Assume the picture below:


From inlet port In1 and In2 we can have electromagnetic or acoustic energy entrace single sine wave with same amplitude and frequency. Assume we have aligned the waves to completely destroy each other After two wave summing in summing point (assume summing point has been created approximately perfect with no reflection). Then we sense zero energy after summing point, where the waves have steady state. This rejects energy conservation since it means we after that "summing point steady state output", we will never be able to access that energy. Then have removed some energy from our word at least apparently, though in theory the still present and is bounded together.

What is wrong on my argument? Maybe energy conservation is just applicable for masses and not the waves? I'm electrical engineer and not the physician sorry for weak physical understanding.

I've heard collision of particles can create positive and negative particles, like matter and anti-matter.

  • 1
    $\begingroup$ Possibly relevant: physics.stackexchange.com/questions/306845/… $\endgroup$
    – Andrew
    Jan 24, 2023 at 12:15
  • $\begingroup$ Please note a simple real example similar to your depicted one. Consider a tube where you have a plane wave travelling to the right. At some point you introduce a cancelling wave with a loudspeaker. To the right of the speaker there is complete cancellation (you can do even with trial and error). Does that violate energy conservation? Nope, of course not (as nicael suggested). We haven't considered what happens to the left of the loudspeaker. The pressure is doubled!!! This is where all the energy goes. So in effect you created an impedance mismatch at that point in order to reflect (cont.) $\endgroup$
    – ZaellixA
    Jan 26, 2023 at 1:00
  • $\begingroup$ (cont.ed) the incoming wave energy. Of course this is not the only way to setup an ANC system. Pressure minimisation and power minimisation are, probably, the two most well known approaches to (generic) ANC systems, but they are not the only ones. I strongly suggest you take a look at a very nice "review" by Nelson and Elliot "Active Noise Control" (here: researchgate.net/profile/Philip-Nelson/publication/…) $\endgroup$
    – ZaellixA
    Jan 26, 2023 at 1:05
  • $\begingroup$ @ZaellixA Since speakers are dipole radiator, we have also left wave of both In1 and In2. But we can control them through the tube in such a way to cancel those too. $\endgroup$ Jan 26, 2023 at 21:51
  • $\begingroup$ Yes, you are right, but I was conceptualizing a system where the "cancelling" radiator (as well as the "primary" one) would be at the end points of the tube, or if the cancelling would not be, then only one of its side would be firing inside the tube. This way, the "back side" of the radiators wouldn't comprise part of the system so they wouldn't take part in the energy conservation principle. $\endgroup$
    – ZaellixA
    Jan 27, 2023 at 10:31

1 Answer 1



Of course, the conservation of energy is not violated here. This is a common misconception related to Active Noise Control (ANC). In this 1D problem (I consider this to be a 1D problem since we assume only plane waves and we don’t take into account the angle of incidence at the junction) is shown below why this is the case.

What follows is based on the basic derivations provided in ”Signal Processing for Active Control” by Steven Elliot, which is a highly recommended textbook for those who would like to get started with ANC.


I will assume that the reader is familiar with the plane wave equation which is given below

$$ p \left( x, t \right) = A e^{j \left( \omega t - k x \right)} \tag{1} \label{1} $$

where $x$ is the spatial coordinate (a scalar in this 1D case), $t$ is the time dimention, $A$ is the amplitude of the wave, $\omega$ the radial frequency for which $\omega = 2 \pi f$, with $f$ the temporal frequency is true, $k$ is the wavenumber (scalar in this 1D case) for which $k = \frac{\omega}{c} = \frac{2 \pi}{\lambda}$ is true with $c$ the speed of propagation and $\lambda$ the wavelength.

It is easier to drop the time dependence and work in the wavenumber domain, in which equation \eqref{1} is written as

$$ p \left( x \right) = A e^{- j k x} \tag{2} \label{2} $$

Now, I will use a simpler system than the one you provide and then extend to what you have shown in your image. So, below you see an equivalent system (source: ”Signal Processing for Active Control” by Steven Elliot)

1D Active Noise Control problem

In the image, we see three propagating waves where $p_{p+} \left( x \right)$ is the “noise” (I’ll use this term to denote the signal we care to cancel, in the literature this is found as the primary source) propagating on the positive $x$ direction, $p_{s+} \left( x \right)$ and $p_{s-} \left( x \right)$ waves which are the propagating waves in the positive and negative directions respectively due to the “control” source (in the literature this is found as secondary source).

Note that $u$ is the signal of the “control” source and it is in the same frequency $\omega$ (hence same wavenumber $k$) as the “noise” source signal. The waves due to the “control” signal can be written as

$$ \begin{align} p_{s+} \left( x \right) = B e^{-j k x}, ~~~ x > 0 \tag{3.a} \label{3.a}\\ p_{s-} \left( x \right) = B e^{+j k x}, ~~~ x < 0 \tag{3.b} \label{3.b} \end{align}$$

where again here $B$ is the amplitude of the waves.

The condition we want to impose is

$$ p_{p+} \left( x \right) + p_{s+} \left( x \right) = 0 \implies p_{s+} \left( x \right) = -p_{p+} \left( x \right) \implies B e^{-j k x} = -A e^{-j k x} \implies B = -A \tag{4} \label{4} $$

By setting the amplitude of the “control” source waves equal to the negative of the “noise” wave we can achieve perfect (in theory only) noise reduction downstream.

Now, what about upstream, towards the direction of the “noise” source? Well, we can calculate the resulting sound field as the summation of the two fields in this region. This is

$$ p_{p+} \left( x \right) + p_{s-} \left( x \right) = A e^{-j k x} - A e^{+j k x} = -A \left( e^{j k x} - e^{-j k x} \right) = -2 j A \sin \left(k x \right) \tag{5} \label{5} $$

where in the last step the known equality $e^{j x} - e^{-j x} = 2 j \sin \left( x \right)$ has been used. Notice that equation \eqref{5} describes a standing wave, which is the superposition of a positive-going with a negative-going wave. The nodes (pressure nodes in this case) are at $x = 0, \frac{\lambda}{2}, \lambda, \ldots, n \frac{\lambda}{2}, ~~~ n \in \mathbb{N}$ (this is where the function $\sin \left( k x \right)$ is zero). Similarly, at positions $x = \frac{\lambda}{4}, \frac{3 \lambda}{4}, \ldots, \frac{\left(2 n + 1 \right) \lambda}{4}, ~~~ n \in \mathbb{N}$ the amplitude of the resulting wave is $2 A$.

Thus, the energy that was moving towards the right (positive $x$ direction) has now been diverted to the left (negative $x$ direction). This constitutes a realisation of the energy conservation principle.

Extention to your problem

Please note that the above situation is completely equivalent to your system. That is because the phenomena at the junction (you describe it as the ”Summing point”) are exactly the same as those described in the previous section (Derivation) if we assume (lossless) plane wave propagation in an infinite domain. The back-propagating wave from the “control” source you mention is not part of the system and thus it plays absolutely no role in the conservation of energy in the system. The energy in this other system (the propagation of the “back-propagating” wave of the “control” source) is conserved by definition since there are no other sources and the system is lossless by definition.


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