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# 电子工程代写|三维成像代写Three-Dimensional Imaging代考|ARTD3005 Integral Harmonic Feedback

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## 电子工程代写|三维成像代写Three-Dimensional Imaging代考|Integral Harmonic Feedback

Integral Harmonic Feedback In first- or second-harmonic feedback, described above, the correction signal is linearly proportional to the amplitude of the first or second harmonic with amplification constants $A$ and $B$, respectively. If there is a continuous phase shift drift during the exposition $\left(\varphi_N\right.$ increases continuously during the exposure), the PZT-supported mirror must be shifted continuously to compensate for the fringe drift. In this case, as the amplification is not infinity, $\varphi$ will shift further away continuously from its initial value of zero or $\frac{1}{2} \pi$ to provide the necessary voltage for the PZTsupported mirror. To correct this phase shift, $V_0$ must be manually changed during the exposure to compensate for the drift.

In this case, more efficient performance of the stabilization system may be obtained if an integral feedback is used by introducing a simple integrator device between the lock-in amplifier output and the PZT power supply. The effect of this integrator in the feedback loop may be understood by substituting the linear time-independent terms $A \sin (\varphi)$ and $B \cos (\varphi)$ in Eqs. $3.13$ and 3.17, respectively, by the integral voltages
\begin{aligned} &\frac{A}{\tau_i} \int_0^t \sin (\varphi) d t \ &\frac{B}{\tau_i} \int_0^t \cos (\varphi) d t \end{aligned}
with $\tau_i$ being the time constant of the integrator device and the $t$ the time elapsed since the instant that the loop is closed $(t=0)$.

In the integral feedback the correction signal may be much greater than the error signal, keeping the phase $\varphi$ close to its initial value.

The necessary amplification may be achieved by increasing the amplification factors $A$ or $B$ or by decreasing the integration time $\tau_i$.

This integral feedback is particularly interesting for compensating large and slowly varying perturbations like temperature drift and air current drafts, allowing the stabilization of nonstationary holograms.

## 电子工程代写|三维成像代写Three-Dimensional Imaging代考|Fringe Lock with Arbitrary Phase

Fringe Lock with Arbitrary Phase $\varphi$ To record self-stabilized stationary holograms in photorefractive crystals in the presence of an external electrical field [8], or even to self-stabilize holograms using the reflected waves [7], feedback using single-harmonic signals does not work because the phase shift $\varphi$ between the interfering waves must be different from $0, \pi$, or $\pm \frac{1}{2} \pi$.
By adequately processing and combining the first- and second-harmonic signals, before feedback of the PZT-supported mirror, it is possible also to lock the fringe pattern with an arbitrary phase shift $\varphi$. A discussion of this processing, described in detail in a recent paper [9], is presented below.
The electrical signal measured directly by the photodetector contains all harmonics of the dither frequency $\Omega$. The second-harmonic electrical signal may be represented by
$$V_{2 \Omega}(t)=4 k_2 J_2\left(\varphi_d\right) \sqrt{\eta_0 \eta_1 I_1 I_2} \cos (\varphi) \cos (2 \Omega t)$$
which is directly proportional to the light intensity $I_{2 \Omega}(t)$ of Eq. 3.12.
The first-harmonic signal, proportional to light intensity $I_{\Omega}(t)$ of Eq. 3.11, is separated from $I_{\mathrm{R}}$ (Eq. 3.9) using a bandpass filter. After the bandpass filter, the first-harmonic signal is frequency doubled, phase shifted in relation to the second-harmonic signal (Eq. 3.22), and amplified to generate a new electrical second-harmonic signal:
$$V_{\Omega 2}(t)=-4 k_1 J_1\left(\varphi_d\right) \sqrt{\eta_0 \eta_1 I_1 I_2} \sin (\varphi) \sin (2 \Omega t+\delta)$$

## 电子工程代写|三维成像代写三维成像代考|积分谐波反馈

\begin{aligned} &\frac{A}{\tau_i} \int_0^t \sin (\varphi) d t \ &\frac{B}{\tau_i} \int_0^t \cos (\varphi) d t \end{aligned}
，其中$\tau_i$是积分器的时间常数，$t$是自环路关闭的瞬间$(t=0)$经过的时间

## 电子工程代写|三维成像代写三维成像代考|条纹锁与任意相位

，与光强成正比 $I_{2 \Omega}(t)$ 式3.12的。

$$V_{\Omega 2}(t)=-4 k_1 J_1\left(\varphi_d\right) \sqrt{\eta_0 \eta_1 I_1 I_2} \sin (\varphi) \sin (2 \Omega t+\delta)$$

## MATLAB代写

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