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Consider the signals $e(t)$ and $h(t)$ with units $\left[\frac{V}{m}\right]$ and $\left[\frac{A}{m}\right]$ and their Fourier transforms $E(\omega)$ and $H(\omega)$ with units $\left[\frac{Vs}{m}\right]$ and $\left[\frac{As}{m}\right]$. The time avaverage P of the product of the two signals (~power spectral density), $e(t)h(t)$, in function of the Fourier transforms $E(\omega)$ and $H(\omega)$ can be written as \begin{align} P &= <P'(t)> = \lim\limits_{T\to \infty} \frac{1}{T} \int\limits_{-T}^{T} e(t)h(t) dt \\ &= \lim\limits_{T\to \infty} \frac{1}{T} \int\limits_{-T}^{T} \int\limits_{-\infty}^{\infty} E(\omega)e^{i\omega t} d\omega \int\limits_{-\infty}^{\infty} H(\omega')e^{i\omega' t} d\omega' dt \\ &= \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty}E(\omega) H(\omega')\left[ \lim\limits_{T\to \infty} \frac{1}{T} \int\limits_{-T}^{T}e^{i(\omega+\omega')t} dt \right] d\omega d\omega' \\ &= 2 \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty}E(\omega) H(\omega')\delta(\omega'+\omega) d\omega d\omega' \\ &= 2 \int\limits_{-\infty}^{\infty} E(\omega) H(-\omega)d\omega \\ \end{align}

The question: the unit of the time average $P$ should be $\left[\frac{VA}{m^2}\right]$ whereas the unit of the final result is $\left[\frac{VA}{m}s\right]$. How can this be and how to interpret it?

Example: if you know the Fourier transforms $E(\omega)$ and $H(\omega)$, and you want to calculate the average $P$, then you will use $P=2 \int\limits_{-\infty}^{\infty} E(\omega) H(-\omega)d\omega$ but this will give another unit as expected?! So, then you have the wrong result?

What is unit of time average of product of two signals?

It's the product of the units of the signal. In my example is $V \cdot A = W$ which is power and in yours it's $V/m \cdot A/m = W/m^2$ which is Intensity.

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The spectrum of the product and the product of the spectra are completely different things.

Let's look at a simple example. Let's look at a cirucit with a voltage source and single impedance $Z$. For simplicity we will use the complex admittance $Y = 1/Z = Y_m \cdot e^{j \varphi}$ where $Y_m$ is the magnitude and $\varphi$ is the phase of the admittance.

Let's drive this with a voltage $u(t) = u_0 \cdot cos(\omega t)$. The current waveform will then be $$i(t) = u_0 \cdot Y_m \cdot cos(\omega t + \varphi) $$

The instantaneous power is $$p(t) = u(t) \cdot i(t) = u_0^2 \cdot Y_m\cdot [cos(\omega t) \cdot cos(\omega t + \varphi)]$$

Using the trig identitiy we get $$p(t) = u_0^2\cdot Y_m/2 \cdot [cos(\varphi)+cos(2 \cdot \omega \cdot t + \varphi)]$$

The time averaged power would simply be $$<p(t)> = u_0^2\cdot Y_m/2 \cdot cos(\varphi)$$

The spectrum of the instantaneous power has two frequency compontents: One at $0Hz$ and one at $2 \omega$. The $0Hz$ component is the time averaged power and the actual physical power that would be dissipated in the impedance. It obviously depends on the phase of the impedance. For example for a capacitor or inductor, the dissipated power is 0 and the instantaneous power spectrum ONLY has a reactive componet at $2 \cdot \omega$.

Note that the spectrum of the instantaneous power is a VERY different from the power spectrum of both voltage and the current , i.e. only one component at $\omega$

The assumption that you can get the spectrum of the power my muliplying the spectra of the field quantities is just plain wrong. The math doesn't work out either: multiplication in the time domain is CONVOLUTION in the frequency domain, not multiplication. If you convolve the units work out as well

$$P(\omega ) = \int_{s}^{}U(s) \cdot I(s-\omega)ds$$

which has the correct units of $W/Hz$.

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  • $\begingroup$ Thank you for this explanation, it is very useful! You also mention that I assume that "you can get the spectrum of the power by multiplying the spectra of the field quantities" but I do not see where I assume this...? Or do I implicitly assume it somewhere? Is there a mistake in my mathematical derivation? I want to express the time average intensity in function of the Fourier transforms of the $e$ and $h$ field. $\endgroup$ – Frederic Oct 29 '20 at 13:17
  • $\begingroup$ I do not understand the expression for time averaged power. $\endgroup$ – jomegaA Oct 29 '20 at 13:37
  • $\begingroup$ @jomegaA, I use the general definition of the average of a function, $<f(t)>=\lim\limits_{T\to \infty} \frac{1}{2T}\int\limits_{-T}^{T} f(t) dt$, and apply it to the function $e(t)h(t)$. $\endgroup$ – Frederic Oct 29 '20 at 15:09
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Your derivation is wrong. Your assumption about the limit

$$\lim_{T\to\infty}\frac{1}{T}\int_{-T}^Te^{j\omega t}dt\tag{1}$$

is not correct. Clearly, for $\omega\neq 0$ the limit $(1)$ equals zero. For $\omega=0$ you get

$$\lim_{T\to\infty}\frac{1}{T}\int_{-T}^Tdt=2\tag{2}$$

So there's no Dirac delta impulse involved here. Note that a Dirac impulse doesn't have a finite value when its argument is zero.

Also note that in the definition of the time average you should either divide by $2T$ or integrate from $-T/2$ to $T/2$.

Furthermore, the limit you use in the definition of the average is applicable to power signals, which don't have a Fourier transform. Yet you assume the existence of the Fourier transforms of the individual signals. So there are other things to worry about than the units.

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  • $\begingroup$ The limit only becomes 2 if $\omega+\omega'=0$ and is 0 otherwise. So, this corresponds to the Dirac function $2\delta(\omega'-\omega)$. I do not see the mistake? $\endgroup$ – Frederic Oct 29 '20 at 15:04
  • $\begingroup$ @Frederic: No, the Dirac delta does not equal unity for zero argument. $\endgroup$ – Matt L. Oct 29 '20 at 15:35

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