北邮通信原理课后习题答案
《通信原理习题解答》第四章模拟通信系统4.1将模拟信号m(t)= \sin 2 \pi f_{m}t与载波c(t)=A_{c} \sin 2 \pi f_{c}t相乘得到双边带抑制载波调幅(DSB-SC)信号,设f_{c}=6f_{m}(1)画出DSB-SC的信号波形图;(2)写出DSB-SC信号的傅式频谱,画出它的振幅频谱图;(3)画出解调框图。解:)s(t)=m(t)c(t)=A_{c} \sin(2 \pi f_{m}t)\sin(2 \pi f_{c}t)=A_{c} \sin(2 \pi f_{m}t)\sin(12 \pi1/5m(2)s(t)可化为s(t)= \frac {A_{c}}{2}(\cos 10 \pi f_{m}t- \cos 14 \pi f_{m}t)于是S(f)= \frac {A_{c}}{4}[ \delta(f-5f_{m})+\delta(f+5f_{m})- \delta(f-7f_{m})- \delta(f+7f_{m})s(1)A_{c}/4fits-7-5057(3)接收信号低通滤波器输出sin12πfm14.2已知s(t)= \cos(2 \pi \times 10^{4}t)+4 \cos(2.2 \pi \times 10^{4}t)+\cos(2.4 \pi \times 10^{4}t)是某个AM已调信号的展开式(1)写出该信号的傅式频谱,画出它的振幅频谱图;(2)写出s(t)的复包络s2(t);(3)求出调幅系数和调制信号频率;(4)画出该信号的解调框图。解:(1)s(t)的频谱为1/25《通信原理习题解答》S(f)= \frac {1}{2} \Bigg l[ \delta \Bigg l(f+10 \times 10^{3} \Bigg r)+\delta \Bigg l(f-10 \times 10^{3} \Bigg r)\Bigg r]+2 \Bigg l[ \delta \bigl(f+\frac