Carsons rule, which bits go where?

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The discussion focuses on calculating frequency deviation using Carson's rule in the context of an FM signal. The input analogue signal is 1.2 kHz, and the minimum bandwidth to avoid over-sampling is 3.6 kHz, while the signal is transmitted over 1.1 km with a 12 kHz bandwidth on an FM carrier. The carrier frequency is specified at 875 MHz, which is in the GSM reverse band. The formula for bandwidth requirement (CBR) is given as CBR = 2(Δf + fm), leading to the equation Δf = 0.5CBR - fm. Clarification is sought on the correct application of CBR and fm in the context of the problem.
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Homework Statement


calculate the frequency deviation using an approximate technique


Homework Equations


input analogue signal is 1.2kHz
min bandwidth to avoid over-sampling is 3.6kHz
signal transmitted over 1.1km with a 12kHz bandwidth on a FM carrier
Carrier freq is in the GSM reverse band at 875MHz


The Attempt at a Solution



Bandwidth requirement (CBR) =2(Δf + fm)
0.5CBR - fm= Δf = freq deviation

does CBR=12kHz and fm=875MHz?
Don't know what goes where
 
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Mickydawg25 said:
Bandwidth requirement (CBR) =2(Δf + fm)

CBR is the total bw estimate of the FM signal (12kHz)

fm is the bandwidth of the message signal

Δf is peak frequency deviation of the instantaneous FM frequency and depends on the amplitude of the message signal.
 
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