# 数学代写|傅里叶分析代写Fourier analysis代考|Half-Wave Symmetry of Periodic Waveforms

## 数学代写|傅里叶分析代写Fourier analysis代考|Half-Wave Symmetry of Periodic Waveforms

A physical understanding of the basics of the DFT algorithms can be obtained using the half-wave symmetry of periodic waveforms. If a given periodic function $x(n)$ with period $N$ satisfies the condition
$$x\left(n \pm \frac{N}{2}\right)=x(n),$$
then it is said to be even half-wave symmetric.
The samples of the function over any half period are the same as those in the succeeding or preceding half period. In effect, the period is $N / 2$. If the DFT is computed over the period $N$, then the odd-indexed DFT coefficients will be zero. That is, the function is composed of even-indexed frequency components only.
If a given periodic function $x(n)$ with period $N$ satisfies the condition
$$x\left(n \pm \frac{N}{2}\right)=-x(n),$$
then it is said to be odd half-wave symmetric. The samples of the function over any half period are the negatives of those in the succeeding or preceding half period. If the DFT is computed over the period $N$, then the even-indexed DFT coefficients will be zero. That is, the function is composed of odd-indexed frequency components only. It is due to the fact that any periodic function can be uniquely decomposed into even half-wave and odd half-wave symmetric components. If the even halfwave symmetric component is composed of the even-indexed frequency components, then the odd half-wave symmetric component must be composed of the odd-indexed frequency components. Therefore, if an arbitrary function is decomposed into its even half-wave and odd half-wave symmetric components, then we have divided the original problem of finding the $N$ frequency coefficients into two problems, each of them being the determination of $N / 2$ frequency coefficients. First, let us go through an example of decomposing a periodic function into its even half-wave and odd half-wave symmetric components.

## 数学代写|傅里叶分析代写Fourier analysis代考|The PM DIF DFT Algorithm

A given waveform $x(n)$ can be expressed as the sum of $N$ frequency components (e.g., with $N=8$ ),
\begin{aligned} & x(n)=X(0) e^{j 0 \frac{2 \pi}{8} n}+X(1) e^{j 1 \frac{2 \pi}{8} n}+X(2) e^{j \frac{2 \pi}{8} n}+X(3) e^{j 3 \frac{2 \pi}{8} n} \ & \quad+X(4) e^{j 4 \frac{2 \pi}{8} n}+X(5) e^{j 5 \frac{2 \pi}{8} n}+X(6) e^{j 6 \frac{2 \pi}{8} n}+X(7) e^{j 7 \frac{2 \pi}{8} n}, n=0,1, \ldots, 7 \end{aligned}

For the most compact algorithms, the input sequence $x(n)$ has to be expressed as 2-element vectors
$$a^0(n)=\left{a_0^0(n), a_1^0(n)\right}=2\left{x_{e h}(n), x_{o h}(n), n=0,1, \ldots, \frac{N}{2}-1\right}$$
The first and second elements of the vectors are, respectively, the scaled even and odd half-wave symmetric components $x_{e h}(n)$ and $x_{o h}(n)$ of $x(n)$. That is, the DFT expression is reformulated as
\begin{aligned} X(k) & =\sum_{n=0}^{N-1} x(n) e^{-j \frac{2 \pi}{N} k n}, k=0,1, \ldots, N-1 \ & =\left{\begin{array}{l} \sum_{n=0}^{(N / 2)-1}\left(x(n)+x\left(n+\frac{N}{2}\right)\right) e^{-j \frac{2 \pi}{N} k n}=\sum_{n=0}^{(N / 2)-1} a_0^0(n) e^{-j \frac{2 \pi}{N} k n}, k \text { even } \ \sum_{n=0}^{(N / 2)-1}\left(x(n)-x\left(n+\frac{N}{2}\right)\right) e^{-j \frac{2 \pi}{N} k n}=\sum_{n=0}^{(N / 2)-1} a_1^0(n) e^{-j \frac{2 \pi}{N} k n}, k \text { odd } \end{array}\right. \end{aligned}

# 傅里叶分析代写

## 数学代写|傅里叶分析代写Fourier analysis代考|Half-Wave Symmetry of Periodic Waveforms

$$x\left(n \pm \frac{N}{2}\right)=x(n)$$

$$x\left(n \pm \frac{N}{2}\right)=-x(n)$$

## 数学代写|傅里叶分析代写Fourier analysis代考|The PM DIF DFT Algorithm

$$x(n)=X(0) e^{j 0 \frac{2 \pi}{8} n}+X(1) e^{j \frac{2 \pi}{8} n}+X(2) e^{j \frac{2 \pi}{8} n}+X(3) e^{j \frac{2 \pi}{8} n} \quad+X(4) e^{j 4 \frac{2 \pi}{8} n}+$$

$a^{\wedge} 0(n)=\backslash$ left $\left{a_{-} 0^{\wedge} 0(n), a_{-} 1^{\wedge} 0(n) \backslash\right.$ right $}=2 \backslash$ left $\left{x_{_}{e h}(n), x_{-}{0 h}(n), n=0,1, \backslash \mid d o t s, \backslash f r a c{N}{2}-1 \backslash\right.$ right $}$

$\$ \\backslash begin{aligned } X(\mathrm{k}) \&=\backslash sum_{n=0 }^{\wedge}{\mathrm{N}-1} \times(n) \mathrm{e}^{\wedge}{-\mathrm{j} \backslash frac {2 \backslash pi }{\mathrm{N}} \mathrm{kn}}, \mathrm{k}=0,1, \backslash \backslash dots, N-1 \backslash 8=\backslash left { \sum_{n=0}^{(N / 2)-1}\left(x(n)+x\left(n+\frac{N}{2}\right)\right) e^{-j \frac{2 \pi}{N} k n}=\sum_{n=0}^{(N / 2)-1} a_0^0(n) e^{-j \frac{2 \pi}{N} k n}, k even \sum_{n=0}^{(N / 2)-1} 佂确的。 〈结束{对齐} \ \

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