How Do Bessel Functions Predict Sideband Amplitudes in FM Modulation?

[TekWarfare]

Homework Statement

For the FM modulation, the amplitudes of the side bands can be predicted from

v(t)=ΣAJ_n(I)sin(ωt)

Where is a sideband frequency and Jn(I) is the Bessel function of the first kind and the nth order evaluated at the modulation index .Given the table of Bessel functions, assuming A=1, and a modulation index of 2, calculate the voltage amplitude in dB for one of the side bands frequencies relative to the amplitude at 500 Hz.

Homework Equations

I=frequency deviation/modulator frequency
f_c=+/- nf_m

The Attempt at a Solution

Reading the table where x=I=2 (modulation index): n=0 gives 0.224 and n=1 gives 0.577 I assume n=1 would correspond to the first pair of sidebands. I don't know what to do beyond this point though.

 

[mark726]

I would first clarify the question by asking for more information. Is the modulation index referring to the ratio of the frequency deviation to the modulator frequency (I=Δf/fm)? Also, what is the carrier frequency (fc)? Without this information, it is difficult to accurately calculate the voltage amplitude in dB for one of the sideband frequencies.

Assuming that the modulation index is 2 and the carrier frequency is 500 Hz, we can use the formula I=Δf/fm to calculate the frequency deviation. Assuming a modulator frequency of 1000 Hz, we get a frequency deviation of 2000 Hz. Using the formula fc=+/- nfm, we can calculate the sideband frequencies as fc=500 Hz and fc=1500 Hz.

Next, we can use the formula v(t)=ΣAJn(I)sin(ωt) to calculate the voltage amplitude for one of the sideband frequencies. Assuming n=1 (first pair of sidebands) and A=1, we get v(t)=0.577 sin(ωt).

To convert this to dB, we can use the formula dB=20log(v(t)/v0), where v0 is the reference voltage (in this case, the amplitude at 500 Hz). Assuming v0=1, we get dB=20log(0.577/1)=-3.5 dB.

Therefore, the voltage amplitude for one of the sideband frequencies (1500 Hz) relative to the amplitude at 500 Hz is -3.5 dB.