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How can we determine what which Bessel function amplitudes contain the majority of the the energy? Similar to the Carson bandwidth rule, I want to determine which sidebands help make up the 95% of total energy.

For example, a Bessel function $J_n(x)$ has the following amplitudes: (from n = 0 to n = 5)

$$0.895406$$ $$0.310493$$ $$0.051825$$ $$0.005714$$ $$0.000471$$ $$3.1e-5$$

I had expected that the sum of these would be equal to 1 but unfortunately it does not work out that way. Obviously in this case, the carrier band would be one of the main components, how would I know which sidebands contribute to 95% of the energy? Furthermore, in some cases, the carrier band is smaller than the sidebands, how would I go about in that case?

Lastly, what happens in the case where one of the amplitudes is negative? Thanks.

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How does the Bessel function amplitude carry energy? I don't understand any of your questions. –  user02138 May 21 '12 at 1:01
@user02138 This is related to frequency modulation. Typically we use bessel functions to represent the amplitude of the sideband and carrier frequencies of a signal undergoing FM. –  suzu May 21 '12 at 1:14
It might be a good idea to explain these concepts thoroughly when asking your question to a typically pure mathematical audience.... –  user02138 May 21 '12 at 3:59
In general, energy should be proportional to the square of the amplitude. If your amplitude for $n$ is also the amplitude for $-n$ by symmetry, the total is $$0.895406^2 + 2 \times 0.310493^2 + 2 \times 0.051825^2 + 2 \times 0.005714^2 + 2 \times 0.000471^2 + 2 \times 0.000031^2$$ which is very close to $1$ (the slight discrepancy probably due to roundoff error). –  Robert Israel May 21 '12 at 5:56
@suzu I don't think that the wording is at all inappropriate for the site. There are many mathematicians here who lean toward the applied side, and many who would describe themselves as 'pure' but have a wide knowledge of applied mathematics. You can't expect to understand every question asked on the site (I steer clear of analysis, for example). The fact that you don't understand the questions is a reflection on your knowledge and background, not necessarily an indication of a poorly asked question. –  Chris Taylor May 21 '12 at 22:51

1 Answer 1

up vote 5 down vote accepted

Using the Jacobi-Anger expansion and some trigonometric identities, it is possible to expand the equation describing an FM wave of the following form: e = $A\sin(\omega_ct +\delta\sin(\omega_mt))$ into a product of Bessel functions and differences of sines:

$e = J_0(\delta)A\sin(\omega_ct)) + J_1(\delta)A[\sin(\omega_c + \omega_m)t -\sin(\omega_c - \omega_m)t] + J_2(\delta)A[\sin(\omega_c + 2\omega_m)t -\sin(\omega_c - 2\omega_m)t]...$

Where $\omega_c$ is the carrier frequency, $\omega_m$ is the modulation frequency, and $\delta$ is a parameter called the "deviation ratio," defined as $\frac{\omega_d}{\omega_m}$, where $\omega_d$ is how far the carrier frequency "swings" from its mean value under modulation.

So the power spectrum of an FM wave will have the components of its amplitude determined by the magnitude of $J_n(\delta)$, and the frequencies where these spectral lines occur will be at multiples of the modulating frequency $\omega_m$, mirrored about the y axis defined by $\omega_c$ as shown here. Since it can be shown that $\sum_{n=-\infty}^\infty J_n^{2} = 1$, by Parseval's theorem this implies that the total power of the modulated signal is the same as the power of the unmodulated carrier (unlike ordinary amplitude modulation).

If I understand your question correctly, you're basically looking for a way to find how many sidebands you need, in terms of multiples of $\omega_m$, to account for 95% of the energy contained in the FM signal. By looking at a plot for various values of $\delta$, one can see that the spectral density varies nonlinearly with increasing $\delta$; qualitatively speaking the spectral density starts out concentrated close to the carrier frequency, as $\delta$ is increased the amount of energy concentrated around the fundamental drops and the "peaks" occur at higher sidebands. The signal power is proportional to $P = \frac{A^2}{2}$, so for a given A and a given $\delta$ all you would need to do is calculate $[J_0^2(\delta) + \sum_{n=1} 2J_n^{2}(\delta)]*P$ and stop summing and check the value of n when you go over $0.95*P$.

Of course this approach only works for a single value of $\delta$. A formula that gives the power spectral density for an arbitrary modulation index is more complicated; I haven't read through the derivation here fully yet but it looks promising.

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