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](/uploads/image/z/11502013-13-3.jpg?t=%E5%B7%B2%E7%9F%A5%CF%85%28t%29+%3D+500cos%282%CF%80%C3%97108t+%2B+20sin2%CF%80%C3%97103t%29+%28mV%29%2C%EF%BC%881%EF%BC%89%E8%8B%A5%E4%B8%BA%E8%B0%83%E9%A2%91%E6%B3%A2%2C%E8%AF%95%E6%B1%82%E8%BD%BD%E6%B3%A2%E9%A2%91%E7%8E%87%26%23402%3BC%E3%80%81%E8%B0%83%E5%88%B6%E9%A2%91%E7%8E%87F%E3%80%81%E8%B0%83%E9%A2%91%E6%8C%87%E6%95%B0Mf%E3%80%81%E6%9C%80%E5%A4%A7%E9%A2%91%E5%81%8F%CE%94%26%23402%3Bm%E3%80%81%E6%9C%89%E6%95%88%E9%A2%91%E8%B0%B1%E5%AE%BD%E5%BA%A6BWCR%E5%92%8C%E5%B9%B3%E5%9D%87%E5%8A%9F%E7%8E%87Pav%EF%BC%88%E8%AE%BE%E8%B4%9F%E8%BD%BD%E7%94%B5%E9%98%BBRL+%3D+50%CE%A9%EF%BC%89.%EF%BC%882%EF%BC%89%E8%8B%A5%E4%B8%BA%E8%B0%83)
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