I85PREP
⌘ K
GATE EC study materials
GATE EC study resources

GATE EC practice questions

One question at a time. Build your GATE EC understanding.

First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Electronics and Communication Engineering (GATE paper)

A silicon pn junction at 300 K has doping N_A = 10^16 cm^-3 on the p-side and N_D = 10^16 cm^-3 on the n-side, with intrinsic carrier concentration n_i = 1.5×10^10 cm^-3. What is the built-in potential?

  1. About 1.44 V
  2. About 0.72 V Correct answer.
  3. About 0.18 V
  4. About 0.36 V

Explanation

The built-in potential is V_bi = V_T ln(N_A N_D / n_i^2). With V_T ≈ 0.0259 V and N_A N_D / n_i^2 ≈ (10^16/1.5×10^10)^2 ≈ 4.4×10^11, the logarithm is about 26.8, giving roughly 0.69–0.72 V.

Question 2Electronics and Communication Engineering (GATE paper)

In a common-emitter BJT amplifier biased in the forward-active region, the small-signal output resistance looking into the collector (ignoring Early effect) is best described as:

  1. Equal to 1/gm
  2. Equal to beta times r_pi
  3. Infinite in the ideal hybrid-pi model without Early effect Correct answer.
  4. Equal to r_pi

Explanation

Without the Early effect the collector current is independent of collector-emitter voltage, so the output characteristic is flat and the small-signal output resistance is infinite. Finite output resistance appears only when the Early voltage is included.

Question 3Electronics and Communication Engineering (GATE paper)

An ideal op-amp is configured as a non-inverting amplifier with R_f = 9 kΩ and R_1 = 1 kΩ. A 0.5 V input is applied. What is the output voltage?

  1. 10.0 V
  2. 0.5 V
  3. 5.0 V Correct answer.
  4. 4.5 V

Explanation

The non-inverting gain is 1 + R_f/R_1 = 1 + 9 = 10. Multiplying the 0.5 V input by this gain yields 5.0 V at the output.

Question 4Electronics and Communication Engineering (GATE paper)

A first-order RC low-pass filter has R = 1 kΩ and C = 1 µF. What is the approximate -3 dB cutoff frequency?

  1. About 1 kHz
  2. About 159 Hz Correct answer.
  3. About 16 Hz
  4. About 1.59 MHz

Explanation

The cutoff frequency is f_c = 1/(2πRC). With RC = 10^-3 s, f_c ≈ 1/(6.283×10^-3) ≈ 159 Hz, which is the frequency where the response falls 3 dB below the passband.

Question 5Electronics and Communication Engineering (GATE paper)

For a discrete-time LTI system with impulse response h[n] = (0.5)^n u[n], what is the region of convergence of the z-transform H(z)?

  1. |z| > 0.5 Correct answer.
  2. The entire z-plane
  3. |z| < 0.5
  4. 0.5 < |z| < 1

Explanation

A right-sided causal sequence has a z-transform that converges outside the circle whose radius equals the pole magnitude. Here the pole is at z = 0.5, so the ROC is |z| > 0.5, which also includes z = infinity.

Question 6Electronics and Communication Engineering (GATE paper)

A periodic signal has fundamental period T = 2 ms. What is the frequency spacing between adjacent spectral lines in its Fourier series?

  1. 2000 Hz
  2. 500 Hz Correct answer.
  3. 2 kHz per line spacing of 1 Hz
  4. 2 Hz

Explanation

The Fourier series of a periodic signal has harmonics at integer multiples of the fundamental frequency f_0 = 1/T. With T = 2 ms, f_0 = 500 Hz, which is the spacing between adjacent lines.

Question 7Electronics and Communication Engineering (GATE paper)

In a 4-bit successive-approximation ADC with a 4.096 V full-scale reference (LSB = 256 mV), the input is 3.0 V. After the first three approximation steps, what is the trial code?

  1. 101
  2. 111
  3. 100
  4. 110 Correct answer.

Explanation

Step 1 sets MSB: 2048 mV < 3000 mV, keep 1. Step 2 adds 1024 mV giving 3072 mV > 3000 mV, so that bit is cleared. Step 3 adds 512 mV to 2048 mV giving 2560 mV < 3000 mV, so it is kept, producing 110.

Question 8Electronics and Communication Engineering (GATE paper)

A 4-to-1 multiplexer has select lines S1 (MSB) and S0 (LSB). Which data input is routed to the output when S1 = 1 and S0 = 0?

  1. D0
  2. D3
  3. D2 Correct answer.
  4. D1

Explanation

The selected input index equals the binary value of the select lines, S1 S0 = 10, which is decimal 2. Therefore input D2 is connected to the output.

Question 9Electronics and Communication Engineering (GATE paper)

A Boolean function is F = A'B + AB' + AB. Which simplified expression is equivalent?

  1. A ⊕ B
  2. AB
  3. A + B Correct answer.
  4. A'B'

Explanation

The terms A'B + AB' form A ⊕ B, and adding AB fills the remaining minterm, so the function covers all cases where A or B is 1. Hence F = A + B by the consensus/absorption simplification.

Question 10Electronics and Communication Engineering (GATE paper)

A JK flip-flop has J = 1 and K = 1 and is clocked repeatedly. What does the output do?

  1. It sets to 1 and stays there
  2. It holds its previous state
  3. It toggles on every active clock edge Correct answer.
  4. It resets to 0 and stays there

Explanation

With both J and K asserted the JK flip-flop operates in toggle mode, so the output flips on each active clock edge. This is the standard behavior that distinguishes it from the SR latch's invalid state.

Question 11Electronics and Communication Engineering (GATE paper)

A 50 Ω lossless transmission line is terminated in a purely resistive 100 Ω load. What is the magnitude of the reflection coefficient at the load?

  1. 2/3
  2. 1/3 Correct answer.
  3. 1
  4. 1/2

Explanation

The reflection coefficient is Γ = (Z_L − Z_0)/(Z_L + Z_0). Substituting gives (100 − 50)/(100 + 50) = 50/150 = 1/3, a positive real value indicating a partially reflected wave.

Question 12Electronics and Communication Engineering (GATE paper)

The characteristic impedance of a lossless transmission line depends on which set of primary parameters?

  1. Per-unit-length inductance L and capacitance C Correct answer.
  2. Only the load impedance
  3. Per-unit-length resistance R and conductance G
  4. Only the physical length of the line

Explanation

For a lossless line Z_0 = sqrt(L/C), where L and C are the per-unit-length inductance and capacitance determined by the geometry and dielectric. Resistance and conductance cause loss and do not set the lossless characteristic impedance.

Question 13Electronics and Communication Engineering (GATE paper)

An antenna has a directivity of 10 (10 dBi) and a radiation efficiency of 80%. What is its gain in dBi?

  1. About 12.5 dBi
  2. About 10 dBi
  3. About 9 dBi Correct answer.
  4. About 8 dBi

Explanation

Gain equals directivity multiplied by radiation efficiency, so G = 10 × 0.8 = 8 in linear terms, which is about 9 dBi. The 1 dB reduction relative to directivity reflects the 80% efficiency.

Question 14Electronics and Communication Engineering (GATE paper)

A sinusoidal carrier is amplitude-modulated by a single tone with modulation index m = 1. What fraction of the total transmitted power resides in the two sidebands?

  1. Two thirds
  2. One half
  3. One quarter
  4. One third Correct answer.

Explanation

For tone AM the total power is P_c(1 + m^2/2) and the sideband power is P_c m^2/2. With m = 1 the ratio is (0.5)/(1.5) = 1/3, so one third of the power is in the sidebands.

Question 15Electronics and Communication Engineering (GATE paper)

A source emits symbols with probabilities 0.5, 0.25, 0.125, and 0.125. What is the source entropy in bits per symbol?

  1. 2.0
  2. 1.5
  3. 1.75 Correct answer.
  4. 1.0

Explanation

Entropy is the sum of −p log2 p over the symbols. The terms are 0.5, 0.5, 0.375, and 0.375, which sum to 1.75 bits per symbol, below the maximum of 2 bits for four equiprobable symbols.

Question 16Electronics and Communication Engineering (GATE paper)

A memoryless binary symmetric channel has crossover probability p = 0.1. What is the channel capacity in bits per channel use?

  1. About 0.900
  2. About 0.469
  3. About 0.100
  4. About 0.531 Correct answer.

Explanation

The capacity of a BSC is C = 1 − H(p), where H(p) is the binary entropy function. With p = 0.1, H(0.1) ≈ 0.469, so C ≈ 1 − 0.469 = 0.531 bits per channel use.

Question 17Electronics and Communication Engineering (GATE paper)

A signal x(t) = cos(2π·100t) is sampled at 250 Hz. What is the resulting discrete-time frequency in radians per sample?

  1. 0.2π
  2. 1.6π
  3. 0.4π
  4. 0.8π Correct answer.

Explanation

Discrete frequency is ω = 2π f / f_s = 2π(100)/250 = 0.8π rad/sample. Since this lies between 0 and π, no aliasing occurs.

Question 18Electronics and Communication Engineering (GATE paper)

A discrete-time LTI system has all poles strictly inside the unit circle and is causal. Which statement about the system is correct?

  1. It is stable and its impulse response decays to zero Correct answer.
  2. It is stable only if the input is bounded
  3. It is unstable because it is causal
  4. It is marginally stable because poles are nonzero

Explanation

For a causal system, stability requires all poles to lie strictly inside the unit circle. Under that condition the impulse response is absolutely summable and decays with n, so a bounded input always gives a bounded output.

Question 19Electronics and Communication Engineering (GATE paper)

A silicon diode with a reverse saturation current of 1 nA is forward biased at 0.6 V at room temperature (V_T ≈ 26 mV). What is the approximate diode current?

  1. About 0.1 mA Correct answer.
  2. About 1 µA
  3. About 10 mA
  4. About 1 nA

Explanation

Using the Shockley equation I = I_s(exp(V/V_T) − 1), the exponent is 0.6/0.026 ≈ 23.1, so exp(23.1) ≈ 1.1×10^10. Multiplying by 1 nA gives roughly 0.1 mA.

Question 20Electronics and Communication Engineering (GATE paper)

A MOSFET in saturation has I_D = 0.5 mA at V_GS = 2 V and I_D = 2 mA at V_GS = 3 V, with the same V_DS. Assuming square-law behavior, what is the threshold voltage?

  1. 2.0 V
  2. 0.5 V
  3. 1.5 V
  4. 1.0 V Correct answer.

Explanation

In saturation I_D is proportional to (V_GS − V_T)^2, so the ratio 2/0.5 = 4 means (3 − V_T)/(2 − V_T) = 2. Solving gives 3 − V_T = 4 − 2V_T, hence V_T = 1.0 V.

Question 21Electronics and Communication Engineering (GATE paper)

In a CMOS inverter, both transistors are minimum-sized and the supply is V_DD. During the input transition when both devices conduct, the current that flows is called:

  1. Gate tunneling current
  2. Short-circuit (crowbar) current Correct answer.
  3. Junction reverse leakage
  4. Subthreshold leakage

Explanation

When the input sits between the rails, both the NMOS and PMOS are simultaneously on, creating a direct path from V_DD to ground. This crowbar or short-circuit current is a major dynamic power component in CMOS logic.

Question 22Electronics and Communication Engineering (GATE paper)

An ideal transformer has a turns ratio of 10:1 (primary to secondary). A 100 Ω load is connected to the secondary. What impedance is seen at the primary?

  1. 1 MΩ
  2. 1 kΩ
  3. 100 Ω
  4. 10 kΩ Correct answer.

Explanation

Impedance transforms by the square of the turns ratio, so Z_primary = n^2 Z_secondary = 10^2 × 100 Ω = 10 kΩ. This is the basis of impedance matching with transformers.

Question 23Electronics and Communication Engineering (GATE paper)

A 3-bit flash (parallel) ADC requires how many comparators for a full unipolar conversion?

  1. 8
  2. 7 Correct answer.
  3. 3
  4. 4

Explanation

An n-bit flash ADC uses 2^n − 1 comparators, since the reference ladder has 2^n levels and adjacent levels define 2^n − 1 thresholds. For n = 3 this gives 7 comparators.

Question 24Electronics and Communication Engineering (GATE paper)

A uniform plane wave propagates in free space with electric field amplitude 10 V/m. What is the magnitude of the time-averaged Poynting vector?

  1. About 26.5 W/m^2
  2. About 377 W/m^2
  3. About 0.133 W/m^2 Correct answer.
  4. About 0.265 W/m^2

Explanation

The time-averaged power density is S = E_0^2/(2η_0), where η_0 ≈ 377 Ω is the intrinsic impedance of free space. Thus S = 100/(2×377) ≈ 0.133 W/m^2.

Question 25Electronics and Communication Engineering (GATE paper)

A CMOS gate's dynamic power dissipation is dominated by which mechanism?

  1. Reverse leakage of the source junctions
  2. Charging and discharging the load capacitance Correct answer.
  3. Static bias current in the output stage
  4. Ohmic heating in the interconnect only

Explanation

Dynamic power in CMOS is P = α C_L V_DD^2 f, arising from charging and discharging the load capacitance on each switching event. Static bias current is negligible in a complementary CMOS gate because one device is always off in steady state.

Question 26Electronics and Communication Engineering (GATE paper)

In a common-emitter BJT amplifier, the transistor is biased in the forward-active region. If the collector current is doubled while the thermal voltage is held constant, how does the small-signal transconductance g_m change?

  1. It halves
  2. It stays the same
  3. It increases by a factor of four
  4. It doubles Correct answer.

Explanation

In the forward-active region the transconductance is g_m = I_C/V_T, so it is directly proportional to the collector current. Doubling I_C at fixed V_T therefore doubles g_m. The base-emitter resistance r_pi = beta/g_m falls by half, which is consistent with the same result.

Question 27Electronics and Communication Engineering (GATE paper)

An ideal op-amp is connected as a non-inverting amplifier with R_f = 9 kOhm and R_1 = 1 kOhm. A 0.5 V DC source drives the non-inverting input. What is the output voltage?

  1. 4.5 V
  2. 0.5 V
  3. 5.0 V Correct answer.
  4. 10 V

Explanation

The closed-loop gain of a non-inverting amplifier is 1 + R_f/R_1 = 1 + 9 = 10. Multiplying the 0.5 V input by this gain gives 5.0 V, assuming the op-amp output can swing that far within its supply rails.

Question 28Electronics and Communication Engineering (GATE paper)

A silicon pn junction diode is reverse biased well beyond the breakdown voltage in a circuit with a series current-limiting resistor. Which statement describes the resulting behaviour?

  1. The diode blocks all current and behaves as an open circuit
  2. The diode conducts heavily in reverse with an almost fixed terminal voltage Correct answer.
  3. The diode conducts heavily in forward direction with a 0.7 V drop
  4. The diode capacitance increases without limit and blocks conduction

Explanation

Beyond breakdown the reverse current rises sharply while the terminal voltage stays nearly constant at the breakdown value, so the device acts like a voltage reference. The series resistor sets the current and prevents destruction. This is the basis of Zener-diode regulator design.

Question 29Electronics and Communication Engineering (GATE paper)

A first-order RC low-pass filter has R = 1 kOhm and C = 1 microfarad. What is its -3 dB corner frequency?

  1. 15.9 kHz
  2. 159 Hz Correct answer.
  3. 100 Hz
  4. 1 kHz

Explanation

The corner frequency of a first-order RC low-pass filter is f_c = 1/(2*pi*R*C). With R = 1000 Ohm and C = 1e-6 F, the time constant is 1 ms, giving f_c = 1/(2*pi*1e-3) which is about 159 Hz.

Question 30Electronics and Communication Engineering (GATE paper)

A signal x(t) has the Fourier transform X(jw) that is zero for |w| > 2000*pi rad/s. It is sampled at a rate of 2.5 kHz. What happens to the spectrum of the sampled signal?

  1. The spectrum is shifted to baseband and scaled by 2.5 kHz
  2. The spectrum becomes continuous and no replicas appear
  3. Aliasing occurs because the sampling rate is below Nyquist Correct answer.
  4. The spectrum is replicated at 2.5 kHz intervals without overlap

Explanation

The signal bandwidth is 2000*pi rad/s, that is 1 kHz, so the Nyquist rate is 2 kHz. Sampling at 2.5 kHz is above that, so the replicas are separated and no aliasing occurs.

Question 31Electronics and Communication Engineering (GATE paper)

A continuous-time signal x(t) has the Laplace transform X(s) = 1/((s+2)(s+3)) with region of convergence -3 < Re(s) < -2. Which statement about the corresponding time-domain signal is true?

  1. It is right-sided and grows without bound
  2. It is two-sided and decays to zero
  3. It is two-sided and grows without bound Correct answer.
  4. It is left-sided and grows without bound

Explanation

Poles at -2 and -3 with a vertical strip between them give a two-sided signal containing both a right-sided term e^{-2t}u(t) and a left-sided term e^{-3t}u(-t). Because the strip lies in the right half-plane, at least one exponential grows as t increases, so the signal is unbounded.

Question 32Electronics and Communication Engineering (GATE paper)

The z-transform of a causal sequence is X(z) = z/(z - 0.5). What is the value of x[0] using the initial-value theorem?

  1. 0
  2. 2
  3. 0.5
  4. 1 Correct answer.

Explanation

For a causal sequence, x[0] equals the limit of X(z) as z tends to infinity. Here z/(z-0.5) tends to 1 as z grows without bound, so x[0] = 1. This matches the inverse transform (0.5)^n u[n] evaluated at n = 0.

Question 33Electronics and Communication Engineering (GATE paper)

A discrete-time LTI system is described by y[n] = x[n] + 2x[n-1] + x[n-2]. Which statement about its frequency response is correct?

  1. It has a zero at z = 1 and rejects DC
  2. It has a double pole at z = 1 and is unstable
  3. It has a zero at z = -1 and passes high frequencies
  4. It has a double zero at z = -1 and rejects the Nyquist frequency Correct answer.

Explanation

The transfer function is H(z) = 1 + 2z^{-1} + z^{-2} = (1 + z^{-1})^2, which has a double zero at z = -1. At the Nyquist frequency w = pi, z = -1, so the response is exactly zero and that frequency is completely rejected.

Question 34Electronics and Communication Engineering (GATE paper)

A linear time-invariant system has an impulse response h(t) = e^{-2t}u(t). A unit step u(t) is applied at the input. What is the steady-state value of the output as t tends to infinity?

  1. 1
  2. 0
  3. Infinity
  4. 0.5 Correct answer.

Explanation

The step response is the integral of the impulse response, giving (1 - e^{-2t})/2 for t >= 0. As t tends to infinity the exponential vanishes and the output settles at 1/2. This equals the DC gain H(0) = 1/2 of the system.

Question 35Electronics and Communication Engineering (GATE paper)

A 3-input CMOS NAND gate is to be built with a standard static CMOS style. How many NMOS and PMOS transistors are required, and how are they arranged?

  1. Two NMOS in series and two PMOS in parallel
  2. Three NMOS in series and three PMOS in parallel Correct answer.
  3. Six NMOS in series and six PMOS in parallel
  4. Three NMOS in parallel and three PMOS in series

Explanation

In static CMOS the pull-down network must conduct only when all inputs are high, so the three NMOS devices are stacked in series. The pull-up network is the dual and must conduct when any input is low, so the three PMOS devices are placed in parallel.

Question 36Electronics and Communication Engineering (GATE paper)

A negative-edge-triggered D flip-flop has D tied to its own complemented output Q'. The flip-flop is initially cleared. What is the output sequence Q after each successive falling clock edge?

  1. 1, 0, 1, 0, ...
  2. 0, 1, 0, 1, ... Correct answer.
  3. 0, 1, 1, 0, ...
  4. 0, 0, 0, 0, ...

Explanation

With D = Q', each active edge loads the complement of the present state, so the flip-flop toggles on every falling edge. Starting from Q = 0, the outputs after successive edges are 1, 0, 1, 0, and so on. This is the standard divide-by-two configuration.

Question 37Electronics and Communication Engineering (GATE paper)

A 4-bit ripple-carry adder is built from full adders, and the carry propagates through all four stages. If each full adder contributes a carry delay of 2 ns, what limits the maximum operating frequency of the adder?

  1. The sum delay of the last stage only
  2. The clock-to-output delay of the register
  3. The setup time of the first flip-flop
  4. The total carry propagation delay of 8 ns Correct answer.

Explanation

In a ripple-carry adder the critical path is the carry chain, which must settle before the next clock edge. With four stages at 2 ns each, the worst-case carry delay is 8 ns, which caps the clock period and hence the frequency. This is why carry-lookahead structures are used to shorten the path.

Question 38Electronics and Communication Engineering (GATE paper)

A 4-variable Karnaugh map has a group of four adjacent ones forming a 2x2 square. Which statement about the resulting product term is correct?

  1. It contains three literals
  2. It contains two literals Correct answer.
  3. It reduces to the constant 1
  4. It contains four literals

Explanation

A group of 2^k ones eliminates k variables from the product term. A group of four therefore removes two of the four variables, leaving a term with two literals. Larger groups yield simpler terms, which is the purpose of the map.

Question 39Electronics and Communication Engineering (GATE paper)

A uniform lossless transmission line of characteristic impedance 50 Ohm is terminated in a load of 50 + j50 Ohm. What is the magnitude of the reflection coefficient at the load?

  1. 0.447 Correct answer.
  2. 0.5
  3. 1
  4. 0

Explanation

The reflection coefficient is (Z_L - Z_0)/(Z_L + Z_0). Substituting gives j50/(100 + j50), whose magnitude is 50/sqrt(100^2 + 50^2) = 50/111.8, approximately 0.447. The load is mismatched in reactance, so some power is reflected.

Question 40Electronics and Communication Engineering (GATE paper)

A rectangular waveguide with broad wall dimension a = 2.286 cm is air-filled. What is the cutoff frequency of the dominant TE10 mode?

  1. 9.84 GHz
  2. 4.64 GHz
  3. 3.28 GHz
  4. 6.56 GHz Correct answer.

Explanation

For the TE10 mode the cutoff wavelength is twice the broad-wall dimension, so lambda_c = 4.572 cm. The cutoff frequency is c/lambda_c = 3e8/0.04572, which is approximately 6.56 GHz. This corresponds to X-band waveguide.

Question 41Electronics and Communication Engineering (GATE paper)

A lossless half-wave dipole is fed with 1 kW of input power and radiates uniformly over a sphere of radius 10 km in free space. Neglecting losses, what is the power density at that distance?

  1. 7.96 mW/m^2
  2. 0.796 mW/m^2 Correct answer.
  3. 79.6 mW/m^2
  4. 0.0796 mW/m^2

Explanation

For a lossless antenna all input power is radiated, so the average power density at distance r is P/(4*pi*r^2). With P = 1000 W and r = 10^4 m, the denominator is about 1.257e9 m^2, giving roughly 7.96e-4 W/m^2, that is 0.796 mW/m^2.

Question 42Electronics and Communication Engineering (GATE paper)

A superheterodyne receiver uses a local oscillator below the received radio frequency. A strong signal at the image frequency can cause interference. For an RF of 10 MHz and an intermediate frequency of 455 kHz, what is the image frequency?

  1. 10.455 MHz
  2. 9.545 MHz
  3. 10.910 MHz Correct answer.
  4. 9.090 MHz

Explanation

With a low-side local oscillator at f_LO = f_RF - f_IF = 9.545 MHz, the image lies at f_LO - f_IF = 9.545 - 0.455 = 9.090 MHz on the other side, or equivalently f_RF + 2*f_IF = 10 + 0.91 = 10.910 MHz. The image is separated from the RF by twice the intermediate frequency.

Question 43Electronics and Communication Engineering (GATE paper)

An FM signal is described by s(t) = A cos(2*pi*f_c*t + 5 sin(2*pi*f_m*t)). Using Carson's rule, what is the approximate transmission bandwidth?

  1. 5*f_m
  2. 2*f_m
  3. 12*f_m Correct answer.
  4. 6*f_m

Explanation

The modulation index is 5 and the peak frequency deviation is 5*f_m. Carson's rule gives bandwidth approximately 2*(delta_f + f_m) = 2*(5*f_m + f_m) = 12*f_m. This captures over 98 percent of the signal power for a sinusoidal modulating tone.

Question 44Electronics and Communication Engineering (GATE paper)

A PCM system uses uniform quantization with 8-bit encoding over a full-scale range of 2 V. What is the quantization step size, and what is the approximate signal-to-quantization-noise ratio for a full-scale sinusoid?

  1. 7.81 mV and about 44 dB
  2. 3.91 mV and about 56 dB
  3. 7.81 mV and about 50 dB Correct answer.
  4. 15.6 mV and about 44 dB

Explanation

With 8 bits there are 256 levels, so the step is 2 V/256 = 7.81 mV. For a full-scale sinusoid the SQNR is approximately 6.02n + 1.76 dB, giving about 49.9 dB for n = 8. Both values must be consistent for the option to be correct.

Question 45Electronics and Communication Engineering (GATE paper)

A digital communication link uses binary antipodal signalling with equally likely symbols and an additive white Gaussian noise channel. If the bit energy is doubled while the noise power spectral density is unchanged, how does the bit error probability change?

  1. It decreases because the argument of the Q-function grows by a factor of the square root of two Correct answer.
  2. It decreases by exactly a factor of two
  3. It stays the same because the signalling is antipodal
  4. It doubles because the energy is doubled

Explanation

For antipodal binary signalling the error probability is Q(sqrt(2*E_b/N_0)). Doubling E_b raises the argument by the square root of two, so the error probability falls, though not by a simple constant factor. The Q-function is monotonically decreasing in its argument.

Question 46Electronics and Communication Engineering (GATE paper)

A linear block code has a generator matrix in systematic form with 4 message bits and 3 parity bits. What are the code rate and the number of distinct codewords?

  1. 4/7 and 128
  2. 3/7 and 8
  3. 7/4 and 16
  4. 4/7 and 16 Correct answer.

Explanation

The code has n = 7 and k = 4, so the rate is k/n = 4/7. A linear (7,4) code contains 2^k = 16 distinct codewords, one for each message word. The remaining 2^n - 2^k words are non-codewords used for error detection.

Question 47Electronics and Communication Engineering (GATE paper)

A random variable X takes values 0 and 1 with probabilities 0.7 and 0.3. What is the variance of X?

  1. 0.30
  2. 0.21 Correct answer.
  3. 0.70
  4. 0.49

Explanation

For a Bernoulli random variable the variance is p(1-p). With p = 0.3 this gives 0.3*0.7 = 0.21. The mean is 0.3, and the second moment minus the square of the mean gives the same result.

Question 48Electronics and Communication Engineering (GATE paper)

A causal LTI system has the transfer function H(s) = 10/(s^2 + 2s + 10). A sinusoidal input at the undamped natural frequency is applied. What is the magnitude of the steady-state output relative to the input amplitude?

  1. 1 Correct answer.
  2. 2
  3. 10
  4. 0.5

Explanation

The natural frequency is sqrt(10) rad/s and the damping ratio is 2/(2*sqrt(10)), which is less than one. At w = sqrt(10) the denominator magnitude is 2*w = 2*sqrt(10), while the numerator is 10, so the gain is 10/(2*sqrt(10)) = sqrt(10)/2, approximately 1.58. Rechecking with the standard resonance relation gives a peak near one for this damping; the value at w_n is |H| = 1/(2*zeta) with zeta = 1/sqrt(10), approximately 1.58.

Question 49Electronics and Communication Engineering (GATE paper)

An ideal current source of 2 mA drives a parallel combination of a 1 kOhm resistor and a 1 kOhm resistor. What is the voltage across the parallel combination?

  1. 4 V
  2. 2 V
  3. 0.5 V
  4. 1 V Correct answer.

Explanation

Two 1 kOhm resistors in parallel give 500 Ohm. The 2 mA current flowing through this equivalent resistance produces V = I*R = 2 mA * 500 Ohm = 1 V, by Ohm's law.

Question 50Electronics and Communication Engineering (GATE paper)

A two-port network has z-parameters z11 = 10 Ohm, z12 = 2 Ohm, z21 = 2 Ohm and z22 = 5 Ohm. What is the condition for the network to be reciprocal, and is this network reciprocal?

  1. The determinant equals one; yes, it is reciprocal
  2. z12 = -z21; no, it is not reciprocal
  3. z12 = z21; yes, it is reciprocal Correct answer.
  4. z11 = z22; no, it is not reciprocal

Explanation

A two-port is reciprocal when the transfer terms are equal, that is z12 = z21. Here both equal 2 Ohm, so the network is reciprocal. Reciprocity is independent of whether the port impedances z11 and z22 are equal.