Brief Question
I'm trying to figure out a proof for the Complex Power Conservation Theorem, which does not use Nodal Analysis.
Theory I know
In my university this theorem has ben taught with the name "Boucherot's Theorem", but I couldn't find a lot of mentions of it. Example here: Boucherot Eng Wiki page. So this is just a new reference to it (and furthermore shorter names win).
It states, in his briefer version (which I took and summarised from the book "Chua-Desoer-Kuh"):
Boucherot's Theorem (thesis): Consider a linear time invariant circuit, made up of B branches, driven by a number of AC (sinusoidal) independent sources a the same frequency \$ \omega \$, in his sinusoidal steady state. Then, assuming a unique sign convention for power measurement ("passive" or "active"), the following statement (Complex Power Conservation) is true: $$\sum_{k=1}^B S_k=\sum_{k=1}^B V_k \overline{I}_k=0$$ Where \$ S_k, V_k, \overline{I}_k \$ are respectively Complex Power, Voltage phasor, conjugate Current phasor for branch \$ k\$. Absolute values of the phasors here are to be considered rms values.
For the sake of completeness, the nodal analysis proof I know is the following one.
Nodal Analysis Proof: Assuming \$ [A]\$ to be the reduced incidence matrix of the circuit for a datum node (GND), and \$ \{V\}\$, \$ \{I\}\$, \$ \{E\}\$ respectively the columns of branch-voltage, branch-current and node-potential phasors; then for the KLC and KLV respectively: $$ [A]\{I\}=\{0\} \qquad [A]^t\{E\}=\{V\}$$ \$[A]\$ is a real matrix, so taking the conjugate of the KLC matrix equation we obtain \$ [A]\{\overline{I}\}=\{0\}\$. Therefore: $$\sum_{k=1}^B S_k=\sum_{k=1}^B V_k \overline{I}_k=\{V\}^t\{\overline{I}\}=([A]^t\{E\})^t\{\overline{I}\}=\{E\}^t[A]\{\overline{I}\}=\{E\}^t\{0\}=0 \qquad QED$$ (Nodal Analysis uses passive sign convention for each branch)
What I've found, and what I've thought of
Searching the web for a different proof, I stumbled upon this document here: Original Paul Boucherot Writings for 1900 Congrès international d'électricitè. And I think there is something wrong: the math is really unclear.
For first, it starts from the Conservation of Instantaneous Power: assuming a unique sign convention for power measurement ("passive" or "active"): $$\sum_{k=1}^B p_k=\sum_{k=1}^B v_k i_k=0$$ In my studies this is Tellegen's Theorem thesis, for which here no proof are presented. With Nodal Analysis the proof is quite identical to the one I showed before: branch-voltages, branch-current, and node-potential satisfy the same equations in fact. The only differences are that it's true for every circuit at every instant, and obviously that we're not talking about phasors.
At this point I would like to see also a proof for Tellegen's theorem which does not uses Nodal Analysis because, as far as I know, it's a mathematical instrument that came many years after Boucherot's work. I could assume it true using the generic Principle of Conservation of Energy, right, but I find it like cheating.
Anyway, assuming the previous statement true, it seems to me there are still math problems in the writings I linked. For example, why does he write \$\psi\$ and \$\varphi\$, instead of \$\psi_k\$ and \$\varphi_k\$? From the French, I got that \$\psi_k=\angle V_k\$ and \$\varphi_k\equiv\angle V_k - \angle I_k\$ (it's the argument of the impedance when there is one, used for active and reactive powers): these angles depend on the branch for sure.
Please note I use the subscript \$k\$ instead of \$n\$, and \$V_k\$ instead of \$E_n\$ for branch-voltage phasors, to keep a consistent notation, and equivalence symbol \$\equiv\$ to highlight that one can take the equivalent angle usually in the interval \$(-\pi,\pi]\$.
Then, if I name \$\vartheta_k\$ the argument \$\angle I_k\$, knowing that \$\psi_k=\angle V_k\equiv\angle I_k+\varphi_k=\vartheta_k+\varphi_k\$, Tellegen's equation become: $$\sum_{k=1}^B \sqrt{2} |V_k| \sin(\omega t + \psi_k) \sqrt{2} |I_k| \sin(\omega t + \vartheta_k)=0$$ $$\sum_{k=1}^B |V_k| |I_k| \sin(\omega t + \vartheta_k + \varphi_k) \sin(\omega t + \vartheta_k)=0$$ $$\sum_{k=1}^B |V_k| |I_k| \cos(\varphi_k) \sin^2(\omega t + \vartheta_k) + |V_k| |I_k|\sin(\varphi_k) \sin(\omega t + \vartheta_k)\cos(\omega t + \vartheta_k) =0$$ $$\sum_{k=1}^B P_k \sin^2(\omega t + \vartheta_k) +\sum_{k=1}^B Q_k \sin(\omega t + \vartheta_k)\cos(\omega t + \vartheta_k) =0$$
Where \$P_k=|V_k| |I_k| \cos\varphi_k\$ and \$Q_k=|V_k| |I_k| \sin\varphi_k\$ are respectively the active ad reactive powers for branch \$k\$.
He gets a similar equation, but using \$\psi_k\$. How? Anyway, he just proceeds writing down that "hence, the equations": $$\sum_{k=1}^B P_k \sin^2(\omega t + \vartheta_k)=0 \qquad \sum_{k=1}^B Q_k \sin(\omega t + \vartheta_k)\cos(\omega t + \vartheta_k) =0$$ This just seems wrong to me.
Then he talks about integrating the equations, that is ok, but he completely skip over the fact that \$\vartheta_k\$ depends on the branch, so in general every integral inside the sums does.
Anyway, \$\sin^2(\omega t+\vartheta_k)\$ and \$\sin(\omega t+\vartheta_k)\cos(\omega t+\vartheta_k)\$ are periodic in t, with (minimum) period \$\frac{\pi}{\omega}\$, hence integrating the complete equation over a whole number of periods solves the problem of the branch-dependence. With easy substitutions, and using the said property for integrals of periodic functions, I got: $$\int_{-\frac{\pi}{\omega}}^{\frac{\pi}{\omega}} (\sum_{k=1}^B P_k \sin^2(\omega t + \vartheta_k) +\sum_{k=1}^B Q_k \sin(\omega t + \vartheta_k)\cos(\omega t + \vartheta_k) \,dt)=0$$ $$\int_{-\pi}^{\pi} \sin^2x\,dx\cdot\sum_{k=1}^B P_k + \int_{-\pi}^{\pi} \sin x\cos x\,dx\cdot\sum_{k=1}^B Q_k=0$$ Here I noticed that the function \$\sin x \cos x\$ is odd (I choose symmetrical integration interval for this reason), then its integral over \$(-\pi,\pi)\$ is 0. Furthermore \$\sin^2 x\$ is non negative and positive for some intervals, so its integral over the same interval is positive (its value is \$\pi\$). Hence we get: $$\sum_{k=1}^B P_k=0$$
This is half Boucherot's thesis: \$\sum_{k=1}^B Q_k=0\$ is missing. How do I obtain it? The problem I see, is that integrating must be done over periods, in order to isolate the sum, but this lead to 0 coefficient for the sum.
This seems completely ok, because as everyone knows "instantaneous reactive power", i.e. the contribute to \$p_k\$ due to \$Q_k\$, has an average value of zero over a period. In particular, I mean this addend of the \$p_k(t)\$ expression: $$p_{k,r}(t)=2 Q_k \sin(\omega t + \vartheta_k)\cos(\omega t + \vartheta_k)=Q_k \sin(2(\omega t + \vartheta_k))$$
So, my idea initially was to differentiate in \$t\$ the original equation, in order to obtain even functions for the \$Q_k\$ products, and odd ones for \$P_k\$ products, and then integrating back. In other words, it's just taking the original equation (a primitive), evaluating it in \$\frac{\pi}{\omega}\$ and in \$-\frac{\pi}{\omega}\$, and evaluating their difference. Again this lead to having zeroes, so it does not work.
It seems like I have to use a slightly different equation (something like a \$\pm\frac{\pi}{2}\$ phase shift only for voltage phasors) from the beginning, in order to have \$\sin\varphi_k\$ to multiply the \$\sin^2\$.
I must be missing something. Someone knows what? Every idea, and different proof, is welcome.
In the end, another consideration. For this proof, more than once I read online people talking about "transforming Tellegen's equation in its phasor version", i.e. obtaining \$\sum_{k=1}^B V_k I_k=0\$ substituting phasors in \$\sum_{k=1}^B v_k i_k=0\$ (and then saying: \$\overline{I_k}\$ and \$I_k\$ satisfy the same equations, so \$\sum_{k=1}^B V_k \overline{I_k}=0\$ also must be true).
I find this wrong too, because phasors don't work like that with products: \$V_kI_k\$ is not the phasor of \$v_ki_k\$, but of \$|V_k| |I_k| \sin(\omega t + \vartheta_k + \psi_k)\$, which has nothing to do with these calculations. Is this right, or am I missing something here too?
