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My understanding is that the relationship between complex signals and real signals is such that for some complex signal $f(t) = s(t) \cdot e^{j2\pi f_c}$, the actual broadcast signal sent through the physical channel is $\Re{\{f(t)\}}$, such that for a given complex signal magnitude $r(t)$ and phase $\theta(t)$ of $s(t)$, the physical signal is then $r(t) \cdot \cos(2\pi f_c + \theta(t))$.

However, if we use the standard definition of signal power (which is $\frac{1}{b-a}\int_a^b \lvert f^2(t)\rvert \,dt$), we get two different values for, e.g., $s(t) = \frac{1 + j}{\sqrt{2}}$. The power appears as 1 when calculating the complex signal $f(t)$ directly, but only $\frac{1}{2}$ when calculating the power of $\Re\{f(t)\}$.

Why did they define power that way then? Why not have a conversion factor to make complex signal power equal its real corresponding power?

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    $\begingroup$ The imaginary component of the complex signal counts for something, too. Usually $|z| \ge |\Re e\{z\}|$ so I think that $|z|^2 \ge |\Re e\{z\}|^2$. $\endgroup$ Sep 12, 2023 at 6:23

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It is an interesting question, I see it in this way.

First, calculating the energy by integration in the frequency with Rayleigh's theorem: $\int\limits_{-\infty}^{\infty}|V(f)|^2df$ it is easy to generalize that the energy of a complex signal is twice the energy of the real signal if the analytic signal is considered as the complex signal.

Consider $x(t)$ the real signal and $z(t)$ the complex analytic signal, then:

$Z(\omega)=[1+sgn(\omega)]X(\omega)$

That means that the spectrum of $z(t)$ is $0$ in negative frequencies and double magnitude in positive ones compared with the spectrum of $x(t)$. Then:

$E_z=\int\limits_{-\infty}^{\infty}|[1+sgn(\omega)]X(\omega)|^2df=4\int\limits_{0}^{\infty}|X(f)|^2df$

and in the case of $E_x$, using the even symmetry of the module of the spectrum of a real signal:

$E_x=\int\limits_{-\infty}^{\infty}|X(\omega)|^2df=2\int\limits_{0}^{\infty}|X(f)|^2df$

So, $E_z=2E_x$

Second, where does this additional energy come from? If we start from $x(t)$ and construct $z(t)$ we add an imaginary part to $x(t)$. For the analytic signal, this imaginary part is the Hilbert transform of $x(t)$. Since with Hilbert transform you are multiplying negative and positive frequencies by $j$ and $-j$ the module of the spectrum remains equal and according to Rayleigh's theorem that means same energy. Then the energy of $\mathrm{Im}\{z(t)\}$ is the same than $\mathrm{Re}\{z(t)\}$.

Since energy and power aren't linear, there is to consider the cross term. As both parts are shifted $90$ degrees, the term should look like $a^2(t)\cos(\phi(t))\sin(\phi(t))$ what have odd symmetry, and the integration over a period is $0$.

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Why did they define power that way then? Why not have a conversion factor to make complex signal power equal its real corresponding power?

You need to be careful who "they" are. One usual definition of baseband allows for a factor of 2 to compensate. So, your complaints only apply to a specific definition that doesn't have the desired property, for whatever reason (simplicity is a good reason, because real-world signal chains never have unity gain without very much work to achieve that, so things being mathematically equal are inconsequential for the real-world implementation: you'd always create a signal digitally in baseband, then would have to calibrate the gains of all stages together until the antenna to get the power you wanted – a factor of 2 doesn't hurt or help with that).

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