Question

2) The exponential Fourier series of a periodic signal x(t) is given as x(t) = (4 + j3)e-j6t + j3e-j4t + 2 - j3ej4t + (4 - j3

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Answer #1

Solution? nit)=14+j3) e Jo 13 e tj5 e v4+2 -13e jet +4-35)e+jot (in fundamomenteel- frequency F = HCF(6,4) = Radlsec -34 e clC Cas (+3 C): 77H1 =(4+j3) e Jott 14-33) et jot JN A +jse-14t C2 betonta Power of signal = { 1 CM? = (144j51]!+ [64-js.] 24

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