Question

6.5 (8 pts) Determine the magnitude (in dB) and the phase (in degrees) of H(a) at ω-1 if H (a) equals (a) H (ω)-0.6 (b) H(w) = ju/12 10]ω 2 +ju (c) H(a) 6

Answer 1

H(w) = 0.6 therefore H(w) is a constant, independent of frequency w. therefore magnitude Vertical-bar H(w) Vertical-bar = 0.6 and phase angle H(w) 0 degree H(w) = jw/12 magnitude Vertical-bar H(w) Vertical-bar = Vertical-bar w/12 Vertical-bar therefore Vertical-bar H(w) Vertical-bar = 1/12 phase of H(w) = 90 degree or pi/2 rad for all w. therefore angle H(w) = pi/2 rad H(w) = 10 jw/2 + jw magnitude Vertical-bar H(w) Vertical-bar = Vertical-bar 10 w/ squareroot 2^2 + w^2 = 10/squareroot 5 therefore Vertical-bar H(w) Vertical-bar = 2 squareroot 5 phase angle H(w) = 90 degree - tan^-1(w/2) therefore anglel H(w) = 2 squareroot 5 phase angle H(w) = 90 degree - tan^-1 (w/2) therefore angle H(w) = 90 - tan^-1(1/2) = 63.44 degree H(w) = 6/2 + jw + 3/1 + jw = 6 + 6jw + 6 + 3jw/(2 + jw)(1 + jw) = 12 + 9jw/2 - w^2 + 3jw magnitude Vertical-bar H(w) Vertical-bar = squareroot 144 + 81/squareroot 1 + 9 Vertical-bar H(w) Vertical-bar = 4.74 phase angle H(w) = tan^-1(9w/12) - tan^-1(3w/2 - w^2)]] = tan^-1 (9/12) - tan^-1 3 = -34.7 degree