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A purely phase-modulated signal has no power modulation. This is obvious enough if you look at the time series, but I'd like to "see" it in the frequency domain.

In physical terms, if we take a laser beam and apply phase modulation via an electro-optic modulator and then send this beam to a photodiode, we will see only a DC term on the photodiode. I would like to work out how the sidebands created by phase modulation manage to conspire to eliminate any beat frequencies.

The operation of phase modulation at a frequency $\Omega$ consists of multiplication by $\exp\{i\Gamma\sin\Omega t\}$ which is easily seen to have unit amplitude at all times. Via the Jacobi-Anger identity, this can be written in terms of sidebands:

$$\exp\{i\Gamma\sin\Omega t\} = \sum_{n=-\infty}^\infty J_n(\Gamma) \exp\{i n \Omega t\}$$

where $\exp\{i n \Omega t\}$ is interpreted as creating a sideband at a frequency of $n\Omega$ Hz away from the carrier whose amplitude is given by the nth Bessel function $J_n(\Gamma)$ where $\Gamma$ is the modulation depth.

A photodiode will see the modulus squared of this signal. We can write down the result by grouping terms by like powers of $\exp\{i\Omega t\}$.

It's easy to see that the energy is conserved at DC due to the relation $ 1 = \sum_{n=-\infty}^\infty |J_n(x)|^2 $.

Likewise, it's easy to see that there is no signal at frequency $1\Omega$ on the photodiode, because the beat of the $n$th upper sideband with the $(n-1)$st upper sideband always cancels with the beat of the nth lower sideband with the $(n-1)$st lower side band, because of the property of the Bessel functions that $J_{-n}(x)=(-1)^n J_n(x)$.

However, the same sort of nice cancellation does not appear for the $2\Omega$ signal, because the bessel functions always appear in products with two bessel functions of the same parity, i.e. $\cdots + J_{-3}J_{-1} + J_{-2}J_0 + J_{-1}J_1 + J_0 J_2 + J_1 J_3 + \cdots$.

How can we show that there is no beat signal at frequencies $2\Omega$ (and higher)?

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Just to check that we're on the same page:

Multiplication of a constant-amplitude complex signal $s=\exp{\left(i f(t)\right)}$ by another complex function with constant amplitude won't give a time-varying amplitude:

$\exp{\left(i f(t)\right)} \cdot \exp{\left(i g(t)\right)} = \exp{\left(i h(t)\right)}$

Since $f$ and $g$ are real, $h$ is real, and $\frac{d}{dt}\left|\exp{\left(i h(t)\right)}\right| = 0$.

You're interested in the particular case where $g(t) = \Gamma \sin{(\Omega t)}$:

$\exp{\left(i f(t)\right)} \cdot \exp{\left(i \Gamma \sin{(\Omega t)}\right)} = \exp{\left(i h(t)\right)}$

Again, since $f$ and $\Gamma \sin{(\Omega t)}$ are real, $h$ will be real, and the result will have constant amplitude.

But, now you want to see the same truth in the frequency domain.

Starting from the Jacobi-Anger identity:

$\exp{\left(i \Gamma \sin{(\Omega t)}\right)} = \sum_{n=-\infty}^{\infty}J_{n}(\Gamma)\exp{\left(i n \Omega t\right)}$

Multiplying each side by its complex conjugate:

$1 = \sum_{m=-\infty}^{\infty}\sum_{n=-\infty}^{\infty}J_{m}(\Gamma)J_{n}(\Gamma)\exp{\left(i (m-n) \Omega t\right)}$

Now make the change of variables $p=(m-n)/2$, $q=(m+n)/2$:

$1 = \sum_{p=-\infty}^{\infty}\sum_{q=-\infty}^{\infty}J_{q+p}(\Gamma)J_{q-p}(\Gamma)\exp{\left(i 2p \Omega t\right)}$

Rewrite it a little:

$1 = \sum_{p=-\infty}^{\infty}\exp{\left(i 2p \Omega t\right)}\left(\sum_{q=-\infty}^{\infty}J_{q+p}(\Gamma)J_{q-p}(\Gamma)\right)$

Since the complex exponentials are orthogonal, I think the sum in parentheses has to vanish unless $p$ is 0:

$\sum_{q=-\infty}^{\infty}J_{q+p}(\Gamma)J_{q-p}(\Gamma) = \delta(p)$

Am I allowed to assume the Jacobi-Anger identity? If so (and if I haven't made a mistake), I've just proven that this sum in parentheses vanishes. Would you like a separate proof, not based on the Jacobi-Anger identity?

Disclaimer: I'm an experimental physicist. My math is rusty. Please forgive and point out any mistakes.

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Very nice! I suppose my question reduces to asking for some kind of intuitive justification of why $\sum J_{q+p}J_{q-p} = \delta_p$. Unless "due to the magic of the Bessel functions" is all there is. (-: –  nibot Mar 13 '11 at 20:41
    
Maybe try the reduced version on math stack exchange? They seem pretty good over there. –  Andrew Mar 14 '11 at 4:09
    
Asked on math.SE: math.stackexchange.com/q/27294/2191 –  nibot Mar 15 '11 at 23:18
    
I notice you changed the 'reduced version' slightly; I'm not sure I proved the exact version you posted. Your case with $m$ an even number would be my case. Your case might still be true, of course, but I haven't checked it. –  Andrew Mar 15 '11 at 23:23
    
Hm, yes. It does seem to work numerically. –  nibot Mar 15 '11 at 23:57
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