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## 数学代写|傅里叶分析代写Fourier analysis代考|The 2-D DFT and IDFT

It is easier to interpret the 2-DFT as two sets of $1-\mathrm{D}$ DFTs and it is usually computed using 1-D DFTs. Formally, the 2-D DFT of a $N \times N$ signal $x(m, n)$ is defined as
$$X(k, l)=\sum_{m=0}^{N-1} \sum_{n=0}^{N-1} x(m, n) e^{-j \frac{2 \pi}{N}(m k+n l)}, \quad k, l=0,1, \ldots, N-1 .$$
As in the case of the 1-D IDFT, the 2-D IDFT synthesizes the 2-D signal by multiplying the basis signals with the respective coefficients and summing the samples at all points $(m, n)$. The 2-D IDFT is given by
$$x(m, n)=\frac{1}{N^2} \sum_{k=0}^{N-1} \sum_{l=0}^{N-1} X(k, l) e^{j \frac{2 \pi}{N}(m k+n l)}, \quad m, n=0,1, \ldots, N-1$$
In this definition, the DC coefficient $X(0,0)$ is placed in the top left-hand corner of the coefficient matrix. While this format is mostly used for computational purposes, placing $X(0,0)$ in the center of the coefficient matrix is desired for better visualization of the spectrum. Further, it is easier to derive some derivations using this form. This form, called the center-zero format, with $N$ even, is given as
$$X(k, l)=\sum_{m=-\frac{N}{2}}^{\frac{N}{2}-1} \sum_{n=-\frac{N}{2}}^{\frac{N}{2}-1} x(m, n) e^{-j \frac{2 \pi}{N}(m k+n l)}, k, l=-\frac{N}{2},-\frac{N}{2}+1, \ldots, \frac{N}{2}-1$$
The corresponding 2-D IDFT is given by
$$x(m, n)=\frac{1}{N^2} \sum_{k=-\frac{N}{2}}^{\frac{N}{2}-1} \sum_{l=-\frac{N}{2}}^{\frac{N}{2}-1} X(k, l) e^{j \frac{2 \pi}{N}(m k+n l)}, \quad m, n=-\frac{N}{2},-\frac{N}{2}+1, \ldots, \frac{N}{2}-1$$
One format of the signal or the spectrum can be obtained from the other by swapping the quadrants of the signal or spectrum in the other format.
The 2-D DFT and IDFT definitions for a $M \times N$ signal are
$$X(k, l)=\sum_{m=0}^{M-1} \sum_{n=0}^{N-1} x(m, n) e^{-j 2 \pi\left(m \frac{k}{M}+n \frac{l}{N}\right)}, k=0,1, \ldots, M-1, l=0,1, \ldots, N-1$$
$$x(m, n)=\frac{1}{M N} \sum_{k=0}^{M-1} \sum_{l=0}^{N-1} X(k, l) e^{j 2 \pi\left(m \frac{k}{M}+n \frac{1}{N}\right)}, m=0,1, \ldots, M-1, n=0,1, \ldots, N-1$$

## 数学代写|傅里叶分析代写Fourier analysis代考|The 2-D DFT of Real-Valued Signals

The DFT, with complex exponential basis signals, is inherently designed for complexvalued signals. However, in practice, most of the signals are real-valued. For realvalued signals, the DFT coefficients always occur as complex conjugate pairs or real values. Coefficients
$$X(0,0), X\left(\frac{N}{2}, 0\right), X\left(0, \frac{N}{2}\right), X\left(\frac{N}{2}, \frac{N}{2}\right)$$
of a $N \times N$ real-valued signal are real-valued, as the basis functions are of the form 1 and $(-1)^n$. The rest are complex conjugate pairs. For example,
\begin{aligned} 2|X(k, l)| \cos (&\left.\frac{2 \pi}{N}(m k+n l)+\angle(X(k, l))\right) \ &=X(k, l) e^{j \frac{2 \pi}{N}(m k+n l)}+X^(k, l) e^{-j \frac{2 \pi}{N}(m k+n l)} \ &=X(k, l) e^{j \frac{2 \pi}{N}(m k+n l)}+X^(k, l) e^{j \frac{2 \pi}{N}(m(N-k)+n(N-l))} \end{aligned}
With $X(k, l)=X_r(k, l)+j X_i(k, l)$, the magnitude is
$$|X(k, l)|=\sqrt{X_r^2(k, l)+X_i^2(k, l)}$$
and the phase is
$$\angle X(k, l)=\tan ^{-1} \frac{X_i(k, l)}{X_r(k, l)}$$
Using Eq. (4.3), with $N$ even, a real-valued $N \times N$ signal can be expressed as a sum its constituent sinusoidal surfaces.
\begin{aligned} x(m, n)=& \frac{1}{N^2}\left(X(0,0)+X\left(\frac{N}{2}, 0\right) \cos (\pi m)\right.\ &+X\left(0, \frac{N}{2}\right) \cos (\pi n)+X\left(\frac{N}{2}, \frac{N}{2}\right) \cos (\pi(m+n)) \ &+2 \sum_{k=1}^{\frac{N}{2}-1}\left(|X(k, 0)| \cos \left(\frac{2 \pi}{N} m k+\angle(X(k, 0))\right)\right.\ &+2 \sum_{l=1}^{\frac{N}{2}-1}\left(|X(0, l)| \cos \left(\frac{2 \pi}{N} n l+\angle(X(0, l))\right)\right.\ &+2 \sum_{l=1}^{\frac{N}{2}-1}\left(\left|X\left(\frac{N}{2}, l\right)\right| \cos \left(\frac{2 \pi}{N}\left(m \frac{N}{2}+n l\right)+\angle\left(X\left(\frac{N}{2}, l\right)\right)\right)\right. \end{aligned}

$+2 \sum_{k=1}^{\frac{N}{2}-1} \sum_{l=1}^{N-1}\left(|X(k, l)| \cos \left(\frac{2 \pi}{N}(m k+n l)+\angle(X(k, l))\right)\right)$
$m, n=0,1, \ldots, N-1$

# 傅里叶分析代写

## 数学代写|傅里叶分析代写傅立叶分析代考|二维DFT和IDFT

2-DFT更容易解释为两组$1-\mathrm{D}$ dft，它通常使用1-D dft计算。形式上，$N \times N$信号$x(m, n)$的二维DFT定义为
$$X(k, l)=\sum_{m=0}^{N-1} \sum_{n=0}^{N-1} x(m, n) e^{-j \frac{2 \pi}{N}(m k+n l)}, \quad k, l=0,1, \ldots, N-1 .$$

$$x(m, n)=\frac{1}{N^2} \sum_{k=0}^{N-1} \sum_{l=0}^{N-1} X(k, l) e^{j \frac{2 \pi}{N}(m k+n l)}, \quad m, n=0,1, \ldots, N-1$$

$$X(k, l)=\sum_{m=-\frac{N}{2}}^{\frac{N}{2}-1} \sum_{n=-\frac{N}{2}}^{\frac{N}{2}-1} x(m, n) e^{-j \frac{2 \pi}{N}(m k+n l)}, k, l=-\frac{N}{2},-\frac{N}{2}+1, \ldots, \frac{N}{2}-1$$

$$x(m, n)=\frac{1}{N^2} \sum_{k=-\frac{N}{2}}^{\frac{N}{2}-1} \sum_{l=-\frac{N}{2}}^{\frac{N}{2}-1} X(k, l) e^{j \frac{2 \pi}{N}(m k+n l)}, \quad m, n=-\frac{N}{2},-\frac{N}{2}+1, \ldots, \frac{N}{2}-1$$

$$X(k, l)=\sum_{m=0}^{M-1} \sum_{n=0}^{N-1} x(m, n) e^{-j 2 \pi\left(m \frac{k}{M}+n \frac{l}{N}\right)}, k=0,1, \ldots, M-1, l=0,1, \ldots, N-1$$
$$x(m, n)=\frac{1}{M N} \sum_{k=0}^{M-1} \sum_{l=0}^{N-1} X(k, l) e^{j 2 \pi\left(m \frac{k}{M}+n \frac{1}{N}\right)}, m=0,1, \ldots, M-1, n=0,1, \ldots, N-1$$

## 数学代写|傅里叶分析代写傅立叶分析代考|实值信号的二维DFT

$$X(0,0), X\left(\frac{N}{2}, 0\right), X\left(0, \frac{N}{2}\right), X\left(\frac{N}{2}, \frac{N}{2}\right)$$

\begin{aligned} 2|X(k, l)| \cos (&\left.\frac{2 \pi}{N}(m k+n l)+\angle(X(k, l))\right) \ &=X(k, l) e^{j \frac{2 \pi}{N}(m k+n l)}+X^(k, l) e^{-j \frac{2 \pi}{N}(m k+n l)} \ &=X(k, l) e^{j \frac{2 \pi}{N}(m k+n l)}+X^(k, l) e^{j \frac{2 \pi}{N}(m(N-k)+n(N-l))} \end{aligned}

$$|X(k, l)|=\sqrt{X_r^2(k, l)+X_i^2(k, l)}$$
，相位是
$$\angle X(k, l)=\tan ^{-1} \frac{X_i(k, l)}{X_r(k, l)}$$

\begin{aligned} x(m, n)=& \frac{1}{N^2}\left(X(0,0)+X\left(\frac{N}{2}, 0\right) \cos (\pi m)\right.\ &+X\left(0, \frac{N}{2}\right) \cos (\pi n)+X\left(\frac{N}{2}, \frac{N}{2}\right) \cos (\pi(m+n)) \ &+2 \sum_{k=1}^{\frac{N}{2}-1}\left(|X(k, 0)| \cos \left(\frac{2 \pi}{N} m k+\angle(X(k, 0))\right)\right.\ &+2 \sum_{l=1}^{\frac{N}{2}-1}\left(|X(0, l)| \cos \left(\frac{2 \pi}{N} n l+\angle(X(0, l))\right)\right.\ &+2 \sum_{l=1}^{\frac{N}{2}-1}\left(\left|X\left(\frac{N}{2}, l\right)\right| \cos \left(\frac{2 \pi}{N}\left(m \frac{N}{2}+n l\right)+\angle\left(X\left(\frac{N}{2}, l\right)\right)\right)\right. \end{aligned}

$+2 \sum_{k=1}^{\frac{N}{2}-1} \sum_{l=1}^{N-1}\left(|X(k, l)| \cos \left(\frac{2 \pi}{N}(m k+n l)+\angle(X(k, l))\right)\right)$
$m, n=0,1, \ldots, N-1$

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