GATE EC 2014 Official Paper: Shift 2 (Previous Year Paper)

The correct answer is option 2), i.e. both. 

Explanation:

• Both: We use both to refer to two things or people together.
• Here, communication skill and interpersonal skill are two different things, that's we will use both for them.
  

Extra Bites:

Let us try to find the use of words as well- 

• Each: It is used to refer to individual things in a group or a list of two or more things.
• E.g. : Each vehicle was being checked by the police.
• All: It is used to refer to a whole group, with the emphasis that nothing is being left out.
• E.g. : All kids were asked to stand on the ground.
• Either: It is used to refer to anyone object out of the two objects. It is used to expresses choice between the two things. 
• E.g.   : Both these roads go to DB Mall so you can take either way. 
  

Mistake Point:

Although 'all' also looks appropriate here, 'Both' has already been given in one of the options. So we will prefer 'Both'.

The correct answer is option 2), i.e. gets some rest. 

Explanation:

• In the given sentence "if clause" has been used which is a type of conditional clause. Generally, when two actions are given in the Future tense, for the action representing a condition, we use the present indefinite tense, and for the other action, we use future indefinite tense.  "Conditional clause in Present Indefinite tense + Principle clause is in Future indefinite tense (will/shall)"
  

Extra Bites:

• In the given sentence, "She will feel much better" is the principal clause (the action taking place later) which is dependent on the conditional clause.
• A conditional clause has "if" as to show the condition and should be in the present indefinite tense.

The correct answer is option 3), i.e. bear, losing. 

Let us try to find the meaning of words given in the option for better understanding- 

• Bear: (verb) to carry/stand a thought.
• Bare: (verb) uncover a part of the body or other thing and expose it to view.
• Loosing: (gerund) the act of setting free or releasing.
• Losing: (gerund) suffering a loss or failing to keep possession of something.

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Clearly, according to the context of the sentence, the only correct option is option 3), i.e. bear, losing.

For a very large number of throws, the frequency should be same for an unbiased die.

But given frequencies are not same, hence the die is biased

Q, W, Z, B ⇒ Q + 6 → W, W + 3 → Z, Z + 2 → B

B, H, K, M ⇒ B + 6 → H, H + 3 → K, K + 2 → M

W, C, G, J ⇒ W + 6 → C, C + 4 → G, G + 3 → J

M, S, V, X ⇒ M + 6 → S, S + 3 → V, V + 2 → X

All follow the same pattern except "W, C, G, J".

Hence, option 3) is the correct answer.

Let the no. of red lights be equal to X and the no. of green lights be equal to Y.

According to the given condition, there exists a blue light for every red light.

Since there are only two lights on each step of the ladder, we can conclude that for every blue light, there exists only one red light. The same condition can be applied to the pair of green and yellow lights.

Thus, we have, no. of red lights = no. of blue lights = X.

Also, no. of green lights = no. of yellow lights = Y.

Therefore, options 1 and 2 are necessarily correct.

Now, the sum of the red and the green lights = X + Y = the sum of the yellow and the blue lights.

Therefore, option 3 is also necessarily correct.

No. of steps of the ladder that have the blue-red pair may be less than the no. of steps that have the green-yellow pair or vice versa.

Hence, option 4 is not necessarily correct.

Given:

The total quantity of exports of all the items = 500 thousand tonnes

The total revenues = 250 crore rupees

Calculation:

Total revenue for item 1 = 12% ×  250 × 107  = 3 × 108 rupees

Total export for item 1 = 11% × 500 × 10⁶ = 55 × 10⁶ kg

Item1revenue per kg = (3/55) × 10²  = 5.45 rupees

Total revenue for item 2 = 20% × 250 × 107  = 5 × 10⁸  rupees

Total export for item 2 = 20% × 500 × 106 = 10⁸ kg

Item 2 revenue per kg = 5 rupees

Total revenue for item 3 = 23% × 250 × 107  = 5.75 × 10⁸

Total export for item 3 = 19% × 500 × 106 = 95 × 106 

Item 3 revenue per kg = 6.052 rupees

Total revenue for item 4 = 6% × 250 × 107  = 1.5  × 108

Total export for item 4 = 22% × 500 × 106 = 11 × 107

Item 4 revenue per kg = 1.363 rupees

Total revenue for item 5 = 20% × 250 × 107 = 5 × 108  rupees

Total export for item 5 = 12% × 500 × 106 = 6 × 107  rupees

Item 5 revenue per kg = 8.333 rupees

Total revenue for item 6 = 19% × 250 × 107  = 475  × 106  rupees

Total export for item 6 = 16% × 500 × 106 = 8  × 107  rupees

Item 6 revenue per kg = 5.9375  rupees

∴ Item 5 has generated the maximum revenue per kg

Given:

A half-full tank by draining pipe = 30 minutes

Calculation:

A full tank can drain = 60 min.

Let 60 litres can be draining in 60 min.

The efficiency of draining pipe = 1

Half tank filled in 10 min. by both = 30/10 = 3

The efficiency of pump water = 3 + 1 = 4 

∴ Pump water is 4 times the draining rate.

Concept:

The point of maxima or minima is obtained by solving for the derivative of the function and equating to zero.

Then, to check if the point is a point of maxima ‘or’ minima we check the second derivative at that point.

This is explained with the help of the following graph:

If $\frac{{{d^2}f}}{{d{x^2}}} < 0$; the point will be a point of maxima

If $\frac{{{d^2}f}}{{d{x^2}}} > 0$, the point will be a point of minima.

Analysis:

Given f(t) = e^-t – 2e^-2t

f'(t) = -e^-t + 4e^-2t

Solving for f’(t) = 0, we get:

4e^-2t = e^-t

$\frac{4}{{{e^t} \cdot {e^t}}} = \frac{1}{{{e^t}}}$

e^t = 4

t = In 4

f”(t) = e^-t – 8 e^-2t

At t = In 4, we get

f”(In 4) = e^{-In 4} – 8 e^{-2 In 4}

$= \frac{1}{4} - \frac{8}{{16}}$

$= \frac{1}{4} - \frac{1}{2}$

= -0.26

Since, f”(t) < 0 at t = In 4, it is a point of maxima.

Putting x = ∞ to the given limiting function, we get:

${\left( {1 + \frac{1}{\infty }} \right)^\infty } = {1^\infty }$

1^∞ is one of the indeterminant forms.

We can modify the given limits as:

$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{x}} \right)^x} = \mathop {\lim }\limits_{x \to \infty } {e^{ln{{\left( {1 + \frac{1}{x}} \right)}^x}}}$

$= {e^{\mathop {\lim }\limits_{x \to \infty } ln{{\left( {1 + \frac{1}{x}} \right)}^x}}}$

$= {e^{\mathop {\lim }\limits_{x \to \infty } x;ln\left( {1 + \frac{1}{x}} \right)}}$

The above can be written as:

e^f(x)    --(1)

where:

$f\left( x \right) = \mathop {\lim }\limits_{x \to \infty } x\left( {1 + \frac{1}{x}} \right)$       ---(2)

The above can be written as:

$\mathop {\lim }\limits_{x \to \infty } \frac{{ln\left( {1 + \frac{1}{x}} \right)}}{{1/x}}$  

Putting on the limit of x = ∞, we get:

$\mathop {\lim }\limits_{x \to \infty } \frac{{In\left( {1 + \frac{1}{x}} \right)}}{{1/x}} = \frac{0}{0}$

Applying L-Hospitals rule, we get:

$= \mathop {\lim }\limits_{x \to \infty } \frac{{\left( {\frac{1}{{1 + \frac{1}{x}}}} \right)\left( {\frac{{ - 1}}{{{x^2}}}} \right)}}{{ - 1/{x^2}}}$

$= \mathop {\lim }\limits_{x \to \infty } \frac{1}{{1 + \frac{1}{x}}} = 1$

Putting this in equation (1), we can write:

${e^{\mathop {\lim }\limits_{x \to \infty } x;In;\left( {1 + \frac{1}{x}} \right)}} = {e^1}$

$\therefore \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{x}} \right)^x} = e;$

Concept:

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

1. Real, distinct roots, D₁ ≠ D₂

2. Complex Roots, D₁, D₂ = a + ib

3. Double roots, D₁ = D₂ = 0

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

$y = \left( {{C_1} + {C_2}x} \right){e^{{D_1}x}}$

Calculation:

Solving the given differential equation, we get:

D² + 2 α D + 1 = 0

Given the roots are equal, which indicates:

(2α)² – 4(1)(1) = 0

4α² – 4 = 0

α² = 1

α = ± 1

Concept:

• Norton’s Theorem states that “Any linear circuit containing several energy sources and resistances can be replaced by a single constant current source in parallel with a single resistor “.
• It is an analytical method used to change a complex circuit into a simple equivalent circuit consisting of a single resistance in parallel with a current source.

Steps to follow for Norton’s Theorem:

• Calculate Norton’s current source by removing the load resistor from the original circuit and replacing it with a short circuit.
• Calculating the current through a shorted wire.
• Calculating the Norton resistance by removing all power sources in the original circuit (voltage sources shorted and current sources open).
• Calculating total resistance between the open connection points.
• Draw the Norton equivalent circuit, with the Norton current source in parallel with the Norton resistance.
• The load resistor re-attaches between the two open points of the equivalent circuit.

Concept:

For a compensated semiconductor with donor concentration greater than the acceptor concentration, the majority carrier electron concentration is calculated as:

${n_0} = \frac{{{N_d} - {N_a}}}{2} + \sqrt {{{\left( {\frac{{{N_d} - {N_a}}}{2}} \right)}^2} + n_i^2}$

With Nd - Na ≫ ni, the above equation becomes:

n0 ≅ Nd - Na

Also, the charge carriers at thermal equilibrium follow mass-action law, i.e.

${n_0}{p_0} = n_i^2$

Application:

Given:

Nd = 2.25 × 1015 cm-3

ni = 1.5 × 1010 cm-3

Na = 0 cm^-3

Nd - Na ≫ n_i, the majority carrier electron concentration will be:

n0 ≅ Nd - Na = 2.25 × 1015 cm-3

Since the donor concentration is greater than the acceptor concentration, the material will be n-type with the majority carrier electron concentration as:

n0 ≅ Nd - Na

n0 = 2.25 × 1015 cm-3

The minority carrier hole concentration is obtained using mass action law as:

${n_0}{p_0} = n_i^2$

${p_0} = \frac{{n_i^2}}{{{n_0}}}$

${p_0} = \frac{{{{{2.25 \times {{10}^{20}}}}}}}{{2.25 \times {{10}^{15}}}}$

p0 = 10⁵ cm-3

The Base Recombination in BJT increases by the following:

(i) Increase the doping concentration of Base

(ii) Increase the Base-width

I_E = I_B + I_C

Any of the above two processes will increase recombination in the base, which will increase base current, and reduce the collector current.

Reduced collector current will cause a reduction in β, because $\beta = \frac{{{I_C}}}{{{I_B}}}$

Hence common-emitter current gain will decrease.

Hence, option A is wrong.

Power rating = BV_CBO × I_CBO

Power rating = BV_CEO × I_CEO

Since an increase in recombination will cause a decrease in I_CBO. Hence I_CEO will decrease.

This is because I_CEO = (1 + β) I_CBO

But power rating will remain constant, i.e.

\(\begin{array}{*{20}{c}} = &{;B{V_{CEO}}}& \times &{{I_{CEO}}}\ ;& \downarrow &;& \downarrow \ ;&{Increases}&;&{decreases} \end{array}\) 

∴ BV_CEO ( Breakdown voltage) increases.

Hence, option B is correct.

Analyzing option C:

Unity gain cut-off frequency is given by:

${f_T} = \frac{{{g_m}}}{{2\pi \left( {{C_\pi } + {C_\mu }} \right)}}$ 

Since,

${g_m} = \frac{{{I_C}}}{{{V_T}}}$ 

& I_C ↓, g_m ↓, because of which f_T ↓

Hence, option C is incorrect.

Option D is also incorrect because g_m decreases with recombination.

Low energy ion implantation, because it provides independent control of close and depth.

Ion implantation deposits controlled amount of charged species in a specific region of a semiconductor.

In a current series feedback, current is sampled from the output and voltage is feedback to the source.

In the given amplifier circuit, the feedback signal becomes zero by opening the output feedback.

Hence, it is a current series feedback.

Important Points

Procedure to find the type of feedback

Step - 1: Draw AC Model of circuit capacitor-short circuited DC supply-connect to ground.

Step-2: Identify the feed back element in circuit in above circuit R_E is feedback element.

Step-3: If f/b element is already connected to output then it indicate voltage sampling otherwise current sampling. R_E directly not connected to output so current sampling.

Step-4: If f/b element is directly connect to input these it indicate shunt mixing otherwise series mixing. Since R_E is directly not connected to input so series mixing.

Answer - current-series feedback.

Differential voltage Gain (A_d) is given by:

A_d = -g_mR_c

And the Common-mode gain  (A_cm) is given by:

${A_{cm}} = \frac{{ - {R_c}}}{{2{R_E}}}$ 

$CMRR = \left| {\frac{{{A_d}}}{{{A_{cm}}}}} \right|$ 

$CMRR = \frac{{{g_m}{R_c}}}{{\frac{{{R_c}}}{{2{R_E}}}}}$ 

$= 2{g_m}{R_E}$

We can observe that with an increase in R_E,  A_cm decreases and CMRR increases.

In a n variable Boolean function, the maximum number of prime implicant is given by:

${{\rm{P}}_{{\rm{max}}}} = \frac{{{2^{\rm{n}}}}}{2} = {2^{{\rm{n}} - 1}}$

Example with n = 4:

Maximum number of prime applicants = 2n-1 = 24 -1 = 8

The Truth Table for a half subtractor is drawn as:

 

The Boolean expression for difference can be written as:

N = X̅ Y + XY̅

N = X ⊕ Y

And the Boolean expression for the borrow can be written as:

M = X̅ Y

Concept:

Minimum phase system is:

1) Stable and causal: All poles of H(z) are inside the unit circle.

2) Stable and causal Inverse: All poles of 1/H(z) are inside the unit circle and all zeroes of H(z) are inside the unit circle.

Thus, to have a minimum phase system, all poles and zeroes of H(z) must be inside the unit circle.

Maximum Phase system is:

1. Stable and anti causal have a stable and anti causal Inverse.

2. All poles and zeros are outside the unit circle and ROC Includes the unit circle.

Analysis:

$H\left( z \right) = 1 + \frac{7}{2}{z^{ - 1}} + \frac{3}{2}{z^{ - 2}} = 0$ 

2z² + 7z + 3 = 0

2z² + 6z + z + 3 = 0

2z (z + 3) + 1 (z + 3) = 0

(z + 3) (2z + 1) = 0

${z_1} = - \frac{1}{2}$ (Inside the unit circle)

z₂ = -3 (outside the unit circle)

Hence, the given FIR system is a mixed-phase system.

Concept:

Time reversal property of Z-transform

X[n] ↔ X(z)

X[-n] ↔ X(z^-1)

Calculation:

Given that x[n] = x[-n]

⇒ X(z) = X(z^-1)

Zero of X(z) = 1 + j2

Then another zero will be:

$\frac{1}{1+j2}=\frac{1-j2}{5}$

$=0.2-0.4j$

Given block diagram

For X₁(s) = 0 the block diagram can be redrawn as,

$\frac{{Y\left( s \right)}}{{{X_2}\left( s \right)}}$ is 

$\frac{{Y\left( s \right)}}{{{X_2}\left( s \right)}} = \frac{{G\left( s \right)}}{{1 + G\left( s \right)H\left( s \right)}}$

$= \frac{{s + 1}}{{s\left( {s + 2} \right)}}$

Concept: we can remember the relation  $y = \mathop {\lim }\limits_{x \to \infty } x;{\log _2}\left[ {1 + \frac{1}{x}} \right]$ = log2 e = 1.442

Application: It is given that the capacity of Band-limited AWGN is given by:

$C = W{\log _2}\left[ {1 + \frac{P}{{{\sigma ^2}W}}} \right]$

It is given W →∞

$C = \mathop {\lim }\limits_{W \to \infty } W{\log _2}\left[ {1 + \frac{P}{{{\sigma ^2}W}}} \right]$

$C = \mathop {\lim }\limits_{W \to \infty } \frac{{W{\sigma ^2}}}{P}\frac{P}{{{\sigma ^2}}}{\log _2}\left[ {1 + \frac{P}{{{\sigma ^2}W}}} \right]$

$C = \mathop {\lim }\limits_{W \to \infty } \frac{P}{{{\sigma ^2}}}\left( {\frac{{W{\sigma ^2}}}{P}} \right){\log _2}\left[ {1 + \frac{P}{{{\sigma ^2}W}}} \right]$

Let $\frac{{W{\sigma ^2}}}{P} = x$

Since W →∞ then x →∞

$C = \mathop {\lim }\limits_{x \to \infty } \frac{P}{{{\sigma ^2}}};x;lo{g_2}\left[ {1 + \frac{1}{x}} \right]$ 

$C = \frac{P}{{{\sigma ^2}}}\mathop {\lim }\limits_{x \to \infty } {\log _2}{\left[ {1 + \frac{1}{x}} \right]^x}$

Considering:

$y = \mathop {\lim }\limits_{x \to \infty } {\log _2}{\left[ {1 + \frac{1}{x}} \right]^x}$

$y = \mathop {\lim }\limits_{x \to \infty } x;{\log _2}\left[ {1 + \frac{1}{x}} \right]$

$y = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\log }_2}\left[ {1 + \frac{1}{x}} \right]}}{{1/x}}$

Putting x →∞, we are getting $\frac{0}{0}$ form, hence using the L-hospitals rule, we can write:

$y = \mathop {\lim }\limits_{x \to \infty } \frac{1}{{1 + \frac{1}{x}}}\frac{{{{\log }_2}e}}{{\left( {\frac{{ - 1}}{{{x^2}}}} \right)}};\left( {\frac{{ - 1}}{{{x^2}}}} \right)$

y = log₂ e

$C = \frac{P}{{{\sigma ^2}}}{\log _2}e$ 

$C = 1.44\frac{P}{{{\sigma ^2}}}$

C = 1.44 × 1000 bps

C = 1.44 kbps

Concept:

A TEM wave traveling in the given direction must satisfy the Poynting theorem, as it describes the magnitude and direction of the flow of energy in electromagnetic waves.

Mathematically, the Poynting vector is the cross-product of the Electric field vector and the magnetic field vector, i.e.

$P = \vec E × \vec H;Watt/{m^2}$

Analysis:

Option 1:

With E = +8 ŷ and H = -4 ẑ

According to Poynting theorem, the direction of propagation will be:

P̅ = 8 ŷ × (-4ẑ)

P̅ = - 32 x̂

Since the direction of propagation given is +x direction, Option 1 is incorrect.

Option 2:

With E = -2ŷ and H = -3ẑ  

P̅ = -2ŷ × (-3ẑ)

P̅ = 6 x̂

Since the direction of propagation given is +x direction, Option 2 is correct.

Option 3:

With E = +2ẑ and H = +2ŷ

P̅ = 2ẑ × (2ŷ)

P̅ = - 4 x̂

Since the direction of propagation given is +x direction, Option 3 is incorrect.

Option 4:

With E = -3ŷ and H = +4ẑ  

P̅ = -3ŷ × (4ẑ)

P̅ = - 12 x̂

Since the direction of propagation given is +x direction, Option 4 is incorrect.

Concept:

Consider the system of m linear equations

a11 x1 + a12 x2 + … + a1n xn = b1

a21 x1 + a22 x2 + … + a2n xn = b2

…

am1 x1 + am2 x2 + … + amn xn = bm

The above equations containing the n unknowns x1, x2, …, xn. To determine whether the above system of equations is consistent or not, we need to find the rank of following matrices.

\(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}& \ldots &{{a_{1n}}}\ {{a_{21}}}&{{a_{22}}}& \ldots &{{a_{2n}}}\ \ldots & \ldots & \ldots & \ldots \ {{a_{m1}}}&{{a_{m2}}}& \ldots &{{a_{mn}}} \end{array}} \right]\)

\(\left[ {A{\rm{|}}B} \right] = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}& \ldots &{{a_{1n}}}&{{b_1}}\ {{a_{21}}}&{{a_{22}}}& \ldots &{{a_{2n}}}&{{b_2}}\ \ldots & \ldots & \ldots & \ldots & \ldots \ {{a_{m1}}}&{{a_{m2}}}& \ldots &{{a_{mn}}}&{{b_m}} \end{array}} \right]\)

A is the coefficient matrix and [A|B] is called an augmented matrix of the given system of equations.

We can find the consistency of the given system of equations as follows:

(i) If the rank of matrix A is equal to the rank of an augmented matrix and it is equal to the number of unknowns, then the system is consistent and there is a unique solution.

The rank of A = Rank of augmented matrix = n

(ii) If the rank of matrix A is equal to the rank of an augmented matrix and it is less than the number of unknowns, then the system is consistent and there are an infinite number of solutions.

The rank of A = Rank of augmented matrix < n

(iii) If the rank of matrix A is not equal to the rank of the augmented matrix, then the system is inconsistent, and it has no solution.

The rank of A ≠ Rank of an augmented matrix

Calculation:

The given system of equations can be represented in a matrix form as shown below.

\(A = \left[ {\begin{array}{{20}{c}} 2&1&{ 3}\ 3&0&1\ 1&{ 2}&{ 5} \end{array}} \right],X = \left[ {\begin{array}{{20}{c}} a\ b\ c\end{array}} \right],B = \left[ {\begin{array}{*{20}{c}} { 5}\ { -4}\ 14 \end{array}} \right]\)

The Augmented matrix can be written by:

\([A|B] = \left[ {\begin{array}{{20}{c}} 2&1&{ 3}\ 3&0&1\ 1&{ 2}&{ 5} \end{array}{\rm{|}}\begin{array}{{20}{c}} { 5}\ { -4}\ 14 \end{array}} \right]\)

R2 → 2R2 – 3R1 and R3 → 2R3 – R1

\(= \left[ {\begin{array}{{20}{c}} 2&1&{ 9}\ 0&-3&-7\ 0&{ 3}&{ 7} \end{array}{\rm{|}}\begin{array}{{20}{c}} { 5}\ { - 23}\ 23 \end{array}} \right];\)

R3 → R3 + R2

 

\([A|B] = \left[ {\begin{array}{{20}{c}} 2&1&{ 9}\ 0&-3&-7\ 0&{ 0}&{ 0} \end{array}{\rm{|}}\begin{array}{{20}{c}} { 5}\ { - 23}\ 0 \end{array}} \right];\)

The rank of matrix A = 2

The rank of Augmented matrix = 2

Rank of A = Rank of augmented matrix = 2 < n = 3

Hence, the system is consistent and has infinitely many solutions.

Concept:

If f(z) = u + iv is an analytic function, then it satisfies the following:

$\frac{{\partial u}}{{\partial x}} = \frac{{\partial v}}{y};;and;\frac{{\partial u}}{{\partial y}} = - \frac{{\partial v}}{{\partial x}}$

Calculation:

Given: u = e^-y cos x

$\frac{{\partial u}}{{\partial x}} = - {e^{ - y}}\sin x$

$\frac{{\partial u}}{{\partial y}} = - {e^{ - y}}\cos x$

$\frac{{\partial v}}{{\partial y}} = - {e^{ - y}}\sin x$    ---(1)

$\frac{{\partial v}}{{\partial x}} = {e^{ - y}}\cos x$      ---(2)

Integrate equation (1) w.r.t. y, taking x as constant, we get:

v = e^-y sin x

Concept:

A capacitor does not allow a sudden change in voltage, i.e.

V_c(0^-) = V_c(0⁺)

Similarly, an inductor does not allow a sudden change in current across it, i.e.

i_L(0^-) = i_L(0⁺)

Analysis:

For t = 0^- the capacitor has no initial voltage, i.e.

V_c(0^-) = 0

∴ V_c(0⁺) = 0V

The steady-state voltage across the capacitor can be obtained by open-circuiting the capacitor as shown:

Using voltage division rule:

${V_c}\left( \infty \right) = \frac{{5 \times 2}}{{2 + 1}} = \frac{{10}}{3}V$

The voltage across the capacitor will be:

${V_C}\left( t \right) = {V_C}\left( \infty \right) - \left( {{V_C}\left( \infty \right) - {V_C}\left( 0 \right)} \right){e^{\frac{{ - t}}{{{R_{eq}}C}}}}$

$\therefore {V_C}\left( t \right) = \frac{{10}}{3} - \left( {\frac{{10}}{3} - 0} \right){e^{\frac{{ - t}}{{{R_{eq}}C}}}}$

${V_C}\left( t \right) = \frac{{10}}{3}\left( {1 - {e^{\frac{{ - t}}{\tau }}}} \right)$

τ = R_eq C = Time constant.

R_eq = Equivalent impedance across the capacitor which is obtained as:

${R_{eq}} = \frac{{2 \times 1}}{{2 + 1}} = \frac{{2k}}{3}$

$\therefore \tau = \frac{{2k}}{3} \times 1\mu = \frac{2}{3}msec$

Since the capacitor is connected in parallel to the Resistance R₂, the voltage across R₂ will also be V_C(t).

∴ The required current I will be:

$I = \frac{{{V_C}\left( t \right)}}{{{R_2}}}$

$I = \frac{{10}}{{3 \times 2k}}\left( {1 - {e^{\frac{{ - t}}{\tau }}}} \right)$

$I = \frac{5}{3}\left( {1 - {e^{\frac{{ - t}}{\tau }}}} \right)mA$

$\tau = \frac{2}{3}msec$

Concept:

Consider the following p-n junction:

The total depletion with W is the sum of the depletion width on the p-side and n-side, i.e.

W = W_P + W_N

Where W_p is given by:

${W_P} = \frac{{W \cdot {N_D}}}{{{N_A} + {N_D}}}$ 

And W_N is given by:

${W_N} = \frac{{W \cdot {N_A}}}{{{N_A} + {N_A}}}$ 

Calculation:

Given:

N_A = 9 × 10¹⁶/cm³

N_D = 10¹⁶/cm³

W = 3 μm

${W_p} = \frac{{W \cdot {N_D}}}{{{N_A} + {N_D}}}$ 

${W_P} = \frac{{3 \times {{10}^{16}}}}{{9 \times {{10}^{16}} + {{10}^{16}}}}$ 

${W_P} = \frac{3}{{10}}\mu m = 0.3;\mu m$ 

W_p = 0.3 μm

Peak electric field is given by Poisson’s equation as: 

${E_{peak}} = \frac{V}{E}{W_p}{N_A}$ 

$= \frac{{1.6 \times {{10}^{ - 19}}}}{{1.04 \times {{10}^{ - 12}}}} \times 0.3 \times {10^{ - 4}} \times 9 \times {10^{16}}$ 

E_peak = 4.15 × 10⁵ V/cm

If in the question it is mentioned to the find peak electric field using linear approximation then:

${E_{peak}} = \frac{{ - 2;{V_0}}}{W}$ 

Where,

V₀ = potential at junction

W = depletion width

E_peak = peak electric field intensity at the junction.

If nothing is mention go with the Poisson’s equation:

${E_{peak}} = \frac{{ - V}}{E};{W_N}{N_D} = \frac{{ - V}}{E}{W_p}{N_A}$

For the complete negative half cycle, the diode will be OFF as it will be reversed biased.

This is as shown:

∴ The minimum voltage will be the minimum value that the input voltage can reach, i.e. -5V

For the positive half cycle, the diode will be On when the voltage across the diode exceeds 2.7 V.

The maximum voltage will be when the input voltage is at its maxima, i.e. at + 5V

The circuit is drawn as:

Applying KVL across the loop, we get:

5 – I × 1k – I× 1k – 0.7 – 2 = 0

2.3 = 2I × 1k

$I = \frac{{2.3}}{{2k}}A$

Applying KVL from V_i to V₀, we can write:

$5 - \frac{{2.3}}{{2r}} \times 1k = {V_0}$

V₀ = 5 – 2.15 = 3.85

Concept:

Operation in Triode region:

V_DS < V_DS (sat)

‘OR’

V_DS < [V_GS - V_T]

And the equation of current is given by:

\({I_D} = {\mu n};{C{ox}}\frac{W}{L};\left{ {\left( {{V_{GS}} - {V_T}} \right){V_{DS}} - \frac{{V_{DS}^2}}{2}} \right}\) 

Since V_DS < V_GS – V_T at the triode region, the higher power values of V_DS can be neglected, i.e. neglecting, $\frac{{V_{DS}^2}}{2}$, we get:

\({I_D} = {\mu n}{C{ox}}\frac{W}{L}\left( {{V_{GS}} - {V_T}} \right){V_{DS}}\) 

Let, μ_n C_ox = K_N : Process transconductance Parameter

${I_D} = {K_N};\frac{W}{L}\left( {{V_{GS}} - {V_T}} \right){V_{DS}}$ 

Saturation Region:

V_DS ≥ V_DS (sat)

‘OR’

V_DS ≥ [V_GS - V_T]

If V_D = V_G

<sub>

</sub>

It will always be in the saturation region.

If V_D > V_G

<sub>

</sub>

It will always be in the saturation region.

The equation for a MOSFET in saturation is given by:

\({I_D} = \frac{1}{2};{\mu n};{C{ox}}\frac{W}{L}{\left( {{V_{GS}} - {V_{th}}} \right)^2}\left( {1 + \lambda .{V_{DS}}} \right)\)

λ = channel length parameter, expressed in $\frac{1}{{volt}}$ 

Neglecting λ, i.e. with λ = 0, the current equation becomes:

\({I_D} = \frac{1}{2}{\mu n}{C{ox}}\frac{W}{L}{\left( {{V_{GS}} - {V_T}} \right)^2}\) 

${I_D} = \frac{1}{2}{K_N}\frac{W}{L}{\left( {{V_{GS}} - {V_T}} \right)^2}$ 

Calculation:

Given:

V_Th = 0.8 V

${V_{{D_1}}} = 1.6;V,;;{I_{{D_1}}} = 0.5;mA$

${V_{{D_2}}} = 2V,;;{I_{{D_2}}} = ;?$ 

We observe that the given MOS transistor is in the saturation mode as V_DS > V_GS - V_T

${I_{{D_1}}} = {k_n}{\left( {{V_{G{S_1}}} - {V_T}} \right)^2}$ 

${k_n} = \frac{{{I_{{D_1}}}}}{{{{\left( {1.6 - .8} \right)}^2}}}$ 

${k_n} = 0.78;mA/v$ 

Also V_G = V_D

Now,

V_D2 = 2 V. So, V_GS2 = 2 V

${I_{{D_2}}} = {k_n}{\left( {{V_{G{S_2}}} - {V_T}} \right)^2}$ 

${I_{{D_2}}} = 0.78;{\left( {2 - 0.8} \right)^2};$ 

${I_{{D_2}}} = 1.125;mA$

Consider the given circuit:

Let Q₁ and Q₂ be the output of flip flop 1 and flip flop 2.

Please note that flip flop is a negative edge triggered, i.e. the output of flip flop only changes at the negative edge of the clock.

The function of the D-flip flop is to follow the input when the flip flop is triggered.

Here, the flip flop is triggered at the negative edge of the flip flop.

The output y is similar w3.

Hence, option C is correct.

Concept:

The Truth table of JK flip flop is given by:

 

Analysis:

Initially, Q₁ and Q₂ are 0, 0.

From the figure:

J₁ = Q̅₂ , K₁ = Q₂, J₂ = Q₁, K₂ = Q̅₁

 

Hence the sequence generated at Q₁ is 0110…

Let us understand the concept behind the Question:

• For the data, digital input of 8-bit to be transformed in output, we need to enable I/O device.
• To enable I/O device DS̅₁ should be active low & DS₂  must be active high.
• DS̅₁ is active low when output 0 of the decoder 3 × 8 is active low.
• Output 0 is active low When A₂ = A₁ = A₀ = 0
• But to enable decoder, we need

          $\overline {{G_2}A}$ → active low,

          $\overline {{G_2}B}$ → active low

         & G₁ → active high

• To make G₁ active high A₅ = A_r = A₅ = A₆ = A₂ = 1
• $\overline {{G_2}A}$ is active low when IO/M̅ = 0, i.e. memory operation is being performed.
• $\overline {{G_2}B}$ is active low when RD̅ = 0, i.e. read operation is performed.
• Hence, from the above two statements we have conducted memory read operation is performed.

<br>

To make DS₂ high, we need:

 

The address lines in hexadecimal is F8 F8 H

Hence the correct instrument for data transfer is LDA F8 F8 H, which means load accumulator with the content present at address F8 F8H.

Analyzing the given options:

• Instrument IN F8H & OUT F8H is an invalid instruction because memory operation is performed and not I/O operation.
• HVI F8H, means transfer the data F8H into the accumulator, but in question, no information is given about the data lines, all the information is concerning to the address lines.

y[n] = α y[n – 2] + β x[n]

by applying z-transform on both the sides,

Y(z) = αz^-2 Y(z)+ β X(z)

Y(z) [1 - αz^-2] = β X(z)

$\frac{Y\left( z \right)}{X\left( z \right)}=\frac{\beta }{1-\alpha {{z}^{-2}}}$ 

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

The impulse response can be expressed as

\(H\left( z \right)=\underset{0}{\overset{\infty }{\mathop \sum }},h\left[ n \right]{{z}^{-n}}\)

Put z = 1,

\(H\left( 1 \right)=~\underset{0}{\overset{\infty }{\mathop \sum }},h\left[ n \right]\)

H(1) = 4

$\frac{\beta }{1-\alpha }=4$ 

$1-\alpha =\frac{\beta }{4}$

$4\alpha +\beta =4$

Concept:

Solution of state equation is given as:

x(t) = ϕ(t) x(0)

Where,

ϕ(t) = state transition matrix and

ϕ(t) = L^-1 {(sI-A)^-1}

Analysis:

\(A = \left[ {\begin{array}{*{20}{c}} { - 1}&0\ 0&{ - 2} \end{array}} \right]\)

\(sI - A = \left[ {\begin{array}{{20}{c}} s&0\ 0&s \end{array}} \right] - \left[ {\begin{array}{{20}{c}} { - 1}&0\ 0&{ - 2} \end{array}} \right]\)

\( = \left[ {\begin{array}{*{20}{c}} {s + 1}&0\ 0&{s + 2} \end{array}} \right]\)

\({\left( {sI - A} \right)^{ - 1}} = \frac{1}{{\left( {s + 1} \right)\left( {s + 2} \right)}}\left[ {\begin{array}{*{20}{c}} {s + 2}&0\ 0&{s + 1} \end{array}} \right]\)

\( = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{s + 1}}}&0\ 0&{\frac{1}{{s + 2}}} \end{array}} \right]\)

\(\phi \left( t \right) = {L^{ - 1}}\left{ {{{\left( {sI - A} \right)}^{ - 1}}} \right}\)

\( = \left[ {\begin{array}{*{20}{c}} {{e^{ - t}}}&0\ 0&{{e^{ - 2t}}} \end{array}} \right]\)

x(t) = ϕ(t) x(0)

\(x\left( t \right) = \left[ {\begin{array}{{20}{c}} {{e^{ - t}}}&0\ 0&{{e^{2t}}} \end{array}} \right]\left[ {\begin{array}{{20}{c}} 1\ { - 1} \end{array}} \right]\)

\(\left[ {\begin{array}{{20}{c}} {{x_1}\left( t \right)}\ {{x_2}\left( t \right)} \end{array}} \right] = \left[ {\begin{array}{{20}{c}} {{e^{ - t}}}\ { - {e^{ - 2t}}} \end{array}} \right]\)

x₁(t) = e^-t and x₂(t) = -e^-2t

Concept:

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

$\frac{1}{s}$ is an integrator, ẋ₁ = x₂

Where ẋ = differentiation of x₁

Analysis:

The state equation and output equation as:

State equation:

ẋ₁ = x₂

ẋ₂ = x₃

ẋ₃ = a₁ x₁ + a₂x₂ + a₃x₃ + u

Output equation:

y = c₁x₁ + c₂x₂ + c₃x₃

\(\left[ {\begin{array}{{20}{c}} {{{\dot x}_1}}\ {{{\dot x}_2}}\ {{{\dot x}_3}} \end{array}} \right] = \left[ {\begin{array}{{20}{c}} 0&1&0\ 0&0&1\ {{a_1}}&{{a_2}}&{{a_3}} \end{array}} \right]\left[ {\begin{array}{{20}{c}} {{x_1}}\ {{x_2}}\ {{x_3}} \end{array}} \right] + \left[ {\begin{array}{{20}{c}} 0\ 0\ 1 \end{array}} \right]u\) 

\(y = \left[ {{c_1};;{c_2};;{c_3}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\ {{x_2}}\ {{x_3}} \end{array}} \right]\) 

Here,

\(A = \left[ {\begin{array}{*{20}{c}} 0&1&0\ 0&0&1\ {{a_1}}&{{a_2}}&{{a_3}} \end{array}} \right]\)  is state matrix

\(B = \left[ {\begin{array}{*{20}{c}} 0\ 0\ 1 \end{array}} \right]\) is input matrix

And C = [c₁  c₂  c₃] is the output matrix.

Now,

After state modeling the system, let’s check for controllability & observability.

Kalman’s Test:

(i) For controllability:

Q_c = [B  AB  A²B  …]

If |Q_c| ≠ 0 ⇒ Controllable

            = 0 ⇒ Uncontrollable

(ii) \({Q_0} = \left[ {\begin{array}{*{20}{c}} C\ {CA}\ {C{A^2}} \end{array}} \right]\)

|Q₀| ≠ 0 ⇒ Observable

         = 0  ⇒ Unobservable 

\(AB = \left[ {\begin{array}{{20}{c}} 0&1&0\ 0&0&1\ {{a_1}}&{{a_2}}&{{a_3}} \end{array}} \right]\left[ {\begin{array}{{20}{c}} 0\ 0\ 1 \end{array}} \right]\)

 

\(AB = \left[ {\begin{array}{*{20}{c}} 0\ 1\ {{a_3}} \end{array}} \right]\) 

\({A^2} = \left[ {\begin{array}{{20}{c}} 0&1&0\ 0&0&1\ {{a_1}}&{{a_2}}&{{a_3}} \end{array}} \right]\left[ {\begin{array}{{20}{c}} 0&1&0\ 0&0&1\ {{a_1}}&{{a_2}}&{{a_3}} \end{array}} \right]\)

\({A^2} = \left[ {\begin{array}{*{20}{c}} 0&0&1\ {{a_1}}&{{a_2}}&{{a_3}}\ {{a_1}{a_3}}&{{a_1} + {a_2}{a_3}}&{{a_2} + a_3^2} \end{array}} \right]\) 

\({A^2}B = \left[ {\begin{array}{{20}{c}} 0&0&1\ {{a_1}}&{{a_2}}&{{a_3}}\ {{a_1}{a_3}}&{{a_1} + {a_2}{a_3}}&{{a_2} + a_3^2} \end{array}} \right]\left[ {\begin{array}{{20}{c}} 0\ 0\ 1 \end{array}} \right]\) 

\({A^2}B = \left[ {\begin{array}{*{20}{c}} 1\ {{a_3}}\ {{a_2} + a_3^2} \end{array}} \right]\) 

\({Q_c} = \left[ {\begin{array}{*{20}{c}} 0&0&1\ 0&1&{{a_3}}\ 1&{{a_3}}&{{a_2} + a_3^2} \end{array}} \right]\) 

$\left| {{Q_c}} \right| = 0 = 0 + 1\left( {0 - 1} \right)$

$\left| {{Q_c}} \right| = - 1$

Since |Q_c| ≠ 0, hence our system is always controllable.

Since it is MCQ type Question we need not check other options. Option A is correct.

But let’s check for the observability.

\({Q_0} = \left[ {\begin{array}{*{20}{c}} C\ {CA}\ {C{A^2}} \end{array}} \right]\) 

\(CA = \left[ {{C_1};;{C_2};;{C_3}} \right]\left[ {\begin{array}{*{20}{c}} 0&1&0\ 0&0&1\ {{a_1}}&{{a_2}}&{{a_3}} \end{array}} \right]\) 

CA = [a₁c₃  c₁ + a₂c₃  c₂ + c₃a₃]

\(C{A^2} = \left[ {{c_1};;{c_2};;{c_3}} \right]\left[ {\begin{array}{*{20}{c}} 0&0&1\ 0&{{a_2}}&{{a_3}}\ {{a_1}{a_3}}&{{a_1} + {a_2}{a_3}}&{{a_1} + a_3^2} \end{array}} \right]\) 

$C{A^2} = \left[ {{a_1}{a_3}{c_3};;{a_2}{c_2} + {a_1}{c_3} + {a_2}{a_3}{c_3};;{c_1} + {c_2}{a_3} + {a_2}{c_3} + a_3^2{c_3}} \right]$ 

\({Q_0} = \left[ {\begin{array}{*{20}{c}} {{C_1}}&{{c_2}}&{{c_3}}\ {{a_1}{c_3}}&{{c_1} + {a_2}{c_3}}&{{c_2} + {c_3}{a_3}}\ {{a_1}{a_3}{c_3}}&{{a_2}{c_2} + {a_1}{c_3} + {a_2}{a_3}{c_3}}&{{c_1} + {c_2}{a_3} + {a_2}{c_3} + a_3^2{c_3}} \end{array}} \right]\) 

Since we do not know about the nature of a₁, a₂, a_3, and c₁, c₂, c₃ whether they are positive or

negative numbers we cannot comment on |Q₀| & hence we cannot comment on observability.

Similar is the case for stability since for stable system all the roots of characteristic equation must lie

in the left half of s-plane characteristic equation: |SI - A| = 0

\(\left[ {SI - A} \right] = \left[ {\begin{array}{{20}{c}} s&0&0\ 0&s&0\ 0&0&s \end{array}} \right] - \left[ {\begin{array}{{20}{c}} 0&1&0\ 0&0&1\ {{a_1}}&{{a_2}}&{{a_3}} \end{array}} \right]\) 

\(= \left[ {\begin{array}{*{20}{c}} s&{ - 1}&0\ 0&s&{ - 1}\ { - {a_1}}&{ - {a_2}}&{s - {a_3}} \end{array}} \right]\) 

$\left| {SI - A} \right| = S\left( {{s^2} - {a_3}s - {a_2}} \right) + \left( { - {a_1}} \right)$ 

|SI - A| = s³ – a₃s² – a₂ s – q₁ = 0

Now, again we cannot comment on stability because we do not know the nature of a₁, a₂ & a₃

Hence, only option A is correct.

Concept:

FSK Modulation:

In FSK: transmission of 1 is represented as:

s₁(t) = A_c cos 2π f_Ht

Transmission of 0 is represented as:

s₂(t) = A_c cos 2π f_Lt

and the Bit error probability $= Q\left[ {\sqrt {\frac{{{E_d}}}{{2{N_0}}}} } \right]$

Where E_d is the energy of s₁(t) – s₂(t)

\({E_d} = \mathop \smallint \limits_0^{{T_b}} { {s_1}\left( t \right) - {s_2}\left( t \right)}^2;dt;\) 

\({E_d} = \mathop \smallint \limits_0^{{T_b}} s_1^2\left( t \right)dt + \mathop \smallint \limits_0^{{T_b}} s_2^2\left( t \right)dt - 2\mathop \smallint \limits_0^{{T_b}} {s_1}\left( t \right) - {s_2}\left( t \right)dt\) 

Since s₁(t) & s₂(t) are orthogonal, we cn write:

\(\therefore ;\mathop \smallint \limits_0^{{T_b}} {s_1}\left( t \right) - {s_2}\left( t \right)dt = 0\) 

\({E_d} = \mathop \smallint \limits_0^{{T_b}} s_1^2\left( t \right)dt + \mathop \smallint \limits_0^{{T_b}} s_2^2\left( t \right)dt\) 

${E_d} = \frac{{A_c^2{T_b}}}{2} + \frac{{A_c^2{T_b}}}{2} = A_c^3{T_b}$ 

$BER = Q\left( {\sqrt {\frac{{A_c^2{T_b}}}{{2{N_0}}}} } \right)$ 

Analysis:

Given:

A_c = α = 4 mV

$\frac{{{N_0}}}{2} = 0.5 \times {10^{ - 12}};w/Hz$ 

N₀ = 10^-12 w/Hz

${T_b} = \frac{1}{{{R_b}}} = \frac{1}{{800 \times {{10}^3}}}$ 

T_b = 0.2 × 10^-5

T_b = 2 × 10^-6 sec.

$BER = Q\left( {\sqrt {\frac{{A_c^2{T_b}}}{{2{N_0}}}} } \right)$ 

$BER = Q\left( {\sqrt {\frac{{{{\left( {4 \times {{10}^{ - 3}}} \right)}^2} \times 2 \times {{10}^{ - 6}}}}{{2 \times {{10}^{ - 12}}}}} } \right)$ 

$BER = Q\left( {\sqrt {\frac{{16 \times {{10}^{ - 6}} \times 2 \times {{10}^{ - 6}}}}{{2 \times {{10}^{ - 12}}}}} } \right)$ 

$BER = Q\left( {\sqrt {16} } \right)$ 

BER = Q(4)

Concept: 

E_x = E₁ sin (ωt – βz) â_x

E_y = E₂ sin (ωt – βz + δ) â­_y

If δ = 0 linear polarization.

If δ = 90° and E₁ = E₂ circular

If δ ≠ 90°    δ ≠ 0    E₁ ≠ E₂ Elliptical        

Now left or right ??

Keep your left-hand thumb in the direction of propagation and then check the. The direction of the finger. If it follows the direction then it is left circular otherwise right circular.

Application: The polarization of a plane wave is the figure traced by the tip of the electric field vector as a function of time, at a fixed point in space.

Given:

$\vec E\left( {z,;t} \right) = {a_x}3\cos \left( {\omega t - kz + 30^\circ } \right) - {a_y}4\sin \left( {\omega t - kz + 45^\circ } \right)$

Fixing a point, i.e. z = 0 (say)

$\vec E\left( {0,;t} \right) = {a_x}3\cos \left( {\omega t + 30^\circ } \right) - {a_y}4\sin \left( {\omega t + 45^\circ } \right)$

Now, when ωt = 0

$\vec E = {a_x}3\cos 30^\circ - {a_y}4\sin \left( {45^\circ } \right)$

E = 2.6 a_x – 2.82 a_y

Similarly, $at;\omega = \frac{\pi }{6}\left( {30^\circ } \right)$, we get:

E = a_x3 cos (60°) - a_y 4 sin (75°)

E = 1.5 a_x – 3.86 a_y

The trace is plotted as: