# 数学代写|傅里叶分析代写Fourier analysis代考|MATH3969

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## 数学代写|傅里叶分析代写Fourier analysis代考|Circular Time Shifting

Let $x(n) \leftrightarrow X(k)$ with period $N$. Then,
$$x\left(n \pm n_0\right) \leftrightarrow e^{\pm j \frac{2 \pi}{N} k n_0} X(k)$$
where $n_0$ is an arbitrary number of sampling intervals. The shift of a sinusoidal waveform affects only its phase. There is no change in the magnitude. Therefore, the shift of a waveform in the time domain results in adding increments to the phase of the frequency components, which are linearly proportional to the respective frequency indices. The increment in phase is $e^{\pm j \frac{2 \pi}{N} k n_0}$ for a frequency component with frequency index $k$ due to a shift of $\pm n_0$ sampling intervals in the time domain. For example,
$$\begin{gathered} x(n)={\check{1}, 2,1,4} \leftrightarrow X(k)={\check{8}, j 2,-4,-j 2} \ x(n-1)={\check{4}, 1,2,1} \leftrightarrow X(k)={\check{8}, j 2(-j),-4(-1),-j 2(j)}={\check{8}, 2,4,2} \end{gathered}$$
After shifting, the signal becomes even-symmetric and, therefore, its DFT is also even-symmetric. The computational complexity of computing the DFT of $x(n)$ can be reduced by shifting it to get $x s(n)$, computing its DFT $X s(k)$ and, then, deducing the DFT of $x(n)$ from $X s(k)$ using the shift theorem.
With $x(n)$ advanced by 2 sampling intervals, we get
$$x(n+2)={\overline{1}, 4,1,2} \leftrightarrow X(k)={\check{8}, j 2(-1),-4(1),-j 2(-1)}={\check{8},-j 2,-4, j 2}$$

## 数学代写|傅里叶分析代写Fourier analysis代考|Circular Convolution and Correlation

Let $x(n) \leftrightarrow X(k)$ and $h(n) \leftrightarrow H(k)$, both with period $N$. Then, the circular convolution of the sequences is defined as
$$y(n)=\sum_{m=0}^{N-1} x(m) h(n-m)=\sum_{m=0}^{N-1} h(m) x(n-m), n=0,1, \ldots, N-1$$
The major difference between linear and circular convolutions is that, as the sequences are periodic, the limits are changed from $\pm \infty$ to one period. Otherwise, the convolution may become undefined as the sum may not remain finite. Summation over additional periods is unnecessary, as it yields integer multiples of that of over one period. The circular convolution of two 8-point periodic sequences $x(n)$ and $h(n)$ is computed as follows. The sequences are placed on a circle, as shown in Fig. 3.1, with one of the sequences time-reversed, $h(n)$ as shown in (a). The sum of the products of the corresponding values is the convolution output $y(0)$. The timereversed sequence is shifted by one sample, as shown in (b). The sum of the products of the corresponding values is the convolution output $y(1)$. The procedure is repeated to find the rest of the 6 outputs. The output will repeat after 8 shifts.

Let $x(n)=e^{j \frac{\pi}{4} n}$. The samples of one period are ${1, j,-1,-j}$. The DFT $X(k)={0,4,0,0}$. The pointwise product of $X(k)$ with itself is ${0,16,0,0}$, the IDFT of which is $4{1, j,-1,-j}$. We have used the convolution theorem to find the convolution. We could have also used the defining equation of convolution in the time domain, given above, to find the convolution. If we convolve $x(n)$ with $h(n)=e^{j 3 \frac{2 \pi}{4} n}$, the result is zero.

# 傅里叶分析代写

## 数学代写|傅里叶分析代写Fourier analysis代考|Circular Time Shifting

$$x\left(n \pm n_0\right) \leftrightarrow e^{\pm j \frac{2 \pi}{N} k n_0} X(k)$$

$$x(n)=\check{1}, 2,1,4 \leftrightarrow X(k)=\check{8}, j 2,-4,-j 2 x(n-1)$$

$$x(n+2)=\overline{1}, 4,1,2 \leftrightarrow X(k)=\check{8}, j 2(-1),-4(1),-$$

## 数学代写|傅里叶分析代写Fourier analysis代考|Circular Convolution and Correlation

$$y(n)=\sum_{m=0}^{N-1} x(m) h(n-m)=\sum_{m=0}^{N-1} h(m) x(n-m)$$

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