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41 questions with worked solutions.
Q1. If the input x(t) and output y(t) of a system are related as y(t) = max(0, x(t)), then the system is
Answer: non-linear and time-invariant
The system is non-linear because it involves the maximum function, which does not satisfy the principle of superposition, and it is time-invariant since the output does not change if the input is shifted in time.
Answer: δ[n−2] + 2δ[n] + δ[n+2]
The correct option combines the effects of both impulse responses through convolution, resulting in the sum of shifted delta functions. Each impulse response contributes to the overall system output, leading to the presence of the central term with a coefficient of 2, indicating that the system amplifies the response at n=0.
Answer: N_L x (N-1)
The sub-matrix M represents the relationship between the changes in voltage magnitudes at PQ buses and the changes in angles at all buses except the slack bus. Since there are N_L PQ buses and the slack bus is excluded, the dimension of M is N_L rows and (N-1) columns, corresponding to the remaining buses.
Answer: Maxwell bridge
The Maxwell bridge is specifically designed to measure inductance by balancing the bridge circuit, making it suitable for precise inductance measurements in electrical engineering.
Answer: G(s)H(s) = 3
The infinite semicircular arc maps to the value of G(s)H(s) as s approaches infinity: lim (3s+5)/(s-1) = 3. So the arc maps to G(s)H(s)=3, which is option 2, not the stored option 1.
Answer: -1, 1
In a unity-gain negative feedback system with a proportional-integral controller, the controller output stabilizes to a value that compensates for the steady-state error, leading to a final output of -1 for the plant, while the output of the plant reaches 1 due to the nature of the system and the input.
Answer: a-r, b-p, and c-q
The correct option matches running in generator mode (a) with a negative slip range (r), indicating the machine is supplying power back to the grid. Running in motor mode (b) corresponds to a positive slip range (p), where the machine consumes power to operate. Plugging in motor mode (c) involves a higher slip range (q) as it represents a rapid deceleration or reversal of the motor's direction.
Answer: 95 Hz
Synchronous speed Ns=120*50/10=600 rpm, so slip s=(600-540)/600=0.1. The rotor current frequency due to the backward field is (2-s)f=(1.9)(50)=95 Hz.
Answer: 2e⁻³t u(t)
The correct option represents the impulse response derived from the system's differential equation, which can be solved using the Laplace transform. The term 'u(t)' indicates that the response is causal, and the exponential decay factor '2e^(-3t)' reflects the system's dynamics, confirming it matches the form of the impulse response.
Answer: Discrete-time Fourier transform of x[n] converges if R is 2/3 < |z| < 3
The DTFT converges iff the ROC includes the unit circle. Poles are at 1/5, 2/3, and 3; the ROC 2/3<|z|<3 contains |z|=1, so the DTFT converges. A right-sided sequence would have ROC |z|>3, which excludes the unit circle, so that statement is false.
Answer: R3 and R4 are directional overcurrent relays blocking faults towards Line 1 and Line 2, respectively, R7 is directional overcurrent relay blocking faults towards Line 3 and R5 is directional overcurrent relay blocking faults towards bus 2
The correct option ensures that each relay effectively isolates faults in their respective lines while preventing unnecessary disconnections in other parts of the system. By using directional relays R3, R4, R7, and R5, the system can selectively trip only the affected sections, maintaining stability and minimizing disruption.
Answer: P_G1 will decrease and F will increase
Increasing the value of 'a' in the fuel cost function F1(P_G1) raises the quadratic coefficient, making the cost of generating power at P_G1 more expensive. As a result, to minimize costs while maintaining the total load of 290 MW, P_G1 must decrease, leading to an overall increase in the total generation cost F.
Answer: (C)
A discrete time LTI system with a finite duration impulse response means that the system's output will eventually settle to a steady state, ensuring that it does not produce unbounded outputs for bounded inputs, which is the definition of stability.
Answer: (A) Line charging capacitor of finite value is present in all three lines
Option A cannot be true because the negative imaginary components in the Ybus matrix indicate the presence of capacitive elements, and having line charging capacitors in all three lines would lead to an overall positive imaginary part, which contradicts the given Ybus values.
Answer: 2, 4
The correct option is right because it satisfies the conditions for maximum phase lead and gain amplification at the specified frequency. The values of a and β must be chosen such that the compensator's phase lead reaches its peak at 4 rad/s while providing a gain of 6 dB, which is achieved with a = 2 and β = 4.
Answer: 0.5 leading
The power factor is calculated using the formula for real power in a balanced three-phase system, which relates the total power, line current, and phase voltage. Given the negative wattmeter reading indicates a leading power factor, and the calculated value aligns with 0.5, it confirms that the load is indeed operating at a 0.5 leading power factor.
Answer: Bus 1 P - |V| bus Bus 2 P - Q bus Bus 3 Slack bus
Bus 1 is a P - |V| bus because it maintains a constant voltage magnitude while allowing the active power to vary. Bus 2 is a P - Q bus as it can absorb or supply both active and reactive power, and Bus 3 is the slack bus, which balances the system by providing the necessary power to maintain the overall power flow.
Answer: The charge will progress in the direction of y.
The charge will progress in the direction of y because the electric field exerts a force on the positive charge, causing it to accelerate in that direction, while the magnetic field does not exert a force in the y direction when the charge is initially at rest.
Answer: 2
The correct option is 2 because the addition of R0 (0x60) and R1 (0x46) results in a sum that sets the Auxiliary Carry (A) and Overflow (O) flags, while the Carry (C) and Parity (P) flags remain clear. This corresponds to the binary pattern 0010, which is 2 in decimal.
Answer: 2603
With AB=200 and base angles 87.8 degrees, the apex C lies above the midpoint; its distance is 100*tan(87.8)=100*26.03=2603 m. The stored 2063 is wrong; the answer is 2603.
Answer: 96
Ns=120*50/6=1000 rpm, so slip s=(1000-960)/1000=0.04. With stator copper, core, and rotational losses neglected, efficiency equals (1-s)=0.96, i.e. 96%.
Answer: Settling time of System 2 is more than that of System 1.
The settling time of a second-order system is influenced by the natural frequency ( ext{n}) and damping ratio ( ext{z}). System 1, with a higher natural frequency of 3 rad/sec, will settle faster than System 2, which has a lower natural frequency of 1 rad/sec, despite the difference in damping angles.
Answer: X matches with Q and R; Y matches with P and S.
Current Transformers (CT) are designed to measure line current, hence their primary is connected in series (not parallel), and an open-circuited secondary is undesirable as it can lead to high voltage. Potential Transformers (PT), on the other hand, are connected in parallel to the grid, and the secondary burden does affect the primary current, making the correct matches Q and R for CT, and P and S for PT.
Q24. If the following switching devices have similar power ratings, which one of them is the fastest?
Answer: Power MOSFET
Power MOSFETs are known for their high switching speeds due to their voltage-controlled operation, which allows for rapid turn-on and turn-off times compared to other devices like SCRs and GTOs that rely on current control.
Answer: 30°
The minimum triggering angle of 30° is required to ensure that the triac can effectively control the output voltage across the RL load, allowing for sufficient time for the inductor to build up its magnetic field and avoid excessive current spikes.
Answer: T1 and T2 are not BIBO stable.
Both T1 and T2 fail to be BIBO stable because their output can become unbounded for bounded input signals. T1 accumulates all previous inputs, leading to an ever-increasing output, while T2's weighted sum diverges as n increases, indicating instability.
Answer: 0.87 lagging.
The correct option is right because the power factor at the terminal of the generator is determined by the relationship between the voltage, current, and the reactances involved. Given the inductive load and the reactance values, the resulting power factor is calculated to be 0.87 lagging, indicating that the load is drawing reactive power, which is typical for inductive loads.
Answer: [1 1; 1 -1]
The correct matrix S, [1 1; 1 -1], effectively transforms the sequence voltages Vα and Vβ into the phase voltages Vp and Vq by combining them in a way that reflects the characteristics of a two-phase system, where Vp is the sum of both sequence voltages and Vq is their difference.
Answer: Common source amplifier is inverting and common gate amplifier is non-inverting amplifier
The common source amplifier inverts the input signal, producing a 180-degree phase shift, while the common gate amplifier does not invert the signal, maintaining the same phase as the input. This distinction is fundamental to understanding the behavior of these amplifier configurations.
Answer: nullator.
A nullator inherently has both zero voltage and zero current, so when combined in series with a resistor, the entire circuit will still exhibit these nullator characteristics, effectively behaving as a nullator.
Answer: x[n] + x[n - 1] + x[n - 2]
The correct option is right because the output of the LTI system for the input signal x[n] is determined by the system's response to the impulse function, which indicates that the output is a weighted sum of the current and previous input values, specifically x[n], x[n-1], and x[n-2]. This aligns with the system's impulse response.
Answer: Slack bus
The UPS acts as a slack bus because it provides a constant voltage and can adjust its output to balance the power supply and demand during a power failure, ensuring stability in the system.
Answer: O – P – S
The trajectory O – P – S is the most direct path from the initial to the final operating point, minimizing the time taken by avoiding unnecessary intermediate points and maintaining a more efficient transition through the operating region.
Answer: 0.845
Voltage is purely fundamental, so only the fundamental current (10 A peak, lagging 30 deg) delivers real power. Power factor = (I1_rms/I_total_rms) x cos(30) = (10/sqrt(105)) x 0.866 = 0.845. The stored 0.867 is wrong.
Q35. Instrument(s) required to synchronize an alternator to the grid is/are
Answer: Synchroscope
A synchroscope is specifically designed to indicate the phase difference between the voltage of the alternator and the grid, allowing operators to synchronize the two systems effectively.
Answer: 1.5
Integral of y = (integral of x1)(integral of x2). Integral x1 = 1 (on [0,1]) + 0.5 (triangle) = 1.5; integral x2 = 0.5+0.5 = 1. Product = 1.5.
Q37. Let G(s) = 1/((s+1)(s+2)). Then the closed-loop system shown in the figure below is
Answer: stable for all K > 1.
The closed-loop system is stable for all K > 1 because the poles of the characteristic equation move to the left half of the s-plane as K increases, ensuring that the system remains stable.
Answer: 23.28
To reduce the speed of a DC series motor by half, the armature voltage must also be halved, which requires adding external resistance to drop the voltage across the armature. The calculated resistance of approximately 23.28 Ω accounts for the necessary voltage drop while considering the motor's characteristics and the load torque relationship.
Answer: 100 V and 12.5 Hz
At standstill (s=1) maximum torque occurs when total leakage reactance equals R2'=1 ohm. Since Xs+Xr'=4 ohm at 50 Hz, the frequency must drop to 50*(1/4)=12.5 Hz; keeping V/f=400/50=8 gives V=8*12.5=100 V. So 100 V and 12.5 Hz.
Answer: 169.71
In linear sinusoidal PWM the per-phase fundamental peak = ma x Vdc/2 = 0.8 x 300 = 240 V. RMS = 240/sqrt(2) = 169.71 V. Stored 212.13 V is incorrect.
Answer: 20.71
The peak current through the inductor can be calculated using the relationship between the peak voltage and the inductance, considering the firing angle of the SCR. With a firing angle of 0°, the full peak voltage is applied across the inductor, leading to a peak current of approximately 20.71 A.