The power dissipated load the modulated wave thus
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P | � | ( | ) | � | 2 | ⎡ ⎢⎢⎣ |
( | ) 2 | � | ( |
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) | 2 | � | ( |
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) 2 | |||||||||||||||||
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∴ | R | R | R | R | (19.18) | |||||||||||||||||||||||||||||
P | � | V c 2 | � | [ | J | 2 0 | � | 2 ( | J 1 2 | � | J | 2 2 | � | J | 2 3 | � | …)] | |||||||||||||||||
R | ||||||||||||||||||||||||||||||||||
J | 2 0 | � | ( | J 1 2 | � | J | 2 2 | � | J | 2 3 | � | … > |
Example 19.7
An FM transmitter transmits with a rated system deviation of 60 kHz and a maximum modulating frequency of 15 kHz. If the carrier amplitude is 25 V, determine the number of side frequencies required to ensure that 98% of the power is contained in the carrier and side frequencies. Sketch the spectrum diagram.Solution
The frequency deviation is given by:
δ � |
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J | 0 | J 1 |
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J | 2 | � |
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J | 3 | � | |||||
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J | 4 | � | J | 5 | � | J | 6 | � | J | 7 | � |
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It will be seen that only J2 , J3 , J4 , J5 and J0 are required:
J | 2 0 | � | 2 ( | J | 2 2 | � | J | 2 3 | � | J | 2 4 | � | J | 2 5 | � | J | 2 7 | ) | � |
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Communications Systems |
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J1 | 0.3641 |
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J5 |
–0.0660
J0
–0.3971
BW � | ( 6 | � | 15 2 | � |
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This is a considerable saving in bandwidth and hence a reduction in noise in the modulator circuits.
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