GATE ECE 2020 question paper with solutions

GATE ECE 2020 answer key

General Aptitude GA Q.1 to Q.10

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

   General Aptitude › Grammar & vocabulary

   Prepositions are fixed by collocation. "Killed" takes "in" before an event (killed in an accident, a war or a fire). "Die" takes "of" before a disease or internal cause (die of cancer), and "from" before a wound or external cause (die from injuries). The first blank alone is enough to answer, because only one option begins with "in".

   WATCH OUT FOR

   Picking (D) "from, from" because "die from a disease" sounds acceptable in everyday speech. The first blank rules it out: "killed from accidents" is not idiomatic English. Another mistake is picking (C) because "during" sounds formal. "Killed during accidents" gives only the time of the deaths, not their cause.

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

   General Aptitude › Grammar & vocabulary

   Correlative conjunctions come in fixed pairs. Once "not only" appears, the blank must hold its partner "but also", which adds the second item to the first.

   WATCH OUT FOR

   Choosing "rather than" because it also links two items. It replaces the first item with the second, and that clashes with "not only", which keeps both. Some candidates also pick "but even" because it sounds emphatic, but it is not the standard pair and it implies a surprise the sentence does not suggest.

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

   General Aptitude › Reasoning

   Work out how the first pair is related (here, they are opposites) and then find the word that has the same relation to the third word. Go by meaning, not by prefixes: explicit/implicit are opposites, so the blank takes the opposite of express, which is repress.

   WATCH OUT FOR

   Picking (A) Impress because it copies the ex- to im- prefix change of explicit/implicit. That is a spelling pattern, not a relation of meaning. The other trap is (D) Suppress. It can mean holding something back, but its main sense is putting something down by force. The usual opposite of expressed feelings is repressed feelings.

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

   General Aptitude › Reading

   Answer from what the passage actually states. The words "as opposed to" mark the complaint as a contrast between the two languages, and the only contrast offered in the options is length: 15 seconds of English against 5 seconds of French. Under a rule of equal importance, that gap is the grievance.

   WATCH OUT FOR

   The usual mistake is picking (A) because the complaint was "about" the English announcements. The couple did not object to English being used, since the constitution protects both languages. They objected to English getting more time than French. Some candidates also pick (B) because they read "formal" as "clearer", but the passage never mentions clarity.

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

   General Aptitude › Numerical

   Superadditive means the value at a sum is at least the sum of the values. For $e^x$ the value at a sum is a product, $e^{x_1}{e}^{x_2}$, and a product of two numbers larger than 2 beats their sum. Written exactly: $ab\ge a+b⟺(a-1)(b-1)\ge 1$. The other three functions are concave or decreasing on $x>1$, so for them the value at the sum comes out smaller than the sum of the values.

   WATCH OUT FOR

   Reading the inequality backwards and choosing $\sqrt{x}$ because it "grows". For $\sqrt{x}$, $\sqrt{x_1+x_2}<\sqrt{x_1}+\sqrt{x_2}$ always holds, so it is subadditive. A second trap is to forget the condition $x>1$. On $(0,1)$, $e^{x_1}{e}^{x_2}\ge e^{x_1}+e^{x_2}$ can fail: at $x_1=x_2=0.1$ the left side is $e^{0.2}\approx 1.22$ and the right side is $2e^{0.1}\approx 2.21$. The claim for (A) rests on $e^x>e>2$.

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

   General Aptitude › Reading

   Turn every causal phrase in the passage into an arrow: "because of … following" gives East Asian → sub-prime, "led to" gives sub-prime → banking, and "started with" gives sub-prime → global. Then rule out every option that reverses any one of those arrows. Only (A) keeps all three.

   WATCH OUT FOR

   Ordering the events by where they appear in the passage (the global crisis is mentioned first and the East Asian crisis last) instead of by cause and effect. The East Asian crisis is mentioned last, but it is the earliest cause. Another slip is reading "started with" backwards and putting the global crisis before the sub-prime crisis, as (D) does.

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

   General Aptitude › Numerical

   The hour hand keeps moving between the hour marks at $0.5^{\circ }$ per minute, while the minute hand moves at $6^{\circ }$ per minute. So at 3:15 the minute hand is on the 3 but the hour hand has moved $15\times 0.5^{\circ }=7.5^{\circ }$ past it. In general, $\theta =\left|30H-\frac{11}{2}M\right|$, and if that comes out above $180^{\circ }$ you take $360^{\circ }-\theta$.

   WATCH OUT FOR

   Picking $0^{\circ }$ (A) because both hands seem to point at the 3. That treats the hour hand as fixed on the hour mark, but in 15 minutes it has moved a quarter of the $30^{\circ }$ gap between 3 and 4, which is $7.5^{\circ }$.

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

   General Aptitude › Geometry

   A rectangle inscribed in a circle always has a diameter ($2a$) as its diagonal. For a fixed diagonal, $x^2+y^2=4a^2$ is fixed, and the product $xy$ is largest when $x=y$. So the maximum-area inscribed rectangle is a square of area $2a^2$, and the shaded area is $\pi a^2-2a^2$.

   WATCH OUT FOR

   Taking the measurements from the figure. The rectangle is drawn wider than it is tall, but "maximum possible area" forces a square, so the drawing is not to scale. Another common slip is to treat $a$ as the half-side or as the whole diagonal instead of the radius. That gives a square area of $a^2$ or $\sqrt{2}{a}^2$ and leads to options (A) or (B).

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

   General Aptitude › Numerical

   For the double root $\beta$ of $a{x}^2-bx+c=0$, the sum of the roots gives $2\beta =b/a$ and the product gives ${\beta }^2=c/a$. Multiplying the two gives ${\beta }^3=bc/(2a^2)$ directly.

   WATCH OUT FOR

   Forgetting that the two equal roots add up to $2\beta$, not $\beta$, which leads to (A). Also: taking the sum as $-b/a$ without noticing that the equation is written with $-bx$ (this flips the sign of $\beta$), or writing the product of the roots as $ac$ instead of $c/a$, which leads to (B).

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

    General Aptitude › Data interpretation

    Both averages are taken over the same 5 years, so the ratio of the averages equals the ratio of the totals: $\sum P:\sum (Q-P)=23:(31-23)=23:8$. Since Q is higher than P every year, the sum of the yearly differences is simply $\sum Q-\sum P$.

    WATCH OUT FOR

    Using the average of School Q ($31/5$) in place of the average difference, which gives $23:31$ (option C), or turning the ratio upside down to get $8:23$ (option A). The question asks for P's average divided by the average difference, in that order.

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Electronics and Communication Engineering Q.1 to Q.55

1. Q.1MCQ · 1 markOfficial answer: C

   Engineering Mathematics › Linear algebra

   In an $n$-dimensional space: more than $n$ vectors are always dependent, a spanning set needs at least $n$ vectors, and a set of exactly $n$ vectors is a basis as soon as it spans or is independent (either one implies the other). Having more than $n$ vectors says nothing about whether a particular subset of $n$ of them is independent.

   WATCH OUT FOR

   Confusing "some four of the vectors form a basis" (possible, if the six span ${\mathbb{R}}^4$) with "any four form a basis" (false in general), and so picking (C) as true. A second trap is reading (A) as "the vectors do not span ${\mathbb{R}}^4$". It only says spanning is not guaranteed, which is true.

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2. Q.2MCQ · 1 markOfficial answer: C

   Engineering Mathematics › Vector calculus

   Each field condition belongs to one operator. Solenoidal means $\nabla \cdot \vec{A}=0$ (divergence). Irrotational means $\nabla \times \vec{A}=\vec{0}$ (curl). Harmonic means ${\nabla }^2\vec{A}=0$ (Laplacian). The curl-of-curl identity $\nabla \times (\nabla \times \vec{A})=\nabla (\nabla \cdot \vec{A})-{\nabla }^2\vec{A}$ is standard. Neither of the last two conditions implies the other: $y\hat{i}$ is harmonic but not irrotational, and $3x^2\hat{i}$ is irrotational but not harmonic.

   WATCH OUT FOR

   Taking ${\nabla }^2\vec{A}=0$ for the irrotational condition because it looks like it involves the curl through the identity in (D). It doesn't: ${\nabla }^2\vec{A}=0$ is Laplace's equation, and $\nabla \times \vec{A}=\vec{0}$ is what makes a field irrotational. Another trap is picking (D) because of its sign. The minus sign in front of ${\nabla }^2\vec{A}$ is correct.

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3. Q.3MCQ · 1 markOfficial answer: B

   Engineering Mathematics › Calculus

   In a partial derivative with respect to $x$, treat $y$ and $z$ as constants. The chain rule gives the exponential term the factor $-\cos y$, and the second term is linear in $x$, so its derivative is just its coefficient $z{e}^{-1/(1+y^2)}$. Differentiate in general first, then substitute the point.

   WATCH OUT FOR

   Leaving out the chain-rule factor $-\cos y$ on the first term. That gives $e^0+e\cdot e^{-1}=2$, which is not an option, or, with the sign slipped, a wrong answer such as $1$. The other common slip is substituting $x=1$ before differentiating, which removes the $x$-dependence you are meant to differentiate.

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4. Q.4MCQ · 1 markOfficial answer: C

   Engineering Mathematics › Differential equations

   The auxiliary equation $m^2-6m+9=(m-3{)}^2=0$ has the repeated root $m=3$. A repeated root $m$ gives the pair $e^{mx}$ and $x{e}^{mx}$, so the general solution is $(C_1+C_{2}x)e^{3x}$.

   WATCH OUT FOR

   Two mistakes are common. One is getting the sign of the root wrong: the middle term $-6dy/dx$ gives the root $m=+3$, not $-3$, so (B) is wrong. The other is writing $C_1{e}^{3x}+C_2{e}^{3x}$ for a repeated root. That is really a single constant times $e^{3x}$ (the same as option D), so it cannot be the general solution of a second-order equation. The second solution must carry the factor $x$.

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5. Q.5MCQ · 1 markOfficial answer: C

   Signals and Systems › LTI & convolution

   $y[n]$ is a running maximum of $|x[k]|$ over all past and present samples, so it can only stay the same or rise. With $x[n]=\delta [n]$, the single sample of value $1$ at $k=0$ enters the window at $n=0$ and stays in it for every later $n$. The output jumps from $0$ to $1$ at $n=0$ and holds there, which is $u[n]$.

   WATCH OUT FOR

   Choosing (D), $\delta [n]$, by treating the system as if it passed the current input straight through. The maximum is taken over the whole past window $-\infty \le k\le n$, not only $k=n$, so the $1$ at $k=0$ is remembered for all $n\ge 0$. A second mistake is to use this $h[n]$ in a convolution: the system is non-linear, so $h[n]$ does not determine its response to other inputs.

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6. Q.6MCQ · 1 markOfficial answer: B

   Electronic Devices › Semiconductors

   For an intrinsic semiconductor, equating $n_i=N_c{e}^{-(E_c-E_i)/kT}$ with $p_i=N_v{e}^{-(E_i-E_v)/kT}$ gives $E_i=\frac{E_c+E_v}{2}+\frac{kT}{2}\ln \left(\frac{N_v}{N_c}\right)$, so a larger $N_v$ relative to $N_c$ pushes $E_i$ above mid-gap, toward $E_c$.

   WATCH OUT FOR

   Getting the magnitude right (using $kT\ln (2)/2$) but reversing the direction — assuming a larger $N_v$ pulls $E_i$ toward the valence band, when the correct derivation shows it pushes $E_i$ above mid-gap, toward the conduction band. Since the options here ask only for magnitude, this sign error does not change which option is correct, but it must not be stated as fact.

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7. Q.7MCQ · 1 markOfficial answer: B

   Electronic Devices › Carrier transport

   In SRH recombination through bulk traps, with equal capture cross-sections (${\tau }_n={\tau }_p=\tau$), the recombination rate $U=\frac{np-n_i^2}{\tau (n+n_1)+\tau (p+p_1)}$ is maximized (for a fixed product $np$, which is constant across the depletion region under bias) at the point where $n\approx p$ — by AM-GM, $n+p$ is minimized when $n=p=\sqrt{np}$. This point of equal carrier concentrations lies in the interior of the space-charge region, not at its edges (where one carrier type dominates), so statement (B) is the false one.

   WATCH OUT FOR

   Confusing "maximum recombination" with "maximum generation/depletion edge effects," or assuming the peak SRH rate tracks the majority-carrier profile (which would incorrectly place it at an edge). Students also sometimes forget that with ${\sigma }_n={\sigma }_p$ the trap becomes an effective midgap-like center for symmetry purposes, which is what makes the $n\approx p$ condition the maximizing condition rather than some edge-biased condition.

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8. Q.8MCQ · 1 markOfficial answer: D

   Analog Electronics › Filters & waveforms

   With positive feedback the op-amp acts as a comparator with hysteresis: the output is always at a rail, and it switches only when the input crosses a threshold. For this non-inverting Schmitt trigger, $V_{+}=\frac{V_i+V_o}{2}$ gives thresholds $V_i=\mp 5\text{V}$. A $\pm 1\text{V}$ input stays inside the hysteresis band, so the output is latched at $+5\text{V}$ or $-5\text{V}$.

   WATCH OUT FOR

   Treating the circuit as a linear amplifier. Assuming a virtual short ($V_{+}=V_{-}$) gives a gain of $-1$ and leads to option (B), and misreading it as a non-inverting amplifier gives option (A); neither is valid under positive feedback. A second trap is picking (C) because "a comparator turns a sine into a square wave". That only happens if the input amplitude exceeds the thresholds (here $\pm 5\text{V}$), and a 1 V sine does not.

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9. Q.9MCQ · 1 markOfficial answer: C

   Networks (Electric Circuits) › Theorems

   With the output open, no current flows in the $4\Omega$ resistor, so $V_{TH}$ is simply the voltage of the node before it. Repeated source transformations reduce the left side to a $3\text{ V}$ source behind $3\Omega$ (or a $1\text{ A}$ source in parallel with $3\Omega$). One KCL equation, or current times parallel resistance, then gives the node voltage.

   WATCH OUT FOR

   Three mistakes are common. (1) Counting a voltage drop across the $4\Omega$ resistor, or adding $4\Omega$ into the node equation, even though it carries no current on open circuit. (2) Subtracting the $2\text{ V}$ source from the transformed $1\text{ V}$ source: both rise in the same direction ($-$ to $+$ going left to right), so they add to $3\text{ V}$. (3) Adding the two current sources directly ($1+2=3\text{ A}$) without first moving the $1\text{ A}$ source past the $2\text{ V}$ source and the $2\Omega$ resistor by source transformation.

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

    Digital Electronics › Combinational

    A MUX output is the sum of each minterm of the select lines ANDed with the data input on that line. Take the enable first: an active-low EN tied to 0 means the MUX is enabled. Then write $F=\sum m_i(S_1,S_0)I_i$ with $S_1=P$ (the MSB) and $S_0=Q$, and simplify.

    WATCH OUT FOR

    Two mistakes are common. The first is reading EN = 0 as 'disabled' (output stuck at 0) and missing the bubble that makes EN active-low. The second is swapping the select lines, taking $Q$ as the MSB. That sends $I_1=0$ and $I_2=R$ to the wrong minterms and gives $F=PQ+\overline{P}R$, which is not among the options.

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

    Control Systems › Nyquist

    By the principle of the argument, when $s$ goes once clockwise round a closed contour, the image $G(s)$ goes clockwise round the origin $N=Z-P$ times. Here $Z$ and $P$ count only the zeros and poles of $G$ inside the contour. $\Gamma$ is clockwise and encloses 3 zeros and 2 poles, so $N=1$ clockwise. A pole-zero map tells you about encirclements of the origin only. Encirclements of $-1+j0$ depend on the zeros of $1+G(s)$, which change with the gain.

    WATCH OUT FOR

    One mistake is to count every pole and zero on the map instead of only those inside $\Gamma$. Another is to read the arrow beside the label $\Gamma$ as a direction arrow. A third is to use $P-Z$ where the rule for clockwise encirclements needs $Z-P$. The last two turn the correct answer (B) into (A). Candidates who go straight to Nyquist habits also pick (C) or (D), because they expect every encirclement question to be about $-1+j0$. That point can't be decided from a pole-zero map with no gain given.

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12. Q.12MCQ · 1 markOfficial answer: D
13. Q.13MCQ · 1 markOfficial answer: A

    Electromagnetics › Transmission lines

    A pure reactance is lossless, so it reflects all the incident power: $|\Gamma |=\left|\frac{jx-1}{jx+1}\right|=1$. On the Smith chart this is the $r=0$ constant-resistance circle, which has centre $\left(\frac{r}{1+r},0\right)=(0,0)$ and radius $\frac{1}{1+r}=1$. That circle is the chart's outer rim.

    WATCH OUT FOR

    Option (D) is the circle for $r=1$: centre $(0.5,0)$, radius $0.5$, the matching circle that passes through the chart's centre. Candidates who mix up the $r=0$ and $r=1$ circles pick it. Option (C) comes from confusing the locus with the real axis, which is the $x=0$ line of purely resistive loads. Option (B) is the single matched point $z=1$ ($\Gamma =0$), not a locus of reactances.

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

    Signals and Systems › z-transform

    For an FIR difference equation, $H(z)$ is just the tap polynomial $\sum h[k]z^{-k}$. Multiply through by $z^{N-1}$ to get $N-1$ poles at the origin and the finite zeros as roots of $z^{N-1}H(z)$. With alternating signs the taps factor neatly: $1-z^{-1}+z^{-2}-z^{-3}=(1-z^{-1})(1+z^{-2})$.

    WATCH OUT FOR

    Each wrong option matches a specific slip. Dropping the $(-1{)}^k$ gives $1+z^{-1}+z^{-2}+z^{-3}=\frac{(z+1)(z^2+1)}{z^3}$, which is option B. Using the geometric-series form $\frac{1-z^{-4}}{1+z^{-1}}$ but forgetting that the denominator cancels the zero at $z=-1$ gives four zeros and a 4th-order pole, which is option D. The sum has four terms ($k=0$ to $3$), so the highest delay is $z^{-3}$ and there are exactly 3 poles at the origin, not 4.

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15. Q.15NAT · 1 markOfficial answer: 48

    Networks (Electric Circuits) › Two-port

    For maximum power, the load must be the conjugate of the Thevenin impedance seen at the output port. For a two-port described by Z-parameters, with a source resistance $R_s$ on port 1, that impedance is $Z_{th}=Z_{22}-\frac{Z_{12}{Z}_{21}}{Z_{11}+R_s}$. You get it by zeroing the source, writing $V_1=-R_s{I}_1$, and eliminating $I_1$.

    WATCH OUT FOR

    The usual errors are leaving out the 10 Ω source resistance, which gives $120-\frac{3600}{40}=30\Omega$, or taking $Z_{22}=120\Omega$ on its own as the output impedance. Both ignore how the port-1 termination couples back through $Z_{12}{Z}_{21}$. A sign slip in $V_1=-10I_1$ is another trap: with $I_1$ defined as flowing into the network, the zeroed source loop forces $V_1=-10I_1$, not $+10I_1$.

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16. Q.16NAT · 1 markOfficial answer: 2.80 to 2.85

    Networks (Electric Circuits) › AC, power & resonance

    In a series RL circuit fed by one sinusoid, $V/I$ gives both the magnitude and the angle of $Z=R+j\omega L$. The phase lag of the current, $\theta$, fixes $\omega L=|Z|\sin \theta$, so $L=\frac{|V_m|}{|I_m|}\cdot \frac{\sin \theta }{\omega }$. You never need the value of $R$.

    WATCH OUT FOR

    Three common slips. (1) Using $|Z|=20\Omega$ as if it were $\omega L$, which gives $L=4\text{H}$. (2) Forgetting to divide by $\omega$ and answering $14.14$. (3) Treating the $\pi /4$ lag as $\pi /4$ degrees, or using $\cos 45^{\circ }$ for the reactive part instead of $\sin 45^{\circ }$. At exactly $45^{\circ }$ the last slip happens to give the right number, but it breaks at any other angle.

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17. Q.17NAT · 1 markOfficial answer: 644 to 657

    Analog Electronics › Diode circuits

    This is a half-wave voltage doubler. $C_1$ with $D_1$ is a clamper that lifts the input so it swings from $0$ to $2V_m$. $D_2$ with $C_2$ is a peak detector that holds that peak. With ideal parts and no load, $V_o=2V_m$, and the given rms value must first be turned into a peak: $V_m=\sqrt{2}{V}_{rms}$.

    WATCH OUT FOR

    Using 230 V as the peak and answering $460\text{V}$. Missing the doubling and answering the single peak, $325.27\text{V}$. Treating the two equal $0.1\mu \text{F}$ capacitors as a capacitive divider that splits the voltage. In fact, with no load the capacitor values affect only how fast the output settles, not the final value.

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18. Q.18NAT · 1 markOfficial answer: 6

    Analog Electronics › Op-amps

    In an ideal op-amp, no current enters the input terminals, but the output terminal can source any current. The output current is the sum of all the currents leaving the output node, which here means the load current plus the feedback-network current. First use the virtual short to find $V_o=4\text{V}$, then check that it lies inside the $\pm 5\text{V}$ rails.

    WATCH OUT FOR

    Counting only the load current, $V_o/R_L=4\text{mA}$, and forgetting that the output also drives the feedback divider ($2\text{mA}$). Another slip is taking the op-amp's zero input current to mean the output current is zero too.

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19. Q.19NAT · 1 markOfficial answer: 14

    Digital Electronics › Processors & memory

    In a byte-addressable memory, $n$ address lines select $2^n$ locations. So the number of address lines is ${\log }_2(\text{number of byte locations})$, with $1\text{K}=2^{10}$. The number of lines depends only on how big the memory bank is. The 8085's total of 16 address lines does not change it.

    WATCH OUT FOR

    Answering 16 because the 8085 has 16 address lines. That is the whole address space ($64\text{K}$), not what a $16\text{K}$ bank needs. Two other mistakes: counting bits instead of bytes ($16\text{K}\times 8$ bits, giving 17), and taking $1\text{K}=1000$ instead of $1024$.

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20. Q.20NAT · 1 markOfficial answer: 3.050 to 3.080

    Digital Electronics › ADC & DAC

    A DAC's output is the decimal value of the input code times one step. "Calibrated over the full range 0 to 10 V" means the largest code, $2^n-1$, gives 10 V, so one step is $\frac{10}{2^n-1}$ V. Convert the hex input to decimal first, then multiply by the step.

    WATCH OUT FOR

    Many students read 13A as decimal 13 or 130, or mis-convert A (it is 10, so 13A in hex is 314 in decimal). Some divide by $2^{10}=1024$ instead of $2^{10}-1=1023$. That gives $314\times \frac{10}{1024}\approx 3.066$ V, which also falls inside the official range of 3.050 to 3.080, so the key accepts both conventions. Under the stated calibration (all-ones code = 10 V), 1023 is the correct divisor.

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21. Q.21NAT · 1 markOfficial answer: 6.25

    Electromagnetics › Transmission lines

    On a lossless line, any length that is an odd multiple of $\lambda /4$ ($\lambda /4,3\lambda /4,5\lambda /4,\dots$) makes $\tan \beta l\to \infty$. The line then acts as an impedance inverter: $Z_{in}=Z_0^2/Z_L$. Input impedance repeats every $\lambda /2$, so only the length modulo $\lambda /2$ matters.

    WATCH OUT FOR

    One common slip is to treat $3\lambda /4$ like a half-wave line and answer $Z_{in}=Z_L=400\Omega$. Only multiples of $\lambda /2$ repeat the load; $3\lambda /4=\lambda /2+\lambda /4$ inverts it. Another is to invert the ratio and write $Z_L^2/Z_0=3200\Omega$. A quick sanity check catches both: a load larger than $Z_0$ must turn into an input impedance smaller than $Z_0$ after an odd number of quarter-waves.

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22. Q.22NAT · 1 markOfficial answer: 0.5
23. Q.23NAT · 1 markOfficial answer: 160

    Control Systems › Routh–Hurwitz

    Build the Routh array of $s^3+10s^2+(16+K)s+11K$. Marginal stability happens at the value of $K$ that turns the $s^1$ row entirely to zero, provided the other first-column entries stay positive. The auxiliary equation from the $s^2$ row then gives the $j\omega$-axis roots and the oscillation frequency.

    WATCH OUT FOR

    Forgetting the zero term $K(s+11)$ when forming the characteristic equation. This drops the $Ks$ term or the $11K$ constant and gives a wrong $s^1$ entry. Another mistake is using the third-order shortcut $a_2{a}_1=a_3{a}_0$ with the coefficients in the wrong order: the correct form is $10(16+K)=1\cdot 11K$, which gives $K=160$. Some students also answer $K=0$. That makes the $s^0$ row zero, which only puts a root at the origin (the open-loop pole at $s=0$), so it does not give the sustained oscillation the question asks about.

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24. Q.24NAT · 1 markOfficial answer: 6
25. Q.25NAT · 1 markOfficial answer: 0.25

    Engineering Mathematics › Probability & statistics

    For 0/1 variables, $\min (M,N)$ equals the product $MN$: it is 1 only when both are 1. So $X$ is Bernoulli with $p=P(M=1)P(N=1)=\frac{1}{4}$, and the mean of a Bernoulli variable is $p$. Here $E(X)=E(MN)=E(M)E(N)=\frac{1}{2}\cdot \frac{1}{2}$, which uses the independence of the tosses.

    WATCH OUT FOR

    Answering 0.75 by finding $P(X=0)$ when the question asks for $E(X)$, or mixing up min with max: $\max (M,N)$ is 1 in three of the four outcomes and gives $E=0.75$. A second mistake is writing $E(\min (M,N))=\min (E(M),E(N))=0.5$. Expectation does not commute with a nonlinear function like min.

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

    Engineering Mathematics › Linear algebra

    A system $Ax=b$ is consistent exactly when $b$ satisfies every linear dependency among the rows of $A$. Here $\text{rank}(A)=2$ with 4 equations, so there are two such dependencies: row 2 equals $2\times$ row 1, and $6R_1-3R_3+R_4=0$. The right-hand side has to obey the same two relations.

    WATCH OUT FOR

    Spotting only the obvious condition $b_2=2b_1$ and then guessing the second one, or getting a sign wrong when eliminating. Solving equations 1 and 3 gives $x_2=b_3-3b_1$ and $x_1=7b_1-2b_3$. Substituting into equation 4 gives $3b_3-6b_1=b_4$, which is $6b_1-3b_3+b_4=0$, not $3b_1-6b_3+b_4=0$. Also watch for option (B), which writes $b_3=2b_1$ in place of $b_2=2b_1$.

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27. Q.27MCQ · 2 marksOfficial answer: A

    Engineering Mathematics › Differential equations

    Separate the variables to get the general family $\ln |y-1|=\frac{1}{2}{x}^2+C$. Then remember that dividing by $y-1$ threw away the case $y=1$. Substituting it back shows that the constant $y=1$ is also a solution.

    WATCH OUT FOR

    There are two traps. One is getting the coefficient wrong when integrating $xdx$ (it gives $x^2/2$, not $2x^2$). The other is picking $y=-1$ as the constant solution. The constant solution is the root of the $y$-factor, $y-1=0$, so it is $y=1$. Put any candidate constant back into the equation: $y=-1$ gives $0=-2x$, which fails.

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

    Networks (Electric Circuits) › AC, power & resonance

    The wire carrying I puts M and J at the same potential. Take that node as the reference. The two sources then fix V_T = 120∠−90° V and V_B = +120∠−30° V directly, and KCL at J involves only the two middle impedances, giving I = −(V_T + V_B)/Z. The right-hand Z sits straight across the sources and never touches node J, so it does not enter the calculation for I.

    WATCH OUT FOR

    Misreading the lower source's polarity. Its + terminal is at the bottom, so V_B = +120∠−30° relative to M, and the two phasors ADD in the numerator. Getting that sign wrong gives the wrong magnitude. A second trap is option (A): treating the two identical Z's as a balanced divider and concluding no current flows in the M–J wire. That would need V_T = −V_B, and the phase difference between the sources rules it out. A third mistake is adding the angles of the numerator and Z instead of subtracting them when dividing phasors.

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

    Signals and Systems › DTFT & DFT

    Sampling at $400\text{Hz}$ gives $x[n]=\cos (\pi n/2)=\cos \left(\frac{2\pi \cdot 2n}{8}\right)$, which completes exactly 2 cycles in the 8-point window. A cosine sitting exactly on a DFT bin $k_0$ puts all its energy into bins $k_0$ and $N-k_0$, each with value $N/2$. Here that means $X[2]=X[6]=4$ and every other bin is zero, with no leakage.

    WATCH OUT FOR

    Picking bin 4 by treating $\pi /2$ as "half of $\pi$" and so half of $N=8$, instead of writing $\pi /2=2\pi k_0/8$, which gives $k_0=2$. The other common mistake is forgetting the negative-frequency half: the $e^{-j\frac{2\pi }{8}2n}$ term lands at $k=8-2=6$, not at $k=-2$. Expecting leakage into every bin (option A) is also wrong here, because an integer number of cycles fits in the window.

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

    Control Systems › State space

    To get a circuit's state equations, write KVL across each inductor to find $L\frac{d{i}_L}{dt}$ and KCL at each capacitor node to find $C\frac{d{v}_C}{dt}$. Then remove every non-state quantity (here the left node voltage $v_A$) using the resistor and source relations. The signs of the diagonal entries (both negative for a passive RLC network) and where the zeros sit in $B$ are quick checks that rule out most options.

    WATCH OUT FOR

    Treating $v_A$ as if it equals $2i_1$ and forgetting that the inductor current also flows out of the left node. That drops the $-4i$ term in $\dot{i}$. The other common error is getting the sign of the coupling terms wrong: current $i$ charges the capacitor, so $\dot{v}$ has $+4i$, while $v$ opposes the inductor current, so $\dot{i}$ has $-2v$. Flipping either sign leads to (B), (C) or (D).

    Full solution →
31. Q.31MCQ · 2 marksOfficial answer: Marks to all

    Electronic Devices › P-N junction

    For a one-sided abrupt junction, $1/C_D^2$ is linear in $V$ and meets the $V$ axis at $V_{bi}$. A single $(V,C_D)$ point gives the slope only if $V_{bi}$ (or a second point) is also known: $m=-\frac{1/C_D^2}{V_{bi}-V}$.

    WATCH OUT FOR

    Assuming a "typical" $V_{bi}$ (0.5 V, 0.7 V, 0.85 V...) to land on an option. The answer changes with that choice, so any single option picked this way is a guess, not a result. Another slip is to ignore $V_{bi}$ and use $4\times 10^{20}/0.2=2\times 10^{21}$, which treats the line as passing through the origin and contradicts the figure.

    The question and its official answer →
32. Q.32MCQ · 2 marksOfficial answer: A

    Electronic Devices › MOS devices

    MOS threshold voltage relation for a p-type substrate: $V_{TH}=V_{FB}+2{\varphi }_F+\frac{Q_{dep}}{C_{ox}^{\prime }}$, where $V_{FB}={\Phi }_m-{\Phi }_s$ is obtained directly from the semiconductor band diagram (electron affinity plus the gap from $E_C$ down to $E_{Fs}$).

    WATCH OUT FOR

    Misreading the band diagram — adding instead of subtracting the 0.2 eV arrow (which points from $E_V$ up to $E_{Fs}$), giving $E_i-E_{Fs}=0.8$ eV instead of $0.3$ eV, or forgetting that the semiconductor work function must include the full gap from $E_C$ to $E_{Fs}$ (not just to $E_i$) when computing ${\Phi }_s$.

    Full solution →
33. Q.33MCQ · 2 marksOfficial answer: Marks to all

    Electronic Devices › BJT

    For a narrow base, $\beta \propto 1/G_B$, where $G_B={\int }_0^W{N}_B(x)dx$ is the base Gummel number, while the emitter-side base current stays fixed. A linear profile's Gummel number is its average doping times $W$: $(10^{17}+10^{14})/2\approx 5\times 10^{16}{\text{cm}}^{-3}$, about half of the uniform $10^{17}{\text{cm}}^{-3}$ base. The more heavily doped base (T2) therefore has about half the gain.

    WATCH OUT FOR

    Inverting the ratio, i.e. computing ${\beta }_{T1}/{\beta }_{T2}=2$ and choosing option (A), when the question asks for T2's gain as a multiple of T1's. Another mistake is picking whichever printed option is nearest to 0.5, such as 0.7. When no option matches a fully justified result, the honest answer is that none is correct.

    The question and its official answer →
34. Q.34MCQ · 2 marksOfficial answer: A

    Electronic Devices › Optoelectronics

    Link the solar cell's efficiency/fill-factor rating to its short-circuit current density ($P_{max}=\eta P_{in}A=FF\cdot V_{oc}{I}_{sc}$), then use $J_{sc}=qGt$ (full carrier collection over the cell thickness) to back out the average generation rate.

    WATCH OUT FOR

    Forgetting to convert the thickness from micrometres to centimetres (or forgetting it at all), and mixing area/current-density units when area is exactly $1{\text{cm}}^2$ — either error shifts the answer by orders of magnitude, landing on option (D) or a value between the given choices.

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

    Analog Electronics › Amplifiers

    The DC base voltage is 0 V because $v_i$ is direct coupled, which fixes $I_E=(0-0.7+10)/20\text{k}\Omega =0.465\text{mA}$ and $g_m=I_C/V_T$. With $C_E$ bypassing $R_E$ and $v_{be}=v_i$, the gain of this common-emitter stage is $-g_m$ times the total AC collector load $R_C\parallel R_L$.

    WATCH OUT FOR

    Leaving out $R_L$: $C_1$ is a short at signal frequencies, so $R_L$ sits in parallel with $R_C$. Using $R_C$ alone gives $-g_m(10\text{k}\Omega )=-178.85$, which is option (C). Other slips are ignoring $V_{BE}$ in the bias ($I_E=10/20\text{k}\Omega =0.5\text{mA}$), and keeping $R_E$ in the gain even though $C_E$ bypasses it.

    Full solution →
36. Q.36MCQ · 2 marksOfficial answer: C

    Digital Electronics › ADC & DAC

    A MOS sampling switch has to be designed for the worst-case input, not the typical one. An n-channel switch conducts while $V_G-V_S\ge V_T$, and its source is whichever terminal is lower. To stay ON for every input, $V_G$ must be $V_T$ above the most positive input: $10+3=13\text{V}$. To stay OFF for every input and every held value, $V_G$ must be no more than $V_T$ above the most negative level: $-10+3=-7\text{V}$.

    WATCH OUT FOR

    Using only the threshold (answer A, $\pm 3\text{V}$) and forgetting that the source moves with the signal. Another error is checking both conditions at the same input level, e.g. taking $V_I=+10\text{V}$ for holding and getting $V_G\le 13\text{V}$, or subtracting $V_T$ instead of adding it (answer D). For holding, the worst case is the most NEGATIVE channel voltage, because that is where a channel forms most easily.

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

    Analog Electronics › Amplifiers

    Looking into the source of a MOSFET whose gate is at AC ground, the drain-side resistance $r_{ds}+R$ is divided by $(1+g_m{r}_{ds})$, which is about the device's intrinsic gain. So the source is a low-resistance node, close to $1/g_m$ unless the drain load $R$ is comparable to $g_m{r}_{ds}$ times $1/g_m$. This is the dual of looking into the drain, where the source-side resistance is multiplied up by the same factor.

    WATCH OUT FOR

    Picking (A), $r_{ds}+R+g_m{r}_{ds}R$. That has the form of the resistance looking into the DRAIN of a source-degenerated MOSFET, where resistance is multiplied up. Here we look into the SOURCE, and resistance is divided down. Other slips: taking $v_{gs}=+v_x$ instead of $-v_x$, which gives $(r_{ds}+R)/(1-g_m{r}_{ds})$, and ignoring the dependent source altogether, which gives (D), $r_{ds}+R$.

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

    Digital Electronics › Number systems

    The same bit pattern means a different number in each representation. With MSB $=1$: in signed magnitude the magnitude is the remaining bits, in 1's complement it is the complement of all the bits, and in 2's complement the MSB counts as $-2^{n-1}$. To write a negative number in 6-bit 2's complement, take $2^6-|x|$, or complement $+|x|$ and add 1.

    WATCH OUT FOR

    A common slip is reading $1100$ as $-4$ in 1's complement too. Its 1's complement value is $-3$, because the magnitude is $\overline{1100}=0011$, not $100$. Another slip is forming the 6-bit answer from the 4-bit pattern (for example by sign-extending $1100$) instead of converting the sum $-11$. Some students also give $001011$, which is $+11$, or forget the final $+1$ and pick the 1's complement pattern $110100$.

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

    Digital Electronics › Sequential

    In a Mealy diagram, find the one edge that outputs 1 ($S_4\stackrel{0/1}{\to }{S}_3$). The detected sequence is the input path from $S_0$ into $S_4$ (0101) followed by that edge's input (0), which gives 01010.

    WATCH OUT FOR

    Reading the machine as Moore and taking the sequence that merely reaches $S_4$ (0101), then picking 01011 because it also passes through $S_4$. In a Mealy machine the 1 is produced on the transition $S_4\stackrel{0/1}{\to }{S}_3$, so the last bit must be 0. A 1 from $S_4$ resets the machine to $S_0$ with output 0.

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

    Control Systems › Root locus

    First rewrite the characteristic equation as $1+KG(s)H(s)=0$, which gives $G(s)H(s)=\frac{K(s+3)}{s(s+1)(s+2)}$. The real-axis rule shows which segments are on the locus. Then solve $dK/ds=0$ and keep only the roots that sit on a locus segment and give $K>0$.

    WATCH OUT FOR

    Solving $dK/ds=0$ and accepting every root as a break point. The cubic $s^3+6s^2+9s+3=0$ has roots in $(-1,0)$, $(-2,-1)$ and $(-4,-3)$, so a student who skips the real-axis check (or the $K>0$ check) may pick (B) or (D). Another slip is to treat $3K$ as a constant and read the poles off $s^3+3s^2+(K+2)s$ without separating out the $K$ terms.

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

    Analog Electronics › Filters & waveforms

    In an ideal inverting amplifier the inverting input is a virtual ground, so any element connected from that node to ground has zero volts across it and carries no current. Here the $2C$ and $2R$ shunt drop out. The gain is set only by $Z_f/R$ with $Z_f=R\parallel C$, which gives a single pole at $\omega =1/RC$.

    WATCH OUT FOR

    Treating $2C\parallel 2R$ as part of the input network, for example as a voltage divider with the series $R$ or as an added pole at $1/(4RC)$, and so getting a lower cut-off frequency. Those elements sit between virtual ground and real ground and carry no current. A second slip is using ${\omega }_c=1/RC$ directly as the answer in hertz (500) instead of dividing by $2\pi$.

    Full solution →
42. Q.42MCQ · 2 marksOfficial answer: B
43. Q.43MCQ · 2 marksOfficial answer: A

    Electromagnetics › Antennas

    Above a perfect conductor, a vertical current element's image points the same way (in phase), at $z=-h$. So the pattern is $\sin \theta \cos (kh\cos \theta )$. Array-factor maxima need $kh\cos \theta =n\pi$. The $n=0$ lobe always sits at $90^{\circ }$, so the first maximum that can land at $60^{\circ }$ is $n=1$: $\frac{2\pi }{\lambda }h\cdot \frac{1}{2}=\pi$, which gives $h=\lambda$.

    WATCH OUT FOR

    Giving the image the wrong sign. That is the rule for a horizontal element, whose image is reversed. It gives $AF=\sin (kh\cos \theta )$ and the maxima condition $kh\cos \theta =(2n+1)\pi /2$, which leads to $0.5\lambda$. At $h=0.5\lambda$ the correct pattern actually has a null at $60^{\circ }$. A second mistake is using $n=0$ and forgetting that it only gives the $90^{\circ }$ lobe.

    Full solution →
44. Q.44NAT · 2 marksOfficial answer: 800

    Analog Electronics › Diode circuits

    In a series regulator with an error-amplifier transistor, the feedback loop settles where the sampled output at the divider tap equals the reference plus one base-emitter drop: $V_O\frac{R_2}{R_1+R_2}=V_Z+V_{BE}$. This is the same as $V_O=(V_Z+V_{BE})\left(1+\frac{R_1}{R_2}\right)$. $R_3$ and $V_I$ only set the bias current, not the output.

    WATCH OUT FOR

    Setting the divider tap equal to $V_Z=3.3\text{V}$ and forgetting the $V_{BE}$ of $Q_2$. That gives $R_2=3300/5.7\approx 579\Omega$. Other mistakes are putting the $V_{BE}$ of the pass transistor $Q_1$ into the divider equation, or trying to use $V_I=15\text{V}$ and $R_3$ to find $R_2$. Neither of those affects the divider.

    Full solution →
45. Q.45NAT · 2 marksOfficial answer: 1

    Electromagnetics › Plane waves

    A uniform plane wave in a source-free medium is transverse. Because $\nabla \cdot \mathbf{H}=0$, the amplitude vector of $\mathbf{H}$ must be perpendicular to the wave vector $\mathbf{\beta }$. You read $\mathbf{\beta }$ off the phase by writing it as $\omega t-\mathbf{\beta }\cdot \mathbf{r}$.

    WATCH OUT FOR

    Students often take the wave vector as $3{\hat{a}}_x-{\hat{a}}_y-{\hat{a}}_z$ straight from the phase. Writing $\omega t+3x-y-z$ in the form $\omega t-\mathbf{\beta }\cdot \mathbf{r}$ gives $\mathbf{\beta }=-3{\hat{a}}_x+{\hat{a}}_y+{\hat{a}}_z$. For this question the sign does not change $b$, because flipping $\mathbf{\beta }$ still gives $\mathbf{\beta }\cdot {\mathbf{H}}_0=0$. It does reverse the direction of propagation, which matters in related questions. Another error is to pick the coefficients that make the dot product $+1$, or to set $\mathbf{H}$ parallel to $\mathbf{\beta }$, instead of perpendicular to it.

    Full solution →
46. Q.46NAT · 2 marksOfficial answer: 0.8

    Electromagnetics › Transmission lines

    Match port 2 with $Z_0$. The transformer then shows port 1 an impedance of $n^2{Z}_0$, which gives $S_{11}=\frac{n^2-1}{n^2+1}$. The transformer steps the total port-1 voltage $V_1^{+}(1+S_{11})$ down by $n$, and that becomes the outgoing wave at the matched port 2, so $S_{21}=\frac{1+S_{11}}{n}=\frac{2n}{1+n^2}$. The answer depends only on the turns ratio. $Z_0$ cancels out, so the given 10 $\Omega$ does not change the result.

    WATCH OUT FOR

    Taking $S_{21}$ as the plain voltage ratio $V_2/V_1=1/n=0.5$. That ignores the reflection at port 1: $S_{21}$ is referred to the INCIDENT wave $V_1^{+}$, not to the total voltage $V_1=V_1^{+}(1+S_{11})$. Another slip is using $n$ instead of $n^2$ for the impedance transformation, which gives $Z_{in}=20\Omega$ and wrong values for both $S_{11}$ and $S_{21}$. Checking $|S_{11}{|}^2+|S_{21}{|}^2=1$ catches both errors.

    Full solution →
47. Q.47NAT · 2 marksOfficial answer: 2
48. Q.48NAT · 2 marksOfficial answer: 3
49. Q.49NAT · 2 marksOfficial answer: 3.95 to 4.05

    Control Systems › Bode & margins

    In the steady state, a stable LTI system driven by $A\cos \omega t$ gives $A|G(j\omega )|\cos (\omega t+\angle G(j\omega ))$. Matching the output amplitude at $\omega =3$ gives $a$ directly. The printed phase of $-1.892\text{rad}$ then works as an independent check.

    WATCH OUT FOR

    Three slips are common. (1) Leaving out the input amplitude 5 and setting $|G(j3)|=1/\sqrt{10}$. That gives $9+a^2=1$, which has no real solution. (2) Treating $-1.892$ as degrees: it is in radians, so a phase check done in degrees fails. (3) Solving from the phase alone and getting ${\tan }^{-1}(3/a)$ wrong, when the amplitude equation gives the answer in one line.

    Full solution →
50. Q.50NAT · 2 marksOfficial answer: 76 to 77

    Digital Electronics › Sequential

    For each flip-flop-to-flip-flop path, the clock period must cover the launching flip-flop's $t_{pd}$, plus the gate delays on the path, plus the capturing flip-flop's $t_{setup}$. The largest of these sets $T_{min}$. Check hold separately: it cannot be fixed by slowing the clock. Here the path with no gates at all (FF2 to FF1) is the slowest, because FF2's 8 ns propagation delay and FF1's 5 ns setup time outweigh the two 2 ns gates on the other path.

    WATCH OUT FOR

    Only timing the obvious path through the gates (FF1 to XOR to NAND to FF2, 11 ns, which gives about 90.9 MHz) and missing the feedback wire from FF2 back to FF1. Another is reading the hop on FF1's output line as a junction, or pairing the wrong setup time with a path: the setup time always belongs to the flip-flop that CAPTURES the data. Adding hold times into the period is also wrong. Hold is a separate minimum-delay check.

    Full solution →
51. Q.51NAT · 2 marksOfficial answer: 2.25

    Engineering Mathematics › Calculus

    The solid is a right prism: its cross-section at every $x$ is the same triangle, of area $\frac{1}{2}$. So the triple integral of $x$ splits into (area of the cross-section) $\times {\int }_0^{3}xdx$. Put another way, it equals volume $\times$ the $x$-coordinate of the centroid: $\frac{3}{2}\times \frac{3}{2}$.

    WATCH OUT FOR

    Misreading the 3-D sketch. Some students take the triangle to lie in the $xy$- or $xz$-plane with the length 3 along $y$ or $z$, which changes the integrand's weighting and gives the wrong value. Others treat the solid as a full box $[0,3]\times [0,1]\times [0,1]$ and get $4.5$, because they miss the slanted face $y+z=1$. A third mistake is to compute only the volume, $1.5$, and forget the factor $x$ in the integrand.

    Full solution →
52. Q.52NAT · 2 marksOfficial answer: 58.50 to 58.80

    Signals and Systems › Fourier transform

    Parseval's theorem turns an integral over frequency into an energy integral over time: $\int |X(\omega ){|}^{2}d\omega =2\pi \int |x(t){|}^{2}dt$ when $\omega$ is in rad/s. For a piecewise-linear signal, split it at its corners, square each linear piece and integrate. Use symmetry to halve the work.

    WATCH OUT FOR

    Leaving out the factor $2\pi$ gives $28/3\approx 9.33$. That factor belongs to the $\omega$ (rad/s) form of Parseval's theorem; only the $f$ (Hz) form has no $2\pi$. Two other slips are common: squaring the triangle's area instead of integrating $x^2(t)$, and misreading the change of slope at $(0,1)$ and $(2,1)$, which treats the signal as a single triangle from $-1$ to $3$ with peak 3 and gives $E_x=12$.

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

    Signals and Systems › z-transform

    A first-order section has constant magnitude response (all-pass) when its zero is the reciprocal of its pole, $z_0=1/p$. On the unit circle, $|e^{j\omega }-1/p|=|e^{j\omega }-p|/|p|$ for real $p$, so the ratio of the two distances does not depend on $\omega$. Pole at $-0.5$ gives zero at $-2$.

    WATCH OUT FOR

    Answering $\alpha =-0.5$ (pole-zero cancellation). That does give a constant magnitude, but it breaks the condition $|\alpha |>1$. Another common slip is to take the reciprocal as $+2$, putting the zero at the mirror image through the origin instead of at $1/p$. That gives $|H{|}^2\propto (5-4\cos \omega )/(1.25+\cos \omega )$, which changes with frequency.

    Full solution →
54. Q.54NAT · 2 marksOfficial answer: 0.299 to 0.301

    Engineering Mathematics › Probability & statistics

    Turn the condition on $Y$ into a condition on $X$ using the increasing map $Y=2X-6$, which gives $X\le 6.5$. Conditioning a uniform variable on a sub-interval leaves it uniform on that sub-interval, so the conditional probability is just a ratio of lengths: $1.5/5$.

    WATCH OUT FOR

    Dividing by the full length $12$ and not by the length of the conditioning interval $[5,10]$, which gives $1.5/12=0.125$. Other slips: solving $2X-6\le 7$ as $X\le 0.5$ or $X\le 6$; forgetting to cap $X\ge 5$ at the support limit $10$; or computing $P(Y\le 7)$ over the whole range, $8.5/12\approx 0.708$, and never applying the condition.

    Full solution →
55. Q.55NAT · 2 marksOfficial answer: 30

    Control Systems › Time response

    With unity feedback, the ramp error of a type-1 loop is $e_{ss}=1/K_v$, where $K_v={\lim }_{s\to 0}sC(s)G(s)$. The controller zero cancels the plant pole at $s=-1$, which leaves $C(s)G(s)=K/(s(s+3))$. That gives $K_v=K/3$, so $e_{ss}=3/K$. The cancelled mode sits at $s=-1$, in the left half-plane, so it cannot make the loop unstable.

    WATCH OUT FOR

    A common slip is to ignore the controller's pole at $s=-3$ and take $K_v=K$, which gives $K=10$. Another is to use the $(s+1)$ factor after cancelling it, for example writing $K_v=K/(1\cdot 3)$ with the wrong factor or keeping $s+1$ in the denominator, which gives a different value. Some students also apply the final-value theorem without checking stability first. Here the loop is stable for every $K>0$, but you should state that check before using the theorem.

    Full solution →