MSBD5004 Mathematical Methods for Data Analysis Homework 5

5004完结撒花🌸🌸ヽ(°▽°)ノ🌸🌸.

Q1

$$\begin{aligned} &if \ n=0,\ c_0 =\int_{-\frac{1}{2}}^{\frac{1}{2}} t \exp (0) d t=0\\ &if \ n\not=0,\ c_n =\int_{-\frac{1}{2}}^{\frac{1}{2}} t \exp (-2 i \pi k t) d t=\frac{i(\pi n \cos (\pi k)-\sin (\pi n))}{2 \pi^{2} n^{2}}=\frac{(-1)^ni}{2\pi n}\\ \end{aligned}$$ $$So, \ f(t) = \sum_{k\not=0} \frac{(-1)^ni}{2\pi n} e^{2\pi i n t}$$

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Q2

$$\begin{aligned} &if \ n=0,\ c_0 =\int_{-\frac{1}{2}}^{\frac{1}{2}} t^2 \exp (0) d t=\frac{1}{12}\\ &if \ n\not=0,\ c_n =\int_{-\frac{1}{2}}^{\frac{1}{2}} t^{2} \exp (-2 i \pi n t) d t=\frac{\left(\pi^{2} n^{2}-2\right) \sin (\pi n)+2 \pi n \cos (\pi n)}{4 \pi^{3} n^{3}}=\frac{(-1)^n}{2\pi^2 n^2} \\ &So, \ f(t) =\frac{1}{12} +\sum_{n\not=0} \frac{(-1)^n}{2\pi^2 n^2} e^{2\pi i n t}\\ &Use\ fourier\ teansform, \\ &1/80=\int_{-\frac{1}{2}}^{\frac{1}{2}} (t^2)^2dt=\frac{1}{12^2}+\sum_{n\not=0} \frac{1}{4\pi^4 n^4} \\ &\sum_{n=2}^{\infty} \frac{1}{ n^4}=\frac{\pi^4}{90}-1 \end{aligned}$$

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Q3

$$g(t)=\left\{\begin{array}{cc} 1, & |t| < 1 \\ 0, & \text { otherwise } \end{array}\right.$$

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Q4

$$\begin{aligned} \hat{f}(s)&=\int_{-\infty}^{\infty} f(t) e^{-2 \pi i s t} d t=\int_{-1}^{1} f(t)e^{-2 \pi i s t} dt\\ &=\int_{-1}^{0} (1+t)e^{-2 \pi i s t} dt\ +\ \int_{0}^{1} (1-t)e^{-2 \pi i s t} dt \\ &= \frac{\sin^2 (\pi s)}{\pi^2 s^2}\\ &=sinc^2(s) \end{aligned}$$

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Q5

1. Compute the Discrete Fourier Transform of [1 1 2 2]T.

$$\mathbf{X}=\left(\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 1 & -i & -1 & i \\ 1 & -1 & 1 & -1 \\ 1 & i & -1 & -i \end{array}\right)\left(\begin{array}{c} X_{0} \\ X_{1} \\ X_{2} \\ X_{3} \end{array}\right)=\left(\begin{array}{c} 6 \\ -1+i \\ 0 \\ -1-i \end{array}\right)$$

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Q6

$$\begin{aligned} A_{N}(f * g)&=\sum_{m=0}^{N-1}[(f * g)(m)]e^{\frac{-2\pi imn}{N}} \\ &=\sum_{m=0}^{N-1}[\sum_{n=0}^{N-1}f(n)g(m-n)]e^{\frac{-2\pi imn}{N}} \\ &=\sum_{n=0}^{N-1}f(n)\sum_{m=0}^{N-1}g(m-n)e^{\frac{-2\pi i(m-n)n}{N}}e^{\frac{-2\pi in^2}{N}} \\ &=\sum_{n=0}^{N-1}f(n)e^{\frac{-2\pi in^2}{N}}\sum_{m=0}^{N-1}g(m-n)e^{\frac{-2\pi i(m-n)n}{N}} \\ &=\left(A_{N} f\right) \cdot\left(A_{N} g\right)\\ \end{aligned}$$

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Q7

Suppose length of f is N: $f=\{f[0],f[1],…,f[N-1]\}$
Then

$$F(f)_k=\sum_{n=0}^{N-1} f[n] e^{\frac{-j 2 \pi k n}{N}}=\sum_{n=0}^{N-1} f[n] W^{n k}=f[0]+f[1] W^{k}+f[2] W^{2 k}+\cdots+f[N-1] W^{(N-1) k}$$

here $W=-\frac{2\pi i}{N}$.
$\tau(f)$ is a circular right shifted signal by 1 unit. Then $\tau(f)$ is $\tau(f)=\{f[N-1], f[0], f[1], \cdots,f[N-2]\}$
The DFT of $\tau(f)$ by definition is :

$$F(\tau(f))_k=\sum_{n=0}^{N-1} \tau(f)[n]e^{\frac{-j 2 \pi k n}{N}}=\sum_{n=0}^{N-1} \tau(f)[n] W^{n k}=f[N-1]+f[0] W^{k}+f[1] W^{2 k}+\cdots+f[N-2] W^{(N-1) k}$$

Now use the property that $W^kW^{(N-1)k}=1$, and rewrite the $F(\tau(f))_k$

$$\begin{aligned} F(\tau(f))_k&=\sum_{n=0}^{N-1} \tau(f)[n]e^{\frac{-j 2 \pi k n}{N}}\\ &=\sum_{n=0}^{N-1} \tau(f)[n] W^{n k}\\ &=W^kW^{(N-1)k}\left[ f[N-1]+f[0] W^{k}+f[1] W^{2 k}+\cdots+f[N-2] W^{(N-1)k}\right] \\ &=W^k\left[ f[N-1]W^{(N-1)k}+f[0]+f[1] W^{k}+\cdots+f[N-2] W^{(N-2)k}\right]\\ &=W^k\tau(f)=e^{-\frac{2\pi ik}{N}}\tau(f)_k \end{aligned}$$

So the relation is $F(\tau(f))=e^{-\frac{2\pi ik}{N}}\tau(f)$.
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有的地方我省略了，不懂的话私信。