GATE 2026 EC Paper: Worked Solutions to All 65 Questions

Key idea: Nocturnal means active at night; its opposite is active by day.

Official answer: B

Of the four options, only diurnal, active during the day, is about the time of day at all, so it is the antonym. The others (normal, abnormal, exceptional) have nothing to do with time of day.

Key idea: The top four scores all lie between 76 and 100, a spread of 24.

Official answer: A

Take (S1) and (S2) together. The four highest scores all lie between the fourth highest (76) and the highest (100).

So those four students are within 25 marks of each other, which is (S3). Option (A) holds.

The others fail: (S3) can come from any four students, so it fixes neither the highest score nor the fourth highest, and (S1) alone says nothing about the fourth score.

Key idea: Match each segment where it enters and leaves the hidden rectangle.

Official answer: D

Follow the stubs outside the rectangle:

• A segment comes in from the upper left through the top edge and leaves near corner Y, going down to the right.
• A segment comes in from the upper right through the top edge and leaves through the bottom edge near Z, going down to the left.
• A nearly horizontal segment enters low on the left edge and leaves higher on the right edge.

Only tile S has all three: two lines crossing from the top edge to the bottom corners, and one shallow line rising from left to right. Tile R has a line starting at corner X, which the stubs don't show. The answer is S, option (D).

Key idea: Taking the nth root of a base multiplies the logarithm by n.

Official answer: B

Change both logarithms to plain \log:

Since \log_p y \cdot \log_y p = 1, the product is n^2 = 16, so n = 4 (all quantities exceed 1, so n > 0). Option (B).

Key idea: Negating for every ... there exists gives there exists ... for every.

Official answer: B

Apply it to the statement: for every student becomes there exists a student, and there exists another student within 10 marks becomes every other student is more than 10 marks away.

That is option (B). Options (A) and (D) make claims about every student, and (C) says exactly one, neither of which follows from the negation.

Key idea: Solve the clues first, then rearrange the four first letters.

Official answer: A

• Synonym of recognize, 8 letters: IDENTIFY.
• A story long enough for a book, 5 letters: NOVEL.
• Two of something, 6 letters: COUPLE.
• One of two equal parts, 4 letters: HALF.

The first letters I, N, C, H rearrange to CHIN. Option (A).

Key idea: Teammates can't be parent and child, and a match can't hold two parent-child pairs.

Official answer: C

Look at who plays together in each match:

• Match 1 (P and X together): X is not P's parent.
• Match 3 (R and X together, Q and Y together): X is not R's parent, and Y is not Q's parent.

So X's child is Q, and Y's child is P or R.

Rule (ii) allows at most one parent-child pair in a match. Match 3 already has X against Q. If Y were R's parent, match 3 would also have Y against R: two pairs. So Y is P's parent, and R has no parent playing. Option (C).

Key idea: P_k = 4k, so the sum is 4 times the sum of 1 to 10.

Official answer: C

Option (C).

Key idea: Every ring's area is \pi(R^2 - r^2), so \pi cancels in the ratio.

Official answer: C

The boundaries are at radii 2.5, 3.5, 5.5, 7.5, 10.5 and 13.5. Every area is \pi times a difference of squares, so \pi cancels:

The ratio is 78.25 / 104 \approx 0.75. Option (C).

Key idea: The example is a reflection in a horizontal mirror.

Official answer: C

In the example, the two lines swap places and every character is flipped upside down, while each line still runs left to right. That is a reflection in a horizontal mirror.

Applied to the new block, the second line moves to the top and both lines appear upside down but in their original left-to-right order. Tile R shows exactly that. P also swaps the two words of one line, Q swaps the parts of the other line, and S does both, which the example never does. Option (C).

Key idea: Each component evolves on its own, so A is diagonal with the exponents on its diagonal.

Official answer: A

Differentiate each component and compare it with the component itself:

Option (A).

Key idea: The gradient of the surface function is normal to the surface.

Official answer: D

Write the surface as f = 2x^2 - y^2 - z^2 = 0 and take its gradient at (1, 0, \sqrt{2}):

This is a multiple of (\sqrt{2}, 0, -1), whose length is \sqrt{3}, so a unit normal is

Option (D). Options (A) and (C) aren't unit vectors, and (B) points the wrong way.

Key idea: Convolution in time is multiplication of Laplace transforms.

Official answer: D

Transform each factor and multiply:

Option (D).

Key idea: A heavily doped junction is thin enough for band-to-band tunnelling.

Official answer: C

In reverse breakdown, the field across a Zener diode's very thin depletion region is strong enough for electrons to tunnel straight from the valence band on the p side to the conduction band on the n side. That tunnelling carries the current. Option (C).

Avalanche breakdown, in lightly doped junctions at higher voltages, works by impact ionisation instead; the question asks about the Zener mechanism.

Key idea: 50 Hz sampled at 40 Hz aliases onto 10 Hz.

Official answer: D

At F_s = 40 Hz, the sample times are t = n/40:

The 50 Hz signal aliases onto 10 Hz, so every sample is identical. Option (D) is the only statement that holds; (A), (B) and (C) all need one set of samples to be greater.

Key idea: The input's zero cancels one of the system's poles.

Official answer: A

The input cancels the pole at 1/3. The response starts at n = 0 (causal) and decays (stable). Option (A).

Key idea: A non-inverting gain is 1 + R_f/R_g; closing S removes 1 kΩ from R_f.

Official answer: C

Take the two switch positions in turn, with R_g = 1 kΩ.

• Switch open: R_f = 3 kΩ, gain 4, so x = 4V_i.
• Switch closed: the 1 kΩ is shorted, R_f = 2 kΩ, gain 3, so V_o = 3V_i.

V_o = \tfrac{3}{4}x. Option (C).

Key idea: Feedback holds the emitter at V_i, which fixes I_E; then I_B = I_E/(\beta + 1).

Official answer: D

The emitter sits at V_i = 3 V, across the 1 kΩ resistor. So

Check the BJT is active: V_C = 12 - 2\text{ k}\Omega \times 2.97\text{ mA} \approx 6.06 V, above the emitter. Option (D).

Key idea: Two identical blocks driven by the same signal give identical outputs, so the feedback is zero.

Official answer: B

Call the error signal E. Both blocks see E, so their outputs are equal, E/(s+4). The feedback is their difference, which is zero:

Option (B).

Key idea: Series pull-up and parallel pull-down is the NOR structure.

Official answer: A

The pull-up network (PMOS in series) conducts only when A = 0 and B = 0. The pull-down network (NMOS in parallel) conducts when A = 1 or B = 1.

That is NOR. Option (A).

Key idea: A 100 dB path loss is a power ratio of 10^{-10}.

Official answer: B

Here P_R/P_T is -90 - 10 = -100 dB. Set the Friis ratio equal to it:

Option (B).

Key idea: Nyquist bandwidth depends on the symbol rate only: half of it.

Official answer: C

With a symbol rate of R_s = 10^4 symbols per second, the number of levels doesn't enter:

Option (C).

Key idea: An impulse of current delivers its charge instantly, so it only shifts the initial voltage.

Official answer: A

Integrate the equation across t = 0. The impulse delivers charge q in zero time, while m(t) (finite) contributes nothing over an instant:

After that, C\,dV/dt = -m(t). So the impulse is equivalent to starting the source-free equation from V_0 + q/C. Option (A), which writes this initial value at t = 0^- for the equivalent problem.

Key idea: Equal amplitudes 90 degrees apart give circular polarization; the rotation sense fixes the hand.

Official answer: A

At z = 0 the field points along \hat{x} at t = 0 and along \hat{y} a quarter period later, with the same magnitude: it is circular, turning from x towards y.

The wave travels along +z. With the right thumb along +z, the fingers curl from x to y, so the wave is right-handed circularly polarized (IEEE convention). Option (A).

Key idea: An applied bias V splits the quasi-Fermi levels by qV.

Official answer: A

At a forward bias of 2 V the split is q \times 2 V = 2 eV, option (A).

Options (C) and (D) give an energy in volts, which isn't an energy unit, and 1 eV doesn't match the bias. The official key lists (A) only.

Key idea: Write \det M in terms of a and b, then keep the options where it isn't zero.

Official answer: B and D

Expand along the first row:

• (A) a = b = -\tfrac{1}{2}: -\tfrac{1}{2} - \tfrac{5}{2} + 3 = 0.
• (B) a = b = \tfrac{1}{2}: 6 \neq 0.
• (C) a = -3, b = 0: 0.
• (D) a = \tfrac{1}{2}, b = -3: -11.5 \neq 0.

Options (B) and (D).

Key idea: 65 states need 2^n \ge 65.

Official answer: B

Six flip-flops give 2^6 = 64 states, one short of the 65 needed for 0 to 64, so n = 7 (2^7 = 128). Option (B) only.

Key idea: The loss tangent is the in-phase (lossy) current over the quadrature current.

Official answer: A and B

With \epsilon = \epsilon' - j\epsilon'' and conductivity \sigma, Ampère's law gives the total current density

The loss tangent is the ratio of the in-phase (lossy) part to the quadrature part:

That is option (B), and option (A) describes the same ratio in words. Susceptibility over permittivity (C) and P over D (D) measure how polarizable the material is, not its loss.

Key idea: The active-region slope is about I_C/|V_A|: the steeper curve has the smaller Early voltage.

Official answer: C and D

Compare the two active-region slopes at the same I_C:

• BJT 1 is steeper, so |V_{A1}| < |V_{A2}|: option (C).
• BJT 2 still has a nonzero slope, so |V_{A2}| is finite: option (D).
• BJT 1 has the steeper slope, so its Early voltage is certainly not infinite, which rules out (B), and (A) is the reverse of (C).

Key idea: The battery's current leaves one midpoint and enters the other, so it cancels from the sum of the node equations.

Official answer: 0.95 to 1.05

Let the bottom rail be 0 V, the top rail T, and the midpoints L and R (volts, kΩ, mA). Whatever current i the battery carries leaves one midpoint and enters the other, so it cancels when the two midpoint equations are added:

At the top rail, the source current leaves through the two upper resistors:

So V_o = 1.00 V, whatever the battery voltage.

Key idea: The resistor sits directly across the ideal source, so resonance doesn't change its voltage.

Official answer: 71.50 to 72.50

The resistor sees the full 12 V peak, so

Resonance only makes the L and C currents cancel; it doesn't change the resistor's voltage.

Key idea: Minimum passband bandwidth equals the symbol rate.

Official answer: 0.95 to 1.05

16-QAM carries \log_2 16 = 4 bits per symbol, so the symbol rate is 1 Msymbol/s. A passband signal needs at least the symbol rate in bandwidth (Nyquist pulses, both sidebands):

Key idea: f_c = c/(2a) with a the broad dimension.

Official answer: 20.50 to 21.50

Key idea: Power lost inside is 1 - |S_{11}|^2 - |S_{21}|^2.

Official answer: 20.00 to 21.00

At the crossing both magnitudes are -4 dB:

About 20.38% is dissipated. (For the lossless network, 2 \times 10^{-0.3} \approx 1, as it should be.)

Key idea: J = K = 1 toggles Q on every clock edge, halving the frequency.

Official answer: 4.90 to 5.10

The clock has 10 falling edges a second and Q changes on each one, so Q completes 5 cycles a second: 5.00 Hz.

Key idea: Ratio test for the first; compare the second with a geometric series.

Official answer: A

For S_A, use the ratio test:

So S_A converges (its sum is in fact 6).

Every term of S_B is a different power of \tfrac{1}{2} (the exponents run 0, 1, 3, 4, 6, 7, 9, and so on). Its partial sums are therefore bounded by the full geometric series:

A series of positive terms with bounded partial sums converges (its sum is 12/7). Both converge. Option (A).

Key idea: Half-wave symmetry removes the even harmonics.

Official answer: A

A shift by T/2 multiplies the nth coefficient by e^{-j\pi n} = (-1)^n. The condition x(t - T/2) = -x(t) then says, coefficient by coefficient,

For even n the bracket is 2, so a_{2m} = 0. This is half-wave symmetry: only odd harmonics survive. Option (A).

Key idea: Independence and E[N] = 0 remove the noise terms, leaving E[X^2] and E[X^3].

Official answer: A

Here correlation means E[XY] (the correlation coefficient of X and Y is 1/2, which isn't an option). Independence and E[N] = 0 remove the noise terms:

The second is zero because x^3 is odd and X is symmetric about 0. Option (A).

Key idea: Two angles add to 90^\circ when the product of their tangents is 1.

Official answer: B

At 159.15 kHz, \omega = 2\pi f = 10^6 rad/s, so both reactances are 1 Ω:

I_C leads V by \theta_1 and I_L lags it by \theta_2, with

Two angles add to 90° exactly when the product of their tangents is 1:

Option (B).

Key idea: Match the characteristic polynomial to the one the target poles give.

Official answer: B

The closed-loop characteristic equation is 1 + C(s)G(s) = 0:

The required poles give

Matching coefficients: 3 + K = 6, so K = 3; and 2 + 3\alpha = 14, so \alpha = 4. Option (B).

Key idea: The transfer function is C(sI - A)^{-1}B.

Official answer: C

Option (C).

Key idea: Last address = first address + size - 1, with 256 KB = 40000H.

Official answer: C

256 KB is 256 \times 1024 = 2^{18} byte locations, 40000H in hex, so

Option (C).

Key idea: Energy goes with the square of the amplitude and with the duration; shifting or reversing a signal leaves it unchanged.

Official answer: C

Scaling the amplitude by 3 multiplies the energy by 9. Compressing time by a factor of 3 divides it by 3. Shifting and reversing don't change it:

The ratio is 4/12 = 1/3. Option (C).

Key idea: QPSK has the same bit error rate as BPSK: \tfrac{1}{2}\mathrm{erfc}\sqrt{E_b/N_0}.

Official answer: B

For Gray-coded coherent QPSK:

Convert the ratio: E_b/N_0 = 10^{0.84} \approx 6.92, so u = \sqrt{6.92} \approx 2.63. With the given approximation,

The nearest option is 10^{-4}. Option (B).

Key idea: The 10's complement of a two-digit number N is 100 - N.

Official answer: B

For the two-digit number 47:

Equivalently, the 9's complement (52) plus 1. Option (B).

Key idea: Integrate |X(f)|^2 from -B to B and set it to 99% of the total energy.

Official answer: B

The total energy is \int_0^\infty e^{-4t}\,dt = \tfrac{1}{4}. The energy in -B < f < B is

As a fraction of the total, that is \tfrac{2}{\pi}\tan^{-1}(\pi B). Setting it to 0.99:

So B \approx 63.66/\pi \approx 20.3 Hz, between 63/\pi and 64/\pi. Option (B).

Key idea: Any interconnection of LTI blocks is LTI, but causality depends on the blocks.

Official answer: B

Reading the figure, the output is

Every block (the two subsystems, the delays and the adders) is linear and time invariant, and any interconnection of LTI blocks is LTI. In the z-domain the whole system is a single transfer function:

So (B) holds, and (C) and (D) are false. Causality (A) isn't guaranteed: nothing says h_1[n] or h_2[n] is zero for n < 0, and if h_1[n] isn't, neither is the overall impulse response. Option (B).

Key idea: Write the MUX output as a sum of select-line minterms times their inputs, then simplify.

Official answer: C

In the figure, I_2 is wired to the same line as I_0 (logic 1), and the 0 line crosses over it to I_1. The MUX output is

The last step uses \bar{z} + z\,(xy) = \bar{z} + xy. Option (C).

Key idea: The serial input is \bar{Q}, and Q stays 0 until a 1 reaches b_7.

Official answer: A

The serial input to b_0 is Q \oplus 1 = \bar{Q}. On each edge, the register shifts left taking in \bar{Q}, while the flip-flop takes in the old b_7.

b_7 stays 0 for the first eight edges, so Q stays 0 and a 1 enters every time. Option (A).

Key idea: The capacitors are shorts at both frequencies, so both tones see the same gain \approx -R_C/(r_e + 500).

Official answer: A

Bias. The divider gives a Thévenin source of 12 \times 25/125 = 2.4 V behind 100\,\|\,25 = 20 kΩ. With I_B = I_E/100 and 1.5 kΩ in the emitter:

Capacitors. At \omega = 10^5 rad/s, the 10 µF bypass is 1 Ω against 1 kΩ, and the 100 nF input capacitor is 100 Ω against an input resistance of about 14.5 kΩ. At 10^7 rad/s they are smaller still. Both act as shorts at both frequencies, and the output capacitor carries no current into an open output.

Gain. Only the 500 Ω stays in the emitter for signals:

The same gain applies to both tones, with no phase shift between them, so

The nearest is option (A). Option (C), -190.4, is the gain with the 500 Ω bypassed as well, which the circuit doesn't do.

Note: The exact gain is -9.40, not the option's -9.1. The 500 Ω makes the transistor's output resistance far larger than r_o = 100 V/1 mA, so r_o barely matters. The option's value comes from putting r_o straight across R_C: (5\text{ k}\,\|\,100\text{ k})/525 \approx 9.07. Either way, (A) is the nearest option.

Key idea: Find A and C with the output open, and B and D with it shorted.

Official answer: D

For a T network with series arms Z_1, Z_2 and shunt arm Z_3:

With all three at 1 kΩ: A = D = 2, B = 3 \times 10^3 Ω and C = 10^{-3} Ω⁻¹. As a check, AD - BC = 4 - 3 = 1, as it must be for a reciprocal network. Option (D).

Key idea: The opposite point has \Gamma_{in} = -\Gamma_L; then Z = 50\,\frac{1+\Gamma}{1-\Gamma}.

Official answer: B

Rotate \Gamma_L by 180° (a quarter wavelength of line):

Option (B). The load itself is Z_L \approx 97.7 + j65.1 Ω (close to option (D)), and the two satisfy Z_{in}Z_L = 50^2, as a quarter-wave line requires.

Key idea: The virtual ground fixes the feedback current, which fixes N's voltage; then apply KCL at N.

Official answer: C

The inverting input is a virtual ground. The input current 2\text{ V}/1\text{ k}\Omega = 2 mA flows on through the 1 kΩ feedback resistor, so the node at its far end (call it N) sits at

KCL at N: 2 mA arrives through the feedback resistor, and 1 mA arrives from ground through the 2 kΩ load (since N is at -2 V). Both leave through the 2 kΩ towards the op-amp output, in the direction of I_X:

The op-amp output sits at -2 - 3 \times 2 = -8 V. Option (C).

Key idea: An LED emits in a narrow band just above the gap, at \lambda \approx hc/E_g.

Official answer: A, B and D

The peak wavelength is set by the bandgap:

• 1410 nm (0.88 eV) and 1090 nm (1.14 eV) need photons below the gap. No transition gives them, so no emission: (A) and (B).
• 950 ± 20 nm contains 956 nm: the LED emits here.
• 510 nm (2.43 eV) is more than 1 eV above the gap. Carriers relax to the band edges before recombining, so the emission band (a few kT wide) doesn't reach it: (D).

Options (A), (B) and (D).

Key idea: Find the SNR in dB, then C = B\log_2(1 + \text{SNR}).

Official answer: C

Noise power in 1 MHz (60 dB-Hz):

So C > 3B is the only true statement. Option (C).

Key idea: Group the minterms on a K-map; minterm 14 can be covered two ways.

Official answer: A and D

Plotting the nine minterms gives these groups:

• 0, 2, 8, 10 (the four corners): \bar{x}\bar{z}.
• 5, 7, 13, 15 (x = 1, z = 1): xz.
• 14 is left (w = 1, x = 1, y = 1, z = 0). It pairs with 15 as wxy, or with 10 as wy\bar{z}.

These are options (A) and (D). Option (B) has no term covering minterms 0 and 2 (it gives 0 at w = 0, x = 0, z = 0), and option (C) has no term covering 5 and 7. Options (A) and (D).

Key idea: Any narrowband signal around a carrier can be written with a slowly varying envelope and phase.

Official answer: A, B and C

Check whether each signal fits the form A(t)\cos[2\pi f_c t + \theta(t)]:

• PSK: constant A(t), and \theta(t) stepping between the symbol phases. True, (A).
• AM: A(t) = A_c[1 + k_a m(t)] and \theta(t) = 0. True, (B).
• Band-limited Gaussian noise around f_c: its envelope (Rayleigh) and phase (uniform) are random processes, but the form is the same. True, (C).
• Narrowband FM: constant A(t) and \theta(t) = 2\pi k_f \int m(\tau)\,d\tau with a small peak deviation. So x(t) can represent it, and (D), which says it never can, is false.

Options (A), (B) and (C).

Key idea: x_2 is at twice x_1's frequency; orthogonality depends on the interval you integrate over.

Official answer: A, C and D

x_1 has frequency n and x_2 has frequency 2n. Multiply them with the product-to-sum identity

and the product is

• Over 0 \le t < 1/n, both sines run through whole cycles, so the inner product is 0: orthogonal. True, (A).
• Orthonormal also needs unit energy, but \int_0^{1/n}\cos^2(2\pi nt)\,dt = 1/(2n), which is never 1 for a positive integer n (and x_2's energy is 2/n). False, (B).
• x_2's frequency 2n is twice x_1's, so x_2 is the second harmonic of x_1. True, (C).
• Over 0 \le t < 1/(2n), \sin(6\pi nt) runs through 1.5 cycles and \sin(2\pi nt) through half a cycle, so neither integral cancels:

So they are not orthogonal over that interval. True, (D). Options (A), (C) and (D).

Note: Option (C) uses harmonic in its usual sense: a sinusoid at a whole-number multiple of a fundamental frequency. Here x_2, at 2n, is the second harmonic of the fundamental n of x_1, so the key includes (C).

Key idea: For s^3 + as^2 + bs + K, the Routh array hits the stability edge at K = ab.

Official answer: 1386

The characteristic equation is

For a cubic s^3 + a s^2 + b s + c, the Routh array's s^1 row is (ab - c)/a. It reaches zero, the edge of stability, when

At that gain the auxiliary equation 18s^2 + 1386 = 0 gives poles at s = \pm j\sqrt{77} \approx \pm j8.77 rad/s, on the imaginary axis. K = 1386.

Key idea: A diode-connected MOSFET is always in saturation; equate the resistor and drain currents.

Official answer: 2.50 to 2.60

Call the gate voltage V. Since V_{DS} = V_{GS} > V_{GS} - V_{TH}, the transistor is saturated, and the resistor current equals the drain current (mA, V):

The other root is negative, below threshold, so it's rejected. The drain current is 5 - 2.56 = 2.44 mA. V_G \approx 2.56 V.

Key idea: Use the square's polar moment about its centre, then shift the centre to (1, 0).

Official answer: 0.60 to 0.70

The square has diagonals of length 2, so its area is 2 and its side is \sqrt{2}. Its centre is (1, 0).

For a square of area A and side s, \iint (u^2 + v^2) about its centre is A s^2/6, whatever its orientation. Shifting the centre to (1, 0) adds A \times 1^2:

Key idea: Find the T's middle-node voltage from the virtual ground, then apply KCL there.

Official answer: 2.90 to 3.10

The inverting input is a virtual ground, so v_i/R_1 flows through R_2 to the T's middle node M:

KCL at M: v_i/R_1 arrives through R_2, and R_4 draws V_M/R_4 = -v_i/R_4 (it pushes current in). Both flow on through R_3:

In general the gain is -\tfrac{R_2 + R_3 + R_2R_3/R_4}{R_1}, here -3. The magnitude is 3.00.

Key idea: It fails only when two or more of the seven bits are wrong.

Official answer: 0.10 to 0.20

Subtract the chance of zero or one error in 7 bits from 1:

Key idea: J_n = qD_n\,dn/dx, with D_n = \mu_n V_T from Einstein's relation.

Official answer: -36.50 to -34.50, or 34.50 to 36.50

Einstein's relation gives the diffusion coefficient:

The gradient, over 3\ \mum = 3 \times 10^{-4} cm:

Electrons diffuse towards +x, so the conventional current flows towards -x. Its magnitude is about 35.47 A/mm².

Note: The official key accepts either sign: -36.50 to -34.50 or 34.50 to 36.50. The question asks for the current density without fixing a positive direction.

Key idea: On the positive peak D1 conducts and the two 10 kΩ resistors divide between the input and +2 V.

Official answer: 3.70 to 4.30

With no load, the output equals v_i until a diode turns on.

• D1 (anode towards the output, cathode on the +2 V battery) turns on when v_o > 2 V.
• D2 (cathode on the output, anode on the battery's negative side, at -4 V) turns on when v_o < -4 V, so it only clips the negative half.

On the positive peak D1 conducts, and the two 10 kΩ resistors divide between v_i and 2 V: