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傅里叶级数：周期函数的傅里叶级数展开、收敛性和性质等相关知识点。
傅里叶变换：连续信号和离散信号的傅里叶变换、频域和时域的转换等概念和应用。
频谱分析和滤波：信号在频域的特性分析、频率分析和滤波器设计等相关内容。
傅里叶级数展开：将信号展开为傅里叶级数的形式，分析信号的频谱特征和频率成分。
其他相关主题，如：傅里叶分析在物理、计算机科学、数学物理中的应用、时频分析、傅里叶级数的反变换等。
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Exercise: What is the Fourier transform of the function $f(x)=0$ ?
The Fourier transform of the function $f(x)=0$ is also zero. Since the function $f(x)$ is identically zero, its Fourier transform will also be zero at all frequencies. In other words, all the frequency components of the function $f(x)$ are absent, resulting in a zero Fourier transform.
(a) By expanding both sides as Fourier Sin Series, show that the solution to the equation
$$
\frac{d^2 y}{d x^2}+y=2 x
$$
with boundary conditions $y(x=0)=0, y(x=1)=0$ is
$$
y(x)=\frac{4}{\pi} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n\left(1-n^2 \pi^2\right)} \sin (n \pi x) .
$$
(b) Show that the r.m.s. value of $y(x)$ is
$$
\sqrt{\left\langle y^2(x)\right\rangle}=\frac{4}{\pi} \sqrt{\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^2\left(1-n^2 \pi^2\right)^2}}
$$
(a) To find the solution to the given differential equation, we expand both sides as a Fourier sine series. The general form of a Fourier sine series is given by:
$$
y(x) = \sum_{n=1}^{\infty} B_n \sin(n \pi x)
$$
where $B_n$ are the Fourier coefficients.
Taking the second derivative of $y(x)$ with respect to $x$, we have:
$$
\frac{d^2 y}{d x^2} = -\sum_{n=1}^{\infty} B_n (n \pi)^2 \sin(n \pi x)
$$
Substituting the expressions for $\frac{d^2 y}{d x^2}$ and $y$ into the differential equation, we get:
$$
-\sum_{n=1}^{\infty} B_n (n \pi)^2 \sin(n \pi x) + \sum_{n=1}^{\infty} B_n \sin(n \pi x) = 2x
$$
Matching the coefficients of the same sine functions on both sides, we obtain the following equation for the Fourier coefficients $B_n$:
$$
-(n \pi)^2 B_n + B_n = 2 \delta_{n,1}
$$
where $\delta_{n,1}$ is the Kronecker delta.
Simplifying the equation, we find:
$$
B_n = \frac{4 (-1)^{n+1}}{n (1 – n^2 \pi^2)}
$$
Therefore, the solution to the given differential equation with the specified boundary conditions is:
$$
y(x) = \frac{4}{\pi} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n (1 – n^2 \pi^2)} \sin(n \pi x)
$$
(b) The root mean square (r.m.s.) value of $y(x)$ can be calculated as follows:
$$
\sqrt{\left\langle y^2(x)\right\rangle} = \sqrt{\frac{1}{1-0} \int_{0}^{1} y^2(x) dx}
$$
Substituting the expression for $y(x)$ and performing the integration, we obtain:
$$
\sqrt{\left\langle y^2(x)\right\rangle} = \sqrt{\frac{4}{\pi^2} \sum_{n=1}^{\infty} \frac{1}{n^2 (1 – n^2 \pi^2)^2}}
$$
Therefore, the r.m.s. value of $y(x)$ is given by the equation:
$$
\sqrt{\left\langle y^2(x)\right\rangle} = \frac{4}{\pi} \sqrt{\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^2 (1 – n^2 \pi^2)^2}}
$$
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