Superposition theorem when having voltage sources with different frequency - Page 1
Forum section: electronics / beginners

author: nForce
author_rank: Frequent Contributor
timestamp: November 11, 2017, 05:20:42 pm
content: Hello,When calculating power with two or more voltage sources on let's say resistor in linear circuit, the superposition theorem does not apply. It only applies if and only if we have voltage sources with different frequencies.Can anyone explain intuively and with a mathematical proof why is this the case? So if we have a voltage source let say 5cos(2t) V and other one 2cos(3t) V. Where frequencies are different, so it applies.Thanks.

author: Dave
author_rank: Super Contributor
timestamp: November 11, 2017, 05:37:06 pm
content: As long as the circuit is linear, it does apply.Why wouldn't it?

author: IanB
author_rank: Super Contributor
timestamp: November 11, 2017, 05:38:57 pm
content: I thought the superposition theorem applies to linear systems? So unless there are non-linear elements it should apply.I would have to research the application of the theorem to AC voltage sources to say any more.

author: nForce
author_rank: Frequent Contributor
timestamp: November 11, 2017, 06:39:41 pm
content: I am asking for thepower, not voltage or current.https://www.quora.com/Why-superposition-theorem-is-not-valid-for-power-in-a-resistive-circuit

author: IanB
author_rank: Super Contributor
timestamp: November 11, 2017, 06:44:55 pm
content: Because power is proportional to the square of the current or the square of the voltage, so it is nonlinear. So superposition does not apply directly.However, if you calculate the voltages and currents by superposition first, and then calculate the power in any given element afterwards, then that result will be correct.

author: orolo
author_rank: Frequent Contributor
timestamp: November 11, 2017, 06:49:03 pm
content: It think I see what you mean. There are two kinds of linearity in play. For different frequencies you have linear addition of the powers at each frequency. At the same frequency, you have the power of the linear combination of the two waves. It all boils down to the orthogonality relations for trigonometric functions.Consider a voltage wave, \$V \ = \ A\cos\omega_1 t \, + \, B\cos\omega_2 t\$, if you are to compute power, you must integrate its square. Squaring: \$ V^2 \ = \ A^2\cos^2\omega_1 t + 2AB\cos\omega_1t \cos\omega_2t + B^2\cos^2\omega_2t\$If \$\omega_1 \ne\omega_2\$, integrating over the period makes the middle term go away, and you are left with the powr of the individual components. There is linearity.If \$\omega_1 = \omega_2\$, you have \$V^2 \ = \ (A^2 +2AB + B^2)\cos^2\omega_1t \ = \ (A+B)^2\cos^2\omega_1t\$ . The individual waves do not contribute their power, but interfere and you get the power of thelinearly combinedwave.

author: IanB
author_rank: Super Contributor
timestamp: November 11, 2017, 06:52:42 pm
content: By the way, you began by making a statement as a premise for your question:Quote from: nForce on November 11, 2017, 05:20:42 pmHowever, you did not give a source or attribution for that statement. So we have no way of understanding where it came from, or what the context is. Therefore answering your question becomes difficult because we have to guess what has prompted it.If you could give a reference, such as "I read this in textbook XXX and I don't understand it, could you help to explain it?" then it would be much easier to answer the question, since we can then refer to the same source and understand the background.

author: shteii01
author_rank: Frequent Contributor
timestamp: November 12, 2017, 03:54:41 am
content: P=V*IIf you have voltage and current, you will find power.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 12, 2017, 04:54:25 am
content: The proof you're looking for is the proof of Parseval's theorem.If the frequencies are equal, then they interfere and add or subtract. Else, they interfere in and out periodically, to equal extents, and the average power goes as RMS, rather than arithmetic sum.More generally, when detected by a nonlinear process (in this example, p(t) = v(t)^2), correlated signals interfere, while uncorrelated signals do not (the statistical generalization of RMS is the trend for noise going as 1/sqrt(N) for N sample size).Tim

author: MrAl
author_rank: Super Contributor
timestamp: November 12, 2017, 06:03:46 pm
content: Quote from: nForce on November 11, 2017, 05:20:42 pmHi,It is not entirely clear where you are coming from here. You are quoting two time domain waveforms and then trying to apply some theory, but it's not completely clear why you are doing this. I think you would have to go into more detail.I say this because when we calculate power it's different with time domain signals. We either calculate the instantaneous power or the average power, with average power usually being the goal anyway, and this requires an integration.Start with a simpler example. Start with v1=sin(t) and v2=sin(2*t) and two resistors each 1 ohm, where they are joined together in series, and have v1 at one loose end and v2 at the other loose end, and that forms a voltage divider between v1 and v2, and then try to calculate the power in say the resistor connected to v1.From this you can make it clear what you are trying to find out.Show how you WANT to calculate the power.

author: nForce
author_rank: Frequent Contributor
timestamp: January 29, 2018, 09:24:17 pm
content: I have another question about power calculation in superposition.If we have I = Ig*cos(wt)and U = Ug*cos(wt)and omega are different.Why do we take only the amplitudes of cosinuses and compute the product (P = 1/2 U*I )? So power does not change by cos(t)?Thanks

author: ezalys
author_rank: Frequent Contributor
timestamp: January 30, 2018, 03:08:59 am
content: Here's a calculation that shows why

author: nForce
author_rank: Frequent Contributor
timestamp: January 30, 2018, 12:28:48 pm
content: Hmm, ezalys I understand most of the proof but why is it so that for calculating power, if we take the constant V^2 out, we are left with cos(x)^2 and for calculating power over a period we get:https://www.wolframalpha.com/input/?i=integral+from+0+to+2pi+cos(x)%C2%B2Sorry, but I am not that good at math.

author: ezalys
author_rank: Frequent Contributor
timestamp: January 31, 2018, 01:11:55 am
content: For time averaging you need to divide out by the averaging time, so you would wind up with 1/(2*pi) * \int_0^2\pi \cos(x)^2\,dx=1/2. Hence why there's a factor of two in the time averaged power.

author: Brumby
author_rank: Unknown
timestamp: January 31, 2018, 03:10:22 am
content: Interesting discussion. I'm not going to get into the math - but I have the following scenario to offer for people to address their solutions to the question:Say there are two simple sinusoidal AC sources of the same frequency and amplitude being fed into a load - let's just make it resistive. What is the power dissipated in that load when the phase difference between the two sources is:1. 0ºand2. 180º