# Drawing One-sided Magnitude Spectrum of Transmitted Signal in dBW

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

1. ### KinGuest

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.

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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