GATE ECE 2022 question paper with solutions

GATE ECE 2022 answer key

General Aptitude Q.1 to Q.10

1. Q.1MCQ · 1 markOfficial answer: C

   General Aptitude › Grammar & vocabulary

   Correlative conjunctions come in fixed pairs: either goes with or, and neither goes with nor. When the two blanks are split around two nouns joined in this way, check that both halves come from the same pair.

   WATCH OUT FOR

   Mixing the halves of two pairs, as in neither ... or (A) or either ... nor (B). Each one looks nearly right because one word in it is correct. Another mistake is picking (D) because also and but recall not only ... but also, a pair that needs not only in the first blank.

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2. Q.2MCQ · 1 markOfficial answer: D

   General Aptitude › Numerical

   In a ratio split, every share is (its ratio term) × (one common unit $k$). A given difference between two shares fixes $k$ straight away: $R-S=(4-3)k=k=1000$. You do not need the total sum at all.

   WATCH OUT FOR

   Treating the Rs. 1000 difference as if it were the whole sum and dividing it in the ratio (which gives $\frac{2}{14}\times 1000\approx 143$, not an option). Another slip is to stop at $k=1000$ and pick option (B), forgetting that Q's share is $2k$, not $k$.

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3. Q.3MCQ · 1 markOfficial answer: B

   General Aptitude › Geometry

   Take the rectangle between the parallel sides out of the trapezium. The two legs (3 and 4) and the difference of the parallel sides (11 - 6 = 5) make a 3-4-5 right triangle, and that triangle has the same height as the trapezium. Working out its area two ways gives $h=\frac{3\times 4}{5}=2.4\text{cm}$.

   WATCH OUT FOR

   The options include two numbers from the working. 5.76 is $h^2$, so it is what you get if you forget the square root. 1.80 is the end piece PM, not the height. Some students also try the whole length PQ = 11 as the base of the triangle. The base is only the difference 11 - 6 = 5.

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4. Q.4MCQ · 1 markOfficial answer: B

   General Aptitude › Spatial

   For each size, count the squares by where their top-left cell sits, then subtract the ones that cover the hole. A square of side $k$ with top-left cell $(r,c)$ covers the hole $(2,2)$ exactly when $r\le 2\le r+k-1$ and $c\le 2\le c+k-1$. A hole near a corner leaves no room for 3 × 3 squares, because every 3 × 3 square in a 4 × 4 grid covers the two middle rows and the two middle columns.

   WATCH OUT FOR

   Counting the hole itself as a $1\times 1$ square gives $16+5=21$, which is option (C). Another mistake is to allow bigger squares that have the hole on their edge rather than at their centre, or to subtract only the $2\times 2$ squares that contain the hole ($30-4=26$, option (D)). Stopping after the unit squares gives 15, option (A). Because the hole is a missing building block, no square that contains it can be formed at any size.

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5. Q.5MCQ · 1 markOfficial answer: C

   General Aptitude › Spatial

   A guard sees the whole room exactly when the segment to every point of the room stays inside it. This room has only one inward corner, the notch. Cut the room along the line from the notch to the bottom tip and you get two convex triangles. Any point on that shared line sees all of both triangles, and so sees the whole room. More generally, every point below both inner walls, extended as full lines, sees the whole room, and Q and S are well inside that region. A point up in one arm is blocked from the other arm by the empty V-shaped gap.

   WATCH OUT FOR

   Choosing (B) Q alone. Students assume S, near the bottom tip, is too far down or boxed in to see the tips of the arms, but its sight lines to both arms pass below the notch and stay inside the room. The opposite mistake is choosing (A) or (D) because P or R sits inside an arm and seems close to everything. The straight line from one arm to the other passes over the notch through the unshaded gap, so a guard in one arm cannot see the upper part of the other.

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6. Q.6MCQ · 2 marksOfficial answer: D

   General Aptitude › Reading

   A logical inference must keep the passage's level of certainty. \"May have\" stays \"may have\", and \"remains to be seen\" means unknown. Any option that ranks the two methods or turns an unknown into a fact goes beyond the passage.

   WATCH OUT FOR

   Picking (C) because both methods sound risky. The passage never says GM mosquitoes have undesired consequences, only that it is not yet known, and it says chemicals only \"may\" have them. Students also fall for (A) or (B) because they bring in their own opinion on genetic engineering, which the passage does not give.

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7. Q.7MCQ · 2 marksOfficial answer: C

   General Aptitude › Numerical

   Solve each linear inequality on its own, then intersect the two solution sets. "Satisfies both" means the overlap of the sets, not their union. Keep each end strict or non-strict exactly as the original inequality was.

   WATCH OUT FOR

   Picking (B) $-4<x\le 4$, whether from misreading $2x-1>7$ as $x>-4$ or from treating the endpoint 4 as included. At $x=4$, $2(4)-1=7$, which is not greater than 7, so 4 is excluded. Another slip is taking the union of the two sets, which is every real number, instead of their intersection.

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8. Q.8MCQ · 2 marksOfficial answer: C

   General Aptitude › Geometry

   Plot the points first. Two of them lie on a vertical line and two on a horizontal line, and the two segments cross, so they are the diagonals. When a quadrilateral's diagonals are perpendicular, its area is $\frac{1}{2}{d}_1{d}_2$. Here both diagonals are 4 and bisect each other, so the figure is a square of side $2\sqrt{2}$.

   WATCH OUT FOR

   Joining the vertices in letter order P, Q, R, S. Segments PQ and RS cross, so that path makes a self-intersecting bow-tie and not the quadrilateral. Taking PQ and RS as sides of length 4 gives 16, which is not an option. Another slip is to use $d_1{d}_2=16$ without the factor $\frac{1}{2}$. People who get the side $2\sqrt{2}$ right can still pick $4\sqrt{2}$ or $8\sqrt{2}$ by mistake, because those are the square's perimeter $8\sqrt{2}$ or twice the side.

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9. Q.9MCQ · 2 marksOfficial answer: B

   General Aptitude › Reasoning

   Start with the one statement you know is true: S never lies, so S's claim \"Only one of us is telling the truth\" holds. S is that one person, which means P, Q, R and T are all lying. Negate R's statement and you get P as the copier. Then check that each of the other lies stays false.

   WATCH OUT FOR

   Students often trust P's direct accusation (\"R has copied\") and pick (A). They forget that S's true statement forces everyone except S to be lying. Another slip is to read \"Only one of us\" as leaving S out. S says \"us\", so S counts, and S is the one truth-teller.

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10. Q.10MCQ · 2 marksOfficial answer: B

    General Aptitude › Spatial

    A $180^{\circ }$ rotation about a line lying in the square is a flip over that line. Two flips over perpendicular diagonals make a $180^{\circ }$ rotation about the centre. So XY equals ZZ, which makes XYZZ equal ZZZZ (a full turn, the identity).

    WATCH OUT FOR

    Treating X and Y as $180^{\circ }$ turns about the centre T instead of flips over the diagonals. Then XY looks like a $360^{\circ }$ turn and (D) seems right. Another slip is to think XY undoes itself and leaves the square unchanged, which gives (C). Track the four corner labels step by step: XY sends each corner to the opposite one.

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Electronics and Communication Engineering Q.11 to Q.65

1. Q.11MCQ · 1 markOfficial answer: A

   Engineering Mathematics › Vector calculus

   $\vec{F}=x\vec{ı}+y\vec{ȷ}=\nabla \left(\frac{x^2+y^2}{2}\right)$ is a conservative (gradient) field: its 2-D curl $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}$ is zero everywhere. The line integral of a gradient field depends only on where the path starts and ends, so around any closed contour it is $0$. The figure's geometry is a distraction.

   WATCH OUT FOR

   Picking (C), $8+2\pi$. That is the length of $C$ (two lines of length $4$ plus two semicircles of length $\pi$), which is ${\oint }_C|d\vec{r}|$, not the work integral ${\oint }_C\vec{F}\cdot d\vec{r}$. The other trap is working through the four pieces and getting a direction wrong on the top line or an arc, so the pieces fail to cancel. Check the curl first: once it is zero on the whole plane, the closed-path integral is $0$ with no parametrisation needed.

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2. Q.12MCQ · 1 markOfficial answer: C

   Engineering Mathematics › Linear algebra

   $Ax=b$ is consistent exactly when $\rho (A)=\rho ([A\mid b])$. When the coefficient rows are dependent (here $R_2=-R_1$), the right-hand sides must follow the same relation ($b_2=-b_1$). If they do, there are infinitely many solutions. If they don't, as here ($3\ne -1$), there are none.

   WATCH OUT FOR

   Seeing 3 unknowns and only 2 equations and answering (B) 'infinitely many solutions' straight away. Having fewer equations than unknowns rules out a unique solution, but it does not make the system consistent. You still have to compare $\rho (A)$ with $\rho ([A\mid b])$. Another mistake is to check only $\rho (A)$ and forget the augmented column.

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3. Q.13MCQ · 1 markOfficial answer: B

   Networks (Electric Circuits) › Laws & node/mesh

   A resistor in series with an ideal current source has no effect on the rest of the circuit, because the source fixes the branch current at 1 mA whatever the resistor's value. That leaves a single unknown node voltage. One KCL equation at node A gives $V_A=3.5\text{V}$, and $I=(5-V_A)/2\text{k}\Omega$. Superposition gives the same answer: 1.25 mA from the voltage source minus 0.5 mA pushed back by the current source.

   WATCH OUT FOR

   Using only the 5 V source's contribution (5 V across 4 kΩ = 1.25 mA, option A), or only the current source's share (−0.5 mA, option C), instead of adding the two. Another mistake is putting the right-hand 2 kΩ in parallel with the middle resistor. It is in series with the current source, so it carries exactly 1 mA and can be ignored when finding $V_A$.

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4. Q.14MCQ · 1 markOfficial answer: B

   Networks (Electric Circuits) › Laws & node/mesh

   Draw a closed surface around a subcircuit that is joined to the rest of the network by only one wire. KCL says the net current crossing any closed surface is zero, so the current in that single wire must be zero. A current needs a complete loop, and here there is none. The voltages on each side don't matter.

   WATCH OUT FOR

   Assuming the two bottom wires are both ground. Students then work out the open-circuit node voltages (V_A = 2 V, V_B = 1 V relative to each bottom wire) and write I = (2 − 1)/10 = 0.1 A, which is option (C), or −0.1 A, option (D), if they flip the direction. This is wrong for two reasons. The right bottom wire is floating, with no common return path. And even if a common return existed, the source resistances would load the circuit and give 1/(10 + 4/3) ≈ 0.088 A, not 0.1 A.

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5. Q.15MCQ · 1 markOfficial answer: A

   Signals and Systems › Fourier transform

   Write $x(t)$ as $-\frac{1}{2}\frac{d}{dt}\left(\frac{1}{1+t^2}\right)$. The transform of $\frac{1}{1+t^2}$ is $\pi e^{-|\omega |}$, by duality with $e^{-|t|}\leftrightarrow \frac{2}{1+{\omega }^2}$. Differentiating in time multiplies the transform by $j\omega$. A real, odd signal must have a purely imaginary, odd transform, and that check alone picks out (A).

   WATCH OUT FOR

   Getting a sign wrong when converting between $j$ and $\frac{1}{j}$. Since $\frac{1}{j}=-j$, the result $-\frac{j\pi }{2}\omega e^{-|\omega |}$ is exactly $\frac{\pi }{2j}\omega e^{-|\omega |}$. If you drop the minus sign from the derivative ($\frac{d}{dt}\frac{1}{1+t^2}=\frac{-2t}{(1+t^2{)}^2}$), you get the negative of (A) and may still pick (A) for the wrong reason. Other slips: leaving out the $2\pi$ in duality (which changes the constant), or dropping the $j$ and choosing the real option (B).

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6. Q.16MCQ · 1 markOfficial answer: A

   Electronic Devices › Carrier transport

   In a long p-type bar, electrons are minority carriers whose steady-state profile away from the injection point decays exponentially over the electron diffusion length $L_n$: $\Delta n(x)=\Delta n(0)e^{-x/L_n}$.

   WATCH OUT FOR

   Using the hole diffusion length (1 μm) instead of the electron diffusion length (2 μm) since holes are the majority carrier associated with "p-type," or forgetting to check that the equilibrium electron density $n_0=n_i^2/N_A$ is negligible compared to the injected value (option D, $10^3{\text{cm}}^{-3}$, is exactly that distractor).

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7. Q.17MCQ · 1 markOfficial answer: C

   Electronic Devices › Semiconductors

   In non-degenerate (Boltzmann) statistics, $n=N_C\exp \left(-\frac{E_C-E_{Fn}}{kT}\right)$, so $E_C-E_{Fn}=kT\ln (N_C/n)$. Since $N_C$ and $kT$ are unchanged between the two cases, going from $n_1$ to $n_2=n_1/2$ simply adds $kT\ln 2$ to $E_C-E_{Fn}$ — no need to solve for $N_C$ at all.

   WATCH OUT FOR

   Trying to use $n_i$ and the given thermal voltage of 26 meV to compute $E_C-E_{Fn}$ from an intrinsic-level formula (mixing up $E_i$ and $E_C$ relations), or forgetting that halving $n$ only shifts the Fermi level by $kT\ln 2$ relative to the first case — some solvers redo the full $N_C$ calculation and introduce rounding error instead of using the simple shift.

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8. Q.18MCQ · 1 markOfficial answer: C

   Digital Electronics › Logic families

   With gate tied to drain, a MOSFET that conducts is always in saturation. Here both are, so equate the square-law currents ${\mu }_n(W/L{)}_n(V_O-V_{Tn}{)}^2={\mu }_p(W/L{)}_p(V_{DD}-V_O-|V_{Tp}|{)}^2$. With symmetric thresholds, the node sits at $V_{DD}/2$ only when ${\mu }_n(W/L{)}_n={\mu }_p(W/L{)}_p$. Otherwise the stronger device pulls it towards its own rail. Here $300>200$, so the nMOS wins and $V_O<2\text{V}$.

   WATCH OUT FOR

   Seeing $(W/L{)}_p=5$ against $(W/L{)}_n=1$ and concluding the pMOS is stronger, which gives (B). Device strength is $\mu \cdot W/L$, not $W/L$ alone. The electron mobility is 7.5 times the hole mobility, so $k_n:k_p=300:200$. Choosing (D) also comes from ignoring the mobilities, or from assuming a CMOS stack always settles at $V_{DD}/2$. Some students treat the circuit as an inverter with an unknown input and pick (A). The figure shorts gate to drain, so there is no free input.

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9. Q.19MCQ · 1 markOfficial answer: C

   Digital Electronics › Combinational

   When the function's variables drive a MUX's select lines, the data inputs are simply the function's truth-table column: $A_i=f(m_i)$, with $m_i$ indexed by $S_1{S}_0$. For XOR that column is $0,1,1,0$. XOR is symmetric in C and D, so the answer does not change even if you read $S_0$ as the MSB, because swapping the select lines only swaps $A_1$ and $A_2$, and both are 1.

   WATCH OUT FOR

   Picking (A) (0, 0, 1, 1), which is just $C$ (the MSB select line), or (B) (1, 0, 1, 0), which is $\overline{D}$. This happens when the truth table is written against one variable instead of against the combined index $(S_1{S}_0{)}_2$. Picking (D) means confusing XOR with its complement or with $\overline{C}$.

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10. Q.20MCQ · 1 markOfficial answer: C

    Electronic Devices › MOS devices

    For a MOSFET whose substrate is tied to its own source, the terminal acting as "source" is the one with the lower potential for an nMOS (higher potential for a pMOS), and it is defined by the instantaneous circuit state, not by the schematic labels. Conduction (and hence capacitor charging) continues only as long as $|V_{GS}|>|V_t|$ measured with the source taken this way; steady state is reached when either that inequality just fails (channel pinches off) or the drain-source voltage itself goes to zero (no further current can flow into the capacitor).

    WATCH OUT FOR

    Assuming the PMOS turns off once V2 reaches some value related to its own drain voltage (e.g. stopping at 4 V by analogy with the NMOS), instead of recognizing that V_SG = 10 V is fixed by the two constant supplies and always far exceeds |V_tp| = 1 V — so the PMOS keeps conducting until the capacitor is fully charged to the source rail (V2 = 5 V), not until some gate-threshold balance is reached.

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11. Q.21MCQ · 1 markOfficial answer: C

    Control Systems › Nyquist

    When the plot shown is of $G(s)$ but the loop gain is $KG(s)$, shift the critical point rather than rescaling the plot. Count clockwise encirclements of $-1/K$ (here $-0.5$). Then take the direction of every loop from its own arrows and apply $Z=P+N$ for a clockwise contour. Only loops that actually contain the critical point contribute to $N$.

    WATCH OUT FOR

    Counting encirclements of $-1$ instead of $-1/K=-0.5$. Nothing in the figure encloses $-1$, so this gives $N=0$ and the wrong answer (A). A second error is treating the figure-eight as one loop, or letting its two lobes cancel. The counter-clockwise right lobe does not contain $-0.5$, so it cancels nothing, and the clockwise left lobe adds one encirclement to the outer circle's one.

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12. Q.22MCQ · 1 markOfficial answer: A

    Control Systems › Root locus

    If every open-loop pole is stacked at one point and there are no zeros, $1+\frac{K}{(s+a{)}^n}=0$ becomes $(s+a{)}^n=-K$. The whole locus is then $n$ straight rays from $s=-a$ at angles $\frac{(2k+1)180^{\circ }}{n}$. Count the rays to get $n$, and read the common point to get $a$.

    WATCH OUT FOR

    Picking (B). The poles of $s^5+1$ lie at exactly the angles $\pm 36^{\circ },\pm 108^{\circ },180^{\circ }$, the same angles as the rays in the figure, which makes it look right. But those are five separate starting points on the unit circle around the origin, not five directions leaving $s=-1$. The branches of (B) would start at five different places, while every branch in the figure starts at $-1$. A second slip is to overlook that (C) and (D) have six poles, which means six branches, when the figure shows five.

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13. Q.23MCQ · 1 markOfficial answer: C

    Signals and Systems › Fourier transform

    For a WSS input, the output power spectrum depends only on $|H(f){|}^2$, so a flat unit magnitude over $|f|\le \alpha$ passes $S_X(f)$ through unchanged. Keeping the waveform intact (a pure delay) needs more than that: the phase must also be linear through the origin, $\angle H(f)=-2\pi f{t}_0$, and a magnitude plot says nothing about phase.

    WATCH OUT FOR

    Reading a flat unit-magnitude band as "distortionless" and picking (A). This forgets that a pure delay also needs linear phase through the origin: $H(f)=-1$ or a Hilbert-type phase has the same magnitude and is not a delay. A second slip is writing $S_Y=H(f)S_X(f)$ instead of $|H(f){|}^2{S}_X(f)$ and then worrying that the unknown phase makes Statement II false.

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14. Q.24MCQ · 1 markOfficial answer: C

    Electromagnetics › Maxwell's equations

    Maxwell added displacement current so that total current is continuous: $\nabla \cdot ({\mathbf{J}}_c+\partial \mathbf{D}/\partial t)=0$. In a series loop the conduction current in the wire and resistor carries on across the capacitor gap as displacement current $d{\Psi }_D/dt=dq/dt$. The two are equal at every instant, in both amplitude and phase.

    WATCH OUT FOR

    Picking (D) because "no charge flows through the dielectric". That is true of conduction current, but it is exactly why the gap current is all displacement current, and that current is not zero. Another mistake is picking (A) or (B) by carrying over the 90° voltage–current phase shift of a capacitor. That shift is between the capacitor's voltage and its current. The resistor current and the capacitor current in a series loop are the same current and have no phase difference.

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15. Q.25MSQ · 1 markOfficial answer: A, B

    Engineering Mathematics › Differential equations

    With an exponential trial solution $e^{\xi x+\eta y}$, each second partial derivative becomes a multiplication: by ${\xi }^2$ for $x$ and by ${\eta }^2$ for $y$. The PDE then reduces to one algebraic condition, $a{\xi }^2+b{\eta }^2=1$, and each option is checked against it.

    WATCH OUT FOR

    One mistake is to pick (D) by requiring each term to match $f$ on its own ($a{\xi }^2=1$ and $b{\eta }^2=1$). Only the sum has to equal $1$, and (D) makes it $2$. Another is to pick (C) because a zero exponent looks like the "trivial" solution. But $e^0=1$, not the zero function, and the constant $1$ does not satisfy $0=1$.

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16. Q.26MSQ · 1 markOfficial answer: B, C

    Analog Electronics › Filters & waveforms

    Because of the virtual ground at the inverting input, the series $R_1$–$C$ branch sees $V_{in}$ directly, so the gain is $-Z_f/Z_1$. That gives a single pole at $1/(R_{1}C)$ and a zero at the origin, which is a high-pass response. The feedback resistor only scales the gain: it sets the passband level $R_f/R_1$ and does not move the corner.

    WATCH OUT FOR

    Option (D) comes from putting the total resistance $R_1+R_f=3\text{k}\Omega$ into the time constant, which gives $1/(3\times 10^{-3})=1000/3\text{rad/s}$. The virtual ground isolates $R_f$ from the input branch, so only $R_1$ and $C$ set the corner. Another mistake is to see a capacitor and call the circuit a low-pass filter. A series input capacitor blocks low frequencies. Low-pass behaviour would need the capacitor in parallel with $R_f$.

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17. Q.27MSQ · 1 markOfficial answer: B, C

    Digital Electronics › Boolean logic

    Reduce each option with absorption ($x+xy=x$) and the OR-over-AND distributive law ($x+yz=(x+y)(x+z)$). To prove an option is NOT equivalent, give one input where it differs from $x+yz$.

    WATCH OUT FOR

    Rejecting (B) because it looks like a product of sums rather than a sum of products. In Boolean algebra, unlike ordinary algebra, OR distributes over AND, so $(x+y)(x+z)=x+yz$. A second slip: in (C), dropping the $yz$ term along with $xy$ when applying absorption. Only $xy$ is absorbed by $x$.

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18. Q.28MSQ · 1 markOfficial answer: C

    Digital Electronics › Logic families

    A CMOS inverter draws ideally zero static current, but it still uses dynamic power $\alpha C_L{V}_{DD}^{2}f$ whenever it switches. The ON transistor passes no current in steady state, so its $V_{DS}$ is zero and it sits in the linear (triode) region, not saturation. Noise margins and switching speed both depend on ${\mu }_n(W/L{)}_n$ versus ${\mu }_p(W/L{)}_p$. That is why sizing matters, and why electron mobility affects the pull-down delay.

    WATCH OUT FOR

    Students often mark (B) because they mix up static power (ideally zero) with dynamic switching power (never zero). Another trap is to think the ON nMOS is in saturation because $V_{GS}$ is large. Saturation needs $V_{DS}\ge V_{GS}-V_{Tn}$, and here $V_{DS}=0$. Some also mark (A) because they remember the symmetric-inverter case, but the word "always" together with "irrespective of sizing" makes it false.

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19. Q.29MSQ · 1 markOfficial answer: A, B, D
20. Q.30MSQ · 1 markOfficial answer: A, C

    Engineering Mathematics › Differential equations

    Every solution of $f_{tt}=c^2{f}_{xx}$ has the form $g(x-ct)+h(x+ct)$, with $c=100$ here. To test a term, write its argument as a multiple of $x+vt$ and check that $v=\pm 100$. If a term fails, work out the residual of the whole sum: the bad terms' residuals could in principle cancel, so check that they do not.

    WATCH OUT FOR

    In (D), taking the 100 that multiplies $x$ as the wave speed. The argument is $-100x+t=-100(x-t/100)$, so the speed is $1/100$. The mistake runs the other way in (B): reading $x+1000t$ as close enough to $x+100t$ when its speed is $1000$. A third error is to reject a sum only because each of its terms fails. Terms that fail can still cancel, so always compute the residual of the whole expression.

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21. Q.31NAT · 1 markOfficial answer: 4.0

    Engineering Mathematics › Probability & statistics

    For a discrete frequency distribution, add up the frequencies and find the value where the running total first reaches the middle position(s), $N/2$ and $N/2+1$. Here $N=100$, and the running total goes from 45 (up to 3 wickets) to 65 (up to 4 wickets), so the 50th and 51st values are both 4.

    WATCH OUT FOR

    One mistake is using the grouped-data interpolation formula $L+\frac{N/2-CF}{f}h$ with class $3.5$ to $4.5$, which gives $3.5+\frac{50-45}{20}=3.75\approx 3.8$. That formula is for continuous class intervals, not for whole-number counts like wickets. Other slips are answering with the mode (3, the tallest bar), the middle axis value (5), or the median of the printed frequencies themselves.

    Full solution →
22. Q.32NAT · 1 markOfficial answer: 0.50

    Engineering Mathematics › Complex analysis

    Only the poles inside the contour count. The figure puts $z=-1$ inside $C$ and $z=1$ outside, so the integral is $2\pi i$ times the single residue ${\mathrm{Res}}_{z=-1}=\frac{2^{-1}}{-1-1}=-\frac{1}{4}$. That gives $-\frac{i\pi }{2}$, so $A=\frac{1}{2}$.

    WATCH OUT FOR

    Counting both poles. The residue at $z=1$ is $\frac{2^1}{2}=1$, and adding it gives $2\pi i\cdot \frac{3}{4}=\frac{3\pi i}{2}$, so $A=-1.50$. But the contour crosses the real axis at about $0.6$, to the left of $(1,0)$, so $z=1$ is outside. Two other slips give $A=-0.50$: reading the orientation as clockwise, or dropping the sign of $z-1=-2$ at $z=-1$.

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23. Q.33NAT · 1 markOfficial answer: 0.0

    Signals and Systems › LTI & convolution

    $x_2(t)$ is a finite pulse that ends at $t=2$, not a step. After $t=2$ the output is just the decaying tail $e^{-t}(e^2-1)$ of a stable system, so it tends to zero. The final value theorem confirms this: $sY(s)=\frac{1-e^{-2s}}{s+1}$ is $0$ at $s=0$.

    WATCH OUT FOR

    Treating $x_2(t)$ as the step $u(t)$, which gives a limit of 1. Other errors: reporting the peak $y(2)=1-e^{-2}\approx 0.9$, or multiplying the areas ($1\times 2=2$). That product is $\int y(t)dt$, not the value of $y(t)$ as $t\to \infty$.

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24. Q.34NAT · 1 markOfficial answer: 5.5 to 5.7

    Electronic Devices › MOS devices

    In an ideal MOS capacitor, gate voltage divides among flat-band voltage (0, ideal case), surface potential, and the voltage dropped across the oxide due to all semiconductor-side charge: $V_G=V_{FB}+{\varphi }_s+\frac{Q_{dep}+Q_{IN}}{C_{OX}}$. Once strong inversion is reached, ${\varphi }_s$ and the depletion charge $Q_{dep}$ saturate (pinned near $2{\varphi }_F$ and its corresponding maximum value) and stop changing with further increases in $V_G$. So any additional gate voltage beyond the point of strong inversion goes entirely into building up more inversion charge: $\Delta V_G=\frac{\Delta Q_{IN}}{C_{OX}}$.

    WATCH OUT FOR

    Treating the MOS capacitor as a simple linear capacitor and assuming $Q_{IN}$ is proportional to the full gate voltage $V_G$ (i.e., computing $Q_{IN}(4\text{V})=2\times Q_{IN}(2\text{V})=4.4\mu \text{C}/{\text{cm}}^2$) — this ignores that ${\varphi }_s$ and $Q_{dep}$ have already saturated by the time strong inversion is reached, so only the incremental voltage $\Delta V_G$ (not the total $V_G$) maps to incremental inversion charge.

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25. Q.35NAT · 1 markOfficial answer: 2.99 to 3.01
26. Q.36MCQ · 2 marksOfficial answer: B

    Engineering Mathematics › Calculus

    On a closed interval, the absolute minimum is at a critical point or at an endpoint. Here the only critical point, $x=2$, is a maximum, because $f^{\prime \prime }<0$ everywhere. A concave function takes its minimum at an endpoint, so you only have to compare $f(1)=2$ with $f(e)=11-e^2\approx 3.61$.

    WATCH OUT FOR

    Setting $f^{\prime }(x)=0$, getting $x=2$ and picking option (A) without checking which kind of extremum it is. $f^{\prime \prime }(2)=-4<0$, so $x=2$ is where $f$ is largest on the interval. The other trap is picking $x=e$ because it is the right-hand endpoint, without actually comparing $f(1)$ and $f(e)$.

    Full solution →
27. Q.37MCQ · 2 marksOfficial answer: A

    Engineering Mathematics › Linear algebra

    $v_1$ and $v_2$ are orthonormal, so $A=\alpha v_1{v}_1^T+\beta v_2{v}_2^T$ is a spectral decomposition. The matrix $v_i{v}_i^T$ projects onto $v_i$, so $A{v}_1=\alpha v_1$ and $A{v}_2=\beta v_2$. Any vector orthogonal to both is sent to $0$, which gives the third eigenvalue $0$. Checks: trace $=\alpha +\beta$, and rank $\le 2$.

    WATCH OUT FOR

    Adding or combining $\alpha$ and $\beta$ (options B, C, D) as if the two rank-1 terms mix. Because $v_1\perp v_2$, each term acts only on its own direction, so the eigenvalues stay separate as $\alpha$ and $\beta$. Some students also forget the third eigenvalue. A $3\times 3$ matrix of rank at most 2 always has $0$ as an eigenvalue.

    Full solution →
28. Q.38MCQ · 2 marksOfficial answer: A

    Networks (Electric Circuits) › AC, power & resonance

    Read the two ends of the locus. At $\omega =0$ the capacitor is an open circuit, so $Z=R_1+R_2$. As $\omega \to \infty$ it shorts out $R_2$, so $Z=R_1$. The locus of $R_2\parallel C$ is a semicircle of diameter $R_2$, so the arc's height, $R_2/2$, gives an independent check.

    WATCH OUT FOR

    Reading the ends of the locus the wrong way round, so that $R_1=5\text{k}\Omega$ and $R_2=2\text{k}\Omega$ (option B). Another mistake is taking the arc's height of $1.5\text{k}\Omega$ as the radius of the whole locus measured from the origin. A third is treating the right-hand end (5) as $R_2$ on its own rather than $R_1+R_2$, which gives option D. The capacitor removes $R_2$ at high frequency, so the high-frequency end (2) is $R_1$ alone.

    Full solution →
29. Q.39MCQ · 2 marksOfficial answer: B

    Networks (Electric Circuits) › AC, power & resonance

    In a series circuit driven by a sinusoid, the source phasor divided by the current phasor is the total impedance. The two readings come straight off the plot. The ratio of peak voltage to peak current gives |Z_total|. The labelled time of the current's rising zero crossing, multiplied by ω, gives the angle. A rising zero crossing at a positive time means the current lags, so the circuit is inductive. Subtract the known resistor, and what is left is jωL (inductor) or −j/(ωC) (capacitor).

    WATCH OUT FOR

    Reading the rising zero crossing at t = +π/4 as a lead. The current reaches that point of its cycle π/4 s later than the source does, so it lags. Calling it a lead gives the 1 F capacitor (A). A second trap is mixing peak and rms values. 1/√2 A on the plot is the current's PEAK. Dividing the source's rms value (1/√2 V) by that peak gives |Z_total| = 1 Ω, which leaves nothing for Z and matches no option. A third slip is forgetting that the reactance is ωL. Here ω = 1 rad/s, so X = 1 Ω means L = 1 H, but that only works because ω happens to be 1.

    Full solution →
30. Q.40MCQ · 2 marksOfficial answer: C

    Electronic Devices › MOS devices

    A time-varying gate voltage makes the overdrive $V_{ov}(t)$ swing over a range; with $V_{DS}$ fixed, the device moves between saturation and triode as $V_{ov}$ crosses $V_{DS}$, and since $I_D$ keeps rising with $V_{ov}$ in triode too, the true maximum must be checked at the extreme of $V_{ov}$, not assumed to sit in saturation.

    WATCH OUT FOR

    Assuming the device stays in saturation for the whole cycle and blindly maximizing $I_D=\frac{1}{2}{\mu }_n{C}_{ox}(W/L)V_{ov}^2$ at $V_{ov,max}=2\text{V}$, which gives $20\text{mA}$ (option B) — this ignores that for $V_{ov}>V_{DS}=1\text{V}$ the transistor has already left saturation and entered the triode region, where the saturation formula no longer applies.

    Full solution →
31. Q.41MCQ · 2 marksOfficial answer: C

    Analog Electronics › Filters & waveforms

    This is an RC relaxation oscillator: the capacitor on the inverting input charges towards whichever rail the output is at, and the output flips when $V_C$ crosses $v_{+}$. So $V_C$ swings between the two thresholds, not between the rails. The diodes make the thresholds asymmetric. At $+15\text{V}$ output, $D_1$ shorts the $A$–$P$ resistor, so $v_{+}$ is $\frac{1}{3}$ of the output. At $-12\text{V}$ output, $D_2$ shorts the $B$–$V_o$ resistor, so $v_{+}$ is $\frac{2}{3}$ of the output. The thresholds are therefore $+5\text{V}$ and $-8\text{V}$.

    WATCH OUT FOR

    (1) Taking the rails as the capacitor's extremes gives $15-(-12)=27\text{V}$, option (B). The capacitor stops at the thresholds, because the output flips before $V_C$ gets anywhere near a rail. (2) Swapping the diode states, for example by misreading which end is the anode, puts $D_2$ ON at $+15\text{V}$ (giving $+10\text{V}$) and $D_1$ ON at $-12\text{V}$ (giving $-4\text{V}$). That yields $14\text{V}$, option (D). Always check each diode's state against the direction the ladder current actually flows. (3) Ignoring the diodes gives a $4R$ ladder with $2R$ below $P$: thresholds of $+7.5\text{V}$ and $-6\text{V}$, and a swing of $13.5\text{V}$, which is not an option.

    Full solution →
32. Q.42MCQ · 2 marksOfficial answer: A

    Analog Electronics › Filters & waveforms

    The ideal op-amp draws no current, so the input RC is an unloaded first-order low-pass with ${\omega }_c=1/((1\text{k}\Omega )C)$. The non-inverting stage multiplies it by $1+R/1\text{k}\Omega$ and does not move the pole. Read the flat level for $R$ (12 dB is a gain of 4, so $R=3\text{k}\Omega$). Read the point 3 dB below that level for $C$ (9 dB at $\omega =10^3\text{rad/s}$, so $C=1\mu \text{F}$).

    WATCH OUT FOR

    One slip is using the inverting-gain formula $R/1\text{k}\Omega =4$, which gives $R=4\text{k}\Omega$ and lands on option (C). Another is putting the feedback $R$ into the time constant; only the series 1 kΩ and $C$ set ${\omega }_c$. A third is taking the corner from the wrong point: where the curve first visibly bends (about ${\log }_{10}\omega =2.3$), or where it reaches 3 dB absolute (about ${\log }_{10}\omega =3.4$). The corner is the point 3 dB below the flat level, which is 9 dB here.

    Full solution →
33. Q.43MCQ · 2 marksOfficial answer: A

    Digital Electronics › Sequential

    With $J=K$ tied to the counter LSB, the JK flip-flop works as a T flip-flop fed by an $f_0/2$ square wave. It toggles on every second rising edge, which divides the clock frequency by 4. Every transition lands on a rising edge, and rising edges are always one period apart, so the output duty cycle is 50% whatever the input's duty cycle is.

    WATCH OUT FOR

    Three traps. (1) Assuming the clock's 25% duty cycle carries through to the output, which gives option (B). Edge-triggered outputs change only on rising edges, so the clock's high time drops out. (2) Assuming the flip-flop toggles on every clock edge, which gives $f_0/2$, option (C). It toggles only on edges where the LSB is 1. (3) Using the LSB's value just after an edge instead of just before it. On a shared clock the flip-flop sees the pre-edge value.

    Full solution →
34. Q.44MCQ · 2 marksOfficial answer: A

    Control Systems › Routh–Hurwitz

    The roots of an even polynomial are symmetric about the origin, so 'all roots on the $j\omega$ axis' means the same as 'no RHP roots'. In the Routh array the $s^3$ row vanishes and is rebuilt from $dA/ds$. Substituting $x=s^2$ turns the question into one about a quadratic: it needs real roots that are both $\le 0$. That gives $K\le 9/4$ from the discriminant and $K\ge -4$ from the product.

    WATCH OUT FOR

    Checking only that the coefficients are positive ($4+K\ge 0$) and missing the discriminant condition $9-4K\ge 0$. Without it the $s^2$-roots become complex and the roots of $p(s)$ leave the axis in mirror pairs. A second common error is to drop the endpoints. At $K=-4$ there is a double root at $s=0$, which is still on the imaginary axis. At $K=9/4$ the roots are the repeated pair $\pm j\sqrt{5/2}$. A third is to give up when the $s^3$ row comes out all zero, instead of replacing it with the derivative of the auxiliary polynomial.

    Full solution →
35. Q.45MSQ · 2 marksOfficial answer: B, D

    Engineering Mathematics › Calculus

    For $c>0$, the ratio test gives $\lim a_{n+1}/a_n=1/c$ whatever $d$ is. The exponential decides: the series converges for $c>1$ and diverges for $0<c<1$. Only at $c=1$ does $d$ matter, where the series is the p-series $\sum 1/n^{-d}$, which converges only when $d<-1$.

    WATCH OUT FOR

    Picking (A) because $1/n\to 0$. Terms going to zero is necessary but not sufficient, and the harmonic series ($p=1$) diverges. The other trap is picking (C) because the power $n^{-10}$ looks strongly decaying. With $c=0.5$ the factor $1/c^n=2^n$ grows exponentially and swamps any power of $n$, so the terms blow up.

    Full solution →
36. Q.46MSQ · 2 marksOfficial answer: C, D

    Signals and Systems › LTI & convolution

    A sinusoid is an eigenfunction of every LTI system, so an LTI output for $\sin (t)$ can only be $\sin (t)$ scaled and phase-shifted: the same frequency, no new harmonics and no DC. A single input-output pair can prove a system is not LTI, but it can never prove that a system is LTI.

    WATCH OUT FOR

    Marking $S_1$ because $\sin (-t)$ looks like time reversal. In fact $\sin (-t)=-\sin (t)$, which the LTI system $y=-x$ produces. Marking $S_2$ because $x(t+1)$ is non-causal is the same kind of mistake: causality is not part of being LTI. Another error is reading the question as asking which systems ARE LTI, when one input can only show whether an LTI system is possible.

    Full solution →
37. Q.47MSQ · 2 marksOfficial answer: A, C

    Electronic Devices › Carrier transport

    Apply the four separate physical facts: intrinsic carrier symmetry, standard BJT doping hierarchy (emitter heaviest, collector lightest), KCL/continuity in steady state, and phonon-scattering-limited mobility above ~300 K in Si.

    WATCH OUT FOR

    Assuming the collector must be more heavily doped than the base because it is physically larger, or assuming mobility keeps rising with temperature by analogy with intrinsic carrier concentration (which does rise with T) — mobility and carrier concentration have opposite temperature dependences above ~300 K in Si.

    Full solution →
38. Q.48MSQ · 2 marksOfficial answer: A, C

    Engineering Mathematics › Probability & statistics

    In a Markov chain's state diagram, the probabilities on the arrows leaving a state (its self-loop included) add up to 1. Only statements that follow from these per-state equations are universally true. To rule out any other statement, one valid set of numbers that breaks it is enough.

    WATCH OUT FOR

    Summing the arrows coming INTO a state instead of the ones leaving it. That leads to equations like $p_1+p_2=1$ for B, and it makes (B) or (D) look plausible. A second trap is marking (D) because it collects "all the self-loops". The self-loops belong to different states, so nothing forces them to add up to 1.

    Full solution →
39. Q.49MSQ · 2 marksOfficial answer: A, C

    Digital Electronics › Boolean logic

    $D(A,B)=A+\overline{B}$ is the implication gate $B\to A$. On its own it cannot make NOT, because it outputs 1 whenever both inputs are 1. The constant '0' breaks that: $D(0,B)=\overline{B}$. Once you have NOT, $D(A,\overline{B})=A+B$ gives OR, and {OR, NOT} is functionally complete, so every gate can be built.

    WATCH OUT FOR

    Stopping at "D is not a universal gate". Implication with no constants is indeed not complete, and that leads students to tick (B) or (D). The question explicitly supplies '0', which turns D into an inverter and makes the set complete. A second trap is reading (B) and (D) as true statements: both say "cannot", and both are false because OR and AND can be built.

    Full solution →
40. Q.50MSQ · 2 marksOfficial answer: A

    Control Systems › Time response

    For a standard second-order system, percentage overshoot depends only on $\zeta$. The time scale (${\omega }_d$, $t_s$, $t_p$) is set by ${\omega }_n$, and the steady-state value is set by the DC gain $G(0)$. Two systems with the same $\zeta$ but different ${\omega }_n$ and gain therefore share only the percentage overshoot.

    WATCH OUT FOR

    Seeing the same numerator (10) in both transfer functions and assuming the steady-state values match, so marking (B). The steady-state value is $G(0)$, numerator divided by the constant term of the denominator: $10/1=10$ against $10/10=1$. A second slip is thinking a different gain changes the percentage overshoot. It does not, because overshoot is measured relative to the final value.

    Full solution →
41. Q.51MSQ · 2 marksOfficial answer: A
42. Q.52MSQ · 2 marksOfficial answer: Marks to all

    Electromagnetics › Waveguides & fibres

    Between parallel plates, the TM${}_n$ cut-off is $f_{c,n}=nc/(2d)$. The $n=0$ mode has $E_z=0$, which makes it TEM, so the lowest true TM mode is TM${}_1$ with $f_c=c/(2d)$. A TM wave propagates only at frequencies above that cut-off.

    WATCH OUT FOR

    Leaving the separation in the wrong unit. Taking $d=10^{-4}\text{m}$ instead of $10^{-6}\text{m}$ gives $f_c=1.5\times 10^{12}\text{Hz}$, which wrongly adds (D). Two other errors: using $f_c=c/d$ instead of $c/(2d)$, or counting the TEM ($n=0$) mode as a TM wave and ticking all four options.

    Full solution →
43. Q.53NAT · 2 marksOfficial answer: 512

    Engineering Mathematics › Calculus

    Read the region off the figure and slice it the easy way. With vertical strips, the triangle bounded by $y=x$, $y=-x$ and $x=4$ becomes $0\le x\le 4$, $-x\le y\le x$, and one iterated integral covers it. Horizontal strips would split it at $y=0$ into two pieces. The integrand is even in $y$ and the region is symmetric about the x-axis, so the inner integral is twice the integral from $0$ to $x$.

    WATCH OUT FOR

    Integrating over only the upper half, $0\le y\le x$, gives 256. So does integrating $x^2+y^2$ over the whole triangle and forgetting the factor 3 in front (that also gives 256, and 256/3 if only one of the two errors is made). Another slip is using ${\int }_0^x{y}^{2}dy=x^2$ in place of $x^3/3$. Some students also try polar coordinates, where $x=4$ becomes $r=4\sec \theta$ for $-\pi /4\le \theta \le \pi /4$. That leads to a harder ${\sec }^4\theta$ integral with more chances to go wrong.

    Full solution →
44. Q.54NAT · 2 marksOfficial answer: 3.9 to 4.1

    Networks (Electric Circuits) › Theorems

    With port 1 held at a fixed voltage, a linear 2-port looks like a plain Thevenin source at port 2. The $V_2$ versus $-I_2$ line gives both parameters: its intercept is $V_{th}=5\text{ V}$ and the magnitude of its slope is $R_{th}=250\Omega$. Once the Thevenin source is joined to the new load (1 kΩ in series with 10 V), the circuit is a single KVL loop: $I_2=(10-5)/(1000+250)$.

    WATCH OUT FOR

    The usual mistake is getting the direction of $I_2$ wrong. $I_2$ is drawn into the network at A, so the given point $I_2=-4\text{ mA}$ means current flows out of the network into $R$. If you take the port relation as $V_2=5-250I_2$, or add the load battery the wrong way round (treating the sources as series-aiding, for example $(10+5)/1250=12\text{ mA}$), you get the wrong sign or the wrong magnitude. A second mistake is to write $R_{th}=4\text{ V}/4\text{ mA}=1\text{ k}\Omega$ from the second point alone. That is the load resistance at that point, not the Thevenin resistance. $R_{th}$ is the slope, $(5-4)/4\text{ mA}=250\Omega$.

    Full solution →
45. Q.55NAT · 2 marksOfficial answer: 7.9 to 8.1

    Signals and Systems › DTFT & DFT

    Taking the $N$-point DFT twice gives $Nx[(-n{)}_N]$, the input scaled by $N$ and circularly reversed. Index 0 does not move under that reversal, so $y[0]=Nx[0]$. Also, the DFT at index 0 is simply the sum of the input samples.

    WATCH OUT FOR

    Two errors are common. The first is to assume that $DFT(DFT(x))$ gives back $x$, which leads to the answer 1. That confuses the forward DFT with the inverse DFT, which has the opposite sign in the exponent and a $1/N$ factor. The second is to forget the factor $N$ and use the time-reversal alone. Some students also add $x[4]$ in as well, getting 3 or 24, when $x[4]$ actually lands at $y[4]=16$ and not at $y[0]$.

    Full solution →
46. Q.56NAT · 2 marksOfficial answer: 0.57 to 0.61

    Electronic Devices › Carrier transport

    In steady state with no field, the continuity equation says $d{J}_n/dx=q(R-G)$. Because the profiles are symmetric, $J_n(0)=0$. So the current leaving through $x=2l_n$ is $q$ times the carriers generated but not recombined between $0$ and $2l_n$, per unit area per second. You get it by integrating $G-R$; you never need to solve the diffusion equation for $n^{\prime }(x)$.

    WATCH OUT FOR

    The usual error is assuming $R=10^{20}{e}^{-|x|/l_n}$ continues to $x=\pm 2l_n$. The figure shows $R$ stepping down to zero at $x=\pm l_n$. Integrating $R$ out to $2l_n$ gives $e^{-2}$ in place of $e^{-1}$ and the wrong answer $0.22{\text{mA/cm}}^2$. A second trap is using $q=-1.6\times 10^{-19}$ C with its sign and reporting a negative value: the question asks for the magnitude.

    Full solution →
47. Q.57NAT · 2 marksOfficial answer: 0.70 to 0.80

    Analog Electronics › Diode circuits

    With a piecewise-linear diode, the small-signal gain is constant within each region. Below 0.7 V the diode is an open ($r_d=\infty$). On the vertical part of the curve it is a short ($r_d=0$). Each case reduces to a plain resistive divider. The fixed 0.7 V drop only sets where the diode switches (at $V_{in}=2.1\text{V}$) and has no effect on either gain.

    WATCH OUT FOR

    Taking $V_{out}$ at node B (across the bottom resistor only) gives gains $\frac{1}{3}$ and $\frac{1}{2}$ and a ratio of 0.67, which is outside the key. The printed $V_{out}$ terminal is at node A, the diode's anode. Other slips are inverting the ratio (max/min = 1.33), or letting the 0.7 V drop enter the incremental gain when it only shifts the operating point.

    Full solution →
48. Q.58NAT · 2 marksOfficial answer: -0.55 to -0.45

    Digital Electronics › ADC & DAC

    This is an R-2R ladder DAC feeding an inverting amplifier. Because the (−) input is a virtual ground, reduce the ladder from its far end. At each node, two equal $2R$ branches combine into a Thévenin source equal to the average of their two voltages, with resistance $R$, and the next series $R$ turns that back into a source behind $2R$. Each stage halves the weight of every bit to its left. The final current, $V_{th}/(R+2R)$, flows entirely through $3R$, so $V_0=-3R\cdot I$.

    WATCH OUT FOR

    The usual mistake is to treat the circuit as a standard DAC in which the last $2R$ goes to ground and the output is taken at node D, then use $V_0=-\frac{R_f}{R}{V}_D$ with the wrong resistance. In this circuit the last $2R$ is in series and goes to the virtual ground, so the current is $0.5/(3\text{k}\Omega )$, not $0.5/(2\text{k}\Omega )$ or $0.5/(1\text{k}\Omega )$. A second mistake is to forget the extra $2R$ to ground at node A. Node A has two shunt branches, so its Thévenin voltage is $0.8\text{V}$, not $1.6\text{V}$. Dropping the minus sign of the inverting amplifier gives $+0.5$, which is outside the key.

    Full solution →
49. Q.59NAT · 2 marksOfficial answer: -2.1 to -1.9

    Analog Electronics › Amplifiers

    With $r_o\to \infty$ and no body effect, one small-signal current $i_d$ flows through both $R_D$ and $R_S$. The drain voltage is $-i_d{R}_D$ (current flows into the node from ac ground) and the source voltage is $+i_d{R}_S$ (current flows out of the node to ground). So the drain-to-source voltage ratio is simply $-R_D/R_S$. You don't need $g_m$, the bias point or $v_i$.

    WATCH OUT FOR

    Answering $+2$ by dropping the sign: the drain (common-source output) is inverted and the source (source-follower output) is not, so the ratio is negative. Another mistake is working out $v_A/v_i$ or $v_B/v_i$ separately and getting stuck because $g_m$ is not given. It cancels in the ratio. Adding an $r_o$ or a body-effect term also breaks the single-current argument, but the question rules both out ('ideal long channel', 'substrate connected to the source').

    Full solution →
50. Q.60NAT · 2 marksOfficial answer: -0.11 to -0.09

    Control Systems › Time response

    Treat the disturbance as the input and find $Y(s)/D(s)=\frac{G_p}{1+G_c{G}_p}$. Whether a step disturbance is rejected depends on integrators placed BEFORE the point where the disturbance enters, not on the system type as a whole. Here the only integrator is in the plant, after the disturbance, so the step is not rejected: the output settles at $1/G_c=1/10$, and because $r=0$ the error is the negative of that, $e(\infty )=-y(\infty )=-0.1$.

    WATCH OUT FOR

    Two errors are common. The first is reporting $+0.10$, which forgets that $e=r-y=-y$ when $r=0$. The second is answering $0$ on the grounds that the loop is type 1. The integrator sits after the disturbance entry point, so it gives zero error for a step REFERENCE but cannot reject a step DISTURBANCE. Only an integrator in the controller, before the disturbance enters, would drive this error to zero.

    Full solution →
51. Q.61NAT · 2 marksOfficial answer: 1.00
52. Q.62NAT · 2 marksOfficial answer: 0.84 to 0.86
53. Q.63NAT · 2 marksOfficial answer: 0.22 to 0.25
54. Q.64NAT · 2 marksOfficial answer: 1.15 to 1.18
55. Q.65NAT · 2 marksOfficial answer: 0.48 to 0.52

    Electromagnetics › Electrostatics

    Enclosed charge is the volume integral of $\nabla \cdot \mathbf{D}$ (Gauss's law in point form plus the divergence theorem). Here $\nabla \cdot \mathbf{D}=3x^2+3y^2$. Over a cube centred at the origin, ${\int }_{-0.5}^{0.5}{x}^{2}dx=1/12$, so each term gives $1/4$ and $Q=0.5\text{C}$.

    WATCH OUT FOR

    Integrating from 0 to 1 instead of from $-0.5$ to $0.5$. That gives ${\int }_0^1{x}^{2}dx=1/3$ and $Q=2\text{C}$, because the cube is centred at the origin, not placed at a corner. Other slips: differentiating $x{y}^2$ with respect to $x$ or $y$ instead of $z$, which adds a spurious term, or using ${\int }_{-0.5}^{0.5}{x}^{2}dx=2(0.5{)}^3=0.25$ and forgetting to divide by 3.

    Full solution →