Question

An input signal: vi (t)-5cos(100mt+20 ) + 5 cos(400π t + 60 ) [V] is applied to a first order high pass RL filter with break frequecyf 100 [Hz/The filter contains one resistor and one inductor connected in series. The output signal will have the form Vout(t)-Vi cos( 100 π t + Φ) + V2 cos(400 π t + Φ) [vj a) Calculate values for the amplitudes and phase angles for each of the components of the output signal, and record values in the spaces provided below. b) Calculate the required value of L if R 5 [k2]

Answer 1

A simple RL filter is given below. (High pass) H (w) = V_o (jw)/V_i (jw) = jwL/R + jwL = jwL/R/jwL/R + 1 = jw/w_o/jw/w_o + 1 Where w_o = R/L = w_b called break frequency in rad/sec. given f_b = 100 Hz Implies w_b = 200 pi rad/sec Therefore H (w) = jw/200 pi/jw/200pi + 1 V_in is passed through H (w) then output V_out (t) is given by. V_out (t) = 5 Vertical-bar H(100pi) Vertical-bar cos (100pi + angle H (100pi)) + + 20 degree 5 Vertical-bar H (400pi) Vertical-bar cos (400pi + angle H (400pi) + 60 degree) [V]. H (100pi) = j100 pi/200 pi/j 100 pi/200 pi + 1 = j/2/j/2 + 1 Implies Vertical-bar H (100 pi) Vertical-bar = 1/2/squareroot (1/2)^2 + 1 = 0.4472 angle H (100 pi) = 90 degree - tan^-1 (100 pi/200 pi) = 63.44 degree & H (400 pi) = j2/j2 + 1 Implies Vertical-bar H (400 pi) Vertical-bar = 0.8944 & angle H (400) = 90 degree - tan^-1 2 = 26.567 degree Therefore V_out (t) = 2.236 cos (100 pit