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501.

The characteristic equation of a closed loop system is ( + 2) ( + 3) + K = 0. The centroid of the asymptotes in root locus is:

Answer» Centroid Poles are -2 and -3.
Discussion
502.

With respect to the differential amplifier configuration shown below, the differential gain A will be = 4.7 KΩ = 2.7 KΩ = 100 = 1 KΩ = 100 Ω

Answer»
Discussion
503.

In the circuit below, if diodes and the moving coil millimeter are assumed to be ideal if mA reading is , find Vm.

Answer» Now, ∴ Im = pIDC = 0.1 mA ∴ Vm = Im x R = 0.1 mA x 10 K Vm = 1 Volt.
Discussion
504.

Transfer function of a certain system is: The approximate 2% settling time for unit step input is

Answer» Pole '-120' will have no effect on settling time.
Discussion
505.

The polar plot for positive frequencies is shown below. The gain margin is

Answer»
Discussion
506.

Input and output impedance with feedback for voltage series feedback having A = -200, R = 10 kΩ, R = 50 kΩ and feedback factor β = -0.2 will be

Answer» In voltage series feedback output impedance decreases, input impedance increases, , Rif = Ri(1 + Aβ).
Discussion
507.

An FM transmitter delivers 80W to a load of 30 W when no modulation is present. The carrier is now frequency modulated by a single sinusoidal signal and the peak frequency deviation is so adjusted to make the amplitude of the second sideband is zero in the given output.J(0, 0) = 1, J(2, 4) = 0, J(3.8) = - 0.4, J(5.1)= - 0.1J(2, 4) = 0.52, J((3.8) = 0, J(5.1) = - 0.33, J(2.4) = 0.1J(3.8)= 0.41, J(5.1) = 0 The power in all the remaining sidebands is :

Answer» Total power of the FM transmitter = 80 W to a load of 50 W second side band amplitude = J2(β) = 0 ∴ β = 5.1 Average Power = (J0(β))2 x 100 = (0.16)2 x 100 = 2.56 W Power in all remaining = 80 - 2.56 = 77.44 .
Discussion
508.

A MOSFET has a threshold voltage of 1 V and oxide thickness of 500 x 10 [ε = 3.9; ε = 8.85 x 10 F/cm, = 1.6 x 10 ]. The region under the gate is ion implanted for threshold voltage tailoring. The base and type of impant required to shift threshold voltage to - 1 V are __________ .

Answer» VT(new) = VT(odd) + = 6.903 x 10-8 ∴ ∴ fB = - 8.6 x 1011 The threshold voltage is always negative for p-channel and hence implant is of p-type.
Discussion
509.

Z = __________

Answer» Output of first MUX = A(B + C) Output of second MUX = C D + A B).
Discussion
510.

A point charge of 6 μC is located at the origin, uniform line charge density of 180 C/ of 8 length lies along the axis and a uniform sheet of charge equal to 25 C/ of area x 4 lies in the = 0 plane. Calculate total electric flux leaving the surface of sphere of 4 radius centered at origin.

Answer» According to Gauss's law total flux leaving the closed surface is equal to the charge enclosed by the closed surface Qencl = (6 x 10-6) + (8 x 180 x 10-6) + (p x 42 x 25 x 10-9) = (6 x 10-6) + (1.44 x 10-6) + (1.257 x 10-6) = 8.697 μ Coulombs.
Discussion
511.

In a radar system, the range of R is achieved at a frequency . Then the range R at frequency R = 8 would be __________ [Neglect effect of 1 on the Beam width.]

Answer» The range of the radar is directly proportional to the square root of the frequency. ∴ R ∝ ∴ R1 ∝ and R2 ∝ ∴ ∴ .
Discussion
512.

Find Y- parameters

Answer» Apply Nodal Analysis At node V ...(i) At node V2 ...(ii) At node V3 ∴ ∴ Put V3 in equation (i) Put V3 in equation (ii) .
Discussion