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50% Duty Cycle 2 1.5 magnitude 0.5 0 0 0.5 1 1.5 N 3 3.5 4 4.5 2.5 time x 103 10% Duty Cycle 2 1.5 magnitude 1 0.5 0 0 0.5 1

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50% Duty cycles The magnitude function will be given by , 0:5X103 flate) 0-5x10% t <10-3 2 khet O<t < 0.5X1 { The besiod ofP an 2 Jo f(t) las (nt) at whore I 2 0:5X103 51103 2 cos (211541097) at an = 2x100 0:5X10-3 4X103 Jo cos (ant x1 ott) at 0.512 not [1 - (-0] from © MI and IV) I flat) ft zo casentret £ 3. [1-1-0] Sin(2ndTALOG + n TT = 1 +2 I that 1-60n Sincan 19410Ool X103 10% Duty cycle in 2. f (st) م 6 okt 201x103 OnX10 Be it < 1.08103 T ao = 2. fit) et Jo 2 at = 2.8103 JOS4100 So = 4x01 XLO bn = 2x102 Jonze tos sin(entrecoet) of 0.18 no = 422x1g (- cos anterxeot 10 g antxio - (-63 (0-21TTA +1) nu 7 3 [I- co

I have done the calculation little fast. In case you have any doubt please comment below.

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