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Drawing One-sided Magnitude Spectrum of Transmitted Signal in dBW

Discussion in 'Physics' started by Kin, Oct 8, 2018.

  1. Kin

    Kin Guest

    Question

    The voltage waveform measured at the 50ohm antenna terminal of a 11750-kHz AM (DSB-LC) short-wave broadcasting transmitter is shown in Fig. Q3.

    [​IMG]

    How to draw the one-sided magnitude spectrum of the transmitted signal in dBW?

    My Work

    Now we know that \$f_c = 11759*10^3\$Hz and \$f_m = \frac{1}{T} = \frac{1}{0.0025} = 400\$Hz.

    For Signal,

    \$V_{m(rms)}=400/\sqrt{2}=200\sqrt{2}\$V

    \$P_m=V^2/R=(200\sqrt{2})^2/50=1600\$W

    \$P_{m(dBW)}=10\log(1600)=32.04\$dBW

    For Carrier,

    \$V_{c(rms)}=1000/\sqrt{2}=500\sqrt{2}\$V

    \$P_m=V^2/R=(1000\sqrt{2})^2/50=10000\$W

    \$P_{c(dBW)}=10\log(10000)=40\$dBW


    a. I wonder that : The modulation index should be only 0.4. The ratio \$V_{c(rms)}\$ over \$P_{c(dBW)}\$ is too high. My calculation should be wrong?

    b. The term "One-sided Magnitude Spectrum" means that I only have to draw either USB or LSB?


    Thank you for your help.

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