2
$\begingroup$

Suppose there are 2 sinusoidal signals $\cos(\omega t)$ and $\cos(\omega t)$, which implies that their individual power is $1/2$ unit and total power is $1$ unit. When added, the signal $2\cos(\omega t)$, however, has power $2$ unit.

Obviously,

$$\rm Energy_{signal} = Power_{signal}\times Time_{period}$$

So, how does the principle of energy conservation hold in this case?

Improve this question
$\endgroup$
0
$\begingroup$

Power is homogeneous to a second degree quantity. When the data is doubled, the power is quadrupled.

The energy here (classicaly) is defined in an Hilbert space as a squared $L_2$ norm, deriving from some inner product $\langle\cdot,\cdot\rangle$, assuming a real Hilbert space, $\|x\|^2 = \langle x,x\rangle$. Hence: $$\|x+y\|^2 = \langle x+y,x+y\rangle = \|x\|^2 + 2\langle x,y\rangle+ \|y\|^2 \,.$$

Now set $y=x$, you gracefully get $\|x+x\|^2 = 4\|x\|^2$.

Improve this answer
$\endgroup$
9
  • $\begingroup$ True. Power is proportional to the square of amplitude. I understand that. My question is regarding how energy is conserved in this case, That is, we had 2 individual cosine waves before. After passing them through a summer, the energy is more than the sum of their energies before passing them through the summer. How is this possible? $\endgroup$ – Curiosity Mar 7 '18 at 19:14
  • $\begingroup$ @Curiosity The adder is consuming power and delivering it in the form of the new signal. $\endgroup$ – MBaz Mar 7 '18 at 19:20
  • $\begingroup$ Energy is not a simple conservative additive quantity. Imagine two signals $\cos \omega t$ and $-\cos \omega t$. Their sum will be zero energy. Why are you ressorting to an energy conservation principle here? $\endgroup$ – Laurent Duval Mar 7 '18 at 19:22
  • $\begingroup$ Well, yes, that's my question. I mean, that's the part I don't get. Why should energy not be conserved here? $\endgroup$ – Curiosity Mar 7 '18 at 19:25
  • $\begingroup$ I mean, what is the process that consumes the energy? $\endgroup$ – Curiosity Mar 7 '18 at 19:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.