Talk:Linear circuit

Simulation

Removed the line "These can give solutions for any specific circuit, but not much insight into the operation of the circuit in general with different component values or inputs."

This is misleading. In general, running extensive Spice simulations is often the only way to gain insight into the true operation of complex circuits. However, this needs to be done intelligently. I would refer to the understanding of rotating colliding black holes in General Relativity, where the equations are also so complex that they cannot be fully understood without simulations. Kevin Aylward 13:30, 3 August 2013 (UTC) — Preceding unsigned comment added by Kevin aylward (talk • contribs)

Manufactures Linear Designation

Removed “In their linear components, manufacturers work to reduce nonlinear behavior to a minimum, to make the real component conform as closely as possible to the 'ideal' model used in circuit theory.” This is not true. Added correct explanation of what linear means to IC manufactures. Kevin Aylward 13:42, 3 August 2013 (UTC) — Preceding unsigned comment added by Kevin aylward (talk • contribs)

Incomplete definition of a linear circuit

The definition given in the very first paragraph of this article is partial, i.e. it needs more information. The current definition is "A linear circuit is an electronic circuit in which, for a sinusoidal input voltage of frequency f, any steady-state output of the circuit (the current through any component, or the voltage between any two points) is also sinusoidal with frequency f." However, this is only valid for sinusoidal AC circuits. This definition does not include the rest of AC circuits (e.g. square wave, triangular wave, sawtooth wave, and others), and it doesn't even include DC circuits!

By the way, this definition is also not the most fundamental definition. I mean, according to this definition, a circuit is linear by definition if the output signal has the same frequency as the input signal. However, there is a more broad definition by which we can proove the stated in the current definition.

--Alej27 (talk) 05:14, 27 March 2019 (UTC)

No, the definition is true for all circuits. If a circuit is linear, and the only time varying source applied to it is sinusoidal (a sine wave), then after the transient response of the circuit dies away, all the time varying currents and voltages anywhere in the circuit will also be sinusoidal, with the same frequency. In order to get a square wave, triangle wave, or sawtooth wave out of a circuit that only has a sine wave input, the circuit must have nonlinear components.

However, I think you are right that this definition is not the most fundamental definition. I believe the most common definition in textbooks is that a linear circuit is one that obeys the superposition principle. This should probably be the lead definition. --Chetvorno^TALK 11:25, 28 March 2019 (UTC)

I agree that superposition principle should be first. I read Alej27's complaint as coming from the perspective that the driver of the circuit (the oscillating voltage source etc.) is part of the "circuit", as opposed to a separate agent that acts on a circuit. From that perspective, it's a confusing and probably invalid definition. I think it should be possible to tweak the wording to make it clearer. It would also help to have a diagram in which there was a blank space where the input voltage would go (i.e., two open circles at the left side). --Steve (talk) 14:43, 28 March 2019 (UTC)

What diagram? There is no diagram in the article. If you are suggesting a diagram, then what of? SpinningSpark 19:12, 28 March 2019 (UTC)

I think the article could use a diagram illustrating how superposition works. The output of a multisource circuit is equal to the sum of the outputs when each source is applied individually, with all other sources brought to zero. --Chetvorno^TALK 21:00, 28 March 2019 (UTC)

@Alej27: @Sbyrnes321: No, if you read the definitions in the introduction (which are now adequately supported by sources) linearity and nonlinearity are properties of the transfer function (relation of output to input) of a component or circuit, its response to arbitrary driving sources. So the linear or nonlinear circuit itself does not include the "drivers", the independent source or sources of current applied to the circuit. A linear circuit with a square wave applied to it is still a linear circuit. I agree about the need for a diagram though. --Chetvorno^TALK 21:00, 28 March 2019 (UTC)

Yes I agree that Alej27 is wrong, I was saying that the root cause of his/her error is misunderstanding what the word "circuit" is referring to in "linear circuit". To non-specialists (e.g. imagine a grade school science class), all the so-called "circuits" they've ever seen in their lives have at least one battery or other active component, so that they actually do something. Take a disconnected resistor. That's a great example of a linear circuit. But can't you see how it's weird to call it a "circuit" in the first place? It's just a resistor sitting on a table, not connected to anything yet! It doesn't look like a "circuit", as the term is used by the general public. Can you see how some readers might get confused? The diagram proposal I was (poorly) describing is something vaguely like File:Two-port_parameters.PNG. Just anything to show visually that we are talking about so-called "circuits" that are meant to be connected to a (separate) battery or other driver that does not count as part of the circuit itself. I like the recent article edits BTW. --Steve (talk) 01:32, 29 March 2019 (UTC)

Okay, I see what you mean; it is confusing for nontechnical readers. Thanks! I agree a diagram could help - or even several diagrams, showing the different aspects of superposition. If no one else wants to do it, I can probably draw up some SVG diagrams. --Chetvorno^TALK 04:55, 29 March 2019 (UTC)

@Chetvorno:, with respect to the first paragraph of your first reply, you talked about sinusoidal AC inputs, but not about e.g. DC inputs. A DC circuit can in fact be linear (of course, as long as the elements/components of it are linear). This is what I'm trying to say, that a circuit can be linear even if the input signal is not a sinusoid.

Also, you said "In order to get a square wave, triangle wave, or sawtooth wave out of a circuit that only has a sine wave input, the circuit must have nonlinear components", which is true, but that's considering the square/sawtooth/triangle wave as output, and I meant as input; and a circuit that has those kind of waveforms as inputs can be linear (as long as they obey the superposition theorem.) --Alej27 (talk) 01:06, 30 March 2019 (UTC)

@Alej27: I agree with everything you said above. --Chetvorno^TALK 01:48, 30 March 2019 (UTC)

@Sbyrnes321: (Steve), could you please describe the definition of circuit which implies a resistor alone can be considered a circuit? Furthermore, please specify in which textbook/etc. you got that defintion. --Alej27 (talk) 01:14, 30 March 2019 (UTC)

Pedantically, it is not a circuit because there is not a complete current path. On the other hand, that will also apply to most networks we want to discuss; in the majority of cases we are discussing two-port networks without specifying what is connected to the ports. If you are bothered by this, we should rename the article Linear network, but it does not seem to bother a lot of textbook authors who are split roughly 50/50 between "circuit" and "network" in naming their books. On the other hand, the word circuit, has the advantage of immediately implying an electrical subject to the reader. A surprisingly large number of authors (nearly all of them) fail to give a definition of linear circuit/network at all. Ghosh gives the superposition definition but talks of linear systems rather than circuits/networks. Lakshmanan & Murali define a linear circuit as one made up of linear elements, which in turn they define as elements having a linear $v/i$, $v/q$, or $i/\varphi$ curve, so a lone resistor would fit this definition of linear circuit. SpinningSpark 07:32, 30 March 2019 (UTC)

Include yet another definition of linear circuit/device/system: if it satisfies the additivity and the homogeneity properties

As some of you may know, there's another definition for a linear system, and would in fact be an introduction to the first in the article (namely, that a linear circuit satisfies superposition theorem). The defintion I'm talking about is if the device/circuit/system satisfies the additivity property, which happens if $x_{1}(t)\rightarrow y_{1}(t)$ and $x_{2}(t)\rightarrow y_{2}(t)$ then $x_{1}(t)+x_{2}(t)\rightarrow y_{1}(t)+y_{2}(t)$ for all time $t$ and for any inputs $x_{1}(t)$ and $x_{2}(t)$; and if it also satisfies the homogeneity property, which happens if $x_{1}(t)\rightarrow y_{1}(t)$ then $a\,x_{2}(t)\rightarrow a\,y_{1}(t)$ for all time $t$ and any input $x_{1}(t)$ and any constant real numbers $a$. If either or none of these two properties is satisfied, the circuit/device/system is not linear.

Of course, if you/we are going to include this definition, it'd be useful for the reader to explain what each property means. For example, the homogeneity property is satisfied if a scaled input produces a scaled output by the same factor, and the additivity property is satisfied if a sum of inputs produces a sum of individual outputs, you get the point.

Then, combining these two properties or criteria, we get the so-called superposition theorem or principle of superposition: $a_{1}\,x_{1}(t)+a_{2}\,x_{2}(t)\rightarrow a_{1}\,y_{1}(t)+a_{2}\,y_{2}(t)$.

In the context of a two-terminal (i.e. one-port or single-phase) electrical device, the input would be the instantaneous conventional current $i(t)$ through it and the output would be the instantaneous voltage $v(t)$ across it, or vice versa.

Various textbooks in Signals and Systems, and in Control Systems, as well as some in Circuit Theory/Analysis (e.g. Alexander and Sadiku's, or Suresh Kumar's), use the definition of linear circuit based on additivity and homogeneity properties. Just to provide a reference, the one I talked about is extracted from Signals, Systems, and Transforms (4th edition) by Charles L. Phillips, John M. Parr and Eve A. Riskin, on page 74.

--Alej27 (talk) 22:51, 22 March 2021 (UTC)

Isn't that just the definition of superposition rather than something completely different? SpinningSpark 14:44, 5 August 2021 (UTC)

Should we clarify various common misconceptions about linear circuits?

Should we clarify the following common misconceptions about linear circuits?

1. That linear circuits don't have to obey Ohm's law. Sure, devices satisfying Ohm's law ($i(t)\propto v(t)$) also satisfy the superposition principle and are therefore linear, but ideal constant-inductance inductors ($v(t)\propto i'(t)$) and ideal constant-capacitance capacitors ($i(t)\propto v'(t)$) don't satisfy Ohm's law, yet still satisfy the superposition principle and are therefore still linear.
2. That linear circuits don't have to produce a sinusoidal output current when excited by a sinusoidal input voltage. Certainly, devices producing a sinusoidal current due to a sinusoidal voltage also satisfy the superposition principle and are thus linear (as far as I know, those are linear time-invariant devices), but linear time-variant devices, such as inductors and capacitors with inductance/capacitance depending on time but not on magnetic/electric quantities, in general don't produce a sinusoidal current due to a sinusoidal voltage, yet still satisfy the superposition principle and are thus still linear. For example, a rheostat whose knob is being turned by a machine, having a resistance of $R(t)=[10+2.5\cos {(20\pi t)}]{\text{ }}\Omega$, won't produce a sinusoidal current when excited by a sinusoidal voltage, but such rheostat/variable resistor still satisfies superposition and so it's linear.
3. That the characteristics plot (i.e. parametric plot of current versus voltage) of linear circuits doesn't have to be a straight line through the origin. For example, consider an ideal constant-inductance inductor or ideal constant-capacitance capacitor excited by a sinusoidal voltage and operating in steady-state, so the current will also be sinusoidal of same frequency as the voltage, but in general with different amplitude, and the phase shift in time between the voltage and current will be 90°, hence the current-voltage plot of those devices will be an ellipse centered at the origin, not a straight line through the origin, however those devices still satisfy the superposition principle and are therefore linear.

What are your thoughts? We may not have to include the examples above, I just provided them so you can see what I say is true. ​​​​ --Alej27 (talk) 23:50, 29 January 2022 (UTC)

I'm not sure what difficulties you are trying to address here. On the first point, Ohm's law is not mentioned at all in the article, and the inclusion of capacitors and inductors in the list of linear components plainly implies that Ohm's law isn't required. The same thing could be said about your third point due to the inclusion of differentiators and itegrators, although I concede that it might be useful to point out that the meaning of linearity here is different from the common meaning of a straight-line constitutive relationship (that is linear map is meant, not linear function (calculus)). In fact a straight line with a constant offset is not strictly speaking linear, although its deviation from linear under superposition of two signals is only in the DC offset.

The example you give in the second point amounts to a frequency mixer with one of its inputs mechanical instead of electrical. Mixers are not considered linear (its article explicitly says they are nonlinear) because the output is not a linear superposition of its inputs. This of course raises the question of what exactly do we mean by linearity when it comes to multiport devices. Multiports are not currently covered in this article at all, nor does the linked article linear function help very much. SpinningSpark 09:25, 30 January 2022 (UTC)

@Spinningspark: Thanks for your reply. I'll divide this comment for each reply.

• "I'm not sure what difficulties you are trying to address here." // I'd like to add in this article that linear circuits don't have to obey Ohm's law, that they don't have to produce a sinusoidal output current when excited by a sinusoidal input voltage, and that their characteristics curve doesn't have to be a straight line through the origin. Why? Because I've seen many people believe this. For example, in this Quora question many people say that a circuit is linear if it obeys Ohm's law or if its characteristics curve is a straight line through the origin (which is true, but they don't clarify that those are sufficient but not necessary conditions, so many readers leave thinking those are necessary conditions, which is false). As another example, even the IEEE standard 519-1992 defines a non-linear load as a load that produces a non-sinusoidal current when excited by a sinusoidal voltage, which is not consistent with the definition of a linear circuit of this article (and many textbooks).
• "On the first point, Ohm's law is not mentioned [...]. The same thing could be said about your third point [...]" // You're correct. Listing capacitors/inductors and including differentiators/integrators as linear devices already shows my point #1 and #3. I just thought explicitly indicating what I said would be useful.
• "The example you give in the second point [...]" // I would have to disagree that my example is a non-linear device (which what I think you meant), because such device/resistor satisfies the superposition principle. The definition of a linear circuit doesn't require for the output to be sinusoidal when the input is sinusoidal. Just to make it clear, I'll quickly prove such resistor satisfies the superposition principle. Let's consider the voltage as input and the current as output. The voltage-current relation of such resistor is:

$i(t)={\dfrac {v(t)}{R(t)}}={\dfrac {v(t)}{10+2.5\cos {(20\pi t)}}}\quad {\text{(1)}}$

An input voltage $v_{1}(t)$ produces an output current $i_{1}(t)$ given by:

$i_{1}(t)={\dfrac {v_{1}(t)}{10+2.5\cos {(20\pi t)}}}\quad {\text{(2)}}$

An input voltage $v_{2}(t)$ produces an output current $i_{2}(t)$ given by:

$i_{2}(t)={\dfrac {v_{2}(t)}{10+2.5\cos {(20\pi t)}}}\quad {\text{(3)}}$

An input voltage $v_{3}(t)$ given by $v_{3}(t)=a\,v_{1}(t)+b\,v_{2}(t)$ produces an output current $i_{3}(t)$ given by:

${\begin{aligned}i_{3}(t)&={\dfrac {v_{3}(t)}{10+2.5\cos {(20\pi t)}}}\\&={\dfrac {a\,v_{1}(t)+b\,v_{2}(t)}{10+2.5\cos {(20\pi t)}}}\\&=a{\dfrac {v_{1}(t)}{10+2.5\cos {(20\pi t)}}}+b{\dfrac {v_{2}(t)}{10+2.5\cos {(20\pi t)}}}\quad {\text{(4)}}\end{aligned}}$

Substituting eqs. (2) and (3) into (4), we get:

$i_{3}(t)=a\,i_{1}(t)+b\,i_{2}(t)\quad {\text{(5)}}$

Therefore, the voltage-current relation of the resistor $R(t)=10+2.5\cos {(20\pi t)}$ satisfies the superposition principle, and is thus linear according to the present definition of linear circuit of this article. So it is incorrect to say that such resistor is non-linear. (We can also prove it is a time-variant resistor, but this is not relevant to this article.)

Anyways, the main point of this discussion is what I highlighted in italics text. I think from your comment that we don't have to clarify my first point, while it would be useful to clarify my third point, right? Also, now that you've seen the variable resistor of my example is indeed linear, yet I think you said it was non-linear, maybe we should also clarify my second point? --Alej27 (talk) 18:17, 30 January 2022 (UTC)

As I said, in all essentials this is a three port device with two inputs. Your calculation is only working because you have held one port at a constant amplitude and constant frequency. If one sets $v(t)=0$ then $i(t)=0$ whatever the mechanical port is doing. If one then sets $\omega =0$ with a nonzero $v(t)$ then $i(t)=v(t)/12.5$. If one now applies both at once, that is nonzero $v(t)$ and nonzero $\omega =20\pi$ the output is patently not given by superposition, $i(t)\neq 0+v(t)/12.5$. The same is true for an ideal mixer. If one port is held at a steady fixed frequency (a common arrangement in radio, one port will be the carrier frequency) then superposition applies at the other port. Even more simply, an ideal mixer is a multiplier function; with inputs a and b the output is ab. If one holds b=1 then the output is a and any superposition at the a input is tautologically equal to the output. Nevertheless, mixers, even ideal mixers, are considered nonlinear and do not obey superposition in the general case. If you want to say something different, then you need to cough up some sources backing it up. SpinningSpark 15:04, 31 January 2022 (UTC)

@Spinningspark: Oh, I hadn't considered the rotating machine as an input, I just mentioned it to give some practical application of my example. But it's true, considering the circuit as having one electrical input (the voltage), one mechanical output (the rotational speed of the machine, which changes the resistance) and one electrical output (the current), it's indeed non-linear. So let's ignore the point #2 in my first comment. For the point #1, you said it wasn't necessary because we're already including inductors/capacitors. For the point #3, you said we could add it to the article, so I think I'll add it since it may be useful for the reader to know.

Wikipedia is an encyclopedia rather than a textbook. There is no imperative to address misconceptions, common or otherwise. But sometimes the articles do address that. The three items mentioned are not what I would consider common or important misconceptions. An important misconception would be a misconception that hinders the progression of people’s understanding. The ohm’s law misconception is analogous to the misconception that you can assemble an automobile using only a screw-driver. If one has that misconception, one quickly learns that it is false. It doesn't hold anybody back. Constant314 (talk) 02:13, 1 February 2022 (UTC)