Electronics and Communication Engineering - Exam Questions Papers

This is the Root test. Since the signal magnitude is raised to the power of 1/n.

GH = - tan^{- 1} ω - tan^{- 1} 2ω - tan^{- 1} 3ω

= - p

tan^{- 1} ω + tan^{- 1} 2ω = p - tan^{- 1} 3ω

tan^{- l} = p - tan^{- 1} 3ω

= - 3ω

ω = 1 rad/sec.

P = 11101101 = - 25

Q = 11100110 = - 18

Thus P - Q = - 25 - (- 18) = - 7

Thus '- 7's two complement representation is 11111001.

.

Since g = ac + bc

i.e. when c = 0, g = b and c = 1, g = a

Now we evaluate

f = (E)(X) + (0)(X) = EX

f = (X)(Y) + (EX)(Y) = XY + EXY.