EE261 Problem Set 5

课程主页：https://see.stanford.edu/Course/EE261

这次回顾Problem Set 5。

Problem 1

(a)注意到

$$\mathcal F \Lambda = \text{sinc}^2$$

所以

$$\begin{aligned} \int_{-\infty}^{\infty} \operatorname{sinc}^{4}(t) d t &= \int_{-\infty}^{\infty} (\mathcal F \Lambda (t))^2 dt\\ &= \int_{-\infty}^{\infty} \Lambda ^2 (s) ds\\ &= 2 \int_{0}^{1} (1-s)^2 ds\\ &=\frac 2 3 \end{aligned}$$

(b)设

$$g(t)=e^{-a|t|}$$

那么

$$\mathcal F g(s) =\frac{2 a}{a^{2}+4 \pi^{2} s^{2}}$$

如果$a=1$，那么

$$\mathcal F g(s) =\frac{2}{1+4 \pi^{2} s^{2}}$$

另一方面，我们有

$$\mathcal F\Pi (s) =\operatorname{sinc}(s)$$

设

$$f(t)= \Pi \left(\frac t 2 \right)$$

所以

$$\mathcal F f(s) =2\operatorname{sinc}(2s)$$

因此

$$\begin{aligned} \int_{-\infty}^{\infty} \frac{2}{1+(2 \pi t)^{2}} \operatorname{sinc}(2 t) d t &=\frac 12 \int_{-\infty}^{\infty}\mathcal F g(t) \mathcal F f(t) dt\\ &=\frac 12 \int_{-\infty}^{\infty}\mathcal F g(t)\overline{\mathcal F f(t)} dt\\ &=\frac 12 \int_{-\infty}^{\infty}g(s)\overline{f(s)} ds\\ &=\frac 12 \int_{-1}^{1} e^{-|s|} ds\\ &=\int_{0}^{1} e^{-s} ds\\ &=1-e^{-1} \end{aligned}$$

(c)注意到

$$(\mathcal F \Lambda ^{(1)})(s) = (2\pi i s) \mathcal F \Lambda(s)$$

所以

$$\begin{aligned} \int_{-\infty}^{\infty} t^{2} \operatorname{sinc}^{4}(t) d t &= \frac{1}{4\pi^2} \int_{-\infty}^{\infty} |(\mathcal F \Lambda ^{(1)})(t) |^2 dt\\ &=\frac{1}{4\pi^2} \int_{-\infty}^{\infty} \left|\Lambda ^{(1)}(s)\right|^2 ds\\ &=\frac{1}{4\pi^2} \times 2\times \int_{0}^{1} 1ds\\ &=\frac 1 {2\pi^2} \end{aligned}$$

Problem 2

(a)该函数为

$$f(t) = \text{sign}(t)\times \left(1 - \Pi\left(\frac t 2\right)\right)$$

取傅里叶变换得到

$$\begin{aligned} \mathcal F f(s) &= \mathcal F \text{sign}(t) * \mathcal F \left(1 - \Pi\left(\frac t 2\right)\right)\\ &= \frac{1}{\pi i t} * \left(\delta(t) -2 \text{sinc} (2t)\right) \end{aligned}$$

设

$$H(x)=\left\{\begin{array}{ll}{0} & {x \leq 0} \\ {1} & {x>0}\end{array}\right.$$

那么

$$\begin{aligned} g(t) &= H (t-1) \\ f(t) &= g(t)-g^{-}(t) \end{aligned}$$

所以

$$\begin{aligned} \mathcal F g(s) &= e^{-2\pi is} \mathcal F H(s)\\ &= e^{-2\pi is} \frac{1}{2}\left(\delta(s)+\frac{1}{\pi i s}\right) \\ \mathcal F g^{-}(s) &= (\mathcal F g(s))^{-} \\ &= e^{2\pi is} \frac{1}{2}\left(\delta(-s)+\frac{1}{\pi i (-s)}\right)\\ &=e^{2\pi is} \frac{1}{2}\left(\delta(s)-\frac{1}{\pi i s}\right)\\ \mathcal F f(s) &= \mathcal F g(s) - \mathcal F g^{-}(s)\\ &=\frac 1 2 \delta(s)\left(e^{-2\pi is}-e^{2\pi is} \right) +\frac 1 2 \frac{1}{\pi i s} \left(e^{2\pi is}+e^{-2\pi is} \right)\\ &= \frac 1 2 \delta(s)(-2i)\sin(2\pi s)+\frac 1 2 \frac{1}{\pi i s} 2\cos (2\pi s)\\ &=\frac{\cos (2\pi s)}{\pi i s} \end{aligned}$$

(b)注意到

$$\sin (2 \pi|t|)= \frac 1 {2i} \left( e^{2\pi i|t|}-e^{-2\pi i|t|}\right)$$

设

$$g(t)=e^{-a|t|}$$

那么

$$\mathcal F g(s) =\frac{2 a}{a^{2}+4 \pi^{2} s^{2}}$$

所以

$$\begin{aligned} \mathcal F f (s) &=\frac 1 {2i} \left(\frac{2 (-2\pi i)}{-4\pi^2+4 \pi^{2} s^{2}} -\frac{2 (2\pi i)}{-4\pi^2+4 \pi^{2} s^{2}} \right)\\ &=\frac{-4\pi}{4\pi^2 (s^2 -1)}\\ &=\frac{1}{\pi (1-s^2)} \end{aligned}$$

Problem 3

(a)因为

$$\mathcal F\cos (2 \pi \nu t) =\frac{1}{2}(\delta(s-\nu)+\delta(s+\nu))$$

注意时域的滤波等价于频域的乘积，所以频域的输出结果为

$$\frac{1}{2}(\delta(s-\nu)+\delta(s+\nu)) H(s)$$

利用

$$g(x) \delta_{a}=g(a) \delta_{a}$$

取傅里叶逆变换可得

$$\begin{aligned} \mathcal F^{-1}\left(\frac{1}{2}(\delta(s-\nu)+\delta(s+\nu)) H(s)\right) &=\mathcal F^{-1}\left(\frac{1}{2}(\delta(s-\nu)+\delta(s+\nu)) H(s)\right)\\ &=\frac 1 2 \left( \mathcal F^{-1} \delta(s-\nu)H(\nu) + \mathcal F^{-1} \delta(s+\nu)H(-\nu) \right)\\ &= H(\nu)\mathcal F^{-1}\left(\frac{1}{2}(\delta(s-\nu)+\delta(s+\nu))\right)\\ &= H(\nu)\cos (2 \pi \nu t) \end{aligned}$$

(b)算法的思路是利用

$$|H(\nu)| = \max_t | H(\nu)\cos (2 \pi \nu t)|$$

%(b)
Nu = 0 : 20;
res = zeros(1, 21);

for nu = 1 : 21
    resolution = 0.001; 
    t = 0:resolution:10;
    input = cos(2*pi*nu*t);
    output = identme(input, resolution);
    res(nu) = max(output);
end
figure(1);
plot(Nu, res);

(c)

%(c)
s = 0: 0.01: 20;
H = transferfcn(s);
figure(2)
plot(s, H);

Problem 4

$$\begin{aligned} \mathcal F f(s) &=\mathcal F \left(1+\Lambda(3 t) * \text{III}_{1 / 3}(t)\right)\\ &=\delta(s) +\frac 13 \text{sinc}^2 \left(\frac s 3 \right) \times 3\text{III}_{3}(s)\\ &=\delta(s) +\sum_{k=-\infty}^{\infty} \text{sinc}^2 \left(k \right) \delta(s-3k) \end{aligned}$$

注意到我们有

$$\text{sinc}^2 \left(k \right) =\begin{cases} 1 & k=0\\ 0 & 其他 \end{cases}$$

所以

$$\begin{aligned} \mathcal F f(s) &=2\delta(s) \end{aligned}$$

取逆变换得到

$$f(s)=2$$

Problem 5

如果在整周期的时刻看电风扇，那么其图像不变；特别的，如果该电风扇有三个叶子，那么在$\frac 13$周期的整数倍看电风扇，其图像依然不变。