考试题目,我要详细解答。
的有关信息介绍如下:解:根据υ(t)表达式,ωc = 2π×108 rad/s,Δψ(t) = 20 sin(2π×103) rad ,求得:
(1)FM波:ƒC =ωc / 2π = 2π×108 / 2π Hz = 100 MHz
F = Ω / 2π = 2π×103 / 2π Hz = 1 kHz
Mf = 20 rad, Δƒm = Mf F = 20 kHz
所以 BWCR = 2(Mf + 1)F = 42 kHz,Pav = Vm2 / RL / 2 = 2.5 mW
(2)PM波:Mp = 20 rad,kp = 5rad/V,V(下标Ωm) = Mp / kp = 4 V
υ(下标Ω)(t) = 4 sin(2π×103) (V),Δƒm = Mp F = 20 kHz