已知υ(t) = 500cos(2π×108t + 20sin2π×103t) (mV),(1)若为调频波,试求载波频率ƒC、调制频率F、调频指数Mf、最大频偏Δƒm、有效频谱宽度BWCR和平均功率Pav(设负载电阻RL = 50Ω).(2)若为调

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已知υ(t) = 500cos(2π×108t + 20sin2π×103t) (mV),(1)若为调频波,试求载波频率ƒC、调制频率F、调频指数Mf、最大频偏Δƒm、有效频谱宽度BWCR和平均功率Pav(设负载电阻RL = 50Ω).(2)若为调
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已知υ(t) = 500cos(2π×108t + 20sin2π×103t) (mV),(1)若为调频波,试求载波频率ƒC、调制频率F、调频指数Mf、最大频偏Δƒm、有效频谱宽度BWCR和平均功率Pav(设负载电阻RL = 50Ω).(2)若为调
已知υ(t) = 500cos(2π×108t + 20sin2π×103t) (mV),(1)若为调频波,试求载波频率ƒC、调制频率F、调频指数Mf、最大频偏Δƒm、有效频谱宽度BWCR和平均功率Pav(设负载电阻RL = 50Ω).(2)若为调相波,试求调相指数Mp,调制信号υ(下标Ω)(t)(设调相灵敏度kp = 5 rad/V),最大频偏Δƒm.

已知υ(t) = 500cos(2π×108t + 20sin2π×103t) (mV),(1)若为调频波,试求载波频率ƒC、调制频率F、调频指数Mf、最大频偏Δƒm、有效频谱宽度BWCR和平均功率Pav(设负载电阻RL = 50Ω).(2)若为调
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