GATE ECE 2023 question paper with solutions

GATE ECE 2023 answer key

General Aptitude Q.1 to Q.10

1. Q.1MCQ · 1 markOfficial answer: C

   General Aptitude › Grammar & vocabulary

   The four options look alike, so check two things: part of speech, then meaning. The blank must be a noun that belongs to one person and can act as the subject of \"will not permit\". Only conscience, a person's inner moral sense, does all of that, and \"my conscience will not permit it\" is a fixed idiom.

   WATCH OUT FOR

   Picking (A) conscious because it sounds like conscience, when it is an adjective. The other trap is (D) consent, because refusing permission seems to fit. But consent is the permission itself, which you give or withhold; it cannot \"permit\" anything. (B) consensus is always a group's agreement, never one person's.

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2. Q.2MCQ · 1 markOfficial answer: A

   General Aptitude › Reasoning

   Name the relation in the complete pair first. Parliament is the institution that forms the Legislature, so the blank must be the organ of government that the courts form, and that is the Judiciary. Of the three organs of government (Legislature, Executive, Judiciary), only the Judiciary works through the courts.

   WATCH OUT FOR

   Picking (D) Legal because courts deal with law. The link then goes from an institution to its subject matter, which is not the institution-to-organ link that Parliament : Legislature sets up, and 'Legal' is an adjective while 'Legislature' is a noun. Some candidates pick (B) Executive only because it is the other organ of government they remember. But the Executive carries out laws. It does not decide cases.

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3. Q.3MCQ · 1 markOfficial answer: C

   General Aptitude › Numerical

   The ten digits 0 to 9 add up to exactly 45, and any 8 distinct digits add up to at most 44. So the number must use all of 1 to 9 exactly once. Leaving out 0 keeps it to 9 digits, and writing the digits in ascending order makes it as small as possible.

   WATCH OUT FOR

   Picking (D) 99999 because it is the shortest number whose digits add up to 45, without noticing that it breaks the distinct-digits condition. Two smaller slips: adding 0 as a tenth digit (the number gets longer, so it gets larger), and choosing (B) without comparing it digit by digit with (C).

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4. Q.4MCQ · 1 markOfficial answer: C

   General Aptitude › Numerical

   Two-set inclusion–exclusion: $n(R\cup C)=n(R)+n(C)-n(R\cap C)$. Get $n(R\cup C)$ by subtracting the 'neither' group from the total. Then the ratio condition leaves one unknown and one equation.

   WATCH OUT FOR

   One slip is to forget to subtract the overlap: $2x+x=70$ gives $x=23.33$, which is not a whole number. Another is to read 'twice' as applying to the students who like only romantic and only comedy movies. That gives $r_{\text{only}}=2c_{\text{only}}$, so $3c_{\text{only}}=50$, which is also not a whole number. Some candidates also mark 40, which is the number who like ONLY romantic movies. The question asks for everyone who likes romantic movies, so the overlap must be counted too.

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5. Q.5MCQ · 1 markOfficial answer: C

   General Aptitude › Spatial

   A rectangle's sides come from two pairs of perpendicular parallel lines. Sort the lines into perpendicular families. A grid of $m$ lines crossing $n$ lines contains $\left(\genfrac{}{}{0ex}{}{m}{2}\right)\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ rectangles. Here that is $\left(\genfrac{}{}{0ex}{}{3}{2}\right)\left(\genfrac{}{}{0ex}{}{3}{2}\right)=9$ for the tilted $2\times 2$ grid, plus the outer square: 10.

   WATCH OUT FOR

   Counting only the 4 small squares, or forgetting that a square is also a rectangle, so the 2x1 rectangles, the diamond $PQRS$ or the outer square $ABCD$ get left out. Answers of 8 or 9 usually come from missing the outer square or the diamond. Another mistake is counting the corner triangles or treating a diagonal as the side of an extra rectangle outside the diamond.

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6. Q.6MCQ · 2 marksOfficial answer: D

   General Aptitude › Reasoning

   From 'All C are F' you can infer 'Some F are C' only if at least one C exists. The stem gives that directly ("There are also creatures that have claws"), so option (D) is guaranteed. Being inside the same larger set (ferocious) says nothing about how animals and clawed creatures relate to each other.

   WATCH OUT FOR

   Choosing (A) by reasoning 'animals are ferocious and clawed creatures are ferocious, so clawed creatures are animals'. That is the undistributed middle, because two groups that share a property need not overlap. A second trap is rejecting (D) on the grounds that a universal statement does not prove anything exists. The stem states outright that clawed creatures exist, so the existence condition is met.

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7. Q.7MCQ · 2 marksOfficial answer: A

   General Aptitude › Numerical

   Match a sketch to a formula by its qualitative features, not by plotting. Here there are three: symmetry (an even function needs $f(-x)=f(x)$), sign (the curve is never negative) and end behaviour (it decays to 0 on both sides). Only $x^2{2}^{-|x|}$ is even, never negative and decaying at both ends, because $2^{-|x|}$ decays whichever way $x$ goes and beats the polynomial factor.

   WATCH OUT FOR

   Looking only at the right half of the graph, where all four options behave alike (zero at 0, a rise, then decay), and then picking (B) or (C). Test a negative value of $x$: (B) and (D) go negative there, and (C) and (D) blow up because $2^{-x}$ grows as $x\to -\infty$. A second trap is rejecting (A) because the sketch shows a sharp point at the origin while $x^2{2}^{-|x|}$ is actually flat there. The figure is only a qualitative sketch.

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8. Q.8MCQ · 2 marksOfficial answer: B

   General Aptitude › Reading

   Look at the tense and the time markers. "When I was a kid" and "used to imagine" place childhood in the past, so the speaker is a grown-up looking back. "Partial to" is an idiom meaning fond of.

   WATCH OUT FOR

   Taking "partial to" literally as "in part" and picking (C). Also common: picking (A) without noticing that "When I was a kid" means the speaker is no longer a child, or picking (D), a moral the passage never states.

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9. Q.9MCQ · 2 marksOfficial answer: A

   General Aptitude › Reasoning

   Total kits = (number of sets) + 5 × (number of positive sets). The pooled tests always cost 200 kits, so the worst case is the grouping with the most positive sets. With 100 positives, that is 100 sets, one positive in each.

   WATCH OUT FOR

   One mistake is to count only the 100 positive people and answer 200 + 100 = 300, or to forget the 200 pooled tests and answer 500. Both miss that a positive pool means retesting all 5 members, and that the 200 pooled tests happen whatever the result. Another is to take the worst case as all 200 sets being positive (200 + 1000 = 1200), which is impossible because there are only 100 positive people.

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

    General Aptitude › Spatial

    Each fold halves the 32 cm side and doubles the layers. After 5 folds the unfolded sheet is 32 strips 1 cm wide, and 32 strips have 31 boundaries between them. You can also count it as $1+2+4+8+16=2^5-1$, because each fold creases every layer that exists at that moment.

    WATCH OUT FOR

    Picking 32 counts the strips (layers), not the creases between them. Picking 5 counts one crease per fold and forgets that a fold creases every layer of the stack. Picking 63 comes from $2^6-1$, which is an off-by-one in the number of folds.

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Electronics and Communication Engineering Q.11 to Q.65

1. Q.11MCQ · 1 markOfficial answer: C

   Engineering Mathematics › Linear algebra

   The error is shortest when it is perpendicular to ${\mathbf{v}}_2$, which makes $\alpha {\mathbf{v}}_2$ the orthogonal projection of ${\mathbf{v}}_1$ onto ${\mathbf{v}}_2$. So $\alpha =\frac{{\mathbf{v}}_1\cdot {\mathbf{v}}_2}{{\mathbf{v}}_2\cdot {\mathbf{v}}_2}$ (least squares with a single unknown).

   WATCH OUT FOR

   Swapping the roles of the vectors and dividing by $\Vert {\mathbf{v}}_1{\Vert }^2=5$ (which gives $4/5$), or inverting the ratio to get $7/2$ (option A). The denominator is the squared length of the vector you are projecting onto, here ${\mathbf{v}}_2$, so it is $14$. A sign slip in the dot product leads to the negative options B or D.

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2. Q.12MCQ · 1 markOfficial answer: B

   Engineering Mathematics › Vector calculus

   The rate of increase in a direction is the directional derivative, $\nabla f\cdot \hat{v}$. You must dot the gradient with the unit vector along $\mathbf{v}$, not with $\mathbf{v}$ as given.

   WATCH OUT FOR

   The usual slip is to dot with $\mathbf{v}=(2,1,2)$ without dividing by $|\mathbf{v}|=3$. That gives $4$, which is option (D). The other slip is to answer with $|\nabla f|=2$ (option C), which is the largest rate of increase over all directions, not the rate along $\mathbf{v}$.

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3. Q.13MCQ · 1 markOfficial answer: A

   Engineering Mathematics › Complex analysis

   The $n$-th roots of $r{e}^{j\theta }$ are $r^{1/n}{e}^{j(\theta +2k\pi )/n}$ for $k=0,1,\dots ,n-1$. They sit on a circle of radius $r^{1/n}$, spaced $2\pi /n$ apart. For $16j$ that means radius $2$, starting angle $\pi /8$ and spacing $\pi /2$. Any option whose angle is not $\pi /8$ plus a multiple of $\pi /2$ is not a root, and raising it to the fourth power confirms this.

   WATCH OUT FOR

   A common mistake is to take the argument of $16j$ as $\pi /4$ instead of $\pi /2$, or to divide only the principal argument by 4 and forget the $2k\pi /4=k\pi /2$ spacing. Either slip gives the wrong set of roots, and then (A), with angle $2\pi /8=\pi /4$, looks like a valid root. Another slip is to read "cannot be" as "can be" and pick (B), the principal root.

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4. Q.14MCQ · 1 markOfficial answer: B

   Engineering Mathematics › Complex analysis

   Only the poles inside the contour count. The circle is centred at $-1+\frac{3}{2}j$ with radius 1, so it contains $-1+j$ (distance $\frac{1}{2}$) but not $-1-j$ (distance $\frac{5}{2}$). The integral is therefore $2\pi j$ times the single residue $\frac{1+j}{2j}$.

   WATCH OUT FOR

   Reading the centre as $1-\frac{3}{2}j$ instead of $-1+\frac{3}{2}j$. The condition $|z+1-\frac{3}{2}j|=1$ means $|z-(-1+\frac{3}{2}j)|=1$, and with the wrong sign the circle encloses neither pole, which gives a wrong value of 0. A second mistake is adding the residues at both poles, which gives $2\pi j$ (the full integral around a large circle) and is not one of the options.

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5. Q.15MCQ · 1 markOfficial answer: C

   Engineering Mathematics › Linear algebra

   Similar matrices have the same characteristic polynomial, so they have the same eigenvalues. Their eigenvectors are different, though: they are linked by the same matrix that links the two matrices. Write $A=PB{P}^{-1}$. If $B\mathbf{v}=\lambda \mathbf{v}$, then $A(P\mathbf{v})=\lambda (P\mathbf{v})$, so each eigenvector of $A$ is $P$ times an eigenvector of $B$.

   WATCH OUT FOR

   The usual slip is getting the direction of $P$ wrong. Students see $B=P^{-1}AP$ and pick $P^{-1}{\mathbf{v}}_k$ (D), when solving for $A$ gives $A=PB{P}^{-1}$ and so the eigenvectors are $P{\mathbf{v}}_k$. The other slip is assuming similar matrices share eigenvectors as well as eigenvalues (B). To check the direction, substitute the answer back: $A(P{\mathbf{v}}_k)=PB{P}^{-1}P{\mathbf{v}}_k=PB{\mathbf{v}}_k={\lambda }_{k}P{\mathbf{v}}_k$.

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6. Q.16MCQ · 1 markOfficial answer: A

   Electronic Devices › Semiconductors

   In a non-degenerate semiconductor, the Fermi level $E_F$ lies within the forbidden energy gap. When the doping (donor or acceptor) concentration becomes very high (typically approaching or exceeding the effective density of states, $N_D\gtrsim N_C$ or $N_A\gtrsim N_V$), the semiconductor becomes degenerate and $E_F$ moves out of the gap into the conduction band (for very heavy donor/n-type doping) or into the valence band (for very heavy acceptor/p-type doping).

   WATCH OUT FOR

   Confusing "degenerate" with mere heavy doping without connecting it to the Fermi level actually crossing into the band edge; or reversing n-type/p-type by forgetting that donor (electron) doping pushes E_F upward toward the conduction band while acceptor doping pushes it downward toward the valence band.

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7. Q.17MCQ · 1 markOfficial answer: A

   Electronic Devices › Semiconductors

   At absolute zero, an intrinsic semiconductor behaves like a perfect insulator: with no thermal energy available, no electron can be excited across the bandgap, so the valence band stays completely full and the conduction band stays completely empty.

   WATCH OUT FOR

   Confusing "intrinsic" with "no free carriers ever" versus the T = 0 K condition, or assuming some residual thermal population exists at absolute zero; also mixing up which band is which (some students reverse valence/conduction in option (B), thinking of a hypothetical population-inverted or superconducting-like state, which never occurs in an ordinary semiconductor).

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8. Q.18MCQ · 1 markOfficial answer: D

   Networks (Electric Circuits) › AC, power & resonance

   For a series RLC circuit, ${\omega }_0=1/\sqrt{LC}$ and $Q={\omega }_{0}L/R=\frac{1}{R}\sqrt{L/C}$. With $L=1\mu \text{H}$ and $C=1\mu \text{F}$, the characteristic impedance $\sqrt{L/C}$ is $1\Omega$, so $Q=1/R$ and $Q=1000$ needs $R=1\text{m}\Omega$.

   WATCH OUT FOR

   Using the parallel RLC formula $Q=R\sqrt{C/L}$ (or $Q={\omega }_{0}RC$). In a series circuit a smaller $R$ gives a larger $Q$. The parallel formula flips that and points to $R=1000\Omega$, which is not an option, or pushes a student to guess (A).

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9. Q.19MCQ · 1 markOfficial answer: B

   Electronic Devices › MOS devices

   In a MOS capacitor, the depletion width under the gate is set by the surface potential ${\varphi }_s$: it is zero in accumulation, grows through depletion as ${\varphi }_s$ increases with $V_g$ (starting from zero exactly at the flat-band point $V_{fb}$), and saturates at its maximum value once strong inversion sets in at $V_g=V_t$ — the inversion charge then screens the semiconductor, so $W_{dep}$ stays pinned at $W_{dep,max}$ for all $V_g>V_t$.

   WATCH OUT FOR

   Picking a curve whose rise or saturation point is offset from the actual $V_{fb}$/$V_t$ markers (option A), assuming $W_{dep}$ continues to grow (or falls) in inversion instead of saturating (options C, D), or confusing the depletion-width-vs-gate-voltage curve with the C–V curve of the same device.

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

    Electromagnetics › Plane waves

    In an unbounded lossless dielectric with constant ${\epsilon }_r$ and ${\mu }_r$, $\beta =\frac{\omega }{c}\sqrt{{\mu }_r{\epsilon }_r}$ is linear in $\omega$. The medium is therefore non-dispersive, and the phase and group velocities are equal: $v_p=v_g=\frac{c}{\sqrt{{\mu }_r{\epsilon }_r}}$. With ${\mu }_r{\epsilon }_r=4>1$, both velocities equal $c/2$, which is less than $c$.

    WATCH OUT FOR

    Using the waveguide result $v_p{v}_g=c^2$ and so picking (C), $v_p>c$ and $v_g<c$. That relation comes from the dispersive cut-off behaviour of an air-filled hollow waveguide. A plane wave in an unbounded lossless dielectric has no cut-off and no dispersion, so $v_p=v_g=c/2$. The words 'narrow band' only make group velocity meaningful. They do not make the medium dispersive.

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

    Analog Electronics › Amplifiers

    Classify an amplifier by its port impedances. A low input impedance means it takes a current in, and a high output impedance means it delivers a current. A common-gate MOSFET has $R_{in}\approx 1/g_m$ at the source, $R_{out}$ of the order of $r_o$ at the drain, and $i_d=i_s$, so it is a current buffer, which is a CCCS.

    WATCH OUT FOR

    Students often classify from the labels instead of the impedances. Because the input is drawn as a voltage $v_{IN}$ and the output is labelled $v_{OUT}$, they pick VCVS, or VCCS by thinking of a transconductance stage. Others count $R_1$ as part of the amplifier's input impedance. $R_1$ belongs to the driving source (Norton current $v_{IN}/R_1$), and the amplifier's own input, the NMOS source, has an impedance of only about $1/g_m$. A third error is to read the NMOS as a source follower, which would give low output impedance. Its gate is at AC ground and the output is at its drain, so it is common-gate.

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12. Q.22MCQ · 1 markOfficial answer: Marks to all

    Analog Electronics › Feedback & oscillators

    The loop gain $L=\beta A$ decides oscillation, not the closed-loop gain. A negative-feedback loop oscillates when $|L|$ is still above 1 at the frequency where the phase lag reaches $180^{\circ }$. Put another way, the phase crossover comes before the gain crossover, which is a negative phase margin. Each single-pole CS stage adds less than $90^{\circ }$ of lag, so at least three stages are needed. A three-stage ring oscillates at ${\omega }_{180}=\sqrt{3}{\omega }_p$ once each stage's gain exceeds 2.

    WATCH OUT FOR

    Taking "closed loop gain" literally. A stable unity-feedback amplifier has a closed-loop gain of about 1, and at oscillation that gain is unbounded, so it is not the test. Also: (1) choosing (C) on the idea that more than $180^{\circ }$ of phase shift is enough on its own, when the loop gain must still exceed 1 where the lag reaches $180^{\circ }$; (2) believing two single-pole stages can reach $180^{\circ }$, when they only approach it as $\omega \to \infty$, where the gain has gone to zero; (3) confusing the start-up condition ($|L|>1$ at $180^{\circ }$) with the Barkhausen condition for a steady amplitude ($|L|=1$ at exactly $180^{\circ }$).

    Full solution →
13. Q.23MCQ · 1 markOfficial answer: A

    Digital Electronics › Combinational

    A $2\times 1$ MUX computes $Y=\overline{S}{I}_0+S{I}_1$. Tie one data input to a constant and feed a variable into the other, and it becomes a basic gate. With $I_0=0$ and $I_1=P$, the output is $Y=SP$. That is an AND gate between the select signal and $P$.

    WATCH OUT FOR

    Getting the select convention backwards, as if $S=1$ picked $I_0$. That gives $Y=\overline{Q}P$, which matches none of the options, and some students then guess OR or NAND. Another slip is reading the grounded input as logic 1 instead of 0.

    Full solution →
14. Q.24MCQ · 1 markOfficial answer: A

    Digital Electronics › Sequential

    In a pipeline driven by one clock, throughput is set by the clock alone: one bit per cycle, so $f_{clk}$ bits/s whatever the number of stages. Latency is set by the depth: $N$ stages take $N$ clock periods. Here that gives $1\text{bit/ns}=1000\text{Mbits/s}$ and $3\times 1\text{ns}=3\text{ns}$.

    WATCH OUT FOR

    Dividing the clock rate by the number of stages (1000/3 = 333.33 Mbits/s, options B and D), as if each bit had to clear the whole register before the next could enter. In fact all three stages hold different bits at once, so a bit leaves on every clock. A second mistake is to take the latency as one clock period (1 ns, option B) when a bit has to pass through all three flip-flops.

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15. Q.25MCQ · 1 markOfficial answer: A

    Control Systems › Bode & margins

    Phase crossover is where $\angle G(j\omega )=-180^{\circ }$, and the gain $k$ has no effect on it. The integrator already gives $-90^{\circ }$, so the two lag angles must add up to $90^{\circ }$. That happens when $\omega T_1\cdot \omega T_2=1$. So ${\omega }_{pc}$ is the geometric mean of the two corner frequencies $1/T_1$ and $1/T_2$.

    WATCH OUT FOR

    Picking (B), $1/(T_1{T}_2)$, by solving $\omega T_1{T}_2=1$ instead of ${\omega }^2{T}_1{T}_2=1$, or by multiplying the corner frequencies instead of taking their geometric mean. Another mistake is to forget the $-90^{\circ }$ from the pole at the origin and require the two lags to reach $180^{\circ }$, which they never do. Students also mix up phase crossover (phase $=-180^{\circ }$) with gain crossover ($|G|=1$, which does depend on $k$).

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16. Q.26MCQ · 1 markOfficial answer: B

    Signals and Systems › LTI & convolution

    When the output is the input evaluated at a warped time $y(t)=x(f(t))$: it is causal only if $f(t)\le t$ for every $t$, and time invariant only if $f(t)=t-c$ (a pure shift). Here $f(t)=e^t$ is greater than $t$ everywhere and is not a pure shift, so the system is non-causal and time varying.

    WATCH OUT FOR

    Delaying the output wrongly by replacing $t$ with $t-t_0$ only in the input and writing $y(t-t_0)=x(e^t-t_0)$, which makes the system look time invariant. Another slip is checking causality only at $t=0$, where $e^0=1>0$ is easy to overlook, or assuming a very negative $t$ makes $e^t$ lie in the past. In fact $e^t>t$ for every $t$.

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17. Q.27MCQ · 1 markOfficial answer: B

    Signals and Systems › Fourier transform

    Multiplying by $\cos {\omega }_{0}t$ splits the spectrum into two half-amplitude copies at $\pm {\omega }_0$. When ${\omega }_0$ exceeds the signal bandwidth the copies do not overlap. Each copy then carries ${\left(\frac{1}{2}\right)}^{2}E=E/4$, so the total is $E/2$. It is the same as saying the average of ${\cos }^2$ is $\frac{1}{2}$.

    WATCH OUT FOR

    Picking $E$ because $|\cos {\omega }_{0}t|\le 1$ and the cosine "only shifts the spectrum", which forgets the factor $\frac{1}{2}$ on each copy. Or picking $E/4$ by squaring that $\frac{1}{2}$ and counting only one of the two copies. The given ${\omega }_0=10B$ is there to guarantee there is no spectral overlap. Without it the cross term would not vanish and the energy would not be exactly $E/2$.

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

    Signals and Systems › Fourier transform

    A Gaussian transforms to a Gaussian. Complete the square in $t^2+j\omega t$ and the leftover term gives the factor $e^{-{\omega }^2/4}$. The remaining integral is the given $\sqrt{\pi }$. In general $e^{-a{t}^2}\leftrightarrow \sqrt{\frac{\pi }{a}}{e}^{-\frac{{\omega }^2}{4a}}$ for $a>0$; this question is the case $a=1$.

    WATCH OUT FOR

    Getting the sign of the completed-square term wrong. Since ${\left(\frac{j\omega }{2}\right)}^2=-\frac{{\omega }^2}{4}$, the leftover is $+\frac{{\omega }^2}{4}$ inside $e^{-(\cdot )}$, which gives a decaying $e^{-{\omega }^2/4}$. Dropping the $j$ produces the growing $e^{+{\omega }^2/4}$ (the trap in (A)). Another slip is to use the Gaussian-density form $e^{-t^2/2}\leftrightarrow \sqrt{2\pi }{e}^{-{\omega }^2/2}$, which leads to the $-\frac{{\omega }^2}{2}$ exponent in (D). Setting $\omega =0$ is a fast check: it must give $\int x(t)dt=\sqrt{\pi }$.

    Full solution →
19. Q.29MCQ · 1 markOfficial answer: B

    Signals and Systems › Fourier transform

    Periodicity and discreteness trade places between time and frequency. A signal that is periodic in one domain is discrete in the other, and one that is discrete in one domain is periodic in the other. So CTFT gives continuous and aperiodic, CTFS gives discrete and aperiodic, DTFT gives continuous and periodic, and DTFS/DFT gives discrete and periodic.

    WATCH OUT FOR

    Picking option (A) by assuming the spectrum keeps the time signal's own properties, e.g. that a continuous periodic signal has a continuous periodic spectrum. The properties swap roles: periodicity in time makes the spectrum discrete (ii → c), and sampling in time makes the spectrum periodic (iii → b). Option (D) wrongly gives the DTFT of an aperiodic sequence a discrete spectrum. The DTFT is a continuous function of $\Omega$; only a periodic sequence has a discrete spectrum.

    Full solution →
20. Q.30MSQ · 1 markOfficial answer: A, B
21. Q.31NAT · 1 markOfficial answer: 10

    Digital Electronics › ADC & DAC

    For an ideal $n$-bit ADC driven by a full-scale sine wave, the signal power is $V_m^2/2$ and the quantisation noise power is ${\Delta }^2/12$ with $\Delta =2V_m/2^n$. That gives ${\text{SNR}}_{\text{dB}}=6.02n+1.76$. Solve it for $n$.

    WATCH OUT FOR

    Using the rough '6 dB per bit' rule and leaving out the 1.76 dB term: $61.96/6.02\approx 10.3$ happens to round to the right answer here, but the method is wrong and fails on other values. Taking the signal power as $V_m^2$ instead of $V_m^2/2$, or the full-scale range as $V_m$ instead of $2V_m$, moves the constant away from 1.76 dB.

    Full solution →
22. Q.32NAT · 1 markOfficial answer: 1.30 to 1.40

    Networks (Electric Circuits) › Laws & node/mesh

    Use repeated source transformation. Everything between node $A$ and ground reduces to $2\text{V}$ behind $1\text{k}\Omega$. The $1\text{mA}$ source across the $1\text{k}\Omega$ becomes a series $1\text{V}$ source whose + terminal faces $B$. That leaves a single loop: $3\text{V}$ across $2.2\text{k}\Omega$. Two-node nodal analysis gives the same result.

    WATCH OUT FOR

    Getting the polarity of the transformed $1\text{mA}\parallel 1\text{k}\Omega$ source wrong. A current source pointing from $A$ to $B$ becomes a voltage source with its + terminal at $B$, so it adds to the $2\text{V}$. Subtracting it gives $\frac{1}{2.2}\approx 0.45\text{mA}$. A second trap is dropping the upward $1\text{mA}$ source when combining the branches at $A$, which gives $\frac{2}{2.2}\approx 0.91\text{mA}$.

    Full solution →
23. Q.33NAT · 1 markOfficial answer: 80

    Networks (Electric Circuits) › Two-port

    For maximum power transfer the load must equal the conjugate of the Thevenin impedance seen at port 2. To find it, set the source to zero, which leaves port 1 closed through the 10 Ω source resistance, so $I_1=-V_1/10$. Put that into the Y equations and read off $V_2/I_2$. In one line: $Y_{out}=y_{22}-\frac{y_{12}{y}_{21}}{y_{11}+Y_s}$.

    WATCH OUT FOR

    Using $1/y_{22}=75\Omega$, which ignores the 10 Ω source resistance that loads port 1 (that is the output impedance with port 1 shorted). A second slip is the sign in $V_1=-10I_1$: $I_1$ is defined as entering the network, so it flows through the resistor toward the port. Writing $V_1=+10I_1$ gives $Y_{out}=\frac{4}{300}-\frac{10^{-4}}{0.02-0.1}$ and $Z_{th}\approx 69\Omega$, which is wrong.

    Full solution →
24. Q.34NAT · 1 markOfficial answer: 02

    Digital Electronics › Sequential

    In a cross-coupled NAND (SR) latch, the worst case is an input change that switches its own gate and then, through the feedback wire, switches the other gate. That makes two gate delays. Once the second output reaches the first gate, it only confirms the value already there, so the loop does not keep going round.

    WATCH OUT FOR

    Answering 1 ns, which counts only the gate the input drives and ignores that $\overline{Q}$ settles one gate delay later. The opposite mistake is thinking the feedback loop adds delays without end (or a third delay). The feedback signal arrives at the first gate without changing its output, so the circuit is settled after 2 ns.

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25. Q.35NAT · 1 markOfficial answer: 04

    Networks (Electric Circuits) › Transients

    The inductor current stays at its pre-switching value (2 A, set only by the 1 Ω resistor because the inductor shorts the 2 Ω). Once the switch opens, that current has nowhere to go except back through the 2 Ω resistor, so the resistor voltage jumps from 0 to 2 A × 2 Ω = 4 V in magnitude and then decays with τ = L/R = 0.5 s.

    WATCH OUT FOR

    Answering 0 V because the 2 Ω resistor carried no current before switching. Resistor voltage can jump; only the inductor current is continuous. A second mistake is using the full 3 Ω series path to find i_L(0⁻) (giving 2/3 A and |V_R| ≈ 1.33 V). In DC steady state the inductor shorts out the 2 Ω, so i_L(0⁻) = 2 A. The sign of V_R(0⁺) is negative, but the question asks for the magnitude.

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26. Q.36MCQ · 2 marksOfficial answer: B

    Engineering Mathematics › Probability & statistics

    When a transformation $Y=h(X)$ is constant at $y_0$ over an interval of $x$ that has non-zero probability, $Y$ has a point mass at $y_0$, and its PDF picks up an impulse $P(\text{interval})\delta (y-y_0)$. Where $h$ is strictly monotonic, the density is continuous: $f_Y(y)=f_X(x)/|h^{\prime }(x)|$. So you get one delta for each flat level of the graph, and they sit at the output values ($y$), not at the $x$ breakpoints.

    WATCH OUT FOR

    There are two traps. First, placing the impulses at the $x$-axis breakpoints $\pm 2$ instead of at the output levels $y=\pm 1$, which leads to (C) or (D). Second, overlooking that the segment lying along the $x$-axis for $-1\le x\le 1$ is also a flat piece. It carries about 68% of the probability, so it produces the impulse $b\delta (y)$ at $y=0$. Missing it leads to (A).

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

    Engineering Mathematics › Vector calculus

    $\nabla \times \vec{F}=\vec{0}$, so $\vec{F}=(z^2,3y^2,2xz)$ is the gradient of $\varphi =x{z}^2+y^3$. The integral therefore depends only on its endpoints: $\varphi (Q)-\varphi (P)=29-5=24$. Saying the path is a straight line only reassures you. It does not change the answer.

    WATCH OUT FOR

    Stopping at $\varphi (Q)=29$ and forgetting to subtract $\varphi (P)=5$ picks the distractor (C). A sign error with $dz=-dt$ when parametrising also goes wrong: $z$ falls from 2 to 1 along the line, so the $2xzdz$ term has to be negative. Another mistake is adding $x{z}^2$ twice when building the potential, once from $z^{2}dx$ and once from $2xzdz$. It is a single term.

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

    Engineering Mathematics › Linear algebra

    The diagonal entries of the outer product $\mathbf{x}{\mathbf{x}}^T$ are $x_i^2$, so its trace is $\sum x_i^2={\mathbf{x}}^T\mathbf{x}$, the squared length. You can also get this from $\mathrm{tr}(AB)=\mathrm{tr}(BA)$: it turns the $n\times n$ outer product into the $1\times 1$ inner product.

    WATCH OUT FOR

    Treating the trace as the length itself and choosing $l$ (option C). That forgets that $l$ is defined as the square root of ${\mathbf{x}}^T\mathbf{x}$. The other trap is mixing up the outer product $\mathbf{x}{\mathbf{x}}^T$ (an $n\times n$ matrix) with the inner product ${\mathbf{x}}^T\mathbf{x}$ (a scalar) and then guessing a scaling factor such as $\frac{1}{2}$ or $\frac{1}{4}$ that is not there.

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

    Analog Electronics › Op-amps

    With the same $R_1$ and $R_2$, the inverting gain $-R_2/R_1$ and the non-inverting gain $1+R_2/R_1$ always add to exactly 1. An equal-weight summer that adds the two outputs therefore sees just $V_{IN}$, and the whole circuit reduces to the summer's inverting gain $-R_4/R_3$.

    WATCH OUT FOR

    One mistake is to read the bottom op-amp as a second inverting stage because it also has an $R_1$ and an $R_2$. In fact $V_{IN}$ drives its + input, so it is non-inverting, and $R_1$ goes to ground. Misread this way, the two outputs add to $-2(R_2/R_1)V_{IN}$ and $R_2/R_1$ does not cancel. Another mistake is to forget that the output stage is an inverting summer (its + input is grounded) and choose (B) $+R_4/R_3$. Treating it as a non-inverting stage gives (C).

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

    Analog Electronics › Diode circuits

    With both diodes off, no current flows in the resistor, so the output sits at the $1\text{V}$ source voltage, not at $0\text{V}$. Each diode switches on once $V_{IN}$ is $0.7\text{V}$ away from that $1\text{V}$ level: $D_1$ at $V_{IN}=1.7\text{V}$ and $D_2$ at $V_{IN}=0.3\text{V}$. The result is a flat segment centred on $1\text{V}$, with the output following the input at unit slope on either side.

    WATCH OUT FOR

    Measuring the diode thresholds from $0\text{V}$ instead of from the $1\text{V}$ bias. That gives a dead zone of $\pm 0.7\text{V}$ around the origin (option B) or a flat region at $\pm 1.7\text{V}$ (option C). The other trap is taking $V_{OUT}=0$ while both diodes are off (option D). With no current in the $1\text{k}\Omega$ resistor, the output equals the source voltage, $1\text{V}$.

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

    Control Systems › Time response

    In a stable type-1 unity-feedback loop, a ramp $R(s)=\alpha /s^2$ leaves a steady-state error $e_{ss}=\alpha /K_v$, where $K_v={\lim }_{s\to 0}sG(s)$. Here the $(s+2)$ factor makes $K_v=k/2$, not $k$.

    WATCH OUT FOR

    Taking $K_v=k$ by ignoring the $(s+2)$ factor, which gives $\alpha /k$ (option B). Another is using $e_{ss}=\alpha K_v$ instead of $\alpha /K_v$, which gives $\alpha /(2k)$ (option C). And the final value theorem should not be used before checking that the closed loop is stable.

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

    Control Systems › Block diagrams & SFG

    Both inputs enter the loop ahead of the single forward block $G$, and nothing but a summing junction separates them from it, so they share one forward path (gain $G$) and one characteristic denominator. Each feedback loop's gain is the product around the loop: $-G\cdot 1$ for the outer loop and $-G\cdot H$ for the inner one. That gives $\Delta =1+G+GH$, and so $G_1=G_2=G/(1+G+GH)$.

    WATCH OUT FOR

    Writing the inner loop's contribution as $H$ in place of the loop gain $GH$. The feedback signal $HY$ goes back through $G$ before it reaches $Y$, so the term is $GH$. Dropping the $G$ gives the $1+G+H$ denominators in options (B), (C) and (D). A second slip is to assume the disturbance sees a different denominator because $D$ enters inside the outer loop. Junction 2 is fed straight from junction 1, so $D$ and the outer feedback act at effectively the same point and both loops still act on it.

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

    Control Systems › State space

    This is modal decomposition. Along an eigenvector, the state transition matrix acts as a scalar: $e^{At}{\mathbf{v}}_i=e^{{\lambda }_{i}t}{\mathbf{v}}_i$. Distinct eigenvalues give independent eigenvectors, so any $\mathbf{x}(0)$ splits as $\sum {\alpha }_i{\mathbf{v}}_i$. Each part then evolves on its own as the mode $e^{{\lambda }_{i}t}{\mathbf{v}}_i$, and the free response is the weighted sum of these modes.

    WATCH OUT FOR

    The usual trap is thinking "second order" puts a 2 into the exponent and choosing (B). The order of the system is the number of modes (two terms in the sum), not a factor on ${\lambda }_i$. The term $e^{2{\lambda }_{i}t}{\mathbf{v}}_i$ is a mode of $\dot{\mathbf{x}}=2A\mathbf{x}$, not of $\dot{\mathbf{x}}=A\mathbf{x}$. Substituting it back leaves an extra factor of 2. A second slip is to take the ${\alpha }_i$ as the components of $\mathbf{x}(0)$. They are its coordinates in the eigenvector basis.

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

    Networks (Electric Circuits) › Transients

    Work out $v_C(0^{-})$ and $i_L(0^{-})$ from the DC steady state, with the capacitor open and the inductor shorted. At $t=0^{+}$, replace the capacitor with a voltage source of $v_C(0^{-})$ and the inductor with a current source of $i_L(0^{-})$, redraw the circuit for the new switch positions, and apply KCL. The sign of $i_c$ is set by the marked arrow direction.

    WATCH OUT FOR

    Answering $+1$ by dropping the sign: the capacitor current actually flows upward, against the marked $i_c$. Other traps: choosing $0.8$ because you forgot the inductor keeps drawing its $0.2\text{A}$ after the switch; choosing $0.2$ because you ignored the $25\Omega$ resistor; still treating the capacitor as open at $t=0^{+}$ and getting $i_c=0$; and thinking the open $S_1$ leaves the $1\text{A}$ source in the circuit.

    Full solution →
35. Q.45MCQ · 2 marksOfficial answer: B
36. Q.46MCQ · 2 marksOfficial answer: Marks to all

    Networks (Electric Circuits) › Two-port

    Write the gain in h-parameters, $A_v=\frac{-h_{21}{R}_L}{h_{11}(1+h_{22}{R}_L)-h_{12}{h}_{21}{R}_L}$. Each wrong option changes exactly one parameter from (A), so compare one parameter at a time. $h_{22}$ is a shunt that steals load current. $h_{12}$ is feedback that caps the gain near $1/h_{12}$. $h_{11}$ sets $i_1=V_s/h_{11}$, the current that the output source multiplies by $h_{21}$.

    WATCH OUT FOR

    Choosing (B) from the rule that a voltage amplifier needs a high input resistance. That rule is about not loading a source that has its own internal resistance. Here $V_s$ is ideal, so there is no loading to avoid: $h_{11}$ only sets $i_1=V_s/h_{11}$, and a larger $h_{11}$ means less gain. A second trap is solving the figure literally with $h_{21}{v}_1$, which makes $h_{11}$ and $h_{12}$ irrelevant, and then quietly switching to $h_{21}{i}_1$ to rule out an option.

    Full solution →
37. Q.47MCQ · 2 marksOfficial answer: A

    Signals and Systems › Fourier series

    A sum of the form ${\sum }_{n=0}^{N-1}x[n]e^{-jk\frac{2\pi }{N}n}$ is just $N{a}_k$, so you never need to compute $x[n]$. Split the sine into two complex exponentials with Euler's formula. Then use $N$-periodicity in $k$ (here $a_{-2}=a_3$) to match each exponential to a coefficient you were given.

    WATCH OUT FOR

    The usual slip is sending $e^{+j\frac{4\pi n}{5}}$ to $a_2$ instead of $a_3$. The analysis kernel carries a minus sign, so $e^{+j\frac{4\pi n}{5}}$ belongs to $k=-2\equiv 3$. Swapping $a_2$ and $a_3$ gives $+10$, option (B). Two other mistakes are leaving out the factor $N=5$ from the analysis equation, which gives $\pm 2$ (options C and D), and forgetting the $\frac{1}{2j}$ from Euler's formula.

    Full solution →
38. Q.48MCQ · 2 marksOfficial answer: C

    Signals and Systems › DTFT & DFT

    When both $X(e^{j\Omega })$ and $H(e^{j\Omega })$ are finite sums of powers of $e^{-j\Omega }$, the output transform is just their polynomial product, and each $e^{-jk\Omega }$ turns back into $\delta [n-k]$. You can check it in the time domain too: $h[n]=\delta [n]-\frac{1}{2}\delta [n-2]$, so $y[n]=x[n]-\frac{1}{2}x[n-2]$ with $x[n]=\delta [n]-\delta [n-1]+2\delta [n-3]$. That gives the same five terms.

    WATCH OUT FOR

    Sign slips in the cross terms. The product $(-e^{-j\Omega })(-\frac{1}{2}{e}^{-j2\Omega })=+\frac{1}{2}{e}^{-j3\Omega }$ adds to $2e^{-j3\Omega }$ to give $+\frac{5}{2}$, not $-\frac{5}{2}$. Also, $(2e^{-j3\Omega })(-\frac{1}{2}{e}^{-j2\Omega })=-e^{-j5\Omega }$ is negative. Getting either sign wrong leads to option (A) or (B).

    Full solution →
39. Q.49MCQ · 2 marksOfficial answer: A

    Signals and Systems › Fourier transform

    A sinusoid through an LTI system comes out as the same sinusoid, scaled by $|H(j{\omega }_0)|$ and shifted by $\angle H(j{\omega }_0)$. Here $\pi {\left(\frac{\sin Wt}{\pi t}\right)}^2$ has a triangular spectrum of height $W$ reaching zero at $\pm 2W$. Multiplying by $\cos 10Wt$ moves half of that triangle to $\pm 10W$. The input at $10.5W$ lies $0.5W$ from the centre of one copy, so $H=\frac{1}{2}\cdot W\left(1-\frac{0.5W}{2W}\right)=\frac{3W}{8}$.

    WATCH OUT FOR

    Two mistakes are common. (1) Leaving out the $\frac{1}{2\pi }$ in the multiplication property, or taking the triangle's peak as $W$ rather than the overlap length $2W$. Either error shifts $G(0)$ away from $W$ and leads to option (B) or (D). (2) Forgetting the factor $\frac{1}{2}$ from the modulation property. That gives $|H|=\frac{3W}{4}$ and an output of $\frac{15W}{2}\cos (10.5Wt)$, which is option (B).

    Full solution →
40. Q.50MCQ · 2 marksOfficial answer: D

    Signals and Systems › Sampling

    Multiplying by $\cos ({\omega }_{c}t)$ shifts a baseband spectrum of bandwidth $B$ so it sits at ${\omega }_c\pm B$. The highest frequency in $y(t)$ is therefore the largest carrier plus $B$: $12\pi \times 10^3+4\pi \times 10^3=16\pi \times 10^3$ rad/s. The top branch fills the band from 0 up to $8\pi \times 10^3$, so there is no empty band that band-pass sampling could use, and the Nyquist rate is simply $2{\omega }_{\max }$.

    WATCH OUT FOR

    Answering $2B=8\pi \times 10^3$ (option C), which ignores the fact that modulation moves the spectrum up to the carrier frequency. Another mistake is taking the DSB width $2B$ as the one-sided extent above the carrier. That gives a highest frequency of $12\pi \times 10^3+8\pi \times 10^3=20\pi \times 10^3$, and doubling it gives $40\pi \times 10^3$ (option B). A third is to apply band-pass sampling to the bottom branch alone. The sum $y(t)$ also contains the top branch, which occupies $0$ to $8\pi \times 10^3$ rad/s.

    Full solution →
41. Q.51MCQ · 2 marksOfficial answer: A

    Electromagnetics › Transmission lines

    Moving the reference planes outward along matched lossless lines of electrical lengths ${\theta }_1$ and ${\theta }_2$ keeps every $|S_{ij}|$ the same and multiplies each entry by the phase lag of the path its signal travels: $S_{ij}^{\prime }=S_{ij}{e}^{-j({\theta }_i+{\theta }_j)}$. A reflection crosses its own line twice, giving $e^{-j2{\theta }_i}$. A transmission crosses each line once, giving $e^{-j({\theta }_1+{\theta }_2)}$.

    WATCH OUT FOR

    Using $e^{+j2\theta }$ on the reflection terms (options B and C). A positive exponent is de-embedding: it moves the reference plane in towards the network, which removes line length. Adding line only ever adds phase lag. A second mistake is to give the transmission terms a different sign from the reflection terms (option D). Every entry picks up delay, because every signal path gets longer.

    Full solution →
42. Q.52MSQ · 2 marksOfficial answer: B, C

    Electromagnetics › Transmission lines

    On a lossless line the impedance at any point depends only on the SWR and where the minima are. It is purely resistive at the extremes of the standing wave: $Z_0/S$ at a voltage minimum and $Z_{0}S$ at a voltage maximum. So any load can be replaced by extending the line to a minimum or a maximum of its own standing-wave pattern and ending it there in that resistance. Because minima repeat every $\lambda /2$, the minimum 10 cm towards the generator has a copy 20 cm beyond the load, and there is a maximum 5 cm beyond the load.

    WATCH OUT FOR

    Using the 10 cm distance itself as $l_m$, or measuring the extension towards the generator instead of past the load. A second trap is pairing $Z_0/S=25\Omega$ with the position of a maximum or $Z_{0}S=100\Omega$ with a minimum (options A and D). Both give $(30.77+j19.99)\Omega$, not $Z_L$. Finally, this is an MSQ: marking only one of B and C loses the marks, because both equivalents are valid.

    Full solution →
43. Q.53MSQ · 2 marksOfficial answer: A, B, D

    Electromagnetics › Plane waves

    For $\mathbf{E}={\hat{a}}_x{C}_{1x}\cos (\omega t-\beta z)+{\hat{a}}_y{C}_{1y}\cos (\omega t-\beta z+\theta )$, travelling along $+z$, $(\mathbf{E}\times d\mathbf{E}/dt)\cdot {\hat{a}}_z=-\omega C_{1x}{C}_{1y}\sin \theta$ sets the handedness. With $\sin \theta >0$ ($E_y$ leads $E_x$) the wave is left-handed in the IEEE convention, and with $\sin \theta <0$ it is right-handed. The polarization is elliptical unless $\theta =0$ or $\pi$ (linear), or the amplitudes are equal and $\theta =\pi /2$ or $3\pi /2$ (circular).

    WATCH OUT FOR

    One mistake is using the optics convention, where the wave is viewed coming towards the observer. That reverses the handedness, so it would pick (C) alone. Another is rejecting (A) because its amplitudes are equal: equal amplitudes give a circle only when $\theta =\pi /2$ or $3\pi /2$, and with $\theta =\pi /4$ the path is a tilted ellipse. It is also a mistake to call (B) circular just because $\theta =\pi /2$, since its amplitudes are unequal.

    Full solution →
44. Q.54MSQ · 2 marksOfficial answer: B, C, D

    Electromagnetics › Transmission lines

    A lumped model of a length $\Delta z$ of line is valid if its series elements add up to $(R+j\omega L)\Delta z$ and its shunt elements add up to $(G+j\omega C)\Delta z$. The layout (L, T or $\pi$, split or unsplit) only changes terms of order $(\Delta z{)}^2$, which disappear when you derive the telegrapher's equations.

    WATCH OUT FOR

    Accepting (A) because it contains all four symbols $L$, $R$, $C$, $G$ and looks like a $\pi$-section. It has only one $C\Delta z/2$ and one $G\Delta z/2$, so the shunt admittance is halved. The opposite mistake is rejecting (B) because it is not symmetric. An unsymmetric L-section still gives the correct telegrapher's equations in the limit $\Delta z\to 0$.

    Full solution →
45. Q.55NAT · 2 marksOfficial answer: 0

    Engineering Mathematics › Calculus

    If a region is symmetric about the $y$-axis and the integrand is odd in $x$ ($f(-x,y)=-f(x,y)$), the double integral is zero. The contribution from each point $(x,y)$ cancels the one from $(-x,y)$. Always check for symmetry before integrating.

    WATCH OUT FOR

    A common error is integrating over the right half only (where $x\ge 0$ and the integrand is positive), which gives $1/3$ and rounds to $0$ by luck, without seeing why the full answer is exactly zero. Another is taking $y$ from $x$ to $x+2$ across the whole square, which mixes up the edges that bound each half. The upper and lower boundaries change at $x=0$, so the integral has to be split there.

    Full solution →
46. Q.56NAT · 2 marksOfficial answer: 2.20 to 2.30

    Electronic Devices › Carrier transport

    In an extrinsic semiconductor, electron and hole concentrations obey the mass-action law $np=n_i^2$, so knowing one concentration (here $p=1.5n_i$) fixes the other. Equal drift current densities for electrons and holes (same applied field $E$) then give a direct relation between $n$, $p$ and the mobility ratio.

    WATCH OUT FOR

    Inverting the ratio — computing ${\mu }_p/{\mu }_n$ instead of ${\mu }_n/{\mu }_p$ — or forgetting to apply the mass-action law and instead assuming $n=p$, which would (incorrectly) give a ratio of 1.

    Full solution →
47. Q.57NAT · 2 marksOfficial answer: 4

    Control Systems › Bode & margins

    In an asymptotic Bode magnitude plot, the starting slope is $-20\text{dB/decade}$ for each pole at the origin. After that, the slope rises by $+20\text{dB/decade}$ for each zero at its corner and falls by $-20\text{dB/decade}$ for each pole at its corner. So the power of each factor is the change in slope at its corner divided by 20. Minimum phase is what lets the magnitude plot fix the transfer function uniquely.

    WATCH OUT FOR

    Reading the final slope of $-40\text{dB/decade}$ as the change at ${\omega }_2$ and adding the origin pole again, which gives $c=3$ or $b=2$. The change at ${\omega }_2$ is measured from the flat $0\text{dB/decade}$ segment, so it is $-40$ and $c=2$. A second mistake is taking the flat segment to mean there is no zero. Going from $-20$ to $0$ is a $+20\text{dB/decade}$ change, so there is one zero ($a=1$).

    Full solution →
48. Q.58NAT · 2 marksOfficial answer: 15

    Signals and Systems › LTI & convolution

    The area under a convolution equals the product of the areas of the two signals, because $Y(0)=X_1(0)X_2(0)$. You never need to work out $y(t)$ itself. An impulse adds its strength (the number printed on the arrow) to the area.

    WATCH OUT FOR

    Doing the convolution by hand, and slipping on the overlapping pieces as the pulse slides past the impulses. Also: counting the impulses as zero area because they have zero width, reading the impulse at $t=2$ as strength 1 instead of 2, or taking the width of $x_1$ as 1.5 instead of 3.

    Full solution →
49. Q.59NAT · 2 marksOfficial answer: 0.24 to 0.26
50. Q.60NAT · 2 marksOfficial answer: 0.12 to 0.14

    Electromagnetics › Plane waves

    An anti-reflection coating is a quarter-wave transformer. Its index must be the geometric mean of the two media it sits between, $n_2=\sqrt{n_1{n}_3}$, and its thickness must be a quarter of the wavelength inside the coating, ${\lambda }_0/(4n_2)$, not a quarter of the free-space wavelength.

    WATCH OUT FOR

    Using the free-space wavelength gives ${\lambda }_0/4=0.1875\mu \text{m}$, which rounds to 0.19. Using the glass index instead of the coating index gives ${\lambda }_0/(4\times 2)\approx 0.094\mu \text{m}$. Both answers fall outside the key. Another slip is taking $n={\epsilon }_r$ (4) rather than $n=\sqrt{{\mu }_r{\epsilon }_r}$ (2).

    Full solution →
51. Q.61NAT · 2 marksOfficial answer: 60.00 to 70.00

    Electronic Devices › Semiconductors

    Hole concentration is $p=N_V{e}^{-(E_F-E_V)/kT}$. Both $N_V$ (through the $T^{3/2}$ density-of-states scaling) and $kT$ change with temperature, while the doping-fixed energy gap $(E_F-E_V)$ is taken as unchanged between 300 K and 400 K.

    WATCH OUT FOR

    Scaling only $N_V$ with temperature but forgetting to also scale $kT$ (i.e. plugging $400\text{K}$ carrier density with the $300\text{K}$ value of $kT=0.026\text{eV}$ in the exponential), which changes the exponent from about $10.10$ to about $13.46$ and gives an answer far outside the accepted range.

    Full solution →
52. Q.62NAT · 2 marksOfficial answer: 8.30 to 8.34

    Digital Electronics › ADC & DAC

    An RC sample-and-hold settles exponentially. The capacitor has settled to $N$-bit accuracy once the leftover error $V_{FS}{e}^{-t/\tau }$ has fallen to one LSB, $V_{FS}/2^N$. That gives $t=\tau \ln (2^N)=N\tau \ln 2$, which does not depend on the full-scale voltage.

    WATCH OUT FOR

    Using a half-LSB error limit, $V_{FS}/2^{13}$, gives $13\ln 2\approx 9.01\mu \text{s}$, which falls outside the key. The key here uses a one-LSB limit. Other slips: taking ${\log }_{10}$ where the natural log is needed ($12{\log }_{10}2\approx 3.61\mu \text{s}$), and setting $e^{-t/\tau }$ equal to $1/12$ when it should be $1/2^{12}$.

    Full solution →
53. Q.63NAT · 2 marksOfficial answer: 250, or 500

    Digital Electronics › Sequential

    Write $Q^{+}=D$ for each flip-flop and step the state table from the given initial state until it repeats. The number of states in the cycle is the division ratio: an output that repeats every $N$ clocks has frequency $f_{clk}/N$. A shift loop with exactly one inversion (a twisted-ring, or Johnson, counter) of $n$ flip-flops has $2n$ states, so here $N=4$.

    WATCH OUT FOR

    Reading the diagonal wire as coming from $Q_1$ instead of $\overline{Q_1}$. With no inversion the loop is a plain ring counter, $10\to 01\to 10$, which gives a 2-state cycle and 500 kHz. (The official key accepts both 250 and 500, but the figure as printed gives 250.) A second slip is to count state changes of $Q_2$ rather than full periods: $Q_2$ changes every 2 clocks, but one full period is 4 clocks.

    Full solution →
54. Q.64NAT · 2 marksOfficial answer: 2.00

    Analog Electronics › Amplifiers

    This is a cascade of current mirrors. When transistors share a gate-source voltage and are in saturation with $\lambda =0$, their drain currents scale with $W/L$. The PMOS mirror multiplies the 1 mA reference to 10 mA and 7 mA. The NMOS mirror then scales the 10 mA to 5 mA. KCL at the output node leaves $7-5=2\text{mA}$ for the 1 kΩ load.

    WATCH OUT FOR

    There are three common slips. The first is taking the NMOS ratio against the 1 mA source instead of the 10 mA branch, which gives $I_{D5}=5/1\times 1=5$ mA by luck or $5/10\times 1=0.5$ mA by mistake. The second is scaling the current through $M_5$ from $M_3$'s 7 mA, when $M_5$ is mirrored from $M_4$'s branch. The third is adding the two currents at the output node (12 mA, so 12 V) when $M_3$ sources current into the node and $M_5$ sinks current out of it.

    Full solution →
55. Q.65NAT · 2 marksOfficial answer: 1.97 to 1.99