## Abstract

We separately measure the higher harmonics vibration patterns of a periodic vibrating object by using time-average TV holography and phase modulation. During measurements the frequency of the phase modulation is adjusted to each harmonic component while the excitation of the object is set low enough to record all components on the linear part of the fringe function. Using acoustical phase stepping and calibration of the fringe function, we compute the amplitude and phase distributions of the frequency component. We measure components up to the 65th harmonic by using square-wave excitation.

© 1994 Optical Society of America

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### Equations (11)

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(1)
$${{J}_{0}}^{2}({a}_{0})={\left\{\frac{1}{T}{\int}_{0}^{T}\text{cos}[{a}_{0}\hspace{0.17em}\text{sin}({\mathrm{\omega}}_{0}t+{\mathrm{\varphi}}_{r})]\text{d}t\right\}}^{2}.$$
(2)
$$x(t)=\sum _{k}{a}_{k}\hspace{0.17em}\text{sin}(k\mathrm{\omega}t+{\mathrm{\varphi}}_{k}).$$
(3)
$$X({a}_{k})={\left\{\frac{1}{T}{\int}_{0}^{T}\text{cos}[x(t)]\text{d}t\right\}}^{2}=\prod _{k}{{J}_{0}}^{2}({a}_{k}).$$
(4)
$${X}_{\text{mod}}(a)=\frac{1}{{{J}_{0}}^{2}({a}_{m})}\left({{J}_{0}}^{2}\{{[{{a}_{m}}^{2}+{{a}_{r}}^{2}-2{a}_{m}{a}_{r}\times \text{cos}({\mathrm{\varphi}}_{m}-{\mathrm{\varphi}}_{r})]}^{1/2}\}\prod _{k}{{J}_{0}}^{2}({a}_{k})\right).$$
(5)
$$\frac{{I}_{m}}{{I}_{0}}=\frac{{{J}_{0}}^{2}{[{{a}_{m}}^{2}+{{a}_{r}}^{2}-2{a}_{m}{a}_{r}\hspace{0.17em}\text{cos}({\mathrm{\varphi}}_{m}-{\mathrm{\varphi}}_{r})]}^{1/2}}{{{J}_{0}}^{2}({a}_{m})}.$$
(6)
$${I}_{0}(x,y)={I}_{b}(x,y)-k(x,y){a}_{0}(x,y)\text{cos}\hspace{0.17em}{\mathrm{\phi}}_{0}(x,y),$$
(7)
$${I}_{90}(x,y)={I}_{b}(x,y)+k(x,y){a}_{0}(x,y)\text{sin}\hspace{0.17em}{\mathrm{\phi}}_{0}(x,y),$$
(8)
$${I}_{180}(x,y)={I}_{b}(x,y)+k(x,y){a}_{0}(x,y)\text{cos}\hspace{0.17em}{\mathrm{\phi}}_{0}(x,y),$$
(9)
$${I}_{270}(x,y)={I}_{b}(x,y)-k(x,y){a}_{0}(x,y)\text{sin}\hspace{0.17em}{\mathrm{\phi}}_{0}(x,y),$$
(10)
$${\mathrm{\phi}}_{0}(x,y)=\text{arctan}\left[\frac{{I}_{90}(x,y)-{I}_{270}(x,y)}{{I}_{180}(x,y)-{I}_{0}(x,y)}\right].$$
(11)
$${a}_{0}(x,y)=\frac{{\{{[{I}_{180}(x,y)-{I}_{0}(x,y)]}^{2}+{[{I}_{90}(x,y)-{I}_{270}(x,y)]}^{2}\}}^{1/2}}{2k(x,y)}.$$