# How does a laser Doppler vibrometer work?

I know almost nothing about optics or lasers, but I understand mathematics. I'm trying to understand the basic principle behind a laser Doppler vibrometer, but I don't really get it. The main thing I don't understand is why, as seems to be implied by the linked Wikipedia article, when a beam reflects off of a vibrating object, its frequency is simply increased by some fixed amount $f_D$ which depends on the vibration of the object.

Instead of a vibrating object, let's imagine the beam is bouncing off of a wall moving away at a speed $v$. At the moment a peak of the wave hits the wall, the next peak has to travel a distance $\lambda$ at a speed (relative to the wall) $c-v$. Thus the frequency with which the peaks hit the wall, and the frequency of the reflected beam, should be $\frac{c-v}{\lambda}=\omega-\frac v\lambda$ where $\omega$ is the frequency of the beam. So for an object moving at constant velocity, I can see how there's this constant additive shift of $-\frac v \lambda$.

But if the wall is vibrating, so that say its speed is $A\sin(\omega_w t)$, then that additive shift is itself sinusoidal, and so I would expect the reflected beam to have some crazy time-dependence like $A_{beam}\sin(\omega - \frac{v(t)}{\lambda})=A_{beam}\sin(\omega - \frac{1}{\lambda}A\sin(\omega_w t))$. This is certainly not just the original wave with a constant added to the frequency.

• The wikipedia article doesn't say that the frequency is "simply increased by some fixed amount $f_D$", it actually says that "$f_d = 2 v(t) cos(α)/λ$, where $v(t)$ is the velocity of the target as a function of time" – dasdingonesin Aug 13 '18 at 9:59

I'm not familiar with the devices directly, but the situation you describe is screaming for a spectral description identical to that of the frequency modulation in FM radio. Your signal has instantaneous frequency $$\omega(t) = \omega_0 - \nu \sin(\omega_vt)$$ so the phase function is $$\varphi(t) = \int_0^t\omega(t') \mathrm dt' = \omega_0t + \frac{\nu}{\omega_v} \cos(\omega_v t)$$ and the signal itself is therefore given by $$f(t) = A_0 \sin(\omega_0 t + \frac{\nu}{\omega_v} \cos(\omega_v t)).$$ Now, when you say
by itself, that's a tricky statement, because $f(t)$ doesn't even have a well-defined frequency to begin with. Thus, to talk about the frequency content of $f(t)$, we need to Fourier transform to the frequency domain, to get it spectral amplitude \begin{align} \tilde f(\Omega) & = \int f(t) e^{+i\Omega t}\mathrm dt \\ & = \int A_0 \sin(\omega_0 t + \frac{\nu}{\omega_v} \cos(\omega_v t)) e^{+i\Omega t}\mathrm dt \\ & = A_0 \sum_\pm \int e^{\pm i (\omega_0 t + \frac{\nu}{\omega_v} \cos(\omega_v t))} e^{+i\Omega t}\mathrm dt . \end{align} Now, this is a slightly tricky integral because it resembles a Dirac comb in that the product $e^{\pm i \omega_0 t } e^{+i\Omega t}$ will produce a Dirac-delta term, but the really important bit is in the other term, $e^{\pm i \frac{\nu}{\omega_v} \cos(\omega_v t)}$, which is the Bessel integrand and will therefore produce a bunch of sidebands.
The easiest way to see that is to split the Bessel integrand into its $(2\pi/\omega_v)$-periodic Fourier series: \begin{align} e^{\pm i \frac{\nu}{\omega_v} \cos(\omega_v t)} & = \sum_{n=-\infty}^\infty c_n e^{i\omega_v t} \end{align} where the coefficients are Bessel integrals \begin{align} c_n & = \frac{1}{2\pi} \int_0^{2\pi} e^{\pm i \frac{\nu}{\omega_v} \cos(\omega_v t)}e^{-in\omega_v t} \mathrm d(\omega_v t) \\ & = i^n J_n(\nu/\omega_v) \end{align} (double-check the constants, though). If you then substitute in this series to the spectral amplitude, you get a clear sideband structure: \begin{align} \tilde f(\Omega) & = A_0 \sum_\pm \sum_{n=-\infty}^\infty i^{\pm n} J_n(\nu/\omega_v) \int e^{\pm i \omega_0 t} e^{\pm i n \omega_v t} e^{+i\Omega t}\mathrm dt , \end{align} or, in other words, the original frequency $\omega_v$ is preserved at a relative amplitude $J_0(\nu/\omega_v)$, and the signal acquires a series of sidebands at $\omega_0\pm n\omega_v$, with amplitudes given by the higher-order Bessel functions $J_n(\nu/\omega_v)$. If the modulation amplitude is small (i.e. if the maximal Doppler shift $\nu = v/c$ is small with respect to the oscillation frequency $\omega_v$, then you will typically only have a single added sideband, but of course as the modulation depth increases, you get more and more weight in the higher-order ones.