模拟调制解调技术

前言:由于低频的信号不容易通过天线发射出去,所以调制解调的意义在于将信号放到高频载波上,从而实现无线远程通信

调幅AM

$$
载波 c(t) = \cos (\omega_c t)
\\
基带信号m(t)均值为0
\\ \Rightarrow
已调信号s_{AM} = (A_0 + m(t))\cos (\omega_c t)
$$

AM

则如上图,显然有频谱关系:

$$
S_{AM} (\omega) = \pi A_0 [\delta(\omega + \omega_c) + \delta(\omega - \omega_c)] + \frac{1}{2} [M(\omega + \omega_c) + M(\omega - \omega_c)]
$$

注意:调制加入的直流偏量要大于原信号的最大值,否则会出现“过调幅”,不能使用包络检波

  • 带宽
    $$
    B_{AM} = 2f_H
    $$

  • 功率
    $$
    P_{AM} = P_c + P_s
    \\
    载波功率P_c = \frac{A^2_0}{2},
    边带功率P_s = \frac{\overline{m^2(t)}}{2}
    $$

  • 调制效率
    $$
    \eta_{AM} = \frac{P_s}{P_{AM}}
    $$

    所以,当调制信号幅度达到最大(直流偏置)时,效率才1/3

  • 包络检波滤波器要求
    $$
    f_H \ll \frac{1}{RC} \ll f_c
    $$

  • 抗噪能力

    解调器输入信号(带噪声)可表示为
    $$
    s_m(t) + n_i(t) = [A_0 + m(t) + n_c(t)]\cos \omega_c t - n_s(t)\sin\omega_ct
    \\
    = E(t)\cos[\omega_c t +\Phi(t)]
    \\
    包络E(t)=\sqrt{[A_0 + m(t) + n_c(t)]^2 + n_s^2(t)}
    \\
    \Phi(t) = \arctan\frac{n_s(t)}{A_0 + m(t) + n_c(t)}
    $$

    • 大信噪比情况
      $$
      G_{AM} = \frac{2\overline{m^2(t)}}{A_0^2 + \overline{m^2(t)}}
      \\
      G_{AM}|_{max} = \frac{2}{3}
      $$
    • 小信噪比情况

      产生门限效应,完全白给了

双边带调制DSB

在AM调制中,低调制效率的原因就是因为加了直流偏量,所以如果不用“包络检波”,则可以去掉偏量,即:

$$
s_{DSB}(t) = m(t)\cdot \cos \omega_c t
\\
S_{DSB}(\omega) = \frac{1}{2} [M(\omega+\omega_c)+M(\omega - \omega_c)]
$$

DSB

延用AM中的计算方法:

  • 调制效率
    $$
    \eta_{DSB} = 100%
    $$
  • 抗噪能力
    $$
    S_i = \frac{1}{2}\overline{m^2(t)}
    \\
    N_i = n_0 B
    \\
    S_o = \frac{1}{4}\overline{m^2(t)}
    \\
    N_o = \frac{1}{4}\overline{n_c^2(t)} = \frac{1}{4} n_0 B
    \\
    \Rightarrow G_{DSB} =\frac{S_o/N_o}{S_i/N_i} = 2
    $$

单边带调制SSB

滤除双边带的一边以达到减小带宽的目的

$$
S_{SSB}(\omega) = S_{DSB}\cdot H(\omega)
$$

SSB

时域表示:

$$
s_{SSB}(t) = \frac{1}{2} m(t) \cos \omega_c t \mp \frac{1}{2} \hat m(t) \sin \omega_c t
$$

注意使用了希尔伯特变换(任意分量移相90°)
$$ \hat{M}(\omega) = -jM(\omega) \cdot \text{sgn}(\omega) $$
如下所示为移相组成的SSB调制器

SSB-module

  • 抗噪能力
    $$
    G_{SSB} = 1
    $$

    虽然理论上求的是IO信噪比不变,但其实是和DSB一样都是2。因为求解的时候这是单边带,少了一半的原信号功率

残留边带调制VSB

如图,这个方法不要求理想陡峭的单边滤波器

VSB

得到的调制信号为

$$
S_{VSB}(\omega) = S_{DSB}(\omega)\cdot H(\omega)
$$

采用相干解调法,则混频后

$$
s_p(t) = 2s_{VSB}\cos\omega_c t
\\
\Rightarrow
S_p(\omega) = S_{VSB}(\omega +\omega_c) + S_{VSB}(\omega - \omega_c)
$$

VSB-demodule

则经过低通滤波器后

$$
S_d(\omega) = \frac{1}{2} M(\omega) [H(\omega+\omega_c) + H(\omega - \omega_c)]
$$

所以VSB的精髓之处就在于
$$ H(\omega+\omega_c) + H(\omega - \omega_c) = C$$
如图VSB-demodule

调频FM

$$
s_{FM}(t) = A\cos [\omega_c t + K_f \int m(\tau) d\tau]
$$

  • 调频指数
    $$
    \because m(t) = A_m\cos \omega_m t
    \\ \therefore
    s_{FM} (t) = A\cos[\omega_c t + K_fA_m\int \cos \omega_m \tau d\tau]
    \\ \Rightarrow
    m_f = \frac{K_f A_m}{\omega_m} = \frac{\Delta \omega}{\omega_m} = \frac{\Delta f}{f_m}
    $$
  • 有效带宽(卡森公式)
    $$
    B_{FM} = 2(m_f+1)f_m
    $$
  • 输入信噪比
    $$
    \frac{s_i}{n_i} = \frac{A^2}{2n_0B_{FM}}
    $$
  • 输出信噪比
    $$
    \frac{s_o}{n_o} = \frac{3A^2K_f^2 \overline{m^2(t)}}{8\pi^2n_0f^3_m}
    \\
    G_{FM} = 3m_f^3
    $$

窄带调频NBFM

$$
K_f \int m(\tau) d\tau \ll \frac{\pi}{6}
$$

此时显然频谱宽度较小

$$
s_{NBFM}(t) \simeq A\cos \omega_c t - [AK_f \int m(\tau) d\tau]\sin \omega_c t
\\
S_{NBFM}(\omega) = \pi A[\delta(\omega + \omega_c) + \delta(\omega - \omega_c)] + \frac{AK_f}{2} [\frac{M(\omega -\omega_c)}{\omega - \omega_c} - \frac{M(\omega +\omega_c)}{\omega + \omega_c}]
$$

宽带调频WBFM

非窄带

$$
S_{FM}(t) = \pi A \sum^\infty_{-\infty} J_n(m_f) [\delta(\omega -\omega_c -n\omega_m)+ \delta(\omega +\omega_c +n\omega_m)]
$$

调相PM

$$
s_{PM}(t) = A\cos [ \omega_c t + K_p m(t)]
$$

Attention:FM和PM之间有微积分关系,而FM的实现相对简单(使用VCO就行),所以一般FM用的多

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