GATE EC 2020 Official Paper (Previous Year Paper)

The correct answer is Option 1) i.e. in, of

Let's take the second blank first, we have two choices die of/from, traditionally, 'die of' refers to dying as the result of a disease, and 'die from' refers to death as a result of an external cause such as an accident or a disaster.

For example:

• My uncle died of cancer.
• She died of old age.
• He died from starvation.
  

Therefore, 'of' is the best fit for the second blank and as a result, we can easily negate option 3 and 4. Now we are left with two choices for the first blank. 'In' here is the most appropriate as per the context of the sentence, 'from' is used to indicate the source or provenance of someone or something and 'accident' is neither a source or a medium it is an unfortunate event.

Complete Sentence: The untimely loss of life is a cause of serious global concern as thousands of people get killed in accidents every year while many other die of diseases like cardiovascular disease, cancer, etc

The correct answer is Option 2) i.e. but also

"Not only.....But also" is correlative conjunction which are used in a pair. Correlative conjunctions are pairs such as neither . . . nor, not . . . only, and but . . . also. These conjunctions connect two balanced clauses, phrases, or words. The two elements that correlative conjunctions connect are usually similar in length and grammatical structure.

This Conjunction pair (not only....but also) is used to present two related pieces of information. Both pieces of information are being presented by the writer as surprising or unexpected, with the second one being even more surprising than the first. The first piece of information(accused of theft) will be followed by the second one (conspiracy).

Complete Sentence: He was not only accused of theft but also of conspiracy.

The correct answer is option 2) i.e. Repress.

From the first part of the analogy, it becomes clear that the two words are opposite in meaning i.e. 'Explicit' is the antonym of 'Implicit'. Similarly, the antonym of 'Express' will be ''Repress'.

The meaning of the words are given below:

• Explicit: fully revealed or expressed without vagueness, implication, or ambiguity : leaving no question as to meaning or intent.
• Implicit: involved in the nature or essence of something though not revealed, expressed, or developed.
• Express: to make known the opinions or feelings of (oneself).
• Repress:  to prevent the natural or normal expression(feelings), activity, or development of.

The correct answer is option 3) i.e. the English announcements being longer than the French ones.

After reading the passage we can easily comprehend that the french-speaking couple filed a lawsuit against Air Canada because they were upset at the English announcements being longer than the French ones. It has been mentioned clearly in the passage "formal in-flight announcements in English lasting 15 seconds, as opposed to informal 5-second messages in French.".

Given that a super additive function satisfies the following property:

f(x₁ + x₂) ≥ f(x₁) + f(x₂)

Checking for each option, i.e.

Option 1:

Let us take x₁ = 2, and x₂ = 4

e^x1+x2 = e ²⁺⁴ = e⁶ = 403.42

${e^{{x_1}}} + {e^{{x_2}}} = {e^2} + {e^4} = 61.98$ 

Since ${e^{{x_1} + {x_2}}} > {e^{{x_1}}} + {e^{{x_2}}}$

e^x satisfies the superadditive property.

Option 2:

If $\sqrt x$ is to be a super-additive function, then it must also satisfy:

$\sqrt {{x_1} + {x_2}} \ge \sqrt {{x_1}} + \sqrt {{x_2}}$

Again for x₁ = 2 and x₂ = 4, we get:

$\sqrt {2 + 4} = \sqrt 6 = 2.44$

$\sqrt 2 + \sqrt 4 = 1.414 + 2 = 3.414$

$\sqrt {{x_1} + {x_2}} < \sqrt {{x_1}} + \sqrt {{x_2}}$

∴ √x does not follow the super-additive property.

Option 3: 1/x

For x₁ = 2 and x₂ = 4

$\frac{1}{{{x_1} + {x_2}}} = \frac{1}{6} = 0.166$

$\frac{1}{{{x_1}}} + \frac{1}{{{x_2}}} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4} = 0.75$

Since, $\frac{1}{{{x_1} + {x_2}}} < \frac{1}{{{x_1}}} + \frac{1}{{{x_2}}}$

1/x does not satisfy the superadditive property.

Option 4:

Checking for e^-x

For x₁ = 2, x₂ = 4

${e^{ - \left( {{x_1} + {x_2}} \right)}} = {e^{ - 6}} = 0.0024$

${e^{ - {x_1}}} = 0.135$

${e^{ - {x_2}}} = 0.028$

Since, ${e^{ - \left( {{x_1} + {x_2}} \right)}} < {e^{ - {x_1}}} + {e^{ - {x_2}}}$

e^-x does not satisfy the superadditive property.

The correct answer is option 1) i.e.East Asian crisis →  subprime lending crisis → banking crisis → global financial crisis.

After reading the passage and going through the options (cursory look), we can easily comprehend that we need to select the precedence order on the basis of time( the event that happened first will be on the leftmost side). Let's see the following lines given in the passage "The global financial crisis in 2008 is considered to be the most serious worldwide financial crisis,", so it is clear that 'global financial crisis' is the final result that occurred because of some series of event that took place in the past. Let's connect these chain of events- Excess liquidity resulted in East Asian crises and to compensate this crisis 'sub-prime lending' came into play (provision of loans to those borrowers who may have difficulties in repaying loans), the sub-prime lending crisis led to the banking crisis in 2008 with the collapse of Lehman Brothers in 2008, finally all this resulted in the global financial crisis.

Concept:

The angle between the hour and the minute hand is given by the formula:

$θ=|\frac{11m}{2}-30h|$

h = Hours (≤ 12)

m = minutes (≤ 60)

Calculation:

The time quarter past 3 is 3:15, i.e.

h = 3

m = 15

The angle between the hour and minute hand will be:

$θ=|\frac{11\times15}{2}-30\times3|$

θ = 7.5°

Concept:

Area of a circle is given by:

A = πr²

r = radius

Area of a rectangle is given by:

A = a × b

a = length of the rectangle

b = breadth of the rectangle

For the rectangle to have a maximum area, the two sides must be of equal length, i.e.

For a = b, the area will be maximum.

Application:

The area of the rectangle will be maximum when both the sides are of equal length, as shown:

Applying Pythagoras theorem, we can write:

x² + x² = (2a)²

2x² = 4a²

$x = \sqrt 2 a$

The maximum area of the rectangle wll be:

A = x² = 2a²

Area of circle = πa²

The area of the shaded portion will be the difference between the two, i.e.

Area of the circle – Area of rectangle

= πa² – 2a²

Concept:

For a quadratic equation of the form:

ax² – bx + c = 0, the roots will be given by:

$x = \frac{{b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ 

For the roots to be equal, b² – 4ac should equal 0, i.e.

b² = 4ac     ---(1)

Given that the value of the equal root is β, i.e.

$\beta = \frac{b}{{2a}}$         ---(2)

From the above condition, we conclude that option (1) is not correct.

Squaring equation (2), we get:

${\beta ^2} = \frac{{{b^2}}}{{{{\left( {2a} \right)}^2}}}$ 

Putting b² from equation (1) in the above, we get:

${\beta ^2} = \frac{{4ac}}{{{{\left( {2a} \right)}^2}}} = \frac{{4ac}}{{4{a^2}}}$

${\beta ^2} = \frac{c}{a}$

∴ option (2) is also incorrect.

Cubing equation (2), we can also write:

${\beta ^3} = \frac{{{b^3}}}{{{{\left( {2a} \right)}^3}}}$ 

With b² = 4ac

b³ = 4abc

$\therefore {\beta ^3} = \frac{{4abc}}{{{{\left( {2a} \right)}^3}}}$

${\beta ^3} = \frac{{bc}}{{2{a^2}}}$

∴ Option (3) is correct.

Option (4) cannot be correct, because for the root to be equal, b² = 4ac must satisfy.

The average number of students enrolled in P will be:

${A_p} = \frac{{3k; + ;5k ;+ ;5k ;+; 6k; + ;4k}}{5}$

A_p = 4600

Similarly, The average number of students enrolled in Q will be:

${A_Q} = \frac{{4k; +; 7k ;+; 8k ;+ ;7k; + ;5k}}{5}$

A_Q = 6200

Average of the difference in the number of students enrolled in P and Q will be:

[(4k - 3k) + (7k - 5k) + (8k - 5k) + (7k - 6k) + (5k - 4k)]/5 = 8k/5 = 1600

The required ratio will now be:

$= \frac{{{A_p}}}{{{A_Q} - {A_p}}}$

$= \frac{{4600}}{{1600}} = \frac{{46}}{{16}}$

= 23 : 8

Concept:

Basis: it is defined as a subset of vectors within the space that are linearly independent i.e. it can’t be defined as a set of a linear combination of any other vectors.

Span: It is defined as the linear combination of linearly independent vectors within the space.

R⁴: It shows the space of 4 linearly independent vectors.

Analysis:

1. R⁴ means that there are only 4 linearly independent vectors out of 6 vectors. So it is not necessary that these vectors span R⁴. (TRUE STATEMENT)

2. All these vectors are not linearly independent because there are only 4 linearly independent vectors out of 6 vectors. (TRUE STATEMENT)

3. Any four of these vectors can’t form a basis for R⁴ because only linearly independent vectors can form the basis for R⁴. Out of 6 vectors, only 4 vectors are linearly independent. So, this statement is wrong. (WRONG STATEMENT)

4) If {v₁, v₃, v₅, v₆} span R^4, it means that vectors {v₁, v₃, v₅, v₆} are linearly independent, because span shows the linear combination of linearly independent vectors. Hence, it forms a basis for R⁴. (TRUE STATEMENT)

Explanation:

If a vector field $\vec A$ is solenoidal, it indicates that the divergence of the vector field is zero, i.e.

$\nabla \cdot \vec A = 0$

If a vector field $\vec A$ is irrotational, it represents that the curl of the vector field is zero, i.e.

$\nabla \times \vec A=0$

If a field is scalar A then ${\nabla ^2}\vec A = 0$ is a Laplacian function.

Important Vector identities:

$\nabla \times \left( {\nabla \times \vec A} \right) = \nabla \left( {\nabla \cdot \vec A} \right) - {\nabla ^2}\vec A$

$\nabla .\left( {\nabla \times \vec A} \right) = 0$ (Divergence of a vector field is always zero)

$\nabla \times\left( {\nabla\vec A} \right) = 0$ (Curl of a gradient is always zero)

Concept:

When the input of a function is made up of multiple variables, partial derivative is used to understand how the function changes as we let just one of the variable changes, keeping all the other variables as constant.

Application:

Given $f\left( {x,;y,;z} \right) = {e^{1 - x\cos y}} + xz{e^{ - \frac{1}{{1 + {y^2}}}}}$

Differentiating the above with respect to x and treating y and z as a constant, we get:

$\frac{{\partial f}}{{\partial x}} = {e^{(1 - x\cos y)}} \cdot \frac{d}{{dx}}\left( {1 - x\cos y} \right) + z{e^{\frac{{ - 1}}{{1 + {y^2}}}}} \cdot \frac{\partial }{{\partial x}}\left( x \right)$

$\frac{{\partial f}}{{\partial x}} = {e^{\left( {1 - x\cos y} \right)}} \cdot \left( { - \cos y} \right) + z{e^{\frac{{ - 1}}{{{y^2} + 1}}}}$

$\frac{{\partial f}}{{\partial x}} = - \cos y{e^{(1 - x\cos y)}} + z{e^{ - \frac{1}{{{y^2} + 1}}}}$

At point (1, 0, e), the partial derivative gives a value:

\(\frac{{\partial f}}{{\partial x}}|_\left( {1,0,e} \right); = - \cos \left( 0 \right);{e^{(1 - \cos 0)}} + e.{e^{ - \frac{1}{1}}}\)

= -1e⁰ + e.e^-1

= -1 + 1

\(\frac{{\partial f}}{{\partial x}}|_\left( {1,0,e} \right); =0\)

Concept:

For a second-order differential function, the characteristic equation will have two roots D₁ and D₂. The roots can have three possible forms i.e.

• Real, distinct roots, D₁ ≠ D₂
• Complex roots, D₁, D₂ = a ± ib
• Double roots D₁ = D₂ = 0

The solution for the second-order differential equation with equal roots of the characteristic equation is given by:

$y = \left( {{C_1} + {C_2}x} \right){e^{{D_1x}}}$

Application:

Given: $\frac{{{d^2}y}}{{d{x^2}}} - \frac{{6dy}}{{dx}} + 9y = 0$

The characteristic equation will be:

D² – 6D + 9 = 0

(D - 3)² = 0

D = 3, 3

Since the roots are real and equal, the general solution of the given differential equation will be:

$y = \left( {{C_1} + {C_2}x} \right){e^{3x}}$

Important Point:

If two roots are real and distinct (i.e. D₁ ≠ D₂), the general solution will be:

$y = {C_1}{e^{{D_1}t}} + {C_2}{e^{{D_2}t}}$

If the roots of the characteristic equation are complex (i.e. D_{1, 2} = a ± ib), the general solution to the differential equation will be:

$y = {C_1}{e^{at}}\cos \left( {bt} \right) + {C_2}{e^{at}}\sin \left( {bt} \right)$

Concept:

The unit impulse response of a system is the response when the input is a unit impulse.

This is explained as:

y(n) = x(n) ⊕ h(n)

For a unit impulse input, the output will be:

y(n) = h (n)

Calculation:

Given:

$y\left[ n \right] = \mathop {\max }\limits_{ - \infty \le k \le n} \left| {x\left[ k \right]} \right|$

To find the impulse response of a system, the input will be an impulse, i.e.

x(k) = δ(k)

So,$y\left[ n \right] = \mathop {\max }\limits_{ - \infty \le k \le n} \left| {\delta\left[ k \right]} \right|$

For n < 0:

$y\left[ n \right] = \mathop {\max }\limits_{ - \infty \le k \le n} \left| {\delta\left[ k \right]} \right|=0$

For n = 0:

y[n] = δ(0) = 1

For n > 0:

$y\left[ n \right] = \mathop {\max }\limits_{ - \infty \le k \le n} \left| {\delta\left[ k \right]} \right|=1$

\(\therefore y\left( n \right) = \left[ {\begin{array}{*{20}{c}} {0;}&{ - \infty < n < 0}\ {1;}&{0 \le n < \infty } \end{array}} \right]\)

i.e. y[n] = u[n]

Concept:

For a given position of Fermi-level, the concentration of electrons and holes are given by:

${n_0} = {N_c}{e^{ - \frac{{\left( {{E_c} - {E_F}} \right)}}{{kT}}}}$

${p_0} = {N_v}{e^{ - \frac{{\left( {{E_F} - {E_V}} \right)}}{{kT}}}}$

N­c and Nv are the effective density of states.

We can now calculate the intrinsic Fermi level position.

For the intrinsic semiconductor, the electron and hole concentrations are equal.

∴ we can write:

${N_c};{e^{ - \left( {\frac{{{E_c} - {E_{Fi}}}}{{kT}}} \right)}} = {N_v};{e^{ - \left( {\frac{{{E_{Fi}} - {E_v}}}{{kT}}} \right)}}$

Taking the natural log of both sides of the above equation, we get:

${E_{{F_i}}} = \frac{1}{2}\left( {{E_c} + {E_v}} \right) + \frac{1}{2}kT;In;\left( {\frac{{{N_v}}}{{{N_c}}}} \right)$

Since, $\frac{1}{2}\left( {{E_c} + {E_v}} \right) = {E_{midgap}}$

${E_{{F_i}}} = {E_{midgap}} + \frac{1}{2}kT;In\left( {\frac{{{N_v}}}{{{N_c}}}} \right)$

${E_{Fi}} - {E_{midgap}} = \frac{1}{2}kT;In;\left( {\frac{{{N_v}}}{{{N_c}}}} \right)$   ---(1)

Calculation:

Given V_T = 26 mV and N_V = 2 N_C

Using Equation (1), we can write:

${E_{{F_i}}} - {E_{mid}} = \frac{1}{2}kT;ln\left( {\frac{{{N_V}}}{{{N_C}}}} \right)$

${E_{{F_i}}} - {E_{mid}} = \frac{{26;m}}{2};ln\left( {\frac{{2{N_C}}}{{{N_C}}}} \right)$

${E_{{F_i}}} - {E_{mid}} = 13 \times {10^{ - 3}};ln\left( 2 \right)$

= 9.01 meV

∴ The intrinsic Fermi level $\left( {{E_{{F_i}}}} \right)$ is shifted from the mid-bandgap energy level by 9.01 meV.

Concept:

A forward-biased p-n homojunction diode is shown below:

The maximum recombination rate (U_max) is given by:

${U_{max}} = \frac{1}{2}{n_i}{V_{TH}}{\sigma _0}{{\rm{e}}^{\left( {V/2{V_T}} \right)}}.{N_t}$

Where,

n_i: intrinsic concentration.

V_TH: thermal velocity of carriers.

σ­₀ = Capture cross-section.

V: applied forward voltage.

V_T: Thermal voltage.

N_t: probability density of energy state E_t which is present between E_C and E_v

Observations:

• U_max ∝ n_i, so option 1 is correct.
• Maximum recombination (U_max) occurs at the center of the depletion region. ∴ Option 2 is wrong.
• U_max ∝ e^V, ∴ Option 3 is correct.
• U_max ∝ V_TH, ∴ Option 4 is correct.

Analysis:

The given circuit is a Schmitt trigger circuit. It is redrawn as:

Applying KCL at the non-investing terminal, we get:

$\frac{{{V^ + } - {V_i}}}{{1k}} + \frac{{{V^ + } - {V_0}}}{{1k{\rm{\Omega }}}} = 0$

V⁺ - V_i + V⁺ - V₀ = 0

${V^ + } = \frac{{{V_i} + {V_0}}}{2}$      ---(1)

Case 1:

When V₀ = 5V (+V_sat)

From equation (1), we can write:

${V^ + } = \frac{{5 + {V_0}}}{2} = 2.5 + \frac{{{V_0}}}{2}$

For a positive peak of +1V for V_i, V₊ will be:

V⁺ = 2.5 + 0.5

V⁺ = 3V

Since V^- = 0 V

The difference in the non-inverting and inverting terminal of Op-Amp is:

V_d = V⁺ - V^-

V_d = 3 V

Since the Op-Amp amplifies the difference in the input voltage, the output will remain constant at +V_sat (5V) as V_d = + 3V (Positive)

Now for a negative peak of -1V for V_in, V⁺ will be:

V⁺ = 2.5 – 0.5

V⁺ = 2V

V_d = 2 – 0

V_d = 2V

The difference is positive for both the positive and negative peak of V_i,

∴ The output will remain at + 5V (+V_sat), irrespective of the input.

Case 2:

When V₀ = -5V (-V_sat)

Again apply KCL at the non-investing terminal, we get:

$\frac{{{V^ + } - \left( { - 5} \right)}}{{1k}} + \frac{{{V^ + } - {V_i}}}{{1k}} = 0$

${V^ + } = \frac{{{V_i} - 5}}{2}$

${V^ + } = ; - 2.5 + \frac{{{V_i}}}{2}$

Now, for a positive peak of V_i at 1V, V⁺ will be:

V⁺ = -2.5 + 0.5

V⁺ = -2V

V_d = V⁺ - V^-

V_d = -2V

Since the difference is negative, the output will remain at -V_sat

Similarly, for a negative peak of V_i at -1V, V⁺, will be,

V⁺ = -2.5 – 0.5

V⁺ = -3V

V_d = -3V

Source the difference is negative the output will remain at -V_sat

∴ We can conclude that the output will either be +V_sat or -V_sat

Concept:

According to Thevenin’s theorem, any linear circuit across a load can be replaced by an equivalent circuit consisting of a voltage source Vth in series with a resistor Rth as shown:

Vth = Open circuit Voltage at a – b (by removing the load), i.e.

Application:

Since V_th is simply the open-circuit voltage, we use the substitution theorem to simplify the circuit as shown:

Let V_x be Node Voltage

Applying KCL at node x, we get:

$\frac{{V_x - 2 - 1}}{3} - 2; + \frac{V_x}{2} = 0$

$\frac{{V_x - 3}}{3} + \frac{V_x}{2} = 2$

2V_x – 6 + 3V_x = 12

5V_x = 18

$V_x=18/5=3.6 V$

Since no current flows through the 4Ω resistance, we can write:

V_x = V_th

∴ V_th = 3.6 V

Concept:

For a general 4 to 1 multiplier as shown, the output expression is written as:

F = S̅₁S̅₀I₀ + S̅₁S₀I₁ + S₁S̅₀I₂ + S₁S₀I₃  

i.e. I₀ will be the output when the select inputs are 00, I₁ will be the output when the select Inputs are 01, and so on.

Application:

For the given MUX, the enable input is activated (Active low enable).

∴ The output expression will be:

F = P̅Q̅R + P̅Q(0) + PQ̅R + PQ(1)

F = P̅Q̅R  + PQ̅R + PQ

F = Q̅R(P + P̅) + PQ

F = Q̅R + PQ

Concept:

Cauchy principles argument states that the closed contour Γ is mapped into the G(s)-plane will encircle the origin as many times as the difference between the number of poles (P) and zeros (Z) of the open-loop transfer function G(s) that are encircled by the S – plane locus Γ, i.e.

No. of encirclement is given by:

N = P – Z

Calculation:

The closed contour Γ of a pole-zero map of a rational function G(s) contains 2 poles and 3 zeros.

So, the number of encirclement will be:

N = P – Z

N = 2 – 3 = -1

Hence,

It encircles the origin once in the clockwise direction.

Another method to solve:

The closed contour Γ of a pole-zero map of a rational function G(s) is encircling 2 poles and 3 zeros in a clockwise direction, hence the corresponding G(s) plane contour encircles origin 2 times in anti-clockwise direction and 3 times in clockwise direction.

Hence, Effectively it encircles origin once in the clockwise direction.

Special note:

• If we discuss the stability of the open-loop transfer function then we take encirclement around the origin.
• If we discuss the stability of closed-loop transfer function then we take encirclement around

 -1 + j0. (∴ Option 3 and 4 are incorrect)

Concept:

Size of the block = N bits

The received block will be erroneous when any one of the bit is wrongly detected.

Application:

Since all the bits are independent of each other, the probability that all the bits are received

correctly (P_C) will be:

${P_C} = {\left( {1 - \alpha } \right)_1} \times {\left( {1 - \alpha } \right)_2} \times \ldots {\left( {1 - \alpha } \right)_N}$

P_C = (1 - α)^N

where α is the probability of error in detection.

Now, the probability that the received block is erroneous will be:

P_e = 1 – P_C

P_e = 1 – (1 - α)^N

Concept:

For drawing a smith chart:

First, we calculate normalized impedance (z)

$\frac{{{Z_L}}}{{{Z_O}}} = R + jX$     …1)

Then we solve for constant R circles and constant X-circles by using the following formulas:

Const. R circles:

${\left( {{{\rm{\Gamma }}_R} - \frac{R}{{R + 1}}} \right)^2} + {\rm{\Gamma }}_I^2 = {\left( {\frac{1}{{R + 1}}} \right)^2}$     …2)

Const. X circles:

${\left( {{{\rm{\Gamma }}_R} - 1} \right)^2} + {\left[ {{{\rm{\Gamma }}_I} - \frac{1}{x}} \right]^2} = {\left[ {\frac{1}{x}} \right]^2}$     …3)

Where, Γ_R = Real part of reflection coefficients.

Γ_I = I_maj part of reflection coefficients.

Calculation:

Given: Z = j X

Compare this with equation (1), we can write:

R = 0

After putting this value in equation (2), we get:

(Γ_R)² + (Γ_I)² = 1     …4)

Equation (4) represents the equation of a unit circle with center (0, 0). Hence option (1) is correct.

Concept:

To determine Pole-zero plot, we have to first find the transfer function of LTI System, i.e.

$H\left( z \right) = \frac{{Y\left( z \right)}}{{X\left( z \right)}}$

Calculation:

Input-output difference equation is given as:

\(y\left( n \right) = \mathop \sum \limits_{K = 0}^3 {\left( { - 1} \right)^k}x\left( {n - K} \right)\)

y[n] can be represented as:

y(n) = x(n) – x(n - 1) + x(n - 2) – x(n - 3)

Taking the Z-transform, we get:

Y(z) = X(z) – z^-1 X(z) + z^-2 X(z) – z^-3 X(z)

$H\left( z \right) = \frac{{Y\left( z \right)}}{{X\left( z \right)}} = 1 - {z^{ - 1}} + {z^{ - 2}} - {z^{ - 3}}$

$H\left( z \right) = \frac{{{z^3} - {z^2} + z - 1}}{{{z^3}}} = \frac{{\left( {z - 1} \right)\left( {{z^2} + 1} \right)}}{{{z^3}}}$

Hence, Pole-zero plot is:

Given:

x1 + 2x2 = b1 ....1)

2x1 + 4x2 = b2  ....2)

3x1 + 7x2 = b3  ....3)

3x1 + 9x2 = b4  ....4)

 From equations (1) and (2), it is clear that:

b₂ = 2b₁

So options (2) and (4) are incorrect.

Evaluating option (3) now, we are given:

3b₁ – 6b₃ + b ₄ = 0

Using Equation 1, 3 and 4, we can write:

3(x₁ + 2x₂) - 6(3x₁ + 7x₂) + 3x₁ + 9x₂ ≠ 0

Hence option 3 is also incorrect.

Analysing option 1, we are given:

6b₁ – 3b₃ + b₄ = 0

Put the value from equation 1, 3 and 4

6(x1 + 2x2) - 3(3x1 + 7x2) + 3x1 + 9x2 = 0

 

ence option 1 is correct.

$\frac{{dy}}{{dx}} = \left( {y - 1} \right)x$     ---(1)

The given equation can be solved using the variable separable method as:

$\frac{{dy}}{{y - 1}} = x;dx$

Integrating both sides, we get:

$\smallint \frac{{dy}}{{y - 1}} = \smallint x;dx$

$ln;\left( {y - 1} \right) = \frac{{{x^2}}}{2} + C$     ---(2)

Now, from equation (1), we get:

$\frac{{dy}}{{dx}} = 0$ for y = 1

∴ y = constant = 1 is also a solution to the given differential equation.

Since the voltage sources are in parallel with Z, I₁ and I₂ will be calculated as:

${I_1} = \frac{{120\angle - 90^\circ }}{Z}$ 

Z = 80 – j35

In polar form, this can be written as:

Z = 87.32 ∠-23.63

${I_1} = \frac{{120\angle 90^\circ }}{{87.32\angle - 23.63}}$

I₁ = 1.374 ∠-90 + 23.63°

I₁ = 1.374 ∠-66.37°

Similarly,

${I_2} = \frac{{120\angle - 30^\circ }}{{87.32\angle - 23.63}}$ 

I₂ = 1.374 ∠-6.37°

Applying KCL at node A, we get:

I + I₁ + I₂ = 0

I = -(I₁ + I₂)

I = -(1.374 < -66.37 + 1.374 ∠-6.37°)

Converting back to coordinate axis:

I = -(0.55 – 1.258 j + 1.365 – 0.252 j)

I = (2.915 – 1.41 j)

I = -2.915 + 1.47 j

$I = \sqrt {{{\left( { - 1.916} \right)}^2} + {{\left( {1.41} \right)}^2}} \angle {\tan ^{ - 1}}\left( {\frac{{1.41}}{{ - 1.915}}} \right)$

I = 2.38 ∠-36.36

I = 2.38 ∠180 – 36.36

I = 2.38 ∠143.63

Concept:

Sampling theorem is the bridge between continuous-time signals and discrete-time signals i.e., we convert continuous signals to discrete signals using sampling theorem.

y[n]= x(nTs)

Ts = Sampling interval

Calculation:

It is given that a finite duration discrete signal x(n) is obtained by sampling x(t) = cos(200πt) with T_s = n/400

x(t) = cos 200 π t

Since the sampling instant is t = n/400, x(t) can be written as:

$x\left( n \right) = \cos \left[ {200\pi \cdot \frac{n}{{400}}} \right]$

$x\left( n \right) = \cos \left( {\frac{\pi }{2}n} \right)$ n = 0, 1, 2 …7

\(x\left( n \right) = \left{ {\cos 0,;\cos \frac{\pi }{2},\cos \pi ,\cos \frac{{3\pi }}{2}, \ldots ,\cos \frac{{7\pi }}{2}} \right}\)

x(n) = {+1, 0, -1, 0, +1, 0, -1, 0}

The discrete Fourier Transform of x[n] will be X[k].

Assuming, y(n) = [1, -1, 1, -1], i.e.

$x\left( n \right) = y\left( {\frac{n}{2}} \right)$

Then its 4-point DFT will be:

\(Y\left( K \right) = \left[ {\begin{array}{{20}{c}} 1&1&1&1\ 1&{ - j}&{ - 1}&j\ 1&{ - 1}&1&{ - 1}\ 1&j&{ - 1}&{ - j} \end{array}} \right]\left[ {\begin{array}{{20}{c}} 1\ { - 1}\ 1\ { - 1} \end{array}} \right]\)

After solving we’ll get,

Y(K) \( = \left[ {\begin{array}{*{20}{c}} 0&0&4&0 \end{array}} \right]\)

We know that,

If y(n) ↔ Y(K)

Then $y\left( {\frac{n}{m}} \right)$ ↔ [Y(K), Y(K), Y(K) … m times]

So, $y\left( {\frac{n}{2}} \right)$ ↔ [Y(K), Y(K)]

Since, Y(K) = [0, 0, 4, 0]

\(x\left( n \right) = y\left( {\frac{n}{2}} \right) \leftrightarrow \left[ {\begin{array}{*{20}{c}} {0,}&{0,}&{\underbrace {4,}{X(2)}}&{0,}&{0,}&{0,}&{\underbrace {4,}{X(6)}}&0 \end{array}} \right]\)

So, only X(2) & X(6) are non zero.

After applying source transformation, the circuit is redrawn as:

Applying KVL for the given loop, we get:

$2{i_1} - 2i - 0.5\frac{{di}}{{dt}} - V = 0$

$0.5\frac{{di}}{{dt}} = 2{i_1} - 2i - V$

$\frac{{di}}{{dt}} = - 4i - 2V + 4{i_1}$       ---(1)

Applying KCL at node V, we get:

$i = 0.25\frac{{dV}}{{dt}} + \frac{{V - {i_2}}}{1}$

$\frac{1}{4}\frac{{dV}}{{dt}} = i + {i_2} - V$

$\frac{{dV}}{{dt}} = - 4V + 4i + 4{i_2}$     ---(2)

In Matrix from, equation (1) and (2) can be written as:

\(\frac{d}{{dt}}\left[ {\begin{array}{{20}{c}} V\ i \end{array}} \right] = \left[ {\begin{array}{{20}{c}} { - 4}&4\ { - 2}&{ - 4} \end{array}} \right]\left[ {\begin{array}{{20}{c}} V\ i \end{array}} \right] + \left[ {\begin{array}{{20}{c}} 0&4\ 4&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{i_1}}\ {{i_2}} \end{array}} \right]\)

Concept:

Depletion capacitance is calculated as:

${C_D} = \frac{{\varepsilon A}}{W}$    ---(1)

W = Width of the depletion region given by:

$W = {\left[ {\left[ {\frac{{2\varepsilon }}{q}\left( {\frac{1}{{{N_A}}} + \frac{1}{{{N_D}}}} \right)} \right]\left( {{V_{bi}} - V} \right)} \right]^{1/2}}$      ---(2)

V_bi = built-in potential

V = Anode to cathode applied potential

Calculation:

Given: C_D = 50 pF, V = 0.2 V

From equation 2) & equation 1)

${C_D} = \frac{{\varepsilon A}}{{{{\left[ {\frac{{2\varepsilon }}{q}\left( {\frac{1}{{{N_A}}} + \frac{1}{{{N_D}}}} \right)\left( {{V_{bi}} - V} \right)} \right]}^{1/2}}}}$

${C_D} = \frac{K}{{{{\left( {{V_{bi}} - V} \right)}^{1/2}}}}$

$K = \frac{{\varepsilon A}}{{\sqrt {\frac{{2\varepsilon }}{q}\left( {\frac{1}{{{N_A}}} + \frac{1}{{{N_D}}}} \right)} }}$

$\frac{1}{{{C_{{D^2}}}}} = \frac{1}{{{K^2}}}\left( {{V_{bi}} - V} \right)$       ---(3)

$\because \frac{1}{{{C_{{D^2}}}}} = m\left( {{V_{bi}} - V} \right)$

Comparing this with y = mx + c, we can write:

Slope (m) $= \frac{1}{{{K^2}}}$    

Putting the respective values, we get:

$\frac{1}{{{{\left( {50 \times {{10}^{ - 12}}} \right)}^2}}} = m\left( {{V_{bi}} - 0.2} \right)$

Hence, $m = \frac{{{{10}^{24}}}}{{2500\left( {{V_{bi}} - 0.2} \right)}}$

We need V_bi to find the slope of the plot, V_bi is not given so the data is insufficient and it cannot be determined.

Concept:

The Threshold voltage for a capacitor is calculated as:

\({V_T} = {\phi {ms}} - \frac{{{Q{ox}}}}{{{C_{ox}}}} - \frac{{Q_d'}}{{{C_{ox}}}} + 2{\phi _{fp}}\)        ---(1)

ϕ_ms = Metal – semiconductor work function difference

Q_ox = Oxide charge

Q_d’ = Depletion charge per unit area

C_on = Oxide capacitance

ϕ_fp = Energy gap between the intrinsic gap and the Fermi level.

Application:

Given: ϕ_ms = 3.87 eV

ϕ_s = Energy required to move an electron from the semiconductor Fermi level to the vaccum.

From the given Energy-bad diagram, we get:

ϕ_s = 4 eV + (1 – 0.2) eV

ϕ_s = 4.8 eV

∴ ϕ_ms = 3.87 eV – 4.8 eV

ϕ_ms = -0.93 eV

ϕ_fp = E_i - E_fs

ϕ_fp = (1 – 0.5 – 0.2) eV

ϕ_fp = 0.3 eV

Also, as there is no-charge within the oxide, Q_ox = 0

Substituting these values in equation (1), we get:

$0.16 = - 0.93;eV - \frac{{Q_d'}}{{\left( {100} \right); \times ;{{10}^{ - 9}}}} + 2;\left( {0.3;eV} \right)$

$\frac{{Q_d'}}{{100; \times ;{{10}^{ - 9}}}} = ; - 0.17$

$Q_d' = ; - 1.7 \times {10^{ - 8}};C/c{m^2}$

∴ The magnitude of depletion charge per unit area will be 1.7 × 10^-8 C/m²

Concept:

• ​The doping concentration at the base affects the emitter gain (β), i.e. β is dependent on the doping concentration.
• If the doping at the base increases, the recombination of the carriers from the emitter region increases. This reduces the emitter current gain, as fewer carriers from the emitter finally reach the collector.
• Similarly, less the doping concentration at the base, less will be the recombination of the carriers at the base and more carriers will reach the collector resulting in more emitter gain.

 

Mathematically this can be written as:

$\beta \propto\frac{1}{Base~Doping}$

Analysis:

For transistor T1, we can write:

$\beta \propto\frac{1}{N_{B1}}$   ---(1)

For transistor T2, we can write:

$\beta_2\propto\frac{1}{N_{B2}}$   ---(2)

Dividing both the equations, we get:

$\frac{\beta_1}{\beta_2}=\frac{N_{B2}}{N_{B1}}$   ---(3)

NB = Net concentration at the base also called the dose.

$N_B=∫(Doping~ Concentration)dx$

Calculation:

For transistor T1, we get:

\({N_{B1}} = \mathop \smallint \limits_0^W {N_{B1}}\left( x \right)dx\)

From the given doping profile, we get:

${N_{B1}}\left( x \right) = \left( {\frac{{{{10}^{17}} - {{10}^{14}}}}{{0 - W}}} \right)x + {N_{B1}}\left( 0 \right)$

${N_{B1}}\left( x \right) \approx \left( { - \frac{{{{10}^{17}}}}{W}} \right)x + {10^{17}}$

The net concentration or dose at the base of transistor T2 will be:

\({N_{B2}} = \mathop \smallint \nolimits \left( {\left( { - \frac{{{{10}^{17}}}}{W}} \right)x + {{10}^{17}}} \right)dx\)

\({N_{B2}} = \mathop \smallint \nolimits \left( {\left( { - \frac{{{{10}^{17}}}}{W}} \right)\left( {\frac{{{x^2}}}{2}} \right) + {{10}^{17}}x} \right)_0^W\)

${N_{B2}} = \left( { - \frac{{{{10}^{17}}}}{W}} \right)\left( {\frac{{{W^2}}}{2}} \right) + {10^{17}}W$

${N_{B2}} = {10^{17}}W - \frac{{{{10}^{17}}}}{2}W$

${N_{B2}} = \frac{{{{10}^{17}}}}{2}W$

The ratio of the two emitter gains, using Equation (3) will be:

$\frac{{{\beta _1}}}{{{\beta _2}}} = \frac{{{{10}^{17}}W}}{{\frac{{{{10}^{17}}W}}{2}}}$

$\frac{{{\beta _1}}}{{{\beta _2}}} = 2$

β1 = 2 β2

Concept:

The fill factor of a solar cell is defined as:

$FF = \frac{{{P_{out}}}}{{{V_{oc}}; \times ;{I_{sc}}}}$

V_oc = Open-circuit voltage

P_out = FF × V_oc × I_sc

Efficiency is the ratio of the output power to the input power, i.e.,

$\eta = \frac{{{P_{out}}}}{{{P_{in}}}}$

$\eta = \frac{{F.F; \times; {V_{oc}}; \times ;{I_{sc}}}}{{{P_{in}}}}$       ---(1)

Also, the optical generation rate (G_L) will be:

${G_L} = \frac{{{I_{sc}}}}{{q; \times ;Area; \times ;Thickness}}$       ---(2)

 Calculation:

Given η = 0.25, FF = 0.8, V_oc = 0.7 and thickness = 200 μm.

${P_{in}} = \left( {100\frac{{mW}}{{c{m^2}}}} \right) \times \left( {1;c{m^2}} \right)$ 

I_sc = 0.026 A

The optical generation rate (using equation (2)) will be:

${G_L} = \frac{{0.026}}{{1.6; \times ;{{10}^{ - 19}}; \times ;1; \times ;100; \times ;{{100}^{ - 4}}}}$

= 8.37 × 10¹⁸

= 0.837 × 10¹⁹ cm^-3s^-1

Concept:

The small-signal (low-frequency) voltage gain of a common emitter, with bypassed emitter resistor is given by:

A_v = -g_m (R_c||R_L)

g_m = Trans conductance parameter defined as:

${g_m} = \frac{{{I_c}}}{{{V_T}}}$

I_c = DC collector current obtained by doing the DC analysis of the amplifier

Application:

For DC analysis all the capacitor are open-circuited as shown:

 

Given that the base current is negligible, we can write:

I_c ≈ I_E

Applying KVL for the above loop, we get:

0 - V_BE - I_C (20 k) + 10 = 0

${I_C} = \frac{{10 - 0.7}}{{20k}}mA$

I_C = 0.465 mA

${g_m} = \frac{{{I_c}}}{{{V_T}}} = \frac{{0.465;m}}{{26;m}}$

g_m = 0.01788 A/V

The small signal model is drawn as:

The small-signal gain will be:

A_v = -g_m (R_c||R_L)

Putting on the respective values, we get:

A_v = -(0.0178) (10k||10k)

A_v = -0.0178 × 5k

A_v = -89.42

Concept:

Sample and hold circuit is an analog device that samples the voltage of a continuously varying signal and holds its signal value at a constant level for a specified period.

For proper sampling operation, the MOSFET should be on i.e.

${V_{GS}} > {V_T}$

For proper holding operation, the MOSFET should be off i.e.

${V_{GS}} < {V_T}$

Calculation:

Given: V_T = 3 V, V_(I)min= -10V, V_(I)max = 10V

The MOSFET should be on for sampling operation, i.e.

${V_{GS}} > {V_T}$

${V_G} - {V_{\left( I \right)max}} > {V_T}$

${V_G} - 10 > 3$

${V_G} > 13V$ 

The MOSFET should be off for holding operation, i.e.

${V_{GS}} < {V_T}$ 

${V_G} - {V_{\left( I \right)min}} < {V_T}$

${V_G} - \left( { - 10} \right) < 3$

${V_G} < - 7;V$

Hence, the minimum and the maximum values of V_G required for proper sampling and holding respectively, are 13V and -7V respectively.

AC equivalent circuit is redrawn as:

Since there is a dependent source here, we will use a test battery V_x at the source to evaluate the Norton equivalent resistance as shown:

So, ${R_N} = \frac{{{V_x}}}{{{I_x}}}$

Applying KVL in loop 1, we get:

V_x = -V_gs

Above circuit can be replaced by:

Apply KVL we get,

-V_x + I_x (r_ds + R) – g_mV_xr_ds = 0

$\frac{{{V_x}}}{{{I_x}}} = \frac{{{r_{ds}}; + ;R}}{{1; + ;{g_m}{r_{ds}}}} = {R_N}$

Concept:

1. Signed magnitude representation uses the most significant bit (MSB) a sign bit.

• If the sign bit is ‘0’ then the number is positive.
• If the sign bit is ‘1’ then the number is negative.

The remaining bits represent the magnitude of the binary number.

2. 1’s complement representation:

It is a representation of a binary number obtained by toggling all bits in it i.e. transforming the 0 bit to 1 and the 1 bit to 0.

3. 2’s complement representation:

It is obtained by simply adding 1 to the 1’s complement of that binary number.

Calculation:

Given: 4-bit binary number = 1100

Signed magnitude representation (P): - 4

1’s complement representation (Q): - 3

2’s complement representation (R): - 4

So,

P + Q + R = (-4) + (-3) + (-4) = -11

(-11)₁₀ is represented in 2’s complement as:

-(11)₁₀ = 10101

Since the Options are in 6 bits so, we copy sign bit once towards left.

So 6-bit representation of (-11)₁₀ is 110101

Concept:

In general,

This state diagram shows that the state will transit from S₁ to S₂ when the Input is 0 and at the end of the transition, it will produce output as 0.

Calculation:

The state diagram of a sequence detector is given where S₀ is the initial state.

If the output is 1, we then need to obtain an input sequence.

The above-dotted line shows the desired sequence to get output 1. So the state transition is as follows:

Corresponding Input sequence: 0, 1, 0, 1, 0

Corresponding Output sequence: 0, 0, 0, 0, 1

Hence, Detected Output sequence = 0 1 0 1 0.

Concept:

A  breakaway point is a point on a real axis segment of the root locus between two real poles, where the two real closed-loop poles meet and diverge to become complex conjugates.

The breakaway/ break-in/ saddle point is calculated from the solution of:

$\frac{{dK}}{{dS}} = 0$ .

Calculation:

Given:

${s^3} + 3{s^2} + \left( {K + 2} \right)s + 3K = 0$

${s^3} + 3{s^2} + Ks + 2s + 3K = 0$

$K = \frac{{ - \left( {{s^3} + 3{s^2} + 2s} \right)}}{{s + 3}}$

$\frac{{dK}}{{ds}} = \frac{{ - \left[ {\left( {s + 3} \right)\left( {3{s^2} + 6s + 2} \right) - \left( {{s^3} + 3{s^2} + 2s} \right)} \right]}}{{\left( {s + 3} \right)^2}} = 0$

On solving we’ll get:

$2{s^3} + 12{s^2} + 18s + 6 = 0$

Hence,

$s = - 0.46,; - 3.87, - 1.65$ 

So, break away point lie in between (-1,0).

This is explained as shown:

Note:

Breakaway point is formed when an underdamped system is converted into an overdamped system, i.e. it is the point at which the system becomes critically stable.

Concept:

The cut-off frequency is the frequency at which the gain is $\frac{1}{{\sqrt 2 }}$ times of the maximum gain.

The given circuit is s-domain is as shown:

Applying KCL at node A, we can write:

$\frac{{{V^ - } - {V_i}}}{R} + \frac{{{V^ - }}}{{1/2s}} + \frac{{{V^ - }}}{{2R}} + \frac{{{V^ - } - {V_0}}}{R} + \frac{{{V^ - } - {V_0}}}{{2/sC}} = 0$ 

Since V⁺ = V^- (Virtual short circuit), the above equation becomes:

$\frac{{ - {V_i}}}{R} + 0 - \frac{{{V_0}}}{R} - {V_0}sC = 0$

$\frac{{ - {V_i}}}{R} = {V_0}\left( {\frac{1}{R} + sC} \right)$

$\frac{{ - {V_i}}}{R} = \frac{{{V_0}}}{R}\left( {1 + sCR} \right)$

-V_i = V₀ (1 + sCR)

$\frac{{{V_0}}}{{{V_i}}} = \frac{{ - 1}}{{1 + sCR}}$

In the frequency domain, this can be written as:

$\frac{{{V_0}\left( {j\omega } \right)}}{{{V_i}\left( {j\omega } \right)}} = \frac{{ - 1}}{{1 + j\omega RC}}$

$\frac{{{V_0}\left( {j\omega } \right)}}{{{V_i}\left( {j\omega } \right)}} = \frac{{ - 1}}{{1 + j\left( {\frac{\omega }{{{\omega _0}}}} \right)}}$

Where ${\omega _0} = \frac{1}{{RC}}$

We observe that, when $\omega = {\omega _0} = \frac{1}{{RC}}$, the magnitude of the gain becomes $\frac{1}{{\sqrt 2 }}$ times the maximum gain, which is 1.

∴ $\omega = {\omega _0} = \frac{1}{{RC}}$ is the cut-off frequency.

Calculation:

Given, R = 2kΩ, C = 1μF

${\omega _0} = \frac{1}{{RC}}$

$2\pi {f_0} = \frac{1}{{RC}}$

${f_0} = \frac{1}{{2\pi RC}}$ 

$= \frac{1}{{2\pi \times 2k \times {{10}^{ - 6}}}}$ 

f₀ ≅ 79.6 Hz