物理代写|傅立叶光学代写Fourier optics代考|EE238

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

物理代写|傅立叶光学代写Fourier optics代考|Fourier Series

A periodic function $g(x)$ with period $T$ such that
$$g(x)=g(x+T), \quad-\infty<x<\infty$$
may be represented as a Fourier series:
$$g(x)=\sum_{n=-\infty}^{\infty} G_n \exp (i 2 \pi n x / T) .$$
This is a very important idea as we shall see when studying linear systems. The question of when such an expansion exists is addressed in the Dirichlet sufficiency conditions:

1. The function $g(x)$ must be absolutely integrable over one period.
2. The function $g(x)$ must be piecewise continuous. A finite number of finite discontinuities is allowed.The function $g(x)$ must have finite number of extrema in one period. Something like $\sin (1 / x)$ near $x=0$ is not allowed.
3. The co-efficient $G_n$ may be determined using the following orthogonality relation:
4. 5. \begin{aligned} 6. \int_{-T / 2}^{T / 2} d x \exp [i 2 \pi(m-n) x / T] &=\left[\frac{\exp [i 2 \pi(m-n) x / T]}{i 2 \pi(m-n) / T}\right]{x=-T / 2}^{T / 2} \ &=T \frac{\sin [\pi(m-n)]}{\pi(m-n)} \ &=T \delta{m, n} 7. \end{aligned} 8.
9. The coefficients $G_n$ can therefore be obtained as:
10. $$11. G_n=\frac{1}{T} \int_{-T / 2}^{T / 2} d x g(x) \exp (-i 2 \pi n x / T) . 12.$$
13. If $g(x)$ has a (finite) discontinuity at $x=x_0$, the series expansion converges to:
14. $$15. g\left(x_0\right)=\left[\frac{g\left(x_{0-}\right)+g\left(x_{0+}\right)}{2}\right] . 16.$$

物理代写|傅立叶光学代写Fourier optics代考|Gibbs phenomenon

We note a peculiar phenomenon which arises near the discontinuity of a periodic function that is being represented by means of the Fourier series. We rewrite the Fourier series representation for the square wave considered in the illustration earlier.
\begin{aligned} g(x) &=\frac{1}{2}+\frac{2}{\pi} \sin \left(\frac{2 \pi x}{T}\right)+\frac{2}{3 \pi} \sin \left(\frac{6 \pi x}{T}\right)+\ldots \ &=\frac{1}{2}+\frac{2}{\pi} \sum_{n=0}^{\infty} \frac{1}{(2 n+1)} \sin \left[\frac{2 \pi(2 n+1) x}{T}\right] \end{aligned}
When a finite number of terms is included in the summation above, the left hand side has a discontinuity while the right hand side is a sum of continuous functions. The convergence of the series sum to the periodic square wave is therefore not a point-wise convergence (near the discontinuity one observes undershoot and overshoot) but uniform convergence. In fact the overshoot and undershoot do not die out as the number of terms in the partial series sum increases. This interesting feature is known by the name of Gibbs phenomenon.

The under and overshoot get closer to the discontinuity with increasing number of terms such that the area under them tends to zero. In other words they do not contribute to the energy in the function. This type of convergence is termed as uniform convergence or “almost everywhere” convergence (convergence everywhere except on sets of measure zero). Figure $2.2$ shows a region near the discontinuity of the square wave in Fig. $2.1$ to illustrate the behaviour of the Fourier series representation as the number of terms increases. We may express the uniform convergence property as follows:
$$\lim {N \rightarrow \infty}\left|g(x)-\frac{1}{2}-\sum{n=0}^N \frac{1}{(2 n+1)} \sin \left[\frac{2 \pi(2 n+1) x}{T}\right]\right|^2=0 \text {. (2.14) }$$
The notation above for the L2-norm square is to be understood as:
$$|g(x)|^2=\int_{-\infty}^{\infty} d x|g(x)|^2 .$$

物理代写|傅立叶光学代写傅立叶光学代考|傅立叶级数

$$g(x)=g(x+T), \quad-\infty<x<\infty$$

$$g(x)=\sum_{n=-\infty}^{\infty} G_n \exp (i 2 \pi n x / T) .$$

物理代写|傅立叶光学代写傅里叶光学代考|吉布斯现象

\begin{aligned} g(x) &=\frac{1}{2}+\frac{2}{\pi} \sin \left(\frac{2 \pi x}{T}\right)+\frac{2}{3 \pi} \sin \left(\frac{6 \pi x}{T}\right)+\ldots \ &=\frac{1}{2}+\frac{2}{\pi} \sum_{n=0}^{\infty} \frac{1}{(2 n+1)} \sin \left[\frac{2 \pi(2 n+1) x}{T}\right] \end{aligned}

$$\lim {N \rightarrow \infty}\left|g(x)-\frac{1}{2}-\sum{n=0}^N \frac{1}{(2 n+1)} \sin \left[\frac{2 \pi(2 n+1) x}{T}\right]\right|^2=0 \text {. (2.14) }$$

$$|g(x)|^2=\int_{-\infty}^{\infty} d x|g(x)|^2 .$$

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