GATE Technical: GATE 2019 Electrical Engineering (Set No) questions with solutions

Q1. A system transfer function is H(s) = (a1s² + b1s + c1)/(a2s² + b2s + c2). If a1 = b1 = 0, and all other coefficients are positive, the transfer function represents a

1. low pass filter
2. high pass filter
3. band pass filter
4. notch filter

Answer: low pass filter

When the numerator of the transfer function is a constant (since a1 and b1 are zero), and the denominator is a second-degree polynomial with positive coefficients, the system allows low-frequency signals to pass while attenuating higher frequencies, characteristic of a low pass filter.

Q2. The symbols, a and T, represent positive quantities, and u(t) is the unit step function. Which one of the following impulse responses is NOT the output of a causal linear time-invariant system?

1. e^(−at)u(t)
2. e^(−a(t+T))u(t)
3. 1 + e^(−at)u(t)
4. e^(−a(t−T))u(t)

Answer: 1 + e^(−at)u(t)

The option 1 + e^(−at)u(t) is not the output of a causal linear time-invariant system because it includes a constant term (1) that does not decay over time, violating the principle that the impulse response must be zero for negative time values, which is a requirement for causality.

Q3. A 5 kVA, 50 V/100 V, single-phase transformer has a secondary terminal voltage of 95 V when loaded. The regulation of the transformer is

1. 4.5%
2. 9%
3. 5%
4. 1%

Answer: 5%

Rated secondary voltage is 100 V; under load it drops to 95 V. Voltage regulation = (100 - 95)/100 = 5%. The stored 9% is incorrect.

Q4. The parameter of an equivalent circuit of a three-phase induction motor affected by reducing the rms value of the supply voltage at the rated frequency is

1. rotor resistance
2. rotor leakage reactance
3. magnetizing reactance
4. stator resistance

Answer: magnetizing reactance

Reducing the rms value of the supply voltage affects the magnetizing reactance because it alters the magnetic field strength in the motor, which is directly related to the voltage applied. As the voltage decreases, the magnetizing reactance increases, impacting the motor's ability to generate the necessary magnetic flux for operation.

Q5. A three-phase synchronous motor draws 200 A from the line at unity power factor at rated load. Considering the same line voltage and load, the line current at a power factor of 0.5 leading is

1. 100 A
2. 200 A
3. 300 A
4. 400 A

Answer: 400 A

Real power P = V*I*pf is constant. At unity pf, I=200 A, so P proportional to 200. At pf 0.5, I = 200/0.5 = 400 A. The line current is 400 A.

Q6. The characteristic equation of a linear time-invariant (LTI) system is given by Δ(s)=s⁴+3s³+3s²+s+k=0. The system is BIBO stable if

1. 0 < k < 12/9
2. k > 3
3. 0 < k < 8/9
4. k > 6

Answer: 0 < k < 8/9

The correct option indicates that for the system to be BIBO stable, the roots of the characteristic equation must lie in the left half of the complex plane, which occurs when the coefficients of the polynomial satisfy certain conditions. The range 0 < k < 8/9 ensures that all roots have negative real parts, thus guaranteeing stability.

Q7. Given, Vgs is the gate-source voltage, Vds is the drain source voltage, and Vth is the threshold voltage of an enhancement type NMOS transistor, the conditions for transistor to be biased in saturation are

1. Vgs < Vth; Vds ≥ Vgs − Vth
2. Vgs > Vth; Vds ≥ Vgs − Vth
3. Vgs > Vth; Vds ≤ Vgs − Vth
4. Vgs < Vth; Vds ≤ Vgs − Vth

Answer: Vgs > Vth; Vds ≥ Vgs − Vth

The correct option is right because for an enhancement type NMOS transistor to operate in saturation, the gate-source voltage (Vgs) must exceed the threshold voltage (Vth) to turn the transistor on, and the drain-source voltage (Vds) must be at least equal to the difference between Vgs and Vth to maintain saturation.

Q8. A current controlled current source (CCCS) has an input impedance of 10 Ω and output impedance of 100 kΩ. When this CCCS is used in a negative feedback closed loop with a loop gain of 9, the closed loop output impedance is

1. 10 Ω
2. 100 Ω
3. 100 kΩ
4. 1000 kΩ

Answer: 1000 kΩ

In a negative feedback system, the closed loop output impedance is significantly increased by the loop gain. Given the output impedance of the CCCS is 100 kΩ and the loop gain is 9, the closed loop output impedance becomes 100 kΩ multiplied by (1 + loop gain), resulting in 100 kΩ * 10 = 1000 kΩ.

Q9. A DC-DC buck converter operates in continuous conduction mode. It has 48 V input voltage, and it feeds a resistive load of 24 Ω. The switching frequency of the converter is 250 Hz. If switch-on duration is 1 ms, the load power is

1. 6 W
2. 12 W
3. 24 W
4. 48 W

Answer: 6 W

Duty ratio D = t_on x f = 1 ms x 250 Hz = 0.25, so Vo = 0.25 x 48 = 12 V. Load power = Vo^2/R = 144/24 = 6 W. Stored answer 12 W is incorrect.