You have to convert the magnitude scale into dB
dB = 20*log10(magnitude)
Look at where it is -3 dB. That is your cutoff frequency (fc).
Your bandwidth in this case will be 2*fc. Read more on http://en.wikipedia.org/wiki/Bandwidth_(signal_processing [Broken])
EDIT: Note however, that it is only a convention, that the cutoff frequency is -3 dB. It is something we have defined.
I thought BW is fc in this case. In this case, it looks like the -3dB is about 100KHz. The bandwidth is 100KHz as it is from 0Hz to 100KHz. I don't think it is 2fc = 200KHz.
Yes, I think you are right. The circuit for this graph is a non-inverting amplifier made from a 741 op-amp. Op-amp performance is limited at high frequency; it behaves as a low-pass filter.
"In case of a low-pass filter or baseband signal, the bandwidth is equal to its upper cutoff frequency."
EDIT: I see, the article says it all :) My fault.
You might be right, but I seem to remember that we have to include the imaginary components also.
The frequency response is symmetrical around f = 0, for this case. And using modulation, the signal would be positioned at a frequency f = f0 (where f0 is the carrier wave), with a symmetrical shape like the picture around this frequency. The bandwidth of the signal would then be 2*fc.
In my experience, for a LPF like that, the bandwidth is just fc. If it were a BP filter, then the bandwidth extends to both -3dB frequencies.
Separate names with a comma.