Note #10

Note #10Facilitate the calculation of α m & β mj 2πmft 1 T0/ 20 − j 2πmf0tA) x(t)= ∑ αme=> αm= x(t)e dtT∫ − T0/ 2+∞ +∞∑B) Xt ( ) = α o+ αmcos2mπ ft o + β mcos2mπft om= 1 m=10∑⇒ α o=To /2∫(1/ To) X( t)dt−To/2αTo /2∫(2/ T ) X( t)cos2mπf tdtm = o o−To/2βTo /2∫(2/ T ) X( t)sin 2mπf tdtm = o o−To/21. So, for even function,X(t) = X(-t)α0∫/To /2(2 / To) X()cos2 t mπf tdt (2/ T ) X( t)cos2mπf tdtm = o + o o−To/20Substituting, t =−y inthe first int egral,t = 0⇒ y = 0& t =−To/2 ⇒ y = To/2∫0 To /2α m = (2 / To) X ( −y)cos2 mπ foy ( − dy) + (2 / To) X ( t)cos2 mπfot( dt)To /2 00 To /2=− (2 / To) X ( y)cos 2 mπfoy ( dy) + (2 / To) X ( t)cos 2 mπfot( dt)as, X ( y) = X ( −y)To /2 0To /2 To /2∫∫∫= (2 / To) X ( y) cos 2 mπfoy ( dy) + (2 / To) X ( t) cos 2 mπfot( dt)0 0∫∫∫

So,αTo /2∫2*(2/ T ) X( t)cos2mπf tdtm =o o00 To /2m = (2/ o) ( )sin 2 o + (2/ o) ( )sin 2 o∫β T X t mπ f tdt T X t mπf tdtTo /2 0∫(for y=-t)=0 To /2∫ ∫(2/ To) X( −y)sin−2 πmfoy( − dy) + (2/ To) X( t)sin 2mπfotdtTo /2 0putting X(-y) = X(y)=0 To /2∫ ∫(2/ To) X( y)sin 2 πmfoydy + (2/ To) X( t)sin 2mπfotdtTo /2 0change= -0 To /2∫ ∫(2/ To) X( y)sin 2 πmfoydy + (2/ To) X( t)sin 2mπfotdtTo /2 0= 0

* H/W 1 Problem #2Similarly,If X(t) is odd it implies X(-t) = -X(t)Prove:α m = 0To /2β m = 2*(2/ To) X ( t)sin2πmfo∫tdt0