# Why do complex signals and their real world equivalents have different measurements of power?

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?

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

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).

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 $$0$$.