# Power of an angle-modulated wave

Let $$\phi(t) = A\cos(\omega_c t + km(t)) \tag{1}$$ be an angle-modulated wave(FM or PM). What's the power of $$\phi(t)$$? Intuitively, it seems that the answer is $$P = \frac{A^2}{2} \tag{2}$$ since the amplitude $$A$$ remains constant but I couldn't prove that using the definition which gives $$\lim_{T \to +\infty} \frac{1}{T}\int_{-\frac{T}2}^{\frac{T}2} A^2\cos^2(\omega_ct + km(t))dt \tag{3}$$ It seems we can't evaluate $$(3)$$ in the general case $$m(t)$$.

Let's consider the complex signal

$$\phi_c(t)=Ae^{j(\omega_ct+km(t))}\tag{1}$$

The power of $$(1)$$ is readily computed as

$$P_{\phi_c}=\lim_{T\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}\big|\phi_c(t)\big|^2dt=A^2\tag{2}$$

IF $$(1)$$ were the analytic signal corresponding to the real-valued signal $$\phi(t)$$ then the power of $$\phi(t)$$ would simply be $$P_{\phi_c}/2=A^2/2$$.

In general $$(1)$$ is not analytic. However, in all practical cases of angle modulation, a sufficiently high carrier frequency makes $$(1)$$ analytic for all practical purposes, i.e., there are no measurable negative frequency components. Or, equivalently, a practical angle modulated signal is a bandpass signal and has no components at or close to DC. If this is the case, the power of $$\phi(t)$$ is indeed given by $$A^2/2$$.

In all other (artificial) cases, the power of $$\phi(t)$$ cannot generally be shown to equal $$A^2/2$$.

• Thanks. How can we check that if $\phi_c(t)$ is analytic? Would you give an example, please? Nov 25 '20 at 8:45
• @S.H.W: Compute (numerically) the spectrum of the modulated signal. For realistic message signals and carrier frequencies, the modulated signal is a bandpass signal. This means that $\phi_c(t)$ is analytic (i.e., has no negative frequency components), or, equivalently, $e^{jkm(t)}$ is a lowpass signal with a maximum frequency less than the carrier frequency. Nov 25 '20 at 9:03
• I see. For example take $a(t) = 100\cos[2\pi f_c t + 4\sin 2000\pi t]$ where $f_c = 10 \text{MHz}$. How can we check that if the condition is fulfilled? Nov 25 '20 at 9:13
• @S.H.W: As I said, compute the spectrum. Best choice would be the signal $e^{j4\sin(\omega_mt)}$. This is the corresponding complex lowpass signal. If its maximum frequency is lower than $10MHz$, then shifting it up to $f_c$ will result in an analytic signal. For a purely sinusoidal message signal, you can also do calculations using Bessel functions, but I wouldn't recommend that. Nov 25 '20 at 9:21
• @S.H.W: Thanks, corrected. Nov 26 '20 at 6:32

You can't prove it for the general case, since it doesn't hold in the general case.

Consider $$k \cdot m(t) = \pi/2- \omega_c t$$ In this case the argument of the cosine becomes $$\pi/2$$, the cosine itself becomes zero and so does the power.

You have to make some assumptions around $$m(t)$$. It's probably sufficient to assume that $$m(t)$$ is uncorrelated to $$\omega_c t$$. As long as the argument to the cosine is uniformly distributed on $$[-\pi,\pi]$$ your the power will indeed be $$A^2/2$$

• Thanks. So what are the needed assumptions for $P = A^2/2$? I'm looking for the mathematical reasoning which justifies that assumptions. Nov 24 '20 at 13:01
• @S.H.W as Hilmar said, one sufficient condition is that the argument is uniform mod $2\pi$; that's the case if message signal and the phase of the carrier are uncorrelated. There's infinitely many ways, however, to make that integral take that value, the assumption of uncorrelatedness is just the (most) sensible in this physical context. Nov 24 '20 at 19:02
• @MarcusMüller Would you elaborate on "argument is uniform mod $2\pi$"? I don't understand the meaning of that. And why uncorrelated message signal and the phase of the carrier implies that? Nov 24 '20 at 19:05
• Also, in general, is there any other way to calculate the power of $A\cos(\omega_c t + km(t))$? For instance in the case $m(t)$ is periodic and expressible as Fourier series. Nov 25 '20 at 7:20
• yeah, and that's a wrong statement if you're allowing arbitrary $m(t)$, as Hilmar established. So, you need to think about all "allowed" signals, and then you're suddenly again taking expectation values over all these signals, and that's basically saying you're treating the signals as stochastic processes :) Nov 25 '20 at 10:43