第三章

3-1 三角形式傅里叶级数的系数为

$$ \begin{aligned}&a_{0}=0\\&a_{n}=0\quad(n=1,2,\cdots)\\ \end{aligned} $$

$$ b_{n}=\left\{\begin{aligned}&0&&(n=2,4,\cdots)\\ &\frac{2E}{n\pi}&&(n=1,3,\cdots)\end{aligned}\right. $$

所以

$$ f(t)=\frac{2E}{\pi}\bigg[\sin(\omega_{1}t)+\frac{1}{3}\sin(3\omega_{1}t)+\frac{1}{5}\sin(5\omega_{1}t)+\cdots\bigg]\quad\bigg(\omega_{1}=\frac{2\pi}{T}\bigg) $$

指数形式傅里叶级数的系数为

$$ F_{n}=\left\{\begin{aligned}{}&{{}0\quad(n=0,\pm2,\pm4,\cdots)}\\ {}&{{}-\underbrace{\mathrm{j}E}_{n\pi}\quad(n=\pm1,\pm3,\pm5,\cdots)}\\ \end{aligned}\right. $$

所以

$$ f(t)=-\frac{\mathrm{j}E}{\pi}\mathrm{e}^{\mathrm{j}\omega_{1}t}+\frac{\mathrm{j}E}{\pi}\mathrm{e}^{-\mathrm{j}\omega_{1}t}-\frac{\mathrm{j}E}{3\pi}\mathrm{e}^{\mathrm{j}3\omega_{1}t}+\frac{\mathrm{j}E}{3\pi}\mathrm{e}^{-\mathrm{j}3\omega_{1}t}-\cdots $$

3-2 直流分量为1 V，基波、二次、三次谐波的有效值分别为

$$ \frac{10\sqrt{2}}{\pi}\sin18^{\circ}\approx1.39,\frac{5\sqrt{2}}{\pi}\sin36^{\circ}\approx1.32,\frac{10\sqrt{2}}{3\pi}\sin54^{\circ}\approx1.21 $$

3-3 (1) 1 000 kHz, 2 000 kHz

(2) $ \frac{1\ 000}{3} $ kHz, $ \frac{2\ 000}{3} $ kHz

(3) 1:3

(4) 1:1

3-4

$$ \begin{aligned}&a_{0}=\frac{E}{2}\\&b_{n}=0\\ \end{aligned} $$

$$ a_{n}=\left\{\begin{aligned}&0&(n=2,4,\cdots)\\ &-\frac{4E}{(n\pi)^{2}}&(n=1,3,\cdots)\end{aligned}\right. $$

所以

$$ f(t)=\frac{E}{2}-\frac{4E}{\pi^{2}}\Bigg[\cos(\omega_{1}t)+\frac{1}{3^{2}}\cos(3\omega_{1}t)+\frac{1}{5^{2}}\cos(5\omega_{1}t)+\cdots\Bigg]\quad\left(\omega_{1}=\frac{2\pi}{T}\right) $$

$$ \begin{aligned}&a_{0}=\frac{E}{\pi}\\&b_{n}=0\end{aligned} $$

$$ a_{n}=\frac{2E}{T}\left[\frac{\sin\frac{(n+1)\pi}{2}}{(n+1)\omega_{1}}+\frac{\sin\frac{(n-1)\pi}{2}}{(n-1)\omega_{1}}\right]\quad\left(\omega_{1}=\frac{2\pi}{T}\right) $$

即

$$ a_{n}=\left\{\begin{aligned}&\frac{E}{2}&(n=1)\\ &0&(n=3,5,\cdots)\\ &\frac{2E}{(1-n^{2})\pi}\cos\frac{n\pi}{2}&(n=2,4,\cdots)\end{aligned}\right. $$

所以

$$ f(t)=\frac{E}{\pi}+\frac{E}{2}\bigg[\cos(\omega_{1}t)+\frac{4}{3\pi}\cos(2\omega_{1}t)-\frac{4}{15\pi}\cos(4\omega_{1}t)+\cdots\bigg] $$

3-6 $ F_{0}=\frac{E}{2} $

$$ F_{n}=-\frac{\mathrm{j}E}{2n\pi}\quad(n=\pm1,\pm2,\cdots) $$

所以

$$ \begin{align*}f(t)&=\frac{E}{2}-\frac{\mathrm{j}E}{2\pi}\mathrm{e}^{\mathrm{j}\omega_{1}t}+\frac{\mathrm{j}E}{2\pi}\mathrm{e}^{-\mathrm{j}\omega_{1}t}-\frac{\mathrm{j}E}{4\pi}\mathrm{e}^{\mathrm{j}2\omega_{1}t}+\frac{\mathrm{j}E}{4\pi}\mathrm{e}^{-\mathrm{j}2\omega_{1}t}-\cdots\\&=\frac{E}{2}+\frac{E}{\pi}\left[\sin(\omega_{1}t)+\frac{1}{2}\sin(2\omega_{1}t)+\cdots\right]\end{align*} $$

3-7 (a) 只含有基波和奇次谐波的余弦分量

（b）只含有基波和奇次谐波的正弦分量

(c) 只含有奇次谐波

(d) 只含有正弦分量

(e) 只含有直流和偶次谐波的余弦分量

(f) 只含有直流和偶次谐波的正弦分量

3-8 (a) $ a_{0}=\frac{E}{2} $ $ a_{n}=0 $

$$ b_{n}=\left\{\begin{aligned}&0&&\quad(n=2,4,\cdots)\\ &\frac{4E}{\left(n\pi\right)^{2}}\sin\frac{n\pi}{2}&&\left(n=1,3,\cdots\right)\end{aligned}\right. $$

所以

$$ f(t)=\frac{E}{2}+\frac{4E}{\pi^{2}}\Bigg[\sin(\omega_{1}t)-\frac{1}{3^{2}}\sin(3\omega_{1}t)+\cdots\Bigg] $$

(b)

$$ a_{0}=\frac{3E}{4} $$

$$ b_{n}=0 $$

$$ a_{n}=\frac{-4E}{(n\pi)^{2}}\left(1-\cos\frac{n\pi}{2}\right)\quad(n=1,2,\cdots) $$

所以 $ f(t)=\frac{3E}{4}-\frac{4E}{\pi^{2}}\left[\cos(\omega_{1}t)+\frac{1}{2}\cos(2\omega_{1}t)+\frac{1}{9}\cos(3\omega_{1}t)+\frac{1}{25}\cos(5\omega_{1}t)+\cdots\right] $ $ \left(\omega_{1}=\frac{2\pi}{T}\right) $

$$ I_{0}=\frac{i_{\mathrm{m}}(\sin\theta-\theta\cos\theta)}{\pi(1-\cos\theta)} $$

$$ I_{1}=\frac{i_{\mathrm{m}}(\theta-\sin\theta\cdot\cos\theta)}{\pi(1-\cos\theta)} $$

$$ I_{k}=\frac{2i_{\mathrm{m}}[\sin(k\theta)\cos\theta-k\cos(k\theta)\cdot\sin\theta]}{\pi k(k^{2}-1)(1-\cos\theta)} $$

(2)

$$ I_{0}{\approx}0.22i_{\mathrm{m}}\quad I_{1}{\approx}0.39i_{\mathrm{m}} $$

$$ I_{k}=\frac{2i_{m}\left(\sin\frac{k\pi}{3}-\sqrt{3}k\cos\frac{k\pi}{3}\right)}{\pi k\left(k^{2}-1\right)} $$

(3)

$$ I_{0}=\frac{i_{\mathrm{m}}}{\pi}\quad I_{1}=\frac{i_{\mathrm{m}}}{2}\quad I_{k}=\frac{2i_{\mathrm{m}}\cdot\cos\frac{k\pi}{2}}{\pi\cdot(1-k^{2})} $$

3-11 (a) $ a_{0}=0 $ $ a_{n}=\frac{2}{\pi(4-n^{2})}[1-\cos(n\pi)] $ ，即

$$ a_{n}=\left\{\begin{aligned}&0&&\quad(n=2,4,\cdots)\\ &\frac{4}{\pi(4-n^{2})}&&\quad(n=1,3,\cdots)\end{aligned}\right.\qquad b_{n}=\left\{\begin{aligned}&\frac{1}{2}&&(n=2)\\ &0&&(n\neq2)\end{aligned}\right. $$

$$ \begin{aligned}f(t)&=\frac{4}{\pi}\left[\frac{1}{3}\cos(\omega_{1}t)-\frac{1}{5}\cos(3\omega_{1}t)-\frac{1}{21}\cos(5\omega_{1}t)-\cdots\right]+\frac{1}{2}\sin(2\omega_{1}t)\\&\quad\left(\omega_{1}=\frac{2\pi}{T}=\frac{\pi}{2}\right)\end{aligned} $$

(b)

$$ \begin{aligned}\mathrm{b})\ F_{n}&=\frac{2}{\pi(n^{2}-4)}\sin\frac{n\pi}{2}[\cos(n\pi)-1]\left(\sin\frac{n\pi}{4}+\mathrm{j}\cos\frac{n\pi}{4}\right)\\&\quad\left(\omega_{1}=\frac{2\pi}{T}=\frac{\pi}{2}\right)\end{aligned} $$

3-12 （1）直流 0.25 V，基波幅度 0.305 V，五次谐波幅度 0.018 V

(2) 比值分别为 1.0,0.847,0.303, 此 RC 积分电路是一个低通滤波器, 对高频分量衰减大, 对低频分量衰减少

3-13 （1）频率为100 kHz，幅度为127 V的正弦波

(2) 近于0

（3）频率为100 kHz，幅度为42.4 V的正弦波

3-14 可利用此电路直接选出以下频率成分的正弦信号: $ 100\ kHz $, $ 300\ kHz $

3-15

$$ F(\omega)=\frac{\tau E}{2}\bigg[\mathrm{S a}\bigg(\frac{\omega\tau}{2}-\frac{\pi}{2}\bigg)\quad+\mathrm{S a}\bigg(\frac{\omega\tau}{2}+\frac{\pi}{2}\bigg)\bigg]=\frac{2E\tau\cos\frac{\omega\tau}{2}}{\pi\bigg[1-\bigg(\frac{\omega\tau}{\pi}\bigg)^{2}\bigg]} $$

3-16

(a) $ \mathrm{j}\frac{2E}{\omega}\left[\cos\left(\frac{\omega T}{2}\right)-\mathrm{Sa}\left(\frac{\omega T}{2}\right)\right] $, $ F(0)=0 $

(b) $ \frac{E}{\omega^{2}T}(1-\mathrm{j}\omega T-\mathrm{e}^{-\mathrm{j}\omega T}) $

$$ \frac{E\omega_{1}}{\omega_{1}^{2}-\omega^{2}}(1-\mathrm{e}^{-\mathrm{j}\omega T})=\mathrm{j}\frac{2E\omega_{1}}{\omega_{1}^{2}-\omega^{2}}\sin\left(\frac{\omega T}{2}\right)\mathrm{e}^{-\mathrm{j}\frac{\omega T}{2}},F(\omega_{1})=\frac{E T}{2\mathrm{j}}\quad\left(\omega_{1}=\frac{2\pi}{T}\right) $$

$$ \frac{2E\omega_{1}\sin\left(\frac{\omega T}{2}\right)}{\omega^{2}-\omega_{1}^{2}},F(\omega_{1})=\frac{ET}{2\mathrm{j}}\quad\left(\omega_{1}=\frac{2\pi}{T}\right) $$

3-17 (a) $ \frac{1}{4} $; (b) $ \frac{1}{4} $ (c) $ \frac{1}{4} $ (d) 1 (e) $ \frac{2}{3} $ (f) $ \frac{1}{2} $ (单位均为 MHz)

3-18 $ F(\omega) = E\tau \mathrm{Sa}\left(\frac{\omega\tau}{2}\right)\left[\frac{\cos\left(\frac{k\omega\tau}{2}\right)}{1-\left(\frac{k\omega\tau}{\pi}\right)^{2}}\right] $

3 - 19 (a) $ \frac{A\omega_0}{\pi}\mathrm{Sa}[\omega_0(t+t_0)] $

3-21 $ F_1(-\omega) e^{-j\omega t_0} $

3-22 (1) $ \frac{1}{2\pi}e^{j\omega_{0}t} $ (2) $ \frac{\omega_{0}}{\pi}Sa(\omega_{0}t) $

3-23 $ 2\mathrm{j}E\tau\sin\left(\frac{\omega\tau}{2}\right)\mathrm{Sa}\left(\frac{\omega\tau}{2}\right) $

$$ \frac{\tau_{1}}{4}\left\{\mathrm{S a}^{2}\left[\frac{\left(\omega-\omega_{0}\right)\tau_{1}}{4}\right]+\mathrm{S a}^{2}\left[\frac{\left(\omega+\omega_{0}\right)\tau_{1}}{4}\right]\right\} $$

3-25 (1) $ -\omega $ (2)4 (3) $ 2\pi $ (4)其图形为函数 $ f(t) $之偶分量

3-26 $ \frac{8E}{\omega^{2}(\tau-\tau_{1})}\sin\frac{\omega(\tau+\tau_{1})}{4}\sin\frac{\omega(\tau-\tau_{1})}{4} $

$$ \frac{\omega_{1}E}{\omega_{1}^{2}-\omega^{2}}\left(1+\mathrm{e}^{-\mathrm{j}\frac{\omega T}{2}}\right),\frac{\omega_{1}\omega^{2}E}{\omega^{2}-\omega_{1}^{2}}\left(1+\mathrm{e}^{-\mathrm{j}\frac{\omega T}{2}}\right)\quad\left(\omega_{1}=\frac{2\pi}{T}\right) $$

$$ \frac{1}{(a+\mathrm{j}\omega)^{2}} $$

3-29

$$ \frac{1}{2}\mathrm{j}\frac{\mathrm{d}F\left(\frac{\omega}{2}\right)}{\mathrm{d}\omega} $$

(2) $ \mathrm{j} \frac{\mathrm{d}F(\omega)}{\mathrm{d}\omega} - 2F(\omega) $

(3) $ -F\left(-\frac{\omega}{2}\right)+\frac{\mathrm{j}}{2}\cdot\frac{\mathrm{d}F\left(-\frac{\omega}{2}\right)}{\mathrm{d}\omega} $

(4) $ -F(\omega) - \omega \frac{dF(\omega)}{d\omega} $

(5) $ F(-\omega) e^{-j\omega} $

(6) -j $ \frac{dF(-\omega)}{d\omega}\mathrm{e}^{-j\omega} $

(7) $ \frac{1}{2}F\left(\frac{\omega}{2}\right)\mathrm{e}^{-\mathrm{j}\frac{5}{2}\omega} $

3-31 $ \mathcal{F}[f_1(t) \ast f_2(t)] = E_1 E_2 \tau_1 \tau_2 \mathrm{Sa}\left(\frac{\omega \tau_1}{2}\right) \mathrm{Sa}\left(\frac{\omega \tau_2}{2}\right) $

3-32 $ \mathcal{F}[\cos(\omega_0t)u(t)]=\frac{\pi}{2}[\delta(\omega+\omega_0)+\delta(\omega-\omega_0)]+\frac{\mathrm{j}\omega}{\omega_0^2-\omega^2} $

$$ \mathcal{F}\left[\sin(\omega_{0}t)u(t)\right]={\mathrm{j}}\frac{\pi}{2}\big[\delta(\omega+\omega_{0})-\delta(\omega-\omega_{0})\big]+\frac{\omega_{0}}{\omega_{0}^{2}-\omega^{2}} $$

3-33 $ \frac{E\tau}{4}e^{-j\frac{\omega\tau}{2}}\left\{\mathrm{Sa}^{2}\left[\frac{\left(\omega-\omega_{0}\right)\tau}{4}\right]e^{j\frac{\omega_{0}\tau}{2}}+\mathrm{Sa}^{2}\left[\frac{\left(\omega+\omega_{0}\right)\tau}{4}\right]e^{-j\frac{\omega_{0}\tau}{2}}\right\} $

3-35 $ \sum_{n=-\infty}^{\infty} \frac{\tau_1(-1)^{n+1}}{(2n-1)\pi} \mathrm{Sa}^2 \left\{ \frac{\left[\omega-(2n-1)\frac{\pi}{\tau}\right]\tau_1}{4} \right\} $

3-36 (a) 傅里叶级数 $ f(t) = \sum_{n = -\infty}^{\infty} F_{n} e^{j n \frac{2\pi}{T} t} $

傅里叶变换 $ F(\omega) = 2\pi \sum_{n=-\infty}^{\infty} F_{n}\delta\left(\omega - \frac{2n\pi}{T}\right) $

其中 $ F_{n}=\frac{2ET}{n^{2}\pi^{2}(T-\tau)}\sin\frac{n\pi(T+\tau)}{2T}\sin\frac{n\pi(T-\tau)}{2T} $

(b)傅里叶级数 $ f(t) = \sum_{n = -\infty}^{\infty} F_{n} e^{j n \frac{2\pi}{T} t} $

傅里叶变换 $ F(\omega) = 2\pi \sum_{n=-\infty}^{\infty} F_{n}\delta\left(\omega - \frac{2n\pi}{T}\right) $

其中 $ F_{n} = (-1)^{n} \frac{2E}{\pi (1 - 4n^{2})} $

3-37 (a) $ \frac{8}{\omega^{2}\tau}\sin^{2}\left(\frac{\omega\tau}{4}\right)=\frac{\tau}{2}\mathrm{Sa}^{2}\left(\frac{\omega\tau}{4}\right) $

(b) $ -\frac{4\mathrm{j}}{\omega}\sin^{2}\left(\frac{\omega\tau}{4}\right)=-\mathrm{j}\frac{\omega\tau^{2}}{4}\mathrm{Sa}^{2}\left(\frac{\omega\tau}{4}\right) $

(c) $ \frac{2\tau}{\pi}\frac{\cos\frac{\omega\tau}{2}}{\left[1-\left(\frac{\omega\tau}{\pi}\right)^{2}\right]} $

(d) $ \frac{2}{\omega}\left(\sin\frac{\omega\tau}{4}+\sin\frac{\omega\tau}{2}\right)=\frac{\tau}{2}\mathrm{Sa}\left(\frac{\omega\tau}{4}\right)\left(1+2\cos\frac{\omega\tau}{4}\right) $

3-39

(1) $ \frac{100}{\pi},\frac{\pi}{100} $

(2) $ \frac{200}{\pi},\frac{\pi}{200} $

(3) $ \frac{100}{\pi},\frac{\pi}{100} $

(4) $ \frac{120}{\pi},\frac{\pi}{120} $

3-40

(1) $ \sum_{n=-\infty}^{\infty} a_{n} F(\omega - n\omega_{0}) $

(2) $ \frac{1}{2}\left[F\left(\omega-\frac{1}{2}\right)+F\left(\omega+\frac{1}{2}\right)\right] $

(3) $ \frac{1}{2}[F(\omega-1)+F(\omega+1)] $

(4) $ \frac{1}{2}[F(\omega-2)+F(\omega+2)] $

(5) $ \frac{1}{4}[F(\omega - 1) + F(\omega + 1) - F(\omega - 3) - F(\omega + 3)] $

(6) $ \frac{1}{2}[F(\omega-2)+F(\omega+2)-F(\omega-1)-F(\omega+1)] $

(7) $ \frac{1}{\pi} \sum_{n=-\infty}^{\infty} F(\omega - 2n) $

(8) $ \frac{1}{2\pi} \sum_{n=-\infty}^{\infty} F(\omega - n) $

(9) $ \frac{1}{2\pi}\left[\sum_{n=-\infty}^{\infty}F(\omega-n)-\sum_{n=-\infty}^{\infty}F(\omega-2n)\right] $

(10) $ \frac{1}{3} \sum_{n=-\infty}^{\infty} \frac{\sin(n\pi/3)}{n\pi/3} F(\omega - 2n) $

3-41 (1) $ \frac{1}{3000} $

(2) 梯形周期重复，周期为 $ 6\,000\pi $，幅度为 $ \frac{3}{2} $