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bandwidth at a given frequency

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sharlotte

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If we want to determine the bandwidth at a given frequency f, we should determine the 3dB fractional bandwidth.
according to the Shannon formula, we can calculate the bandwidth to be:
Bandwidth at f =3db_Bandwidth*f

on the other hand when plot the frequency vs received power to determine the 3dB_bandwidth, we have the bandwidth to be twice of the 3dB_bandwidth.

Can these two values be different?
if so what is the difference between the two approaches?
 

your plotting the Nyquist bandwidth, they are not different. The 3dB bandwidth is the full range of frequencies over which the signal is present 3dB below the max.
When you plot depending on how you plot, the plot can soemtiems extend towards negative frequencies for mathematical purposes. All practical purposes it is only the positive side that matters.
 

thank you for your reply.

according to the plot, if the 3db bandwidth is 4khz, the total bandwidth around the frequency (for example : 10Mhz) is 8Khz.
however according to the Shannon theorem, the bandwidth is (3db_Bandwidth * f), which is this case will be: 4khz*10Mhz.

these two values are far too different.

I do not know if I am making a mistake somewhere, but these two values can not be equal.
I am confused!
 

Can you embed the plot image with your post. Your problem will be better understood.
 

View attachment BW.bmp

according to this plot the bandwidth at 10MHz is 2kHz! (where 3dbBW=1kHz)
while according to the Shannon Equation:
the bandwidth is (3db_Bandwidth * f), which in this case will be: 1khz*10Mhz

?!
 

Th 3db BW is 2KHz.

Shannon has a channel capacity theorem: C=B*(1+SNR) but that is only for capacity.
Here bandwidth measurement is a differential measurement i.e. it can be 2KHz centered at f=0Hz.

When you multiply 3db_BW *f you are effectively increasing the BW which is definitely not the signal BW as shown above.

Where as, in Shannons equation: capacity measures throughput which is function of frequency.
 

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