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From: bos1234 on 17 Jun 2010 03:44 when freq.spectrum is plotted for FM signals such as the pic below, how would I go about calculating the power? Is it sum of X(f)^2 (i.e. square each individual compnent and add) The solution has the solution squares each indivial compnents adds it then halves (i.e. 1/2 . A_c^2). Why did they do this?
From: bos1234 on 17 Jun 2010 04:26 >when freq.spectrum is plotted for FM signals such as the pic below, how >would I go about calculating the power? >Is it sum of X(f)^2 (i.e. square each individual compnent and add) > >The solution has the solution squares each indivial compnents adds it then >halves (i.e. 1/2 . A_c^2). Why did they do this? > >[IMG]http://i49.tinypic.com/xmoj6g.jpg[/IMG] >
From: Alexander Sotnikov on 17 Jun 2010 05:59 Given a sinewave with magnitude A and frequency F0 a normalized DFT will contain two components (assuming that the sinewave have an integer number of cycles over N samples, N is the DFT size) at +F0 each having magnitude of A/2. Sum of X(f)^2 will give you 2*((A/2)^2) = (A^2)/2. In your picture only the positive side of the spectrum is plotted. I guess this is where the 1/2 coefficient comes from. Alexander >>when freq.spectrum is plotted for FM signals such as the pic below, how >>would I go about calculating the power? >>Is it sum of X(f)^2 (i.e. square each individual compnent and add) >> >>The solution has the solution squares each indivial compnents adds it >then >>halves (i.e. 1/2 . A_c^2). Why did they do this? >> >>[IMG]http://i49.tinypic.com/xmoj6g.jpg[/IMG] >> >

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