Question

4.15. Use phasor techniques in the following. (a) Find 2 cos(100t - 459) – 3 cos (100t + 60°) (b) Find 50 sin(100t) + (d/dt) cos (100t - 30). Hint: Do not take the derivative in the time domain; replace it by jo in the frequency domain.

Answer 1

Given 2001 (loot - 459) – 3 Col 10067609) For Suppose a Signal is present like this = Acos(wt+$) then the polarform will be = A LA =-27-3598 + = 2 Z-45° - 3460° 2 cos(-45°) + j 2 Sinl-45°) - 3{coscoº +j sin60°) 2x 1/82 + ja (-1/2) - 3/2 - j 38 $3/2 1-414 - 31.414-115 -2.598; 0.086 -0.086 4:012 In polar form No-oºk) + (4ol2 + / eotta ( 1 4.012 _268.77 4.012 {-91-23 4.012 cos( loot-9123) 50 sini lot) + (d/dt) ws(100t -30°) here d/dt = jw where w=100 d/dt = 100 j = 100/90° s go wos (1oot -90) + jioo cos(1006-309) 50 (-90° + (100 1909 x 12-30') 50 (-90° + 100 L60°

-350 + 100609 60+; 100 w S060 -950 + 100 4x + 9100 x 53/2 50 -; 50 + 86.60 j 50 +936.6 (50)+(366)” (tan' ( 35 ) = 61.96 _36-20° 61.96 Coos (100t + 36.29) icts=20 Sini 500t) 20 cos ( 500t 190°) 1909) at ict) = jw x 20 col 5001-909) where w=500 j 500 20 cos (500t -90°) 500 L90° x 20 1-900 10000 ilu) olo j ir respective of time