GATE ECE 2018 question paper with solutions

GATE ECE 2018 answer key

General Aptitude GA Q.1 to Q.10

1. GA Q.1MCQ · 1 markOfficial answer: B

   General Aptitude › Grammar & vocabulary

   Homophones: words that sound alike (piece / peace / peas) are told apart only by meaning. Fill each blank from its context: a "piece" of cake, and lasting "peace" in a house. Checking the first blank on its own already leaves only one option.

   WATCH OUT FOR

   Choosing (C), "peace, piece", by swapping the two homophones: the words look and sound alike, so a hurried reader matches them to the wrong blanks. Also picking (A) or (D) because "peas" sounds right when read aloud; peas is a vegetable and fits neither blank.

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2. GA Q.2MCQ · 1 markOfficial answer: A

   General Aptitude › Grammar & vocabulary

   \"Even though\" signals a concession: the second clause must go against what the first leads you to expect. \"Scope for X\" means potential for something beneficial (improvement), and a potential nobody has acted on is \"neglected\". Also check that each word fits its noun: an area can be neglected, but it cannot be glum or disinterested, because those words describe people.

   WATCH OUT FOR

   Picking (B) because \"rejection\" and \"approved\" are opposites and seem to fit the \"Even though\" contrast. Opposites alone are not enough: each word must also make sense in its own blank, and \"scope for its rejection\" and \"an approved area\" do not. A second trap is (D): matching \"interest\" with \"disinterested\" looks neat, but \"disinterested\" means impartial and describes people, not areas.

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3. GA Q.3MCQ · 1 markOfficial answer: B

   General Aptitude › Numerical

   A number is divisible by 3 exactly when its digit sum is. The known digits add to 22, which leaves remainder 1 when divided by 3, so the missing digit must leave remainder 2 (2, 5 or 8). The smallest of these is 2.

   WATCH OUT FOR

   Using the divisibility rule for 9 instead of 3. The next multiple of 9 after 22 is 27, which gives 5, a trap set as option (C). Another slip is choosing (A) 0 just because it is the smallest whole number, without checking it: a digit sum of 22 is not divisible by 3.

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4. GA Q.4MCQ · 1 markOfficial answer: D

   General Aptitude › Numerical

   If every term is a fixed multiple $r$ of the one before it and $|r|<1$, the infinite geometric series adds up to $\frac{a}{1-r}$, where $a$ is the first term. Find $a$ and $r$ first, then use the formula.

   WATCH OUT FOR

   Using the ratio $\frac{1}{2}$ because the series looks like the familiar $1+\frac{1}{2}+\frac{1}{4}+\cdots =2$. That gives option (A). Another slip is to leave out the leading $1$ and add only $\frac{1}{4}+\frac{1}{16}+\cdots =\frac{1}{3}$. Adding the first few terms also shows the answer is below $1.34$, which rules out (A), (B) and (C).

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5. GA Q.5MCQ · 1 markOfficial answer: B

   General Aptitude › Geometry

   The lamp, the top of the person's head and the tip of the shadow lie on one straight light ray. The lamp post and the person are both vertical, so they form two similar right triangles that share the angle at the shadow's tip. Measure each base from the tip of the shadow: height / base is the same for both triangles.

   WATCH OUT FOR

   Using the 3 m person-to-post distance as the lamp post's base and leaving out the shadow. That gives H/3 = 1.5/3, so H = 1.5 m (option A). The lamp post's base has to run from its foot to the tip of the shadow, 3 + 3 = 6 m. A second slip is taking the shadow as 1.5 m instead of twice her height (3 m).

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6. GA Q.6MCQ · 2 marksOfficial answer: B

   General Aptitude › Numerical

   This is a present-value problem. Undo compound growth by dividing the target amount by ${\left(1+\frac{r}{100}\right)}^n$. Work out $1.1^5$ as $(1.1^2{)}^2\times 1.1=1.61051$ instead of estimating it.

   WATCH OUT FOR

   Using simple interest gives $P(1+5\times 0.1)=1.5P$, so $P\approx 6,66,667$, which is option (C). Another mistake is to take off 10% a year from the target, or to divide by $1.5$ instead of $1.61051$. Under compound interest, (C) is enough money but it is not the minimum.

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7. GA Q.7MCQ · 2 marksOfficial answer: B

   General Aptitude › Numerical

   A ratio has to be turned into fractions of the alloy's mass before alloys are mixed. With equal masses, the gold fraction of C is the plain average of the two gold fractions: $\frac{1}{2}\left(\frac{2}{5}+\frac{3}{10}\right)=\frac{7}{20}$, so copper is $\frac{13}{20}$ and the ratio is $7:13$.

   WATCH OUT FOR

   Adding the ratio terms straight across, $(2+3):(3+7)=5:10$, which is option (A). That only works if you mix 5 units of A with 10 units of B, not equal masses. The two ratios have different part totals (5 and 10), so each has to be scaled to the same alloy mass before the gold and copper amounts are added.

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8. GA Q.8MCQ · 2 marksOfficial answer: C

   General Aptitude › Reading

   Accept a statement only if the passage forces it. 'Finally found a complete captain' after 'many years' tells us such a captain is hard to find. The old picture of a batsman who could not convert aggression into big scores being 'replaced' by a 'successful batsman-captain' tells us he now scores big.

   WATCH OUT FOR

   Reading 'long recognized John's potential as a leader' as proof that he was already considered a good captain as a junior (i). Another is treating 'fans love his aggression, the Board worries about his temper' as a stated disagreement about what each wants in a captain (iii). Neither is said in the passage, so options (A), (B) and (D) all include a statement it does not support.

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9. GA Q.9MCQ · 2 marksOfficial answer: C

   General Aptitude › Statistics & probability

   This is Bayes' theorem. A witness saying "blue" can come from two routes: a blue cab named correctly ($0.15\times 0.8=0.12$) or a green cab named wrongly ($0.85\times 0.2=0.17$). The answer is the share of the first route in the total, $\frac{0.12}{0.12+0.17}$. Green cabs are so common that their 20% misidentifications outnumber the correct "blue" calls.

   WATCH OUT FOR

   Picking 80% (D) because the witness is 80% reliable, which ignores how few cabs are blue. Picking 15% (B) is the opposite mistake: it uses only the proportion of blue cabs and ignores the witness. Picking 12% (A) means stopping at $P(B)P(\text{says blue}\mid B)$ without dividing by $P(\text{says blue})=0.29$.

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10. GA Q.10MCQ · 2 marksOfficial answer: D

    General Aptitude › Reading

    The passage says outright why tourists stay away: poor connectivity (and a lack of accommodation). If the plan removes that stated cause, you can infer that the effect reverses and more tourists come. That inference rests on the passage's own facts. A claim that rests only on what one group fears does not count.

    WATCH OUT FOR

    Picking (B) because the passage mentions crowding and pollution. Those words come from the environmentalists' worry ("are upset that this would lead to"), not from a fact the passage states, and "will become" turns that worry into a certainty. (A) and (C) also fail because they bring in things the passage never mentions: local culture, and reefs in decline.

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Electronics and Communication Engineering Q.1 to Q.55

1. Q.1MCQ · 1 markOfficial answer: C

   Analog Electronics › Amplifiers

   This is a cascode: $M_1$ is common-source and $M_2$, with its gate at AC ground, is common-gate. Looking into the source of $M_2$, node $X$ sees only about $1/g_{m2}$, so all of $M_1$'s signal current $g_{m1}{v}_{gs}$ reaches the output and $g_m\approx g_{m1}$. Looking down from the drain, $M_2$ is a transistor with source degeneration $r_{o1}$. That multiplies the resistance by $M_2$'s intrinsic gain: $r_o=r_{o1}+r_{o2}+g_{m2}{r}_{o2}{r}_{o1}\approx g_{m2}{r}_{o2}{r}_{o1}$.

   WATCH OUT FOR

   Adding the two output resistances in series ($r_o\approx r_{o1}+r_{o2}$, as in (A) and (B)). That treats $M_2$ as a plain resistor. When $v_x$ rises, $v_{gs2}=-v_x$ falls, and $M_2$'s controlled source $g_{m2}{v}_{gs2}$ works against the change in current. That feedback produces the dominant $g_{m2}{r}_{o2}{r}_{o1}$ term. Picking (D), $r_o\approx r_{o2}$, ignores $M_1$ altogether. Writing $g_m\approx g_{m1}+g_{m2}$ treats the devices as if they were in parallel, but they are stacked, carry the same current, and only $M_1$ is driven by the input.

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2. Q.2MCQ · 1 markOfficial answer: C

   Analog Electronics › Op-amps

   A constant 1 mA flows through the feedback network. At first it charges both capacitors in series (0.5 µF), so the output falls at 2 V/ms. Once $C_2$ reaches 2.5 V, the reverse-biased Zener clamps it and carries all the current. From then on only $C_1$ (1 µF) charges, so the output falls at 1 V/ms and carries a fixed 2.5 V offset: $v_{OUT}=-(t_{\text{ms}}+2.5)$ V.

   WATCH OUT FOR

   There are three common slips, each giving one of the wrong options. Treating the circuit as a plain integrator with the series 0.5 µF for the whole time gives 5 ms (B). Using only $C_1$ from $t=0$, or forgetting the 2.5 V held across the clamped $C_2$ in Phase 2, gives 10 ms (D). Stopping at the Zener breakdown time gives 2.5 ms (A). In Phase 2, $C_1$'s voltage keeps growing from its value at $t=0$, not from zero, and the Zener's 2.5 V adds to it.

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3. Q.3MCQ · 1 markOfficial answer: D

   Analog Electronics › Feedback & oscillators

   Let the input and output quantities decide the impedances. Sensing a current needs a low input impedance, because a current divider only passes all of $i_s$ when $R_{in}\to 0$. Delivering a voltage needs a low output impedance, because a voltage divider only passes all of the voltage to the load when $R_{out}\to 0$. Shunt-shunt feedback, the topology used for transimpedance amplifiers, lowers both impedances by the factor $1+A\beta$.

   WATCH OUT FOR

   Going by habit and picking (C), high input and low output impedance. That is the ideal voltage amplifier, which senses a voltage. Option (A) is the ideal current amplifier and option (B) is the ideal transconductance amplifier, which senses a voltage and delivers a current. Some students also read 'impedance' in the name as a hint that the input impedance should be high, but a current input needs a low input impedance.

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4. Q.4MCQ · 1 markOfficial answer: C

   Control Systems › Block diagrams & SFG

   A linear system must obey superposition (additivity plus homogeneity). A necessary consequence is that zero input gives zero output. A constant offset $b\ne 0$ breaks this, so $y=au+b$ is affine, not linear. A constant-coefficient ODE at initial rest and a convolution integral are both linear.

   WATCH OUT FOR

   Calling $y=au+b$ linear because its graph is a straight line. A straight line through the origin is linear; any other straight line is affine. A second trap is doubting (A) because it contains derivatives of $u$. Derivatives are linear operators, and initial rest removes any zero-input response, so (A) is linear.

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5. Q.5MCQ · 1 markOfficial answer: D

   Control Systems › Nyquist

   A Nyquist plot shows $G(j\omega )$ as a set of phasors. With unity feedback, the phasor from $-1+j0$ to any point is $1+G(j\omega )$, so $T(j\omega )=G/(1+G)$ is fixed point by point. This is what the M and N circles are built on. The other three statements are standard facts: both criteria give the stable gain range, Bode's gain-phase relation holds for minimum-phase systems, and Routh needs a polynomial while Nyquist handles $e^{-j\omega T}$ exactly.

   WATCH OUT FOR

   One common mistake is to pick (B), thinking a magnitude plot alone can never give phase. For minimum-phase systems, Bode's gain-phase relation says it does. Another is to pick (C), thinking Routh can handle a delay directly. Routh needs a polynomial characteristic equation, so it works only after a Padé approximation, while Nyquist treats $e^{-j\omega T}$ exactly. Students also forget that the Nyquist plot gives the closed-loop response through $|1+G|$, measured from the $-1$ point.

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6. Q.6MCQ · 1 markOfficial answer: B

   Control Systems › Routh–Hurwitz

   A real quadratic $p^{\prime }(s)$ with no real roots and a positive leading coefficient is positive everywhere. So the cubic $p(s)$ is strictly increasing. It runs from $-\infty$ to $+\infty$, so it crosses the real axis exactly once. The other two roots have to be a complex-conjugate pair.

   WATCH OUT FOR

   Answering 3, from 'a cubic has three roots', without asking which of them are real. Or answering 0, from reading 'p'(s) has no real roots' as a statement about p(s). An odd-degree real polynomial always has at least one real root, so 0 is never possible here. Some students also reach for the Routh array because of the topic tag. Routh counts right-half-plane roots, not real roots, so it does not answer this question.

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7. Q.7MCQ · 1 markOfficial answer: D

   Electronic Devices › P-N junction

   At thermal equilibrium in a p-n junction, drift and diffusion currents cancel separately for holes and for electrons. For each carrier, the drift direction (driven by the built-in field) and the diffusion direction (driven by its own concentration gradient) are opposite to each other — this is exactly what makes the net current zero, and it is the physics tested by option (D).

   WATCH OUT FOR

   Conflating "current direction" with "physical motion direction" for electrons. Because electrons carry negative charge, their conventional current flows opposite to their physical motion, so a careless solver equates "electron drift and diffusion currents are in the same direction" with "electron drift and diffusion motions are in the same direction" — these are two different (and here, oppositely-signed) comparisons, and mixing them up can flip the answer between (B) and (D).

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8. Q.8MCQ · 1 markOfficial answer: D

   Digital Electronics › Logic families

   Find the output from the pull-up network: it conducts, making $f=1$, when every PMOS in some series branch has gate 0. Then check with the pull-down network: it conducts, making $f=0$, when every NMOS in some series branch has gate 1. Both networks here are parallel pairs of series transistors, so this is not the usual series-parallel dual of a static CMOS gate. The two networks are still logically complementary, because the inverters feed each network a different mix of true and complemented inputs.

   WATCH OUT FOR

   The most common slip is reading the pull-down network as if it gave $f$ directly. Its expression $\overline{X}\overline{Y}+XY$ (XNOR) is when the output is pulled LOW, so it gives $\overline{f}$, not $f$. The other common slip is missing the input inverters, especially the right-hand one that feeds $\overline{Y}$ to a PMOS while the plain $Y$ goes to an NMOS. That changes which branch conducts for each input.

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9. Q.9MCQ · 1 markOfficial answer: C

   Digital Electronics › Boolean logic

   For a function of $n$ variables, the minterms and maxterms split the $2^n$ input numbers between them. If $F=\sum m(S)$, then its canonical POS is $\prod M$ over every number not in $S$. Here $F=\sum m(0,2,4)$, so the POS is $\prod M(1,3,5,6,7)$. For a maxterm, a 0 bit gives the plain variable and a 1 bit gives its complement.

   WATCH OUT FOR

   The usual mistake is to keep the same indices and write $\prod M(0,2,4)$, which is option (A). Its factors are 0 exactly where $F$ should be 1. The other is writing each maxterm with the minterm literal rule (a 1 bit as the plain variable). That turns $\prod M(1,3,5,6,7)$ into the wrong-looking option (D) and is a quick way to lose the mark.

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

    Electromagnetics › Transmission lines

    A point on the impedance Smith chart is the reflection coefficient $\Gamma =\frac{z-1}{z+1}$. The centre ($\Gamma =0$) is the matched load $z=1$. The left end of the real axis ($\Gamma =-1$) is $z=0$, a short circuit. The right end ($\Gamma =+1$), where every $r$ circle and $x$ arc meets, is $z\to \infty$, an open circuit.

    WATCH OUT FOR

    Reading the chart as an admittance (Y) Smith chart. That chart is the impedance chart rotated by $180^{\circ }$, so its left end is the open circuit and its right end the short circuit, which leads to option (B). The question says it is a normalized impedance chart, so short circuit is on the left and open circuit on the right. A second slip is calling the centre a short or open circuit: the centre is $\Gamma =0$, which means no reflection, so it is the matched load.

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

    Engineering Mathematics › Linear algebra

    Eigenvectors for distinct eigenvalues are always linearly independent, so distinct eigenvalues are enough for diagonalisability. To knock out the other three options, one counterexample each is enough. $I_4$ shows that repeated eigenvalues can still come with a full set of eigenvectors, and that an invertible matrix can have repeated eigenvalues. $0$ shows that a full set of eigenvectors does not make a matrix invertible.

    WATCH OUT FOR

    Many students read the implication backwards and pick (A), treating "diagonalisable" as if it meant "distinct eigenvalues". $I_4$ is diagonalisable but has one eigenvalue repeated four times. Others mix up diagonalisability with invertibility and pick (B). These are unrelated properties: the zero matrix is diagonalisable (it is already diagonal) but singular.

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12. Q.12MCQ · 1 markOfficial answer: D

    Engineering Mathematics › Calculus

    Write $f$ as $a\frac{x}{y}+b\frac{y}{x}$ first. Each term is then a simple power of $x$ or $y$, so both partial derivatives take one line each and you never need the quotient rule. Put in the point, set the two derivatives equal and solve the linear equation for $a$ in terms of $b$.

    WATCH OUT FOR

    The usual slip is a sign or power when using the quotient rule on the whole fraction. For example, writing $\frac{\partial }{\partial x}\left(\frac{by}{x}\right)=\frac{by}{x^2}$ without the minus sign gives $\frac{a}{2}+2b=-\frac{a}{4}+b$, which leads to $a=-\frac{4b}{3}$, and that matches no option. Swapping which variable is held constant (treating $\frac{ax}{y}$ as $a$ times $\frac{1}{y}$ when differentiating in $x$) produces the distractors $a=\frac{b}{4}$ or $a=\frac{b}{2}$.

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13. Q.13MCQ · 1 markOfficial answer: B

    Signals and Systems › z-transform

    In an all-pass system every pole $p$ has a zero at $1/p^{\ast }$. Stability needs the ROC to contain $|z|=1$. With one pole inside the unit circle ($0.25$) and one outside ($2$), the only ROC that can do that is an annulus, so $h[n]$ must be two-sided.

    WATCH OUT FOR

    Picking (A) by treating the conjugate-reciprocal points $4\angle 0^{\circ }$ and $0.5\angle 30^{\circ }$ as poles, when an all-pass filter places ZEROS there. Picking (C) by confusing the all-pass property (constant magnitude) with constant phase. Also assuming a stable system must be causal: a pole outside the unit circle forces a stable system to be non-causal and two-sided.

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

    Signals and Systems › Fourier series

    The signal and its spectrum do not change when you choose a longer period. With $T^{\prime }=4T$ the new fundamental is ${\omega }_0/4$. The original harmonics sit at every 4th index of the new series, so $b_{4k}=a_k$ and the three indices in between are zero. The nonzero coefficients keep their values, so any sum over them, including $\sum |b_k|$, stays the same.

    WATCH OUT FOR

    Scaling the coefficients by the period ratio. Seeing $\frac{1}{T^{\prime }}=\frac{1}{40}$ in place of $\frac{1}{10}$ and dividing by 4 gives (D) 4. Thinking there are 4 times as many coefficients and multiplying gives (B) 64, and squaring gives (A) 256. Both errors forget two things. First, the integral over $T^{\prime }=40$ covers four copies of the period, and that cancels the extra $\frac{1}{4}$. Second, the new in-between coefficients are exactly zero. This is not a time-scaling of $x(t)$: the function itself is unchanged.

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15. Q.15NAT · 1 markOfficial answer: 0.12 to 0.13
16. Q.16NAT · 1 markOfficial answer: 16
17. Q.17NAT · 1 markOfficial answer: 0.12 to 0.14
18. Q.18NAT · 1 markOfficial answer: 0.40 to 0.43

    Electronic Devices › P-N junction

    For a p-n step junction, the depletion width scales with the square root of the effective barrier potential: $W\propto \sqrt{V_{bi}-V}$, where $V$ is the applied forward bias (taken positive, since forward bias reduces the barrier and hence the width). Equating the ratio of widths at two bias points eliminates the proportionality constant and the junction's physical parameters entirely.

    WATCH OUT FOR

    Treating the forward-bias case with the wrong sign, i.e. writing $W\propto \sqrt{V_{bi}+V}$ instead of $\sqrt{V_{bi}-V}$ — forward bias lowers the built-in barrier, so the depletion width shrinks, not grows, with increasing forward voltage.

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19. Q.19NAT · 1 markOfficial answer: 5

    Digital Electronics › Sequential

    With only a clock as input, the FSM can keep time only by counting clock periods. Its state loop must be as long as the full light cycle measured in clock periods: $150/5=30$ states. $n$ flip-flops give $2^n$ states, so $n=\lceil {\log }_{2}30\rceil =5$.

    WATCH OUT FOR

    Counting only the three colours as three states, which gives 2 flip-flops. That machine cannot hold GREEN for 14 clock periods and RED for 15, because it has no input telling it when to move on. A second mistake is to size for the longest phase alone (15 states, 4 flip-flops) and forget that the count must cover the whole 150 s cycle.

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20. Q.20NAT · 1 markOfficial answer: 0.47 to 0.51
21. Q.21NAT · 1 markOfficial answer: 0.001

    Electromagnetics › Transmission lines

    On a distortionless line ($R/L=G/C$), $Z_0=\sqrt{L/C}=\sqrt{R/G}$ is purely real and $\alpha =\sqrt{RG}$. Eliminating $G$ gives $\alpha =R/Z_0$ (equivalently $\alpha =G{Z}_0$), so only $R$ and $Z_0$ are needed.

    WATCH OUT FOR

    Using the low-loss formula $\alpha =\frac{R}{2Z_0}+\frac{G{Z}_0}{2}$ but keeping only the $R$ term gives $0.0005$. On a distortionless line the $G$ term is not negligible: $G{Z}_0=R/Z_0$, so the two halves are equal and add up to $R/Z_0=0.001$. Another slip is to say $\alpha$ cannot be found because $G$, $L$ and $C$ are not given, when the distortionless condition fixes $G=R/Z_0^2$.

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22. Q.22NAT · 1 markOfficial answer: 2

    Engineering Mathematics › Linear algebra

    A homogeneous square system $\mathbf{Ax}=0$ has non-trivial (and so infinitely many) solutions exactly when $\det \mathbf{A}=0$. Find the real roots of $\det \mathbf{A}$ as a polynomial in $k$ and count each distinct root once.

    WATCH OUT FOR

    Counting roots with multiplicity: $k^2(2-k)=0$ has three roots counted with multiplicity ($0,0,2$), which leads some candidates to answer $3$, but the question asks for DISTINCT values, which gives $2$. A second trap is cancelling a factor of $k$ from the determinant, or dividing a row by $k$, which quietly throws away the root $k=0$ and gives $1$.

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23. Q.23NAT · 1 markOfficial answer: 0.25

    Engineering Mathematics › Probability & statistics

    With $n$ i.i.d. continuous random variables, ties have probability zero and every ordering is equally likely. So any given variable is the minimum (or the maximum) with probability $1/n$, whatever the distribution. Here $n=4$, which gives $1/4$. Zero mean and unit variance play no part in the answer.

    WATCH OUT FOR

    Trying to work with the normal CDF $\Phi$ or normal tables, or thinking the zero mean matters (for example, answering $P(X_4<0)=0.5$ or $(1/2{)}^3=0.125$). Another mistake is to multiply $P(X_4<X_i)=1/2$ over the three other variables to get $1/8$. The events $\{X_4<X_1\}$, $\{X_4<X_2\}$, $\{X_4<X_3\}$ all depend on $X_4$, so they are not independent and you cannot simply multiply their probabilities.

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24. Q.24NAT · 1 markOfficial answer: -0.01 to 0.01

    Engineering Mathematics › Calculus

    You never need to evaluate the integral. Differentiate under the variable upper limit to get $f^{\prime }(x)=e^{-x^2/2}$, then use $a_2=f^{\prime \prime }(0)/2$. You can also see it without calculating: the integral of an even function from $0$ to $x$ is an odd function, and an odd function's Maclaurin series has only odd powers. So every even-indexed coefficient, including $a_2$, is zero.

    WATCH OUT FOR

    Stopping at $f^{\prime }(0)=1$ and reporting it as $a_2$ is the usual slip, and so is forgetting to divide by $2!$. That second error does no harm here only because $f^{\prime \prime }(0)=0$. Another mistake is to read the $t^2/2$ in the exponent as a coefficient in the answer, writing $a_2=-\frac{1}{2}$. That $-\frac{1}{2}$ is the $t^2$ coefficient of the integrand, and integrating moves it to the $x^3$ term of $f$, where it becomes $-\frac{1}{6}$.

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25. Q.25NAT · 1 markOfficial answer: 4.3 to 5.3

    Networks (Electric Circuits) › Two-port

    $B$ is the transfer resistance $V_1/(-I_2)$ with the output shorted ($V_2=0$). With port 2 shorted, find $I_1$ from the input resistance, then use a current divider to get the current through the short. For a T-network, $B=Z_1+Z_2+Z_1{Z}_2/Z_3$ gives the answer in one line.

    WATCH OUT FOR

    Taking $B$ to be the input resistance with port 2 shorted, $V_1/I_1=24/7\approx 3.43\Omega$, which falls outside the key. $B$ divides $V_1$ by the output short-circuit current $-I_2$, not by $I_1$. Some candidates also open-circuit port 2 (that condition defines $A$ and $C$, not $B$), or get the divider backwards and use $2/7$ in place of $5/7$.

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26. Q.26MCQ · 2 marksOfficial answer: B

    Analog Electronics › Diode circuits

    In a Zener shunt regulator, $R$ is set by the worst case for staying in breakdown: lowest input with maximum load current, where $R_{\max }=\frac{V_{I\min }-V_Z}{I_{Z\min }+I_L}$. The diode's power rating is set by the opposite worst case: highest input with minimum load current, where $P_Z=V_Z(I_{R\max }-I_{L\min })$. Choosing the largest workable $R$ keeps that rating as small as possible.

    WATCH OUT FOR

    Sizing $R$ from the nominal $6\text{V}$ or from $V_{I\max }$ instead of $V_{I\min }$. Another mistake is using $I_{Z\min }$ alone and forgetting the $5\text{mA}$ load current, for example $0.7/2\text{mA}$ or similar, which leads towards $186\Omega$. A third is computing the power at $I_{Z\min }$ ($5\times 2=10\text{mW}$) rather than at the maximum Zener current that flows when $V_I=6.3\text{V}$.

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27. Q.27MCQ · 2 marksOfficial answer: D
28. Q.28MCQ · 2 marksOfficial answer: B
29. Q.29MCQ · 2 marksOfficial answer: A

    Control Systems › State space

    For a state model with $D=0$, $G(s)=C(sI-A{)}^{-1}B$. The denominator is $\det (sI-A)$. For a $2\times 2$ system, you get the numerator by multiplying $C\text{adj}(sI-A)B$. When every option has the same denominator, two shortcuts settle it with no inversion: the $s$-coefficient of the numerator equals $CB$, and the DC gain equals $-C{A}^{-1}B$.

    WATCH OUT FOR

    Sign errors when forming $sI-A$: the off-diagonal entries flip sign, giving $+1.5$ and $-4$. Sign errors also happen in the adjugate, where the off-diagonal entries are negated. Get either wrong and you end up with a numerator like $3s-1.875$ (option B). Another mistake is multiplying $C$ and $B$ against $(sI-A)$ instead of its adjugate or inverse. Also note that the DC-gain check alone cannot tell (A) from (D), because both have $G(0)=5/6$. You also need the leading coefficient $CB=3$.

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30. Q.30MCQ · 2 marksOfficial answer: B

    Electronic Devices › Optoelectronics

    A wider band-gap semiconductor has a much smaller intrinsic carrier concentration $n_i$ (since $n_i^2\propto \exp (-E_g/kT)$), and since $V_{bi}=V_T\ln \left(\frac{N_A{N}_D}{n_i^2}\right)$ with the same doping in all three diodes, a smaller $n_i$ (larger $E_g$) gives a larger $V_{bi}$.

    WATCH OUT FOR

    Assuming the built-in voltage is unaffected by band-gap because $N_A$ and $N_D$ are stated to be equal (choosing option C), forgetting that $n_i$ itself is exponentially dependent on $E_g$ and enters the same formula.

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

    Digital Electronics › Combinational

    Work through a MUX cascade one stage at a time. The first MUX's data inputs, 0, 1, 1, 0 for $UV=00,01,10,11$, are the truth table of $U\oplus V$. In the second MUX, $I_0=I_1=Y_1$ and $I_2=I_3=0$, so $F=\overline{W}{Y}_1$ and $X$ does not appear in the answer.

    WATCH OUT FOR

    Reading the inputs 0, 1, 1, 0 as XNOR instead of XOR, which gives (A). A second trap is picking (D): it is the correct function, but the question asks for the minimized form, and $\overline{W}\overline{X}+\overline{W}X$ reduces to $\overline{W}$.

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

    Digital Electronics › Processors & memory

    In a diode ROM, the always-on transistor at the foot of each bit line pulls it low by default. A diode from a word line (anode) to a bit line (cathode) pulls that bit line high when the row is selected. So a diode at a crossing stores $1$ and an empty crossing stores $0$. Read the matrix straight off the diode positions: row $W_0$ has a diode only on $B_0$ and row $W_1$ only on $B_1$.

    WATCH OUT FOR

    Treating a diode as a stored $0$, as if it pulled the line low, which gives option (B). Here the default level comes from the pull-down transistor with its gate tied to $V_{DD}$, so a diode drives the line high and marks a $1$. A second slip is reading the matrix by column instead of by row, or taking the diode's anode position (the $W_1$ diode's anode sits between $B_0$ and $B_1$) as its column. The column is set by the bit line the cathode lands on.

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

    Electromagnetics › Plane waves

    Each skin depth multiplies the amplitude by $1/e$, which is an attenuation of 1 Np, or 8.686 dB. A 20 dB amplitude drop means the amplitude falls by a factor of 10. That takes $\ln 10\approx 2.303$ Np, so the wave must travel $z=\delta \ln 10=2.303\delta$.

    WATCH OUT FOR

    Treating 20 dB as a power ratio ($10{\log }_{10}$, a factor of 100) gives $z=\delta \ln 100=0.46\text{m}$, which is option (C). The skin depth is defined on the field amplitude, so the $20{\log }_{10}$ form applies. A second error is using ${\log }_{10}10=1$ in place of $\ln 10$, which gives $0.1\text{m}$. Multiplying $\delta$ by 23 in place of 2.3 gives option (D).

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

    Engineering Mathematics › Differential equations

    The right-hand side depends on $x$ and $y/x$, so try $y=vx$. The $y/x$ term cancels the $v$ that comes from $\frac{dy}{dx}=v+x\frac{dv}{dx}$, and the $x$ factors also cancel. That leaves the separable equation $\frac{2v}{1+v^2}dv=dx$. It integrates directly to $\ln (1+v^2)$ because the numerator is the derivative of the denominator.

    WATCH OUT FOR

    Integrating $\frac{2v}{1+v^2}$ as $\frac{1}{2}\ln (1+v^2)$, which confuses it with $\int \frac{v}{1+v^2}dv$. That extra half gives option (B). Options (C) and (D) come from putting back $v$ where $v^2$ belongs, i.e. writing $1+y/x$ where the answer has $1+y^2/x^2$.

    Full solution →
35. Q.35MCQ · 2 marksOfficial answer: C

    Networks (Electric Circuits) › AC, power & resonance

    This is a series RC voltage divider. Both $100\text{k}\Omega$ resistors sit in the one loop, so they add to $R=200\text{k}\Omega$ no matter where they are drawn. With $\omega RC=1$ the divider ratio $\frac{1}{1+j\omega RC}$ becomes $\frac{1}{\sqrt{2}}\angle -45^{\circ }$: the amplitude drops by $\sqrt{2}$ and the phase lags by $\pi /4$.

    WATCH OUT FOR

    (1) Leaving out the resistor on the bottom return wire. That uses $R=100\text{k}\Omega$, so $\omega RC=0.5$, giving about $4.47\angle -26.6^{\circ }$, which is not among the options. (2) Converting to rms and then dividing by $\sqrt{2}$ a second time. The rms value of $V_C$ is $2.5\text{V}$; dividing again gives $1.25\sqrt{2}$, which is the magnitude in options (A)/(B). (3) Halving the phase to $-0.125\pi$, or using $\omega =2\pi \times 5$ when the $5$ in $\sin (5t)$ is already $\omega$ in rad/s.

    Full solution →
36. Q.36NAT · 2 marksOfficial answer: 0.284 to 0.348

    Networks (Electric Circuits) › Transients

    A switched series RL loop with a DC source is a first-order circuit. The current starts at zero, because an inductor's current cannot jump. It rises towards $V/R_{\text{total}}$ with time constant $\tau =L/R_{\text{total}}$, and both resistors count because the whole loop is in series. Here 0.5 s is exactly one time constant, so the current has reached $1-e^{-1}\approx 63.2\%$ of its final value.

    WATCH OUT FOR

    Using only one of the 1 Ω resistors. Because the second resistor sits on the far side of the loop, it is easy to miss that it is in series too. With $R=1\Omega$ you get $\tau =1\text{s}$ and $i(0.5)=1-e^{-0.5}\approx 0.393\text{A}$, which is outside the key. A second slip is to use $\tau =RL$ or $\tau =R/L$ instead of $L/R$, or to report the final value $0.5\text{A}$ as if the transient had already died away.

    Full solution →
37. Q.37NAT · 2 marksOfficial answer: 6.2 to 6.6

    Analog Electronics › Diode circuits

    For small signals, a forward-biased diode acts as its dynamic resistance $r_d=\eta V_T/I_D$ in parallel with its junction capacitance. With the dc source shorted, the ac current is the source amplitude divided by the magnitude of the complex loop impedance, which is $(r_d\parallel C_j)$ in series with the resistor.

    WATCH OUT FOR

    Students often ignore $C_j$ and take $5\text{mV}/(1000+100)\Omega \approx 4.5\mu \text{A}$. Another mistake is to add $r_d$ and $X_C$ in series, or to add the magnitudes $|Z_D|+100$ instead of adding the complex impedances first. Some also report only the current through $r_d$ and leave out the current through the capacitance, although the question asks for the total diode current.

    Full solution →
38. Q.38NAT · 2 marksOfficial answer: 0.4 to 0.6

    Analog Electronics › Op-amps

    Always check the linear-mode answer against the supply rails. Here the ideal gain of −31 would need −31 V, so the output clips at −15 V. Once the output is saturated, the virtual short is gone. The input still draws no current, though, so node A is set by a plain voltage divider: 31 kΩ and 1 kΩ between +1 V and −15 V.

    WATCH OUT FOR

    Using the virtual short out of habit and answering V_A = 0 V. The virtual short only holds while the op-amp is in its linear region. The linear answer v_o = −31 V lies outside the ±15 V rails, so the op-amp saturates and V_A becomes 0.5 V. A second slip is to calculate the divider with v_o = −31 V instead of the clipped −15 V, which gives V_A = 0 V again.

    Full solution →
39. Q.39NAT · 2 marksOfficial answer: 8

    Signals and Systems › Fourier transform

    A Hilbert transformer is an all-pass filter, $H(f)=-j\text{sgn}(f)$ with $|H(f)|=1$. It only shifts phase by $\mp 90^{\circ }$, so the output energy equals the input energy. Find the input energy with Parseval: $4\text{sinc}(2t)$ has a flat spectrum of height $2$ over $|f|<1$, so the energy is $2^2\times 2=8$.

    WATCH OUT FOR

    (1) Reading $\text{sinc}(2t)$ as $\frac{\sin 2t}{2t}$ instead of the stated normalised form $\frac{\sin 2\pi t}{2\pi t}$. That gives $16\cdot \frac{\pi }{2}=8\pi \approx 25.13$. (2) Taking the scaled spectrum as $\text{rect}(f/2)$ without the $\frac{1}{2}$ factor, which gives 32. (3) Treating the Hilbert transformer like the analytic-signal operation, which removes negative frequencies, and halving the energy to 4. The Hilbert transformer keeps every frequency and changes only its phase.

    Full solution →
40. Q.40NAT · 2 marksOfficial answer: -0.5
41. Q.41NAT · 2 marksOfficial answer: 14 to 17

    Control Systems › Bode & margins

    For $G(s)=\frac{K}{s(s+a)}$ with unity feedback, the closed loop is a standard second-order system with ${\omega }_n=\sqrt{K}$ and $\zeta =\frac{a}{2\sqrt{K}}$. Setting $M_r=\frac{1}{2\zeta \sqrt{1-{\zeta }^2}}$ gives a quadratic in ${\zeta }^2$. Take only the root with $\zeta <0.707$, because a resonant peak exists only in that range. Then convert $\zeta$ back to $K$.

    WATCH OUT FOR

    Taking the larger root ${\zeta }^2=0.933$, which gives $K\approx 1.07$. That value of $\zeta$ is above 0.707, where the closed-loop response has no resonant peak, so it cannot have $M_r=2$. A second slip is to apply $M_r$ to the open-loop $G(j\omega )$ instead of the closed-loop $T(j\omega )$. A third is to write $\zeta =1/K$ instead of $\zeta =1/\sqrt{K}$ when comparing with $2\zeta {\omega }_n=2$.

    Full solution →
42. Q.42NAT · 2 marksOfficial answer: 0.1

    Control Systems › Bode & margins

    If the open loop is stable ($P=0$) and the phase crosses $-180^{\circ }$ exactly once, the largest stable gain is the gain margin as a plain ratio: $k_0=1/|G(j{\omega }_{pc})|$. Read the magnitude at the phase crossover frequency, not at the gain crossover and not at the peak.

    WATCH OUT FOR

    There are three common slips. (1) Reading the wrong magnitude: the 36 dB DC level (giving $k_0=10^{-36/20}\approx 0.016$), the resonant peak, or the 0 dB gain crossover, instead of the magnitude where the phase is $-180^{\circ }$. (2) Converting 20 dB to 20 instead of 10: magnitude in dB is $20{\log }_{10}|G|$. (3) Forgetting the reciprocal and answering $k_0=10$. The magnitude at ${\omega }_{pc}$ is already greater than 1, so $k$ must shrink the loop gain, not boost it.

    Full solution →
43. Q.43NAT · 2 marksOfficial answer: 0.41 to 0.435

    Analog Electronics › Biasing

    Series transistors carry the same current. M2 has its drain on the 3.3 V rail, so $V_{DS2}-(V_{GS2}-V_T)=2.3\text{ V}>0$ and it is always saturated when on. M1 has $V_{GS1}-V_T=1\text{ V}$, and the requirement $V_{GS2}>V_T$ forces $V_x<1\text{ V}$, so M1 is in triode. Setting the saturation current equal to the triode current gives a quadratic in $V_x$. Keep only the root that satisfies both region assumptions.

    WATCH OUT FOR

    Assuming both transistors are saturated. Equating $k(1-V_x{)}^2$ with $\frac{k}{2}(1{)}^2$ gives $V_x\approx 0.29\text{ V}$, which breaks M1's saturation condition ($V_{DS1}=0.29<1$). Two other mistakes: keeping the root $1.577\text{ V}$, which switches M2 off, and forgetting the factor 2 in $(W/L{)}_2$.

    Full solution →
44. Q.44NAT · 2 marksOfficial answer: 0.59 to 0.63

    Electronic Devices › Optoelectronics

    A solar cell's open-circuit voltage is $V_{oc}=V_T\ln \left(\frac{I_{sc}}{I_0}+1\right)$. Since $I_{sc}/I_0\gg 1$ (verified from the 1-sun data, where $I_{sc}/I_0=e^{0.65/0.026}-1$, an enormous number), this reduces to $V_{oc}\approx V_T\ln (I_{sc}/I_0)$, so a change in illumination (which changes $I_{sc}$ proportionally, with $I_0$ unchanged) shifts $V_{oc}$ by $\Delta V_{oc}=V_T\ln \left(\frac{I_{sc,2}}{I_{sc,1}}\right)$.

    WATCH OUT FOR

    Assuming $V_{oc}$ scales linearly or logarithmically with intensity directly (e.g., $V_{oc,2}=0.2\times 0.65$), instead of recognizing that only $I_{sc}$ is proportional to intensity, and that $V_{oc}$ changes by $V_T\ln (\text{intensity ratio})$ through the diode equation. Also, forgetting to first justify $I_{sc}/I_0\gg 1$ before dropping the "+1" — without that check, the cancellation of $I_0$ between the two conditions is unjustified.

    Full solution →
45. Q.45NAT · 2 marksOfficial answer: 0.11 to 0.13

    Electronic Devices › Semiconductors

    This is a p⁻–p (isotype / high-low) homojunction, not a p-n junction. Since both sides are p-type, the built-in potential is set by the ratio of the majority-carrier (hole) concentrations on the two sides, not by an $n_i^2$/minority-carrier expression: $V_{bi}=V_T\ln \left(\frac{N_{A2}}{N_{A1}}\right)$, with $V_T=kT/q$.

    WATCH OUT FOR

    Treating this as a standard p-n junction and reaching for $V_{bi}=V_T\ln \left(\frac{N_A{N}_D}{n_i^2}\right)$ — that formula does not apply here since there is no donor density and no $n_i$ given (and none is needed): both regions are p-type, so the isotype-junction formula using only the two acceptor densities must be used instead.

    Full solution →
46. Q.46NAT · 2 marksOfficial answer: 0.82 to 0.86

    Digital Electronics › Sequential

    $X=D_{in}\oplus Q$ compares the new data bit with the previous bit held in the flip-flop. $X$ is HIGH only between the moment $D_{in}$ changes ($\Delta T$ after an edge) and the next rising edge, when the flip-flop copies the new bit into $Q$. That window is $T_{CK}-\Delta T$ long and appears only in periods where the data changes. So the average is $V_H\times P(\text{transition})\times (1-\Delta T/T_{CK})$.

    WATCH OUT FOR

    Taking the HIGH window of $X$ as $\Delta T$ instead of $T_{CK}-\Delta T$. That gives $3.3\times 0.3\times 0.15\approx 0.15\text{V}$. The edge samples the OLD bit, because $D_{in}$ changes only after the edge. So $Q$ lags $D_{in}$ for almost the whole period, not just for $\Delta T$. Other slips are leaving out the transition probability (giving $2.81\text{V}$) and multiplying by the clock's 50% duty cycle, which has nothing to do with the output.

    Full solution →
47. Q.47NAT · 2 marksOfficial answer: 8

    Digital Electronics › Boolean logic

    With strong pull-downs and weak pull-ups, a shorted node is the AND of every output tied to it (wired-AND). The node at the OR gate's upper input ties $X_0\cdot (\dots )$ to $\overline{X_0}$, so it is always 0. That leaves $Y=X_3$.

    WATCH OUT FOR

    Treating the shorted nodes as wired-OR, or ignoring the shorts and reading each wire as a single gate's output. A second slip is to feed $\overline{X_0}$ (instead of $X_0$) into the AND gate, or $\overline{X_3}$ (instead of $X_3$) into the OR gate. Both taps are taken before the inverters.

    Full solution →
48. Q.48NAT · 2 marksOfficial answer: 0.7 to 0.8

    Electromagnetics › Waveguides & fibres

    In an air-filled guide, $f_{c,10}=c/2a$ and $f_{c,01}=c/2b$, so the cut-off ratio gives $a=2b$. The guide wavelength is ${\lambda }_g=\lambda /\sqrt{1-(f_c/f{)}^2}$. With $f=1.25f_c$ we get $f_c/f=0.8$ and a denominator of exactly $0.6$. Writing $\lambda =c/f$ in terms of $a$ leaves one unknown.

    WATCH OUT FOR

    One slip is to take $f_c/f=1.25$, or $0.75$ ("25% higher" misread), instead of $1/1.25=0.8$. Another is to set the free-space wavelength, rather than the guide wavelength, equal to 4 cm, which gives $a=2.5$ cm and $b=1.25$ cm. A third is to stop at $a=1.5$ cm and report it, when the question asks for $b=a/2$.

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

    Electromagnetics › Plane waves

    'No reflected wave' for a wave whose $\mathbf{E}$ lies in the plane of incidence (parallel/TM polarisation) means it arrives at the Brewster angle. Between non-magnetic media that angle is $\tan {\theta }_B=\sqrt{{\epsilon }_{r2}/{\epsilon }_{r1}}$. Read ${\theta }_i$ from the wave vector in the phase: $\tan {\theta }_i=k_z/k_x$, measured from the interface normal ${\hat{a}}_x$.

    WATCH OUT FOR

    One mistake is measuring the angle from the interface instead of the normal. That gives $\tan \theta =k_x/k_z=1/\sqrt{2}$ and ${\epsilon }_r=0.5$, which the condition ${\epsilon }_r>1$ rules out. Another is writing $\tan {\theta }_B={\epsilon }_r$ without the square root, which gives ${\epsilon }_r=\sqrt{2}\approx 1.41$. A third is reading the angle off the sketch, which is not to scale: the drawn ray makes about $30^{\circ }$ with the normal. Some students also try a Brewster condition for perpendicular polarisation, but between non-magnetic media none exists. This wave is parallel-polarised because $\mathbf{E}$ has only $x$ and $z$ components.

    Full solution →
50. Q.50NAT · 2 marksOfficial answer: -0.23 to -0.19

    Engineering Mathematics › Differential equations

    Write the equation as $y^{\prime \prime }+y^{\prime }+\frac{5}{4}y=0$. Its auxiliary roots are $-\frac{1}{2}\pm j$, so the motion is a decaying oscillation, $e^{-t/2}(A\cos t+B\sin t)$. Use the two initial conditions to find $A$ and $B$. At $t=\pi$ the sine term is zero, so only $A\cos \pi e^{-\pi /2}$ is left.

    WATCH OUT FOR

    Setting $B=0$ because $y^{\prime }(0)=0$. The decay factor $e^{-t/2}$ also contributes to the derivative, so $y^{\prime }(0)=B-\frac{A}{2}$ and $B=\frac{1}{2}$. At $t=\pi$ this particular slip does no harm, because $\sin \pi =0$. The same slip does change the answer at other times. Another mistake is copying the signs wrongly when moving terms across, for example writing $m^2-m+\frac{5}{4}=0$. That gives a growing solution and $y(\pi )=-e^{\pi /2}\approx -4.81$. Some candidates also forget the negative sign that $\cos \pi =-1$ brings in and enter $+0.21$.

    Full solution →
51. Q.51NAT · 2 marksOfficial answer: 2

    Engineering Mathematics › Complex analysis

    A figure-eight contour winds in opposite senses round its two lobes, so the winding numbers are $+1$ round $z=1$ (counter-clockwise lobe) and $-1$ round $z=-1$ (clockwise lobe). The two residues also have opposite signs ($\frac{1}{2}$ and $-\frac{1}{2}$), so the two sign flips cancel and the contributions add instead of cancelling. Partial fractions give the same result: $\frac{1}{z^2-1}=\frac{1}{2}\left[\frac{1}{z-1}-\frac{1}{z+1}\right]$, so ${\oint }_C=\frac{1}{2}\left[2\pi j(+1)-2\pi j(-1)\right]=2\pi j$.

    WATCH OUT FOR

    Ignoring the arrows and treating the whole contour as counter-clockwise. That adds the residues with the same sign, $\frac{1}{2}+\left(-\frac{1}{2}\right)=0$, and gives the wrong answer $0$. Another slip is reading the direction of only one lobe and assuming the other lobe goes the same way. A single continuous figure-eight always traverses its two lobes in opposite senses.

    Full solution →
52. Q.52NAT · 2 marksOfficial answer: 4.4 to 4.6

    Engineering Mathematics › Calculus

    $z$ is not an independent variable. The constraint makes it a function of $x$ and $y$, so $\frac{\partial r}{\partial x}=\frac{\partial r}{\partial x}{|}_{\text{direct}}+\frac{\partial r}{\partial z}\frac{\partial z}{\partial x}$. Find $\frac{\partial z}{\partial x}$ by implicit differentiation: $\frac{\partial z}{\partial x}=-\frac{F_x}{F_z}$, where $F=z^3-xy+yz+y^3-1$.

    WATCH OUT FOR

    Treating $z$ as a constant gives $\frac{\partial r}{\partial x}=2x=4$, which is outside the key range. The other common slip is a sign error in $-\frac{F_x}{F_z}$: here $F_x=-y$, so $\frac{\partial z}{\partial x}=+\frac{y}{3z^2+y}=-0.5$. Dropping the minus sign gives $+0.5$ and a wrong answer of $3.5$.

    Full solution →
53. Q.53NAT · 2 marksOfficial answer: 7.9 to 8.1

    Networks (Electric Circuits) › Laws & node/mesh

    The network is mirror-symmetric about the line through the two source terminals, so the mirror-image nodes are equipotential. The two $R_2$ bridges then carry no current and can be removed, and the symmetric node pairs can be merged. After that, the network is a plain series–parallel chain: $0.5+(0.5\parallel 1.5)+0.5=1.375\Omega$.

    WATCH OUT FOR

    One mistake is to treat the crossing diagonals as a single junction that splits the $R_3$ resistors differently. Another is to keep the $R_2$ bridges in the calculation and try a full nodal analysis. Both $R_2$ carry zero current by symmetry, and the crossing point is at the upper-pair potential whether or not it is joined, so $R_{eq}=11/8\Omega$. A further slip is combining $0.5\parallel 1.5$ as $0.5+1.5$ or as $1\Omega$ instead of $0.375\Omega$.

    Full solution →
54. Q.54NAT · 2 marksOfficial answer: 13

    Signals and Systems › Sampling

    With a non-ideal reconstruction filter you need two conditions. First, the signal band $W$ must sit inside the flat passband ($W\le f_{\text{pass}}$; here $5\le 6$). Second, the first spectral image must start at or beyond the stopband edge: $f_s-W\ge f_{\text{stop}}$, so $f_s\ge W+f_{\text{stop}}=5+8=13\text{kHz}$.

    WATCH OUT FOR

    Answering the Nyquist rate $2W=10\text{kHz}$ ignores the filter's transition band. Other wrong routes: using the passband edge ($f_s-5\ge 6$, which gives $11\text{kHz}$), which lets the image into the sloped band where $|H(f)|>0$; or taking twice the filter's stopband edge ($2\times 8=16\text{kHz}$), which treats the filter as if it were the signal's bandwidth.

    Full solution →
55. Q.55NAT · 2 marksOfficial answer: 2.90 to 3.10

    Signals and Systems › DTFT & DFT

    For an $N$-point DFT with $N$ even, $X[0]=\sum x[n]$ and $X[N/2]=\sum x[n](-1{)}^n$. Adding them cancels the odd-indexed samples and doubles the even-indexed ones, so the sum of the even samples is $\frac{X[0]+X[N/2]}{2}$. You never need to compute $x[n]$ itself.

    WATCH OUT FOR

    Using $X[k]=k$ in place of $k+1$, which gives $X[0]=0$ and $X[4]=4$ and so the wrong answer 2. Another is forgetting to divide by 2, or confusing this with the IDFT factor $1/N$ and writing $(X[0]+X[4])/8=0.75$. A third is trying to recover all eight samples $x[n]$ through a full IDFT, which takes time and invites arithmetic slips.

    Full solution →