while the plus direction can be debatable, for the minus part we are limited to the age of the universe. \$\endgroup\$ – Sredni Vashtar Apr 13 '18 at 16:14
  • \$\begingroup\$ I am aware that a time limited sine, in effect, has harmonics :) \$\endgroup\$ – rackandboneman Apr 13 '18 at 18:07
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A 'circuit' is usually considered a network of components, with an 'input' and an 'output' port. With network theory, such as Ohms Law, you can derive an equation, the 'transfer function', that describes the output in terms of the input. With 'linear' components, you will always find a 'linear' transfer function.

Let's describe some linear components with functions like output = F(input), output2 = G(input2), etc. Then the combination of such components leads to a combined function like output2 = G(F(input1)). Because both functions are linear, thus of the form y = a * x + b, then those combinations are also linear.

When applying a sinusoidal input signal to the linear network, the output can be amplified by the factor a, and shifted by voltage b. With complex math, or differential equations you can even get 'phase shift', but not a different frequency, because the derivative of a sine has the same frequency.

Do you want this even more formal?

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Either your premise is false or you have not properly articulated the boundary conditions.

Consider a simple passive device such as a diode. It will exhibit a non-linear transfer characteristic resulting in a non-sinusoidal output for a given

Also consider an ideal resonant (LC) circuit with a transfer function resulting in zero output - thus non-sinusoidal.

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    \$\begingroup\$ Yeah, now he's added linear to the question. \$\endgroup\$ – pipe Apr 13 '18 at 14:13
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    \$\begingroup\$ Actually, non-linear devices are more fun. With that simple diode you can e.g. demodulate radio signals (crystal receiver) \$\endgroup\$ – Roland Apr 13 '18 at 15:18
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The eigenfunctions of linear time invariant systems (and passive networks generally are of that kind) are complex exponentials, and their real superpositions are sinoids of arbitrary phase.

An eigenfunction is a function which will only change by a constant (in this case, complex) factor when put through a system. Linear systems are those where the output corresponding to the sum of several inputs corresponds to the sum of the output of the individual inputs, so you can always analyze them by expressing their input as a convenient sum. If this sum can be a sum expressed in an orthogonal eigenfunction basis, things become so much easier.

Hello Fourier analysis.

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