Posted on Categories:Three-Dimensional Imaging, 三维成像, 电子代写

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

## 电子工程代写|三维成像代写Three-Dimensional Imaging代考|MATHEMATICAL ANALYSIS

Assume a point source of strength $A_0$ positioned at a point represented by $\boldsymbol{\rho}=\hat{x} \hat{\xi}+\hat{y} \eta+\hat{z} \zeta$ on the left of a recording plane with coordinates $\mathbf{r}=\hat{x} x+\hat{y} y+\hat{z}$, where the “hat” denotes a unit vector. The complex amplitude distribution on the recording plane (Fig. 2.1) is a spherical wave given by (see, e.g., Ref. 5),
$$u_0(\mathbf{r})=\frac{A_0}{j k|\mathbf{r}-\rho|} \exp (j k|r-\rho|)$$
where $k=2 \pi / \lambda$ is the wave number and $\lambda$ is the wavelength. If the point source moves, the vector $\rho$ is a function of time, which makes the complex amplitude on the observation plane also a function of time. If we wish to record a hologram, we need a reference beam, $u_r$, and expose a recording medium for a time $T$. That is, the recorded intensity pattern will be given by
$$I(\mathbf{r})=\int_0^T\left|u_r+u_o\right|^2 d t=\int_0^T\left(\left|u_r\right|^2+\left|u_o\right|^2+u_r^* u_o+u_r u_o^\right) d t$$ The first two terms constitute the so-called zero-order term, which is of no interest at this point, while the last two terms are responsible for the reconstruction of the recorded object. The fourth term reconstructs a phaseconjugate image that, if properly recorded, is spatially separated from the third term which represents the true image. Therefore, we shall concentrate now on the third term, which has the form $$I_t(\mathbf{r})=\int_0^T u_r^(\mathbf{r}) u_o(\mathbf{r}, t) d t$$
where we noted explicitly that the object wave depends on time while the reference wave is constant in time.

## 电子工程代写|三维成像代写Three-Dimensional Imaging代考|Longitudinal Translation with Constant Velocity

Since the transverse coordinate of the source remains constant in this case, we may choose the $z$ axis along the trajectory such that $\boldsymbol{\rho}_t=0$. Accordingly, Eq. $2.9$ reduces to
$$I_t(\mathbf{r})=C \int_0^T \exp [-j k \zeta(t)] \exp \left[\frac{-j k\left|\mathbf{r}_t\right|^2}{2 \zeta(t)}\right] d t$$
Motion with constant velocity along the $z$ axis can be written as
$$\zeta=\zeta_0+v_z t$$
where $v_z$ is the velocity of the source and $\zeta_0$ is the starting point. Maintaining the paraxial approximation, we may assume $\zeta_0 \gg v_z t$ during the integration time, and then we may write
$$\frac{1}{\zeta(t)}=\frac{1}{\zeta_0+v_z t} \approx \frac{1}{\zeta_0}\left(1-\frac{v_z t}{\zeta_0}\right)$$
Returning to Eq. 2.10, we obtain
$$I_t(\mathbf{r}) \approx C \exp \left[-j k \zeta_0\right] \exp \left[\frac{-j k\left|\mathbf{r}_t\right|^2}{2 \zeta_0}\right] \int_0^T \exp \left[-j k v_z t\left(1-\frac{\left|\mathbf{r}_t\right|^2}{2 \zeta_0^2}\right)\right] d t$$

\begin{aligned} I_t(\mathbf{r}) \approx & \frac{C}{j k v_z\left(1-\left|\mathbf{r}_t\right|^2 / 2 \zeta_0^2\right)} \exp \left[-j k \zeta_0\right] \ & \times \exp \left[\frac{-j k\left|\mathbf{r}_t\right|^2}{2 \zeta_0}\right]\left{1-\exp \left[-j k v_z T\left(1-\frac{\left|\mathbf{r}_t\right|^2}{2 \zeta_0^2}\right)\right]\right} \end{aligned}
With some rearrangement of factors, this can be written in the form
\begin{aligned} I_t(\mathbf{r}) \approx & \frac{C}{j k v_z\left(1-\left|\mathbf{r}_t\right|^2 / 2 \zeta_0^2\right)}\left{\exp \left[-j k \zeta_0\right] \exp \left[\frac{-j k\left|\mathbf{r}_t\right|^2}{2 \zeta_0}\right]\right.\ &\left.-\exp \left[-j k\left(\zeta_0+v_z T\right)\right] \exp \left[-\frac{j k\left|\mathbf{r}_t\right|^2}{2 \zeta_0}\left(1-\frac{v_z T}{\zeta_0}\right)\right]\right} \end{aligned}
Apart from the constant amplitude and phase factors, this expression contains two quadratic phase factors and thus represents two spherical wave fronts. One of these originates at the initial position of the source while the second has a radius of curvature modified to $R=\zeta_0^2 /\left(\zeta_0-v_z T\right)$. This is an equivalent point source at some intermediate position between the starting point and the ending point of the trajectory. There is also an exposure-time and velocity-dependent phase difference between the two sources leading to interference effects that are also exposure and velocity dependent. As a result, the final reconstructed pattern cannot be uniquely predicted from a practical point of view. In any case, the source trajectory cannot be reconstructed unless the total displacement does not exceed a wavelength by much, where “much” is not too well defined.

## 电子工程代写|三维成像代写三维成像代考|数理分析

$$u_0(\mathbf{r})=\frac{A_0}{j k|\mathbf{r}-\rho|} \exp (j k|r-\rho|)$$
，其中$k=2 \pi / \lambda$是波数，$\lambda$是波长。如果点源移动，矢量$\rho$是时间的函数，这使得观测平面上的复振幅也是时间的函数。如果我们希望记录全息图，我们需要一个参考光束$u_r$，并曝光记录介质$T$。也就是说，记录的强度模式将由
$$I(\mathbf{r})=\int_0^T\left|u_r+u_o\right|^2 d t=\int_0^T\left(\left|u_r\right|^2+\left|u_o\right|^2+u_r^* u_o+u_r u_o^\right) d t$$给出。前两项构成所谓的零阶项，这在这里没有意义，而后两项负责对记录的对象进行重构。第四项重建相位共轭图像，如果正确记录，则在空间上与代表真实图像的第三项分离。因此，我们现在将集中于第三项，其形式为$$I_t(\mathbf{r})=\int_0^T u_r^(\mathbf{r}) u_o(\mathbf{r}, t) d t$$
，其中我们明确指出，对象波依赖于时间，而参考波在时间上是恒定的

## 电子工程代写|三维成像代写三维成像代考|纵向平移与恒速

$$I_t(\mathbf{r})=C \int_0^T \exp [-j k \zeta(t)] \exp \left[\frac{-j k\left|\mathbf{r}_t\right|^2}{2 \zeta(t)}\right] d t$$

$$\zeta=\zeta_0+v_z t$$
，其中$v_z$是源的速度，$\zeta_0$是起点。保持近轴近似，我们可以在积分时间内假设$\zeta_0 \gg v_z t$，然后我们可以写
$$\frac{1}{\zeta(t)}=\frac{1}{\zeta_0+v_z t} \approx \frac{1}{\zeta_0}\left(1-\frac{v_z t}{\zeta_0}\right)$$

$$I_t(\mathbf{r}) \approx C \exp \left[-j k \zeta_0\right] \exp \left[\frac{-j k\left|\mathbf{r}_t\right|^2}{2 \zeta_0}\right] \int_0^T \exp \left[-j k v_z t\left(1-\frac{\left|\mathbf{r}_t\right|^2}{2 \zeta_0^2}\right)\right] d t$$

\begin{aligned} I_t(\mathbf{r}) \approx & \frac{C}{j k v_z\left(1-\left|\mathbf{r}_t\right|^2 / 2 \zeta_0^2\right)} \exp \left[-j k \zeta_0\right] \ & \times \exp \left[\frac{-j k\left|\mathbf{r}_t\right|^2}{2 \zeta_0}\right]\left{1-\exp \left[-j k v_z T\left(1-\frac{\left|\mathbf{r}_t\right|^2}{2 \zeta_0^2}\right)\right]\right} \end{aligned}

\begin{aligned} I_t(\mathbf{r}) \approx & \frac{C}{j k v_z\left(1-\left|\mathbf{r}_t\right|^2 / 2 \zeta_0^2\right)}\left{\exp \left[-j k \zeta_0\right] \exp \left[\frac{-j k\left|\mathbf{r}_t\right|^2}{2 \zeta_0}\right]\right.\ &\left.-\exp \left[-j k\left(\zeta_0+v_z T\right)\right] \exp \left[-\frac{j k\left|\mathbf{r}_t\right|^2}{2 \zeta_0}\left(1-\frac{v_z T}{\zeta_0}\right)\right]\right} \end{aligned}

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。