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For the preliminary part of a lab I had to do, I was asked the following question: Question

I get how to do 1.1, but with 1.2, I really don't get what's going on; I really would've thought that the resolution bandwidth would be equal to the bandwidth of the low-pass filter. Furthermore, I'm pretty sure a spectrum analyser has a variable band-pass filter in it, so I really don't get what's going on here.

Any help would be much appreciated. For the record, this isn't part of an assignment or anything like that, it's not worth anything, it's just something that's bugging me.

Thanks.

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This document is seriously awesome. It gave me a good understanding of how spectrum analysers work in an easily understandable way.

http://literature.agilent.com/litweb/pdf/5965-7920E.pdf

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    $\begingroup$ It would be good if you could summarize the relevant portions in your answer. $\endgroup$ – Jim Clay Feb 16 '14 at 23:59
  • $\begingroup$ The link in the answer is broken. $\endgroup$ – jithin Apr 15 at 12:54
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The bandwidth of the lowpass filter is the one-sided bandwidth i.e. $B_{LPF} = f_{max}-0$. If you shift this LPF up to some center frequency, $f_c$, then the bandwidth becomes $B=(f_c + B_{LPF}) - (f_c - B_{LPF}) = 2B_{LPF}$.

When the LPF is centered on DC (0 Hz) the we are only considering the one sided bandwidth of real signals - due to the symmetry of positive and negative frequencies. When the filter is shifted up to some non-zero frequency, then you need to consider the frequencies from the other half of the filter.

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