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If I am not wrong, using Bessel function we can estimate the side bands of a carrier? Is this correct? If so, what are the steps to find these values from the graph?

For example if modulation index is 2 and amplitude of the signal is 2volt what are the first few side bands?

enter image description here

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  • \$\begingroup\$ The Bessel function Jn(x) is the nth sideband for any given modulation index x. The voltage amplitude of the signal is irrelevant. \$\endgroup\$ – user207421 Jun 19 '13 at 12:44
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Amplitude of the signal is irrelevant once you have defined the modulation index. Use this table as an easier guide but remember it applies only to sinewaves as the modulating waveform: -

stolen from wiki

For a mod index of 2.0, your carrier will appear to be 22% of what it was unmodulated and there will be sidebands of amplitude 58%, 35%, 13% and 3% of the original carrier amplitude. The first sideband occurs at a distance equivalent to the modulating frequency away from the original carrier. 2nd s/b at 2 x distance etc..

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  • \$\begingroup\$ Wll only for sine waves means cosines as well? \$\endgroup\$ – Sean87 Jun 19 '13 at 12:55
  • \$\begingroup\$ @Sean87 yes it does; I should really have been more specific - a sinewave of any phase. If you are feeding in data then it's more laborious because your data will have a varying mark-space ratio dependent on what the data is doing and you'll need to take into account the harmonics of the squarewave and you'll need to consider that at some point the squarewave data is band limited for practical reasons of not transmitting massive sidebands BUT Bessel stuff still applies. \$\endgroup\$ – Andy aka Jun 19 '13 at 13:03
  • \$\begingroup\$ Why amplitude of signal is irrelevant? Shouldnt I multiply the values of table with the amplitude? \$\endgroup\$ – Sean87 Jun 19 '13 at 13:37
  • \$\begingroup\$ Multiply it to get what? \$\endgroup\$ – user207421 Jun 19 '13 at 19:07
  • \$\begingroup\$ @Sean87 Mod index is the peak deviation divided by the highest modulating frequency. The deviation is caused by the amplitude of the modulator therefore we don't need to know the modulating signal is 2V. If you meant the amplitude of the carrier then that number is important to you but not important to the answer because it just ratios up everything and doesn't disturb the basic answer. \$\endgroup\$ – Andy aka Jun 19 '13 at 19:38
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Yes. Assuming your modulating function is sinusoidal, you can simply read the amplitudes of each sideband pair from the graph. So for a modulation index of 2, following the yellow curve \$J_2(x)\$, you can see that the amplitude of the second sideband pair (fc ± 2fm) is about 0.3 times the amplitude of the unmodulated carrier.

In general, for a modulation index of \$\beta\$, the amplitude of the nth sideband pair is \$J_n(\beta)\$

Wikipedia has a useful table of Bessel functions here, from which you can see that the amplitude of the second sideband pair is actually about 0.35

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