a amplitude 146 dB phase 625 degrees bamplitude 165 dB pha

a. amplitude = -14.6 dB, phase= -62.5 degrees;;

b.amplitude = -16.5 dB, phase= -73.7 degrees;

c. amplitude = -15.2 dB, phase= -26.5 degrees;

d. amplitude = -17.57 dB, phase= -82.4 degrees;;

a.Zeroes at -0.5 +/- j2.78, poles at 1 +/- j2.64, unstable;

b.Zeroes at -1 +/- j1.63, poles at -2+/- j2.75, stable;

c.Zeroes at 1 +/- j1.63, poles at -2 +/- j2.75, stable;

d.Zeroes at 0.5 +/- j2.78, poles at 1 +/- j2.64, unstable;

Solution

TF= 0.8/(s-0.8)            f=6/2pi = > w= 6

= 0.8/(-0.8+jw) = 0.8/(-0.8+ j6)

magnitude = mod(0.8/(-0.8+j6))

                = 0.8 / sqrt(0.8^2 + 6^2)

                 = 0.132

      in db = 20log 0.132 = -17.57db

phase = tan^-1 (-6/0.8) = -82.4degrees

transfer function is not given in ques 4

if given, poles would be the roots of euqtion in the denominator of transfer function

zeros are the roots of equation in the numerator of the transfer function

system is stable if all the poles and zeros of transfer function lies in the left half of s-plane.if not it is unstable.