第三章作业

3.1

1. $\omega =100\mathrm{rad}/\mathrm{s},T={\displaystyle \frac{\pi }{50}}\mathrm{s}$$\omega = 100\ \mathrm{rad/s}, T = \dfrac{\pi}{50}\ \mathrm{s}$.

2. $\omega ={\displaystyle \frac{\pi }{2}}\mathrm{rad}/\mathrm{s},T=4\mathrm{s}$$\omega = \dfrac{\pi}{2}\ \mathrm{rad/s}, T = 4\ \mathrm s$.

3. $\omega =2\mathrm{rad}/\mathrm{s},T=\pi \mathrm{s}$$\omega = 2\ \mathrm{rad/s}, T = \pi\ \rm s$.

4. $\omega =\pi \mathrm{rad}/\mathrm{s},T=2\mathrm{s}$$\omega = \pi\ \mathrm{rad/s}, T = 2\ \rm s$.

5. $\omega ={\displaystyle \frac{\pi }{4}}\mathrm{rad}/\mathrm{s},T=8\mathrm{s}$$\omega = \dfrac{\pi}{4}\ \mathrm{rad/s}, T = 8\ \mathrm s$.

6. $\omega ={\displaystyle \frac{\pi }{30}}\mathrm{rad}/\mathrm{s},T=60\mathrm{s}$$\omega = \dfrac{\pi}{30}\ \mathrm{rad/s}, T = 60\ \rm s$.

注 展开为三角级数后, 基波频率即所有角频率的最大公约数, 基波周期即所有周期的最小公倍数.

3.2

1. 周期矩形脉冲

   1. 三角形式

      1. $a_0={\displaystyle \frac{1}{4}}{\displaystyle {\int }_{-2}^{2}f(t)\mathrm{d}t={\displaystyle \frac{1}{2}}}$$a_0 = \dfrac{1}{4} \dint_{-2}^2 f(t) \dt = \dfrac{1}{2}$.

      2. $a_n={\displaystyle \frac{2}{4}}{\displaystyle {\int }_{-2}^{2}f(t)\cos \left({\displaystyle \frac{n\pi t}{2}}\right)\mathrm{d}t=\mathrm{Sa}\left({\displaystyle \frac{n\pi }{2}}\right)}$$a_n = \dfrac{2}{4} \dint_{-2}^2 f(t) \cos\pqty{\dfrac{n\pi t}{2}} \dt = \Sa\pqty{\dfrac{n\pi}{2}}$.

      3. $b_n={\displaystyle \frac{2}{4}}{\displaystyle {\int }_{-2}^{2}f(t)\sin \left({\displaystyle \frac{n\pi t}{2}}\right)\mathrm{d}t=0}$$b_n = \dfrac{2}{4} \dint_{-2}^2 f(t) \sin\pqty{\dfrac{n \pi t}{2}} \dt = 0$.

   2. 指数形式 $F_n={\displaystyle \frac{1}{4}}{\displaystyle {\int }_{-2}^{2}f(t){\text{e}}^{-\mathrm{j}n\omega t}\mathrm{d}t={\displaystyle \frac{1}{2}}\mathrm{Sa}\left({\displaystyle \frac{n\pi }{2}}\right)}$$F_n = \dfrac{1}{4} \dint_{-2}^2 f(t) \e^{-\j n\omega t} \dt = \dfrac{1}{2} \Sa\pqty{\dfrac{n\pi}{2}}$.

2. 周期半正弦波

   $$\begin{array}{rl}{a}_0& ={\displaystyle \frac{1}{2}}{\displaystyle {\int }_{-1}^{1}f(t)\mathrm{d}t={\displaystyle \frac{1}{2}}{\displaystyle {\int }_0^1\sin \left(\pi t\right)\mathrm{d}t={{\displaystyle \frac{-\cos \left(\pi t\right)}{2\pi }}\phantom{\int }|}_0^1={\displaystyle \frac{1}{\pi }},}}\\ a_n& ={\displaystyle \frac{2}{2}}{\displaystyle {\int }_{-1}^{1}f(t)\cos \left(n\pi t\right)\mathrm{d}t={\displaystyle {\int }_0^1\sin \left(\pi t\right)\cos \left(n\pi t\right)\mathrm{d}t}}\\ & ={\int }_0^1{\displaystyle \frac{\sin ((n+1)\pi t)-\sin ((n-1)\pi t)}{2}}\mathrm{d}t\\ & ={{\displaystyle \frac{1}{2\pi }}({\displaystyle \frac{\cos ((n-1)\pi t)}{n-1}}-{\displaystyle \frac{\cos ((n+1)\pi t)}{n+1}})\phantom{\int }|}_0^1\\ & ={\displaystyle \frac{1+\cos \left(n\pi \right)}{\pi (1-n^2)}},\\ b_n& ={\displaystyle \frac{2}{2}}{\displaystyle {\int }_{-1}^{1}f(t)\sin \left(n\pi t\right)\mathrm{d}t={\displaystyle \frac{\sin n\pi }{\pi (1-n^2)}},}\\ F_n& ={\displaystyle \frac{1}{2}}{\displaystyle {\int }_{-1}^{1}f(t){\text{e}}^{-\mathrm{j}n\omega t}\mathrm{d}t={\displaystyle \frac{1}{2}}{\displaystyle {\int }_0^1\sin \left(\pi t\right){\text{e}}^{-\mathrm{j}n\pi t}\mathrm{d}t}}\\ & ={{\displaystyle \frac{-\mathrm{j}n\pi \sin \left(\pi t\right)-\pi \cos \left(\pi t\right)}{2{\pi }^2(1-n^2)}}{\text{e}}^{-\mathrm{j}n\omega t}\phantom{\int }|}_0^1={\displaystyle \frac{1+(-1{)}^n}{\pi (1-n^2)}}.\end{array}$$

3.3 不想做, 以后再说.

1. 三角形式

   1. $a_0={\displaystyle \frac{1}{T}}{\displaystyle {\int }_{-\frac{T}{2}}^0{\displaystyle \frac{-2Et}{T}}\mathrm{d}t+{\displaystyle \frac{1}{T}}{\displaystyle {\int }_0^{\frac{T}{2}}{\displaystyle \frac{2Et}{T}}\mathrm{d}t={\displaystyle \frac{1}{2}}}}$$a_0 = \dfrac{1}{T} \dint_{-\tfrac{T}{2}}^0 \dfrac{-2Et}{T} \dt + \dfrac{1}{T} \dint_0^{\tfrac{T}{2}} \dfrac{2Et}{T} \dt = \dfrac{1}{2}$.

   2. $a_n={\displaystyle \frac{2}{T}}{\displaystyle {\int }_{-\frac{T}{2}}^0{\displaystyle \frac{-2Et}{T}}\cos \left(n{\displaystyle \frac{2\pi }{T}}t\right)\mathrm{d}t+{\displaystyle \frac{2}{T}}{\displaystyle {\int }_0^{\frac{T}{2}}{\displaystyle \frac{2Et}{T}}\cos \left(n{\displaystyle \frac{2\pi }{T}}t\right)\mathrm{d}t={\displaystyle \frac{2E[(-1{)}^n-1]}{n^2{\pi }^2}}}}$$a_n = \dfrac{2}{T} \dint_{-\tfrac{T}{2}}^0 \dfrac{-2Et}{T} \cos\pqty{n\dfrac{2\pi}{T}t} \dt + \dfrac{2}{T} \dint_0^{\tfrac{T}{2}} \dfrac{2Et}{T} \cos\pqty{n\dfrac{2\pi}{T} t} \dt = \dfrac{2E\bqty{(-1)^n-1}}{n^2 \pi^2}$.

   3. $b_n=0$$b_n = 0$.

2. 指数形式

   1. $F_0={\displaystyle \frac{1}{T}}{\displaystyle {\int }_{-\frac{T}{2}}^{\frac{T}{2}}f(t)\mathrm{d}t={\displaystyle \frac{1}{2}}}$$F_0 = \dfrac{1}{T} \dint_{-\tfrac{T}{2}}^{\tfrac{T}{2}} f(t) \dt = \dfrac{1}{2}$. (不用展开计算积分)

   2. $$\begin{array}{rl}{F}_n& ={\displaystyle \frac{1}{T}}{\displaystyle {\int }_{-\frac{T}{2}}^0{\displaystyle \frac{-2Et}{T}}{\text{e}}^{-\mathrm{j}n\omega t}\mathrm{d}t+{\displaystyle \frac{1}{T}}{\displaystyle \int }}\end{array}$$

   3.  

   4. $F_n={\displaystyle \frac{(-1{)}^n-1}{n^2{\pi }^2}}$$F_n = \dfrac{(-1)^n - 1}{n^2 \pi^2}$.

3.4 略.

3.5 ⭐️

1. $f(t)=f(-t)$$f(t) = f(-t)$, 故只有余弦分量, 且直流分量为零.

   $f(t)=-f(t+{\displaystyle \frac{T}{2}})$$f(t) =-f\pqty{t + \dfrac{T}{2}}$, 故只有奇次谐波.

   所有只含有奇次谐波的余弦分量. (基波?)

2. 只含有奇次谐波的正弦分量. (基波?)

3. 只含有奇次谐波.

4. 只含有正弦分量.

5. 只含有直流与偶次谐波的余弦分量.

6. 只含有直流与偶次谐波的正弦分量.

3.6

0. $v_i(t)$$v_i(t)$ 的傅里叶展开

   1. $a_0={\displaystyle \frac{1}{T}}{\displaystyle {\int }_0^{\frac{T}{2}}{\displaystyle \frac{2E}{T}}t\mathrm{d}t={\displaystyle \frac{1}{4}}}$$a_0 = \dfrac{1}{T} \dint_0^\tfrac{T}{2} \dfrac{2E}{T} t \dt = \dfrac{1}{4}$.

   2. $a_n={\displaystyle \frac{2((-1{)}^n-1)}{n^2{\pi }^2}}$$a_n = \dfrac{2((-1)^n-1)}{n^2\pi^2}$.

   3. $b_n=-{\displaystyle \frac{2(-1{)}^n}{n\pi }}$$b_n = -\dfrac{2 (-1)^n}{n\pi}$.

1. 稳态时电容两端电压

   1. 直流分量即 $a_0={\displaystyle \frac{1}{4}}\mathrm{V}$$a_0 = \dfrac{1}{4}\ \rm{V}$.

   2. 信号的基波为 ${\displaystyle \frac{-4}{{\pi }^2}}\cos 2\pi {10}^{3}t+{\displaystyle \frac{2}{\pi }}\sin 2\pi {10}^{3}t$$\dfrac{-4}{\pi^2} \cos {\color{red} 2\pi} 10^3 t + \dfrac{2}{\pi} \sin 2\pi 10^3 t$,

      幅度为 0.754679 V, 电容电压幅度为 0.639017 V.

   3. 信号的五次谐波为 ${\displaystyle \frac{-4}{25{\pi }^2}}\cos 10\pi {10}^{3}t+{\displaystyle \frac{2}{5\pi }}\sin 10\pi {10}^{3}t$$\dfrac{-4}{25\pi^2} \cos 10\pi 10^3 t + \dfrac{2}{5\pi} \sin 10\pi 10^3 t$,

      幅度为 0.128352 V, 电容电压幅度为 0.0389310 V.

2. 电容电压与信号电压相应分量的比值

   1. 直流分量: 1.

   2. 基波分量: 0.8467.

   3. 五次谐波: 0.3033.

3. 因此该电路构成低通滤波器, 滤去信号中的高频分量.

注 标答第一问错了吧...

​x
1
f[t_] := Which[
2
    t < 0, 0,
3
    True, t / 500
4
] (* [-T/2, T/2] 的信号, 我们不关心其它处的取值 *)
5

6
Subscript[a, 5] = FourierCosCoefficient[
7
    f[t], t, 5,
8
    FourierParameters -> {1, 2 Pi / 1000}
9
] (* 信号的五次谐波余弦系数 *)
10

11
Subscript[b, 5] = FourierSinCoefficient[
12
    f[t], t, 5,
13
    FourierParameters -> {1, 2 Pi / 1000}
14
] (* 信号的五次谐波正弦系数 *)
15

16
Subscript[u, C, 5] = N[
17
    Sqrt[Subscript[a, 5]^2 + Subscript[b, 5]^2]
18
    10^4/(10 Pi) / Sqrt[10^6 + (10^4/(10 Pi))^2]
19
] (* 电容电压的五次谐波幅度 *)

3.7

1. 思路一 (由定义) $\mathcal{F}[f_1(t)](\omega )={\displaystyle {\int }_0^{\tau }{\text{e}}^{-\mathrm{j}\omega t}\mathrm{d}t={\displaystyle \frac{1-{\text{e}}^{-\mathrm{j}\omega \tau }}{\mathrm{j}\omega }}}$$\cal F[f_1(t)](\omega) = \dint_0^\tau \e^{-\j\omega t} \dt = \dfrac{1 - \e^{-\j\omega\tau}}{\j\omega}$.

   思路二 (由性质) $\mathcal{F}[f_1(t)](\omega )=\tau \mathrm{Sa}\left({\displaystyle \frac{\omega \tau }{2}}\right){\text{e}}^{-\mathrm{j}\frac{\omega \tau }{2}}$$\cal F[f_1(t)](\omega) = \tau \Sa\pqty{\dfrac{\omega\tau}{2}} \e^{-\j\tfrac{\omega\tau}{2}}$.

   这两个结果是一样的.

2. 思路一 (由定义) $\mathcal{F}{f}_2(t)={\displaystyle {\int }_0^{\tau }{\displaystyle \frac{t}{\tau }}{\text{e}}^{-\mathrm{j}\omega \tau }\mathrm{d}t={\displaystyle \frac{{\text{e}}^{-\mathrm{j}\omega \tau }-1}{{\omega }^2\tau }}+{\displaystyle \frac{\mathrm{j}{\text{e}}^{-\mathrm{j}\omega \tau }}{\omega }}}$$\cal F f_2(t) = \dint_0^\tau \dfrac{t}{\tau} \e^{-\j\omega \tau} \dt = \dfrac{\e^{-\j\omega\tau}-1}{\omega^2\tau} + \dfrac{\j\e^{-\j\omega\tau}}{\omega}$.

   思路二 (由性质) $\mathcal{F}{f}_2(t)=\mathcal{F}{\displaystyle \frac{f_1^{(-1)}(t)}{\tau }}-\mathcal{F}u(t-\tau )={\displaystyle \frac{{\text{e}}^{-\mathrm{j}\omega \tau }-1}{{\omega }^2\tau }}+{\displaystyle \frac{\mathrm{j}{\text{e}}^{-\mathrm{j}\omega \tau }}{\omega }}$$\cal F f_2(t) = \cal F \dfrac{f_1^{(-1)}(t)}{\tau} - \cal F u(t-\tau) = \dfrac{\e^{-\j\omega\tau}-1}{\omega^2\tau} + \dfrac{\j\e^{-\j\omega\tau}}{\omega}$.

   评价是不如用定义.

3. $\mathcal{F}{f}_3(t)={\displaystyle {\int }_{-1}^1\cos \left({\displaystyle \frac{\pi }{2}}t\right){\text{e}}^{-\mathrm{j}\omega t}\mathrm{d}t={\displaystyle \frac{\pi \cos \omega }{\frac{{\pi }^2}{4}-{\omega }^2}}}$$\cal F f_3(t) = \dint_{-1}^1 \cos\pqty{\dfrac{\pi}{2} t} \e^{-\j\omega t} \dt = \dfrac{\pi\cos\omega}{\cfrac{\pi^2}{4} - \omega^2}$.

4. $\mathcal{F}{f}_4(t)={\displaystyle {\int }_{-\frac{T}{2}}^{\frac{T}{2}}\sin \left({\omega }_{0}t\right){\text{e}}^{-\mathrm{j}\omega t}\mathrm{d}t={\displaystyle \frac{\frac{4\pi \mathrm{j}}{T}\sin \left(\frac{\omega T}{2}\right)}{{\omega }^2-{\left(\frac{2\pi }{T}\right)}^2}}}$$\cal F f_4(t) = \dint_{-\tfrac{T}{2}}^{\tfrac{T}{2}} \sin(\omega_0 t) \e^{-\j\omega t} \dt = \dfrac{\cfrac{4\pi\j}{T} \sin\pqty{\cfrac{\omega T}{2}}}{\omega^2 - \pqty{\cfrac{2\pi}{T}}^2}$.

3.8

1. $f(t)\leftrightarrow 4\mathrm{Sa}\left(2\omega \right)$$f(t) \rla 4\Sa\pqty{2\omega}$,

   ${\omega }_{\text{c}}={\displaystyle \frac{\pi }{2}}\mathrm{r}\mathrm{a}\mathrm{d}/\mu \mathrm{s}$$\omega_\text c = \dfrac{\pi}{2}\ \rm{rad/\mu s}$.

   $B=f_{\text{c}}={\displaystyle \frac{{\omega }_{\text{c}}}{2\pi }}={\displaystyle \frac{1}{4}}\mathrm{M}\mathrm{H}\mathrm{z}$$B = f_\text c = \dfrac{\omega_\text c}{2\pi} = \dfrac{1}{4}\ \rm{MHz}$.

2. $f(t)\leftrightarrow 10\mathrm{Sa}\left(10\omega \right)-2\mathrm{Sa}\left(\omega \right)$$f(t) \rla 10 \Sa\pqty{10\omega} - 2 \Sa\pqty{\omega}$,

   $B={\displaystyle \frac{1}{10}}\mathrm{MHz}$$B = \dfrac{1}{10}\ \mathrm{MHz}$, 哦不对, 应该是 0.0414 MHz 左右.

3. $f(t)={\displaystyle \frac{1+\cos \frac{\pi }{4}t}{2}}$$f(t) = \dfrac{1 + \cos\cfrac{\pi}{4}t}{2}$,

   $B={\displaystyle \frac{1}{8}}\mathrm{M}\mathrm{H}\mathrm{z}$$B = \dfrac{1}{8}\ \rm{MHz}$.

4. $f(t)\leftrightarrow {\displaystyle \frac{2{\text{e}}^{-\mathrm{j}\omega }(1-\cos \omega )}{{\omega }^2}}$$f(t) \rla \dfrac{2\e^{-\j\omega} \pqty{1 - \cos\omega}}{\omega^2}$.

   $B={\displaystyle \frac{1}{2}}\mathrm{M}\mathrm{H}\mathrm{z}$$B = \dfrac{1}{2}\ \rm{MHz}$.

5.  

注

• 对于矩形信号的叠加, 带宽即时域 $t_{\text{m}}$$t_\text m$ 的两倍.

• 是答案错了, 还是我算错了?

3.9

1. $\tau \mathrm{Sa}\left({\displaystyle \frac{\tau \omega }{2}}\right){\text{e}}^{\frac{\mathrm{j}\omega \tau }{2}}-\tau \mathrm{Sa}\left({\displaystyle \frac{\tau \omega }{2}}\right){\text{e}}^{-\frac{\mathrm{j}\omega \tau }{2}}=2\mathrm{j}\tau \mathrm{Sa}\left({\displaystyle \frac{\tau \omega }{2}}\right)\sin \left({\displaystyle \frac{\omega \tau }{2}}\right)={\displaystyle \frac{4\mathrm{j}}{\omega }}{\sin }^2{\displaystyle \frac{\omega \tau }{2}}$$\tau\Sa\pqty{\dfrac{\tau\omega}{2}} \e^{\tfrac{\j\omega\tau}{2}} - \tau\Sa\pqty{\dfrac{\tau\omega}{2}} \e^{-\tfrac{\j\omega\tau}{2}} = 2\j\tau\Sa\pqty{\dfrac{\tau\omega}{2}} \sin\pqty{\dfrac{\omega\tau}{2}} = \dfrac{4\j}{\omega} \sin^2\dfrac{\omega\tau}{2}$.

2. $6\mathrm{Sa}\left(3\omega \right)+2\mathrm{Sa}\left(\omega \right)={\displaystyle \frac{8\sin \omega {\cos }^2\omega }{\omega }}$$6\Sa\pqty{3\omega} + 2\Sa\pqty{\omega} = \dfrac{8\sin\omega\cos^2\omega}{\omega}$.

3. $f_2(t)={\displaystyle \frac{2}{\tau }}{f}_1^{(-1)}(t)\leftrightarrow {\displaystyle \frac{8}{{\omega }^2}}{\sin }^2{\displaystyle \frac{\omega \tau }{2}}$$f_2(t) = \dfrac{2}{\tau} f_1^{(-1)}(t) \rla \dfrac{8}{\omega^2} \sin^2\dfrac{\omega\tau}{2}$.

4. ${\displaystyle {\int }_{-2\tau }^{2\tau }{\displaystyle \frac{t}{2\tau }}{\text{e}}^{-\mathrm{j}\omega t}\mathrm{d}t={\displaystyle \frac{2\mathrm{j}\cos 2\omega \tau }{\omega }}-{\displaystyle \frac{\mathrm{j}\sin 2\omega \tau }{{\omega }^2\tau }}}$$\dint_{-2\tau}^{2\tau} \dfrac{t}{2\tau} \e^{-\j\omega t} \dt = \dfrac{2\j\cos 2\omega\tau}{\omega} - \dfrac{\j\sin 2\omega\tau}{\omega^2\tau}$. (评价是不如直接用定义).

5. $f_5(t)=\sin \left(6\pi t\right)G_2(t)\leftrightarrow \mathrm{j}[\mathrm{Sa}(\omega +6\pi )-\mathrm{Sa}(\omega -6\pi )]={\displaystyle \frac{12\mathrm{j}\pi \sin \omega }{(6\pi {)}^2-{\omega }^2}}$$f_5(t) = \sin(6\pi t) G_2(t) \rla \j \bqty{\Sa\pqty{\omega + 6\pi} - \Sa\pqty{\omega - 6\pi}} = \dfrac{12\j\pi\sin\omega}{(6\pi)^2 - \omega^2}$.

6. $f_6(t)=\cos \left(10\pi t\right)f_{\mathrm{△}}(t)\leftrightarrow {\displaystyle \frac{1}{2}}{\mathrm{Sa}}^2\left({\displaystyle \frac{\omega +10\pi }{2}}\right)+{\displaystyle \frac{1}{2}}{\mathrm{Sa}}^2\left({\displaystyle \frac{\omega -10\pi }{2}}\right)$$f_6(t) = \cos(10\pi t) f_\triangle(t) \rla \dfrac{1}{2} \Sa^2\pqty{\dfrac{\omega+10\pi}{2}} + \dfrac{1}{2} \Sa^2\pqty{\dfrac{\omega-10\pi}{2}}$.

注

• 答案中将 $\tau$$\tau$ 写成了 $T$$T$.

• 注意从 时域乘积 -> 频域卷积, 需要除以 2π.

3.10

1. 思路一: 由冲激函数的性质, $f(t)={\text{e}}^{-2\mathrm{j}}\delta (t-2)\leftrightarrow {\text{e}}^{-2\mathrm{j}(\omega +1)}$$f(t) = \e^{-2\j} \delta(t-2) \rla \e^{-2\j(\omega+1)}$.

   思路二: 由 $\delta (t)\leftrightarrow 1$$\delta(t) \rla 1$ 先时移后频移, 得到 $f(t)\leftrightarrow {\text{e}}^{-2\mathrm{j}(\omega +1)}$$f(t) \rla \e^{-2\j(\omega+1)}$.

2. 思路一: 由 ${\delta }^{\prime }(t)\leftrightarrow \mathrm{j}\omega$$\delta'(t) \rla \j\omega$ 先频移后时移, 得到 $f(t)\leftrightarrow (\mathrm{j}\omega +3){\text{e}}^{-\mathrm{j}\omega }$$f(t) \rla (\j\omega + 3) \e^{-\j\omega}$.

   思路二: 由冲激函数性质, $f(t)={\delta }^{\prime }(t-1)+3\delta (t-1)\leftrightarrow (\mathrm{j}\omega +3){\text{e}}^{-\mathrm{j}\omega }$$f(t) = \delta'(t-1) + 3\delta(t-1) \rla (\j\omega + 3) \e^{-\j\omega}$.

3. $f(t)=1-2G_6(t)\leftrightarrow 2\pi \delta (\omega )-12\mathrm{Sa}(3\omega )=2\pi \delta (\omega )-{\displaystyle \frac{4\sin 3\omega }{\omega }}$$f(t) = 1 - 2 G_6(t) \rla 2\pi\delta(\omega) - 12 \Sa(3\omega) = 2\pi\delta(\omega) - \dfrac{4\sin 3\omega}{\omega}$.

4. 由 ${\text{e}}^{-\alpha t}u(t)\leftrightarrow {\displaystyle \frac{1}{\alpha +\mathrm{j}\omega }}$$\e^{-\alpha t} u(t) \rla \dfrac{1}{\alpha + \j\omega}$ 时移, 得到 $f(t)\leftrightarrow {\text{e}}^2{\displaystyle \frac{{\text{e}}^{\mathrm{j}\omega }}{2+\mathrm{j}\omega }}={\displaystyle \frac{{\text{e}}^{\mathrm{j}\omega +2}}{2+\mathrm{j}\omega }}$$f(t) \rla \e^2 \dfrac{\e^{\j\omega}}{2 + \j\omega} = \dfrac{\e^{\j\omega + 2}}{2 + \j\omega}$.

5. 由 $u(t)={\delta }^{(-1)}(t)=\pi \delta (\omega )+{\displaystyle \frac{1}{\mathrm{j}\omega }}$$u(t) = \delta^{(-1)}(t) = \pi\delta(\omega) + \dfrac{1}{\j\omega}$ 得到 $f(t)=u(t-2)\leftrightarrow \pi \delta (\omega ){\displaystyle \frac{{\text{e}}^{-2\mathrm{j}\omega }}{\mathrm{j}\omega }}$$f(t) = u(t-2) \rla \pi\delta(\omega) \dfrac{\e^{-2\j\omega}}{\j\omega}$.

3.11

由尺度性质与时移性质, 得到 $f_2(t)=f_1(t_0-t)\leftrightarrow F_1(-\omega ){\text{e}}^{-\mathrm{j}\omega t_0}$$f_2(t) = f_1(t_0 - t) \rla F_1(-\omega) \e^{-\j\omega t_0}$.

3.12

1. 由 $\delta (t+{\omega }_0)\leftrightarrow {\text{e}}^{\mathrm{j}{\omega }_{0}t}$$\delta(t + \omega_0) \rla \e^{\j\omega_0 t}$, 得 ${\displaystyle \frac{{\text{e}}^{\mathrm{j}{\omega }_{0}t}}{2\pi }}\leftrightarrow \delta (-t+{\omega }_0)=\delta (t-{\omega }_0)$$\dfrac{\e^{\j\omega_0 t}}{2\pi} \rla \delta(-t+\omega_0) = \delta(t-\omega_0)$.

2. 由 $G_{2{\omega }_0}(t)\leftrightarrow 2{\omega }_0\mathrm{Sa}\left({\omega }_0\omega \right)$$G_{2\omega_0}(t) \rla 2\omega_0 \Sa\pqty{\omega_0\omega}$, 得 ${\displaystyle \frac{{\omega }_0}{\pi }}\mathrm{Sa}({\omega }_{0}t)\leftrightarrow G_{2{\omega }_0}(t)$$\dfrac{\omega_0}{\pi} \Sa(\omega_0 t) \rla G_{2\omega_0}(t)$.

3. 由上一题, ${\left({\displaystyle \frac{{\omega }_0}{\pi }}\right)}^2\mathrm{Sa}({\omega }_{0}t)\leftrightarrow {\displaystyle \frac{{\omega }_0}{\pi }}{G}_{2{\omega }_0}(t)$$\pqty{\dfrac{\omega_0}{\pi}}^2 \Sa(\omega_0 t) \rla \dfrac{\omega_0}{\pi} G_{2\omega_0}(t)$.

3.13

1. $f_1(t-{\displaystyle \frac{\tau }{2}})=T_{\frac{\tau }{2}}(t-{\displaystyle \frac{\tau }{2}})\leftrightarrow {\displaystyle \frac{E\tau }{2}}{\mathrm{Sa}}^2\left({\displaystyle \frac{\omega \tau }{4}}\right){\text{e}}^{-\mathrm{j}\omega \frac{\tau }{2}}$$f_1\pqty{t - \dfrac{\tau}{2}} = T_{\tfrac{\tau}{2}}\pqty{t - \dfrac{\tau}{2}} \rla \dfrac{E\tau}{2} \Sa^2\pqty{\dfrac{\omega\tau}{4}} \e^{-\j\omega\tfrac{\tau}{2}}$,

2. $\cos \left({\omega }_{0}t\right)={\displaystyle \frac{{\text{e}}^{-\mathrm{j}{\omega }_{0}t}+{\text{e}}^{\mathrm{j}{\omega }_{0}t}}{2}}\leftrightarrow \pi [\delta (\omega +{\omega }_0)+\delta (\omega -{\omega }_0)]$$\cos(\omega_0 t) = \dfrac{\e^{-\j\omega_0 t} + \e^{\j\omega_0 t}}{2} \rla \pi\bqty{\delta(\omega+\omega_0) + \delta(\omega-\omega_0)}$,

3. $f_2(t)\leftrightarrow {\displaystyle \frac{E\tau }{4}}{\text{e}}^{-\mathrm{j}\omega \frac{\tau }{2}}[{\mathrm{Sa}}^2\left({\displaystyle \frac{(\omega -{\omega }_0)\tau }{4}}\right){\text{e}}^{\mathrm{j}{\omega }_0\frac{\tau }{2}}+{\mathrm{Sa}}^2\left({\displaystyle \frac{(\omega +{\omega }_{)}\tau }{4}}\right){\text{e}}^{-\mathrm{j}{\omega }_0\frac{\tau }{2}}]$$f_2(t) \rla \dfrac{E \tau}{4} \e^{ -\j\omega \tfrac{\tau}{2} } \bqty{ \Sa^2 \pqty{ \dfrac{(\omega - \omega_0) \tau}{4} } \e^{\j\omega_0 \tfrac{\tau}{2}} + \Sa^2 \pqty{ \dfrac{(\omega + \omega_) \tau}{4} } \e^{-\j\omega_0 \tfrac{\tau}{2}} }$.

备注

• 门函数 $G_{\tau }(t):=u(t+{\displaystyle \frac{\tau }{2}})-u(t-{\displaystyle \frac{\tau }{2}})$$G_\tau(t) := u\pqty{t+\dfrac{\tau}{2}} - u\pqty{t-\dfrac{\tau}{2}}$, 在 $(-{\displaystyle \frac{\tau }{2}},{\displaystyle \frac{\tau }{2}})$$\pqty{-\dfrac{\tau}{2}, \dfrac{\tau}{2}}$ 非零.

• 三角波 $T_{\tau }(t):=(1-\left|{\displaystyle \frac{t}{\tau }}\right|)G_{2\tau }(t)$$T_\tau(t) := \pqty{1 - \vqty{\dfrac{t}{\tau}}} G_{2\tau}(t)$, 在 $(-\tau ,\tau )$$\pqty{-\tau, \tau}$ 非零.

• 之所以将三角波定义在该区间, 而不是 $(-{\displaystyle \frac{\tau }{2}},{\displaystyle \frac{\tau }{2}})$$\pqty{-\dfrac{\tau}{2}, \dfrac{\tau}{2}}$,

  是因为在卷积上形式简洁, 在傅里叶变换上形式统一:

  1. $T_{\tau }(t)=G_{\tau }(t)\ast G_{\tau }(t)$$T_\tau(t) = G_\tau(t) * G_\tau(t)$.

  2. $G_{\tau }(t)\leftrightarrow \tau \mathrm{Sa}\left({\displaystyle \frac{\tau \omega }{2}}\right),\mathrm{Sa}({\omega }_{0}t)\leftrightarrow {\displaystyle \frac{\pi }{{\omega }_0}}{G}_{2{\omega }_0}(\omega )$$G_\tau(t) \rla \tau \Sa\pqty{\dfrac{\tau\omega}{2}}, \Sa(\omega_0 t) \rla \dfrac{\pi}{\omega_0} G_{2\omega_0}(\omega)$.

  3. $T_{\tau }(t)\leftrightarrow \tau {\mathrm{Sa}}^2\left({\displaystyle \frac{\tau \omega }{2}}\right),{\mathrm{Sa}}^2\left({\omega }_{0}t\right)\leftrightarrow {\displaystyle \frac{\pi }{{\omega }_0}}{T}_{2{\omega }_0}(\omega )$$T_\tau(t) \rla \tau\Sa^2\pqty{\dfrac{\tau\omega}{2}}, \Sa^2\pqty{\omega_0 t} \rla \dfrac{\pi}{\omega_0} T_{2\omega_0}(\omega)$.

• 附门函数与三角波及其傅里叶变换的 Mathematica 绘图代码:

xxxxxxxxxx
21
1
Plot[
2
    {HeavisidePi[t], HeavisideLambda[t]}, {t, -1, 1},
3
    GridLines -> Automatic,
4
    PlotLegends -> LineLegend["Expressions"]
5
] (* 门函数与三角波 *)
6
Plot[
7
    {FourierTransform[
8
        HeavisidePi[t], t, \[Omega],
9
        FourierParameters -> {1, -1}
10
    ], FourierTransform[
11
        HeavisideLambda[t], t, \[Omega],
12
        FourierParameters -> {1, -1}
13
    ]}, {\[Omega], -10 Pi, 10 Pi},
14
    PlotRange -> All,
15
    PlotLegends -> LineLegend[{
16
        "\[ScriptCapitalF][\!\(\*SubscriptBox[\(G\), \
17
        \(\[Tau]\)]\)(t)](\[Omega])",
18
        "\[ScriptCapitalF][\!\(\*SubscriptBox[\(T\), \
19
        \(\[Tau]\)]\)(t)](\[Omega])"
20
    }]
21
] (* 傅里叶变换 *)

3.14

0. $f(t)=2T_2(t-1)\leftrightarrow 4{\mathrm{Sa}}^2\left(\omega \right){\text{e}}^{-\mathrm{j}\omega }$$f(t) = 2 T_2(t-1) \rla 4\Sa^2\pqty{\omega} \e^{-\j\omega}$.

1. $\phi (\omega )=-\omega$$\varphi(\omega) = -\omega$.

2. $F(0)=4$$F(0) = 4$.

3. ${\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}F(\omega )\mathrm{d}\omega =2\pi f(0)=2\pi }$$\inti F(\omega) \dd{\omega} = 2\pi f(0) = 2\pi$.

4. 思路一: 由奇偶性质, 傅里叶变换的实部只含偶分量, 故为 $\mathrm{Re}F(\omega )={\displaystyle \frac{f(t)+f(-t)}{2}}$$\Re F(\omega) = \dfrac{f(t) + f(-t)}{2}$.

   思路二: $F(\omega )=4{\mathrm{Sa}}^2(2\omega ){\text{e}}^{-\mathrm{j}\omega }\leftrightarrow T_2(t+1)+T_2(t-1)$$F(\omega) = 4 \Sa^2(2\omega) \e^{-\j\omega} \rla T_2(t + 1) + T_2(t - 1)$.

注

• 附绘图代码:

xxxxxxxxxx
17
1
f[t_] := 2 HeavisideLambda[(t - 1) / 2]
2

3
Plot[
4
    (f[t] + f[-t])/2, {t, -2, 2},
5
    GridLines -> Automatic
6
] (* f(t) 的偶分量 *)
7

8
Plot[
9
    InverseFourierTransform[
10
        Real[FourierTransform[
11
            f[t], t, \[Omega],
12
            FourierParameters -> {1, -1}
13
        ]], \[Omega], t,
14
        FourierParameters -> {1, -1}
15
    ], {t, -2, 2},
16
    GridLines -> Automatic
17
] (* 第四问; 算了这样计算量有点大, 直接绘 f[t+1] 吧 *)

3.15

1. 思路一: 由微分性质, $tf(2t)\leftrightarrow {\displaystyle \frac{\mathrm{j}}{4}}{F}^{\prime }\left({\displaystyle \frac{\omega }{4}}\right)$$tf(2t) \rla \dfrac{\j}{4} F'\pqty{\dfrac{\omega}{4}}$.

   思路二: 也可以使用频域卷积,

   • 由冲激函数高阶导数的抽样性质, ${\delta }^{(n)}(t)\leftrightarrow (\mathrm{j}\omega {)}^n$$\delta^{(n)}(t) \rla (\j\omega)^n$,

   • 由对称性质, $t^n\leftrightarrow 2\pi {\mathrm{j}}^2{\delta }^{(n)}(\omega )$$t^n \rla 2\pi\j^2 \delta^{(n)}(\omega)$,

   • 由频域卷积, $tf(2t)\leftrightarrow {\displaystyle \frac{1}{2\pi }}\cdot 2\pi \mathrm{j}{\delta }^{\prime }(\omega )\ast {\displaystyle \frac{1}{2}}F\left({\displaystyle \frac{\omega }{2}}\right)={\displaystyle \frac{\mathrm{j}}{4}}{F}^{\prime }\left({\displaystyle \frac{\omega }{4}}\right)$$tf(2t) \rla \dfrac{1}{2\pi} \cdot 2\pi\j\delta'(\omega) * \dfrac{1}{2} F\pqty{\dfrac{\omega}{2}} = \dfrac{\j}{4} F'\pqty{\dfrac{\omega}{4}}$.

2. 类似第一问, $(t-2)f(t)\leftrightarrow \mathrm{j}{F}^{\prime }(\omega )-2F(\omega )$$(t-2) f(t) \rla \j F'(\omega) - 2F(\omega)$.

3. 由尺度性质, $(t-2)f(-2t)\leftrightarrow -{\displaystyle \frac{\mathrm{j}}{4}}{F}^{\prime }(-{\displaystyle \frac{\omega }{2}})-F(-{\displaystyle \frac{\omega }{2}})$$(t-2) f(-2t) \rla -\dfrac{\j}{4} F'\pqty{-\dfrac{\omega}{2}} - F\pqty{-\dfrac{\omega}{2}}$.

4. 由微分性质

   1. 时域微分: $f^{\prime }(t)\leftrightarrow \mathrm{j}\omega F(\omega )$$f'(t) \rla \j\omega F(\omega)$.

   2. 频域微分: $t{f}^{\prime }(t)\leftrightarrow -F(\omega )-\omega F^{\prime }(\omega )$$tf'(t) \rla -F(\omega) - \omega F'(\omega)$.

5. 由尺度性质与时移性质, $f(1-t)\leftrightarrow F(-\omega ){\text{e}}^{-\mathrm{j}\omega }$$f(1-t) \rla F(-\omega) \e^{-\j\omega}$.

6. 由前两问与第五问思路,

   1. 频域微分: $tf(t)\leftrightarrow \mathrm{j}{F}^{\prime }(\omega )$$tf(t) \rla \j F'(\omega)$.

   2. 尺度与时移: $(1-t)f(1-t)\leftrightarrow \mathrm{j}{F}^{\prime }(-\omega ){\text{e}}^{-\mathrm{j}\omega }$$(1-t) f(1-t) \rla \j F'(-\omega) \e^{-\j\omega}$.

7. 由尺度性质与时移性质, $f(2t-5)\leftrightarrow {\displaystyle \frac{1}{2}}F\left({\displaystyle \frac{\omega }{2}}\right){\text{e}}^{-\mathrm{j}\frac{5}{2}\omega }$$f(2t-5) \rla \dfrac{1}{2} F\pqty{\dfrac{\omega}{2}} \e^{-\j\tfrac{5}{2}\omega}$.

注

• 标答第二问复制错了.

• 其它结果与标答一致, 注意到 ${\displaystyle \frac{\mathrm{d}F(a\omega )}{\mathrm{d}\omega }=a{F}^{\prime }(a\omega )}$$\ddv{F(a \omega)}{\omega} = a F'(a\omega)$.

3.16 ⭐️

1. 已知 $\mathrm{sgn}(t)\leftrightarrow {\displaystyle \frac{2}{\mathrm{j}\omega }}$$\sgn(t) \rla \dfrac{2}{\j\omega}$ (可由类似第四问的方式得到)

   以及 $1\leftrightarrow 2\pi \delta (\omega )$$1 \rla 2\pi\delta(\omega)$ (可由对称性质得到),

   于是 $u(t)={\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}}\mathrm{sgn}(t)\leftrightarrow \pi \delta (\omega )+{\displaystyle \frac{1}{\mathrm{j}\omega }}$$u(t) = \dfrac{1}{2} + \dfrac{1}{2} \sgn(t) \rla \pi\delta(\omega) + \dfrac{1}{\j\omega}$.

2. 由定义, $G_{\tau }(t)\leftrightarrow \tau \mathrm{Sa}\left({\displaystyle \frac{\tau \omega }{2}}\right)$$G_\tau(t) \rla \tau\Sa\pqty{\dfrac{\tau\omega}{2}}$,

   由时移性质, $G_{\tau }(t-{\displaystyle \frac{\tau }{2}})\leftrightarrow {\displaystyle \frac{2}{\omega }}\sin \left({\displaystyle \frac{\tau \omega }{2}}\right){\text{e}}^{-\mathrm{j}\omega \frac{\tau }{2}}={\displaystyle \frac{1-{\text{e}}^{-\mathrm{j}\omega \tau }}{\mathrm{j}\omega }}$$G_\tau\pqty{t - \dfrac{\tau}{2}} \rla \dfrac{2}{\omega} \sin\pqty{\dfrac{\tau\omega}{2}} \e^{-\j\omega\tfrac{\tau}{2}} = \dfrac{1 - \e^{-\j\omega\tau}}{\j\omega}$,

   ---

   考虑第二项, 记 $\begin{array}{c}{F}_2(\omega )=\underset{\tau \to \mathrm{\infty }}{lim}{\displaystyle \frac{-{\text{e}}^{-\mathrm{j}\omega \tau }}{\mathrm{j}\omega }}=\{\begin{array}{ll}0,& \omega \ne 0,\\ \mathrm{\infty },& \omega =0.\end{array}\end{array}$$F_2(\omega) = \clim \tau \dfrac{-\e^{-\j\omega\tau}}{\j\omega} = \begin{cases} 0, & \omega \ne 0, \\ \infty, & \omega = 0. \end{cases}$

   如果我们能求出 ${\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{F}_2(\omega )\mathrm{d}\omega }$$\inti F_2(\omega) \dd{\omega}$, 并且值为 $\pi$$\pi$,

   那么我们就可以得到 $u(t)=\underset{\tau \to \mathrm{\infty }}{lim}{G}_{\tau }(t-{\displaystyle \frac{\tau }{2}})\leftrightarrow {\displaystyle \frac{1}{\mathrm{j}\omega }}+\pi \delta (\omega )$$u(t) = \clim \tau G_\tau\pqty{t - \dfrac{\tau}{2}} \rla \dfrac{1}{\j\omega} + \pi\delta(\omega)$.

   一种思路是直接换元, 上式 $={\displaystyle {\int }_{+\mathrm{j}\mathrm{\infty }}^{-\mathrm{j}\mathrm{\infty }}\mathrm{j}{\displaystyle \frac{{\text{e}}^z}{z}}\mathrm{d}z}$$= \dint_{+\j\infty}^{-\j\infty} \j \dfrac{\e^{z}}{z} \dz$, 与 $\tau$$\tau$ 无关,

   但是该积分发散 (不定积分也不可由初等函数表示).

   另一种思路是由 ${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\tau }{\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{F}_2(\omega )\mathrm{d}\omega ={\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\text{e}}^{-\mathrm{j}\omega \tau }\mathrm{d}\omega }}}$$\ddv{\tau} \inti F_2(\omega) \dd{\omega} = \inti \e^{-\j\omega\tau} \dd{\omega}$, 同样可以消去 $\tau$$\tau$,

   但是该积分同样发散.

   ---

   好, 这个路子走不通, 我们另寻他路.

   将原式改写为 $G_{\tau }(t-{\displaystyle \frac{\tau }{2}})={\displaystyle \frac{1-\cos \omega \tau +\mathrm{j}\sin \omega \tau }{\mathrm{j}\omega }}$$G_\tau\pqty{t - \dfrac{\tau}{2}} = \dfrac{1 - \cos\omega\tau + \j \sin\omega \tau}{\j\omega}$, 记

   太乐了, 看仔细点, 这个式子取极限后对于任意非零 $\omega$$\omega$ 都是发散的. 算个锤子.

3. 由 $\delta (t)\leftrightarrow 1$$\delta(t) \rla 1$ 与时域积分, 有 $u(t)\leftrightarrow {\displaystyle \frac{1}{\mathrm{j}\omega }}+\pi \delta (\omega )$$u(t) \rla \dfrac{1}{\j\omega} + \pi\delta(\omega)$.

4. 由 ${\text{e}}^{-\alpha t}u(t)\leftrightarrow {\displaystyle \frac{1}{\alpha +\mathrm{j}\omega }}={\displaystyle \frac{\alpha -\mathrm{j}\omega }{{\alpha }^2+{\omega }^2}}$$\e^{-\alpha t} u(t) \rla \dfrac{1}{\alpha + \j\omega} = \dfrac{\alpha - \j\omega}{\alpha^2 + \omega^2}$,

   1. 思路一:

      1. $\begin{array}{c}\underset{\alpha \to 0_{+}}{lim}{\displaystyle \frac{\alpha }{{\alpha }^2+{\omega }^2}}=\{\begin{array}{ll}0,& \omega \ne 0,\\ \mathrm{\infty },& \omega =0\end{array}\end{array}$$\liml_{\alpha\to0_+} \dfrac{\alpha}{\alpha^2 + \omega^2} = \begin{cases} 0, & \omega \ne 0, \\ \infty, & \omega = 0 \end{cases}$

         ${\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\displaystyle \frac{\alpha }{{\alpha }^2+{\omega }^2}}\mathrm{d}\omega =\pi }$$\inti \dfrac{\alpha}{\alpha^2 + \omega^2} \dd{\omega} = \pi$.

         于是由冲激函数定义, $\underset{\alpha \to 0_{+}}{lim}{\displaystyle \frac{\alpha }{{\alpha }^2+{\omega }^2}}=\pi \delta (\omega )$$\liml_{\alpha\to0_+} \dfrac{\alpha}{\alpha^2 + \omega^2} = \pi\delta(\omega)$.

      2. $\underset{\alpha \to 0_{+}}{lim}{\displaystyle \frac{-\mathrm{j}\omega }{{\alpha }^2+{\omega }^2}}={\displaystyle \frac{1}{\mathrm{j}\omega }}$$\liml_{\alpha \to 0_+} \dfrac{-\j\omega}{\alpha^2 + \omega^2} = \dfrac{1}{\j\omega}$.

      3. 于是 $u(t)\leftrightarrow {\displaystyle \frac{1}{\mathrm{j}\omega }}+\pi \delta (\omega )$$u(t) \rla \dfrac{1}{\j\omega} + \pi\delta(\omega)$.

   2. 思路二: $u(t)={\text{e}}^{-\mathrm{j}(\mathrm{j}\alpha )t}{\text{e}}^{-\alpha t}u(t)\leftrightarrow {\displaystyle \frac{1}{\mathrm{j}\omega }}$$u(t) = \e^{-\j (\j\alpha) t} \e^{-\alpha t} u(t) \rla \dfrac{1}{\j\omega}$.

注

做题前记:

我不喜欢增加参数后计算傅里叶变换, 然后取极限的方法.

不是说技巧性强, 它的技巧很明确, 一般是这两个路子:

1. 通过乘上诸如 ${\text{e}}^{-\alpha t}$$\e^{-\alpha t}$ 的函数, 将不满足绝对可积的函数变为绝对可积. 这与拉普拉斯变换的引入是类似的.

2. 通过乘上诸如 ${\text{e}}^{-\alpha g(t)}$$\e^{-\alpha g(t)}$ 的函数, 对 $\alpha$$\alpha$ 求导后消去信号 $f(t)$$f(t)$ 分母中的 $g(t)$$g(t)$, 从而便于求解积分. 最后对 $\alpha$$\alpha$ 积分即可.

问题是我认为一些情况下, 这样做并不严谨. 比如利用单边指数函数取极限, 我们需要证明含参反常积分一致收敛之后, 才能交换极限与积分的运算次序. 而这实际上是做不到的 (在普通的高等数学的意义下).

诸如冲激函数, 它们本身并不是数学分析中允许存在的函数; 它们在泛函分析中通过广义函数以积分的形式定义. 对于符号函数、直流函数、阶跃函数等, 我们完全可以绕开 (或者说间接地使用) 傅里叶变换中的柯西主值积分, 而只使用冲激函数与傅里叶变换的性质, 去求解它们的傅里叶变换.

这样做不仅有可能 (这么说是因为我没有学过泛函分析与广义函数论) 更加严谨, 而且在推导上更加简便, 即使记不住结论, 也可以轻松推得, 又何乐而不为呢?

做题后记:

作者有使用第二种方法算过吗? 我表示怀疑. 思路看起来没什么问题, 但是难以实践.

第四种方法, 如果使用频移特性, 结果将会出错, 为什么?

如果说条件不适用, 什么条件不适用? $$$% 至少书中没有指出. 谈到广义函数, 书中便丧失了严谨性$

3.17

1. $\cos \left({\omega }_{0}t\right)u(t)\leftrightarrow {\displaystyle \frac{\mathrm{j}\omega }{{\omega }_0^2-{\omega }^2}}+{\displaystyle \frac{\pi }{2}}[\delta (\omega +{\omega }_0)+\delta (\omega -{\omega }_0)]$$\cos(\omega_0 t) u(t) \rla \dfrac{\j\omega}{\omega_0^2 - \omega^2} + \dfrac{\pi}{2} \bqty{\delta(\omega + \omega_0) + \delta(\omega - \omega_0)}$.

2. $\sin \left({\omega }_{0}t\right)u(t)\leftrightarrow {\displaystyle \frac{{\omega }_0}{{\omega }_0^2-{\omega }^2}}+{\displaystyle \frac{\mathrm{j}\pi }{2}}[\delta (\omega +{\omega }_0)-\delta (\omega -{\omega }_0)]$$\sin(\omega_0 t) u(t) \rla \dfrac{\omega_0}{\omega_0^2 - \omega^2} + \dfrac{\j\pi}{2} \bqty{\delta(\omega + \omega_0) - \delta(\omega - \omega_0)}$.

3.18

0. $F(\omega )=T_{{\omega }_2-{\omega }_1}(\omega +{\omega }_0)+T_{{\omega }_2-{\omega }_1}(\omega -{\omega }_0)$$F(\omega) = T_{\omega_2 - \omega_1}(\omega+\omega_0) + T_{\omega_2 - \omega_1}(\omega-\omega_0)$.

1.  

2. $f(t){\text{e}}^{\mathrm{j}{\omega }_{0}t}\leftrightarrow F(\omega -{\omega }_0)=T_{\mathrm{\Delta }\omega }(\omega )+T_{\mathrm{\Delta }\omega }(\omega -2{\omega }_0)$$f(t) \e^{\j\omega_0 t} \rla F(\omega - \omega_0) = T_{\Delta\omega}(\omega) + T_{\Delta\omega}(\omega-2\omega_0)$.

3. 与 1 类似.

图略.

3.19

1. $\mathrm{Sa}\left(100t\right)\leftrightarrow {\displaystyle \frac{\pi }{100}}{G}_{200}(\omega )$$\Sa\pqty{100t} \rla \dfrac{\pi}{100} G_{200}(\omega)$,

   ${\omega }_{\text{m}}=100$$\omega_\text m = 100$.

   $f_{\text{m}}={\displaystyle \frac{2{\omega }_{\text{m}}}{2\pi }}={\displaystyle \frac{100}{\pi }}$$f_\text m = \dfrac{2\omega_\text m}{2\pi} = \dfrac{100}{\pi}$.

   $T_{\text{m}}={\displaystyle \frac{1}{f_{\text{m}}}}={\displaystyle \frac{\pi }{100}}$$T_\text m = \dfrac{1}{f_\text m} = \dfrac{\pi}{100}$.

2. ${\mathrm{Sa}}^2(100t)\leftrightarrow {\displaystyle \frac{\pi }{100}}{T}_{200}(\omega )$$\Sa^2(100t) \rla \dfrac{\pi}{100} T_{200}(\omega)$,

   ${\omega }_{\text{m}}=200$$\omega_\text m = 200$.

   $f_{\text{m}}={\displaystyle \frac{200}{\pi }}$$f_\text m = \dfrac{200}{\pi}$.

   $T_{\text{m}}={\displaystyle \frac{\pi }{200}}$$T_\text m = \dfrac{\pi}{200}$.

3. $f_{\text{m}}={\displaystyle \frac{100}{\pi }},T_{\text{m}}={\displaystyle \frac{\pi }{100}}$$f_\text m = \dfrac{100}{\pi}, T_\text m = \dfrac{\pi}{100}$.

4. $f_{\text{m}}={\displaystyle \frac{120}{\pi }},T_{\text{m}}={\displaystyle \frac{\pi }{120}}$$f_\text m = \dfrac{120}{\pi}, T_\text m = \dfrac{\pi}{120}$.

注 ⭐️

对于时域的理想抽样, 首先傅里叶级数展开抽样脉冲信号,

于是抽样信号的傅里叶变换为

记 $F(\omega )$$F(\omega)$ 仅在 $(-{\omega }_{\text{m}},{\omega }_{\text{m}})$$(-\omega_\text m, \omega_\text m)$ 有限, 于是需要 ${\omega }_{\text{m}}<{\omega }_{\text{s}}-{\omega }_{\text{m}}$$\omega_\text m < \omega_\text s - \omega_\text m$, 即

1. 临界采样角频率 ${\omega }_{\text{s}}=2{\omega }_{\text{m}}$$\omega_\text s = 2\omega_\text m$ (越大越好).

2. 奈奎斯特抽样频率 $f_{\text{s}}={\displaystyle \frac{{\omega }_{\text{s}}}{2\pi }}=2f_{\text{m}}$$f_\text s = \dfrac{\omega_\text s}{2\pi} = 2 f_\text m$.

3. 奈奎斯特抽样间隔 $T_{\text{s}}={\displaystyle \frac{2\pi }{{\omega }_{\text{s}}}}={\displaystyle \frac{1}{2f_{\text{m}}}}$$T_\text s = \dfrac{2\pi}{\omega_\text s} = \dfrac{1}{2 f_\text m}$.

一些常用信号的奈奎斯特抽样频率与间隔:

• $\mathrm{Sa}({\omega }_{0}t)\leftrightarrow {\displaystyle \frac{\pi }{{\omega }_0}}{G}_{2{\omega }_0}(\omega ),{\omega }_{\text{m}}={\omega }_0,f_{\text{m}}={\displaystyle \frac{{\omega }_0}{\pi }},T_{\text{m}}={\displaystyle \frac{\pi }{{\omega }_0}}$$\Sa(\omega_0 t) \rla \dfrac{\pi}{\omega_0} G_{2\omega_0}(\omega), \omega_\text m = \omega_0, f_\text m = \dfrac{\omega_0}{\pi}, T_\text m = \dfrac{\pi}{\omega_0}$.

• ${\mathrm{Sa}}^2({\omega }_{0}t)\leftrightarrow {\displaystyle \frac{\pi }{{\omega }_0}}{T}_{2{\omega }_0}(\omega ),{\omega }_{\text{m}}=2{\omega }_0,f_{\text{m}}={\displaystyle \frac{2{\omega }_0}{\pi }},T_{\text{m}}={\displaystyle \frac{\pi }{2{\omega }_0}}$$\Sa^2(\omega_0 t) \rla \dfrac{\pi}{\omega_0} T_{2\omega_0}(\omega), \omega_\text m = 2\omega_0, f_\text m = \dfrac{2\omega_0}{\pi}, T_\text m = \dfrac{\pi}{2\omega_0}$.

• ${\mathrm{Sa}}^n({\omega }_{0}t),{\omega }_{\text{m}}=n{\omega }_0,f_{\text{m}}={\displaystyle \frac{n{\omega }_0}{\pi }},T_{\text{m}}={\displaystyle \frac{\pi }{n{\omega }_0}}$$\Sa^n(\omega_0 t), \omega_\text m = n \omega_0, f_\text m = \dfrac{n\omega_0}{\pi}, T_\text m = \dfrac{\pi}{n\omega_0}$.

3.20

1. $f(t)=T_{\tau /2}(t)\leftrightarrow {\displaystyle \frac{\tau }{2}}{\mathrm{Sa}}^2\left({\displaystyle \frac{\omega \tau }{4}}\right)$$f(t) = T_{\tau/2}(t) \rla \dfrac{\tau}{2} \Sa^2\pqty{\dfrac{\omega\tau}{4}}$,

2. 略.

3. 略.

xxxxxxxxxx
5
1
Plot[
2
    Sum[
3
        Sinc[(\[Omega] - 8 n Pi)/2]^2, {n, -3, 3}
4
    ], {\[Omega], -16 Pi, 16 Pi}
5
] (* 1; 2、3 略 *)

3.21

1. $f(t)+f(t-t_0)\leftrightarrow F(\omega )+F(\omega ){\text{e}}^{-\mathrm{j}\omega t_0}$$f(t) + f(t-t_0) \rla F(\omega) + F(\omega) \e^{-\j\omega t_0}$, 不改变函数值非零时自变量的区间, 因此仍为 ${\omega }_0$$\omega_0$.

2. $f^{\prime }(t)\leftrightarrow \mathrm{j}\omega F(\omega )$$f'(t) \rla \j\omega F(\omega)$, 同样仍为 ${\omega }_0$$\omega_0$.

3. $f^2(t)\leftrightarrow F(\omega )\ast F(\omega )$$f^2(t) \rla F(\omega) * F(\omega)$, 故为 $2{\omega }_0$$2\omega_0$.

4. $f(t)\cos \left({\omega }_{0}t\right)\leftrightarrow \pi [F(\omega +{\omega }_0)+F(\omega -{\omega }_0)]$$f(t) \cos(\omega_0 t) \rla \pi\bqty{F(\omega + \omega_0) + F(\omega - \omega_0)}$, 于是为 $3{\omega }_0$$3\omega_0$.

注

• 奈奎斯特频率一般用 $f_{\text{s}}$$f_\text s$ 表示; 这里应称呼为奈奎斯特角频率.

• 注意 ${\omega }_{\text{s}}=2{\omega }_{\text{m}}$$\omega_\text s = 2\omega_\text m$, 第四题为 $3{\omega }_0$$3\omega_0$ 而非 $2{\omega }_0$$2\omega_0$.

3.22

1. ${\omega }_{\text{s}}=3000\pi ,T_{max}=T_{\text{s}}={\displaystyle \frac{2\pi }{2{\omega }_{\text{s}}}}={\displaystyle \frac{1}{3000}}$$\omega_\text s = 3000\pi, T_\max = T_\text s = \dfrac{2\pi}{2\omega_\text s} = \dfrac{1}{3000}$.

2. $F(\omega )={\displaystyle \frac{1}{4\pi \cdot {10}^6}}{G}_{2000\pi }(t)\ast G_{4000\pi }(t)$$F(\omega) = \dfrac{1}{4\pi \cdot 10^{6}} G_{2000\pi}(t) * G_{4000\pi}(t)$,

   周期梯形, 周期为 6000 π, 幅度为 1/2000.

注 标答中的幅度似乎有误.

3.23

1. $F(\omega )=14\pi [\delta (\omega +320\pi )+\delta (\omega +240\pi )+\delta (\omega -240\pi )+\delta (\omega -320\pi )]$$F(\omega) = 14\pi \bqty{\delta(\omega + 320 \pi) + \delta(\omega + 240 \pi) + \delta(\omega - 240 \pi) + \delta(\omega - 320 \pi)}$.

   $F_{\text{s}}(\omega )={\displaystyle \sum _{n=-\mathrm{\infty }}^{+\mathrm{\infty }}{\displaystyle \frac{F(\omega -n{\omega }_{\text{s}})}{T_{\text{s}}}}}$$F_\text s(\omega) = \nsuminf \dfrac{F(\omega - n \omega_\text s)}{T_\text s}$.

2. ${\omega }_{\text{m}}=320\pi \mathrm{r}\mathrm{a}\mathrm{d}/\mathrm{s},\mathrm{B}={\mathrm{f}}_{\text{c}}={\displaystyle \frac{{\omega }_{\mathrm{c}}}{2\pi }}={\displaystyle \frac{{\omega }_{\text{m}}}{2\pi }}=160\mathrm{H}\mathrm{z}$$\omega_\text m = 320\pi\ \rm{rad/s}, B = f_\text c = \dfrac{\omega_c}{2\pi} =\dfrac{\omega_\text m}{2\pi} = 160\ \rm{Hz}$.

3. $f_{\text{s}}={\displaystyle \frac{2{\omega }_{\text{m}}}{2\pi }}=320\mathrm{H}\mathrm{z}$$f_\text s = \dfrac{2\omega_\text m}{2\pi} = 320\ \rm{Hz}$.

3.24

1. $x(t):T_{\text{s}}={\displaystyle \frac{\pi }{{\omega }_{\text{m}}}}={\displaystyle \frac{\pi }{8}}$$x(t) \colon T_\text s = \dfrac{\pi}{\omega_\text m} = \dfrac{\pi}{8}$.

   $x(2t):T_{\text{s}}={\displaystyle \frac{\pi }{16}}$$x(2t) \colon T_\text s = \dfrac{\pi}{16}$.

   $x(t/2):T_{\text{s}}={\displaystyle \frac{\pi }{4}}$$x(t/2) \colon T_\text s = \dfrac{\pi}{4}$.

2. 问: 是否会发生混叠?

   答: 当采样周期小于等于奈奎斯特周期时, 不会发生混叠.

   因此 $x_{\text{s}}(2t)$$x_\text s(2t)$ 的频谱会发生混叠, $x_{\text{s}}(t/2)$$x_\text s(t/2)$ 和 $x_{\text{s}}(t)$$x_\text s(t)$ 不会发生混叠.

3. 问: 采样信号的频谱?

   答: 可以由以下代码绘制:

xxxxxxxxxx
11
1
Plot[
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    {
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        Sum[HeavisideLambda[(\[Omega] - 16 n)/4], {n, -5, 5}],
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        Sum[HeavisideLambda[(\[Omega] - 16 n)/8], {n, -5, 5}],
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        Sum[HeavisideLambda[(\[Omega] - 16 n)/16]/2, {n, -5, 5}]
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    }, {\[Omega], -32, 32}, 
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    PlotLegends -> {
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        "\!\(\*SubscriptBox[\(x\), \(s\)]\)(t/2)", 
9
        "\!\(\*SubscriptBox[\(x\), \(s\)]\)(t)", 
10
        "\!\(\*SubscriptBox[\(x\), \(2\)]\)(2t)"}
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]

3.25

1. $G_{\tau }(t)\leftrightarrow \tau \mathrm{Sa}\left({\displaystyle \frac{\tau \omega }{2}}\right)$$G_\tau(t) \rla \tau\Sa\pqty{\dfrac{\tau\omega}{2}}$,

2. ${\omega }_{\text{m}}={\displaystyle \frac{16\pi }{\tau }},f_{\text{s}}={\displaystyle \frac{{\omega }_{\text{m}}}{\pi }}=6.4\mathrm{k}\mathrm{H}\mathrm{z}$$\omega_\text m = \dfrac{16\pi}{\tau}, f_\text s = \dfrac{\omega_\text m}{\pi} = 6.4\ \rm{kHz}$.

3.26

1. ${\omega }_{\text{m}}=5\pi ,T_{\text{s}}={\displaystyle \frac{1}{5}}$$\omega_\text m = 5\pi, T_\text s = \dfrac{1}{5}$, 这里认为会发生混叠.

2. 如果认为低通滤波器滤除的频率包含 $\pi /T=5\pi$$\pi/T = 5\pi$, 那么

   $y(t)={\displaystyle \sum _{n=0}^4{\displaystyle \frac{\sin \left(n\pi t\right)}{2^n}}={\displaystyle \sum _{n=0}^4{\displaystyle \frac{{\text{e}}^{\mathrm{j}n\pi t}-{\text{e}}^{-\mathrm{j}n\pi t}}{2^{n+1}\mathrm{j}}}}}$$y(t) = \dsum_{n=0}^4 \dfrac{\sin(n\pi t)}{2^n} = \dsum_{n=0}^4 \dfrac{\e^{\j n\pi t} - \e^{-\j n\pi t}}{2^{n+1} \j}$.

注

• 第一问中认为临界采样时, 低通滤波器无法保留临界频率的信号, 所以也会发生混叠. ⭐️

• 标答中 ${\text{e}}^{-\mathrm{j}n\pi /t}$$\e^{-\j n\pi /t}$ 改为 ${\text{e}}^{-\mathrm{j}n\pi t}$$\e^{-\j n \pi t}$.

3.27

1. $f_{\text{s}}\ge 2f_1={\displaystyle \frac{{\omega }_1}{\pi }}$$f_\text s \ge 2f_1 = \dfrac{\omega_1}{\pi}$.

2. $X_1(\omega )={\displaystyle \frac{G_{2{\omega }_1}(\omega )+T_{{\omega }_1}(\omega )}{2}}$$X_1(\omega) = \dfrac{G_{2\omega_1}(\omega) + T_{\omega_1}(\omega)}{2}$,

   无需: $x_1(t)={\displaystyle \frac{{\omega }_1}{2\pi }}\mathrm{Sa}({\omega }_{1}t)+{\displaystyle \frac{{\omega }_1}{4\pi }}{\mathrm{Sa}}^2\left({\displaystyle \frac{{\omega }_{1}t}{2}}\right)$$x_1(t) = \dfrac{\omega_1}{2\pi} \Sa(\omega_1 t) + \dfrac{\omega_1}{4\pi} \Sa^2\pqty{\dfrac{\omega_1 t}{2}}$,

   $P(\omega )={\displaystyle \sum _{n=-\mathrm{\infty }}^{+\mathrm{\infty }}{\displaystyle \frac{2\pi }{T_{\text{s}}}}\delta (\omega -n{\omega }_{\text{s}})}$$P(\omega) = \nsuminf \dfrac{2\pi}{T_\text s} \delta(\omega - n \omega_\text s)$,

   $X_{\text{s}}(\omega )={\displaystyle \sum _{n=-\mathrm{\infty }}^{+\mathrm{\infty }}{\displaystyle \frac{\pi }{T_{\text{s}}}}(G_{2{\omega }_1}(\omega -n{\omega }_{\text{s}})+T_{{\omega }_1}(\omega -n{\omega }_{\text{s}}))}$$X_\text s(\omega) = \nsuminf \dfrac{\pi}{T_\text s} \pqty{G_{2\omega_1}(\omega - n \omega_\text s) + T_{\omega_1}(\omega - n \omega_\text s)}$.

3. $H_2(\omega )={\displaystyle \frac{1}{H_1(\omega )}}$$H_2(\omega) = \dfrac{1}{H_1(\omega)}$.