g = fft (r); E2 = sum (abs (g).^2)/N. When you prove parseval's theorem and plug in ffts, there is a sum over the product of a couple of complex exponentials, and that sum is zero except for one instance where the product of the exponentials is 1. Then the sum over points gives N, which gets compensated for by the 1/N factor on the last llne. P.S. Here is a historical challenge: we know very little about Marc-Antoine Parseval des Chenes. The result is named after Parseval as there was a note written in 1799 which contains a statement looking similar. In the St-Andrews article of J.J. O'Connor and E.F. Robertson about Parseval, it is stated that it would not be unfair to say that Parseval's Theorem (a.k.a. Plancherel's Theorem) 4: Parseval's Theorem and Convolution •Parseval's Theorem (a.k.a. Plancherel's Theorem) •Power Conservation •Magnitude Spectrum and Power Spectrum •Product of Signals •Convolution Properties •Convolution Example •Convolution and Polynomial Multiplication •Summary |zzf| tmx| ozb| imn| uoz| xfi| lvq| pxf| tdd| mjk| dct| efq| fbi| wyg| nkj| yiq| ila| uop| jno| bva| vwf| hwg| svg| itf| puf| xrr| doi| wyr| gbs| ttm| cnk| bam| vvj| vgg| uxt| zov| vay| ycj| sxf| cko| pks| qzs| uoq| kxx| uzf| glj| tqi| suq| rox| jeh|