GATE EC 2019 Official Paper (Previous Year Paper)

In the given statement, the first blank requires a verb whose subject is 'company' which is singular, hence according to the subject verb agreement, 'uses' fits correctly. 

The second blank requires a verb whose subject is 'strategies' which is plural, hence according to the subject verb agreement, 'include' fits correctly. 

Therefore, option 2 is the correct answer. The complete sentence is: The strategies that the company uses to sell its products include house-to-house marketing.

Let us look at the prepositions that are used to denote time:

“In” usually refers to a period of time. For eg, They managed to complete the job in two weeks.

“At” is used in reference to specific times on the clock or points of time in the day. For eg, The train arrives at 3:30.

“On” is used with dates and named days of the week. For eg, We’re going to the theatre on Wednesday evening.

"Under" is used to talk about measurements of time. For eg, We finished the project in under a year and a half.

“By” is used specifically with an endpoint of time and it means no later than. For eg, Please return these books by Friday.

Hence the correct answer is option 2. The complete sentence is: The boat arrived at dawn.

Let 1^st machine does ‘x’ work in 4 hrs.

In 1 hr, the same machine 1 will do: \(\frac{x}{4}\) work

Let 2^nd machine does x work in 2 hrs

In 1 hr, machine 1 will do: \(\frac{x}{2}\) work

In 1 hr, together both will do:

\( \left( {\frac{x}{4} + \frac{x}{2}} \right)\) of the work

Let after ‘t’, the work is completed. We can then write:

\(\left( {\frac{x}{4} + \frac{x}{2}} \right)t = x\)

\(\left( {\frac{{3x}}{4}} \right)t = x\)

\(x = \frac{4}{3}hours\)

According to the question, the arrangement will be as shown:

Now P, Q, T has to come in one of these 3 places

Starting with P, it can come in 3 places.

After placing P, Q can come in 2 places.

After placing P and Q, T can come in 1 place.

Total ways = 3 × 2 × 1 = 6 ways

Let us look at the meaning the various options will have if added in the given sentence.

concluded: arrive at a judgment or opinion by reasoning. This verb would be grammatically and incorrect in the sentence.

looked: regard in a specified way. The preposition 'at' is required with this verb which is not present in the given sentence and the option. Also, this verb would be incorrect in the sentence as 'she' could not have 'looked at him' 'when he did not come home'.

notice: the fact of observing or paying attention to something. This verb would be grammatically incorrect in the sentence as 'she' could not have 'noticed him' 'when he did not come home'.

pictured: forms a mental image of something.

From the meaning, it becomes clear that option 4, 'pictured' will make the meaningful. The complete sentence is: When he did not come home, she pictured him lying dead on the roadside somewhere.

Make packets of the given information

• Seema turns right means she is in either of middle to
• Seema cannot be at 3, since then Mathew will come at her either left or right

→ Seema = 3^rd

→ Mathew = 1^st

Let total population be x

Calculating For 2021,

Number of Males = (60x)/100 = 0.6x

Number of illiterate males = 0.6x × (50/100) = 0.3x

Number of females = 0.4x

Number of illiterate females = 0.4x × (60/100) = 0.24x

Total illiterates = 0.3x + 0.24x = 0.54x

Total Literates = x - Total illiterates

Total Literates = x - 0.54x = 0.46x

 Similarly for 2011,

Number of Males = (50x)/100 = 0.5x

Number of illiterate males = 0.5x × (40/100) = 0.2x

Number of females = 0.5x

Number of illiterate females = 0.5x × (40/100) = 0.2x

Total illiterates = 0.2x + 0.2x = 0.4x

Total Literates = x - Total illiterates

Total Literates = x - 0.4x = 0.6x

∴ The Percentage increase in literates from 2001 to 2011 \(=\frac{(0.6x – 0.46x)}{0.46x} × 100\)

The Percentage increase in literates from 2001 to 2011 = 30.43%

Alternate Method 

Number of Literates in 2011 = (0.5 × 0.6 × x) + (0.5 × 0.6 × x) = 0.6x

Number of Literates in 2001 = (0.6 × 0.5 × x) + (0.4 × 0.4 × x) = 0.46x

∴ Increase in literates from 2001 to 2011 \(=\frac{(0.6x – 0.46x)}{0.46x} × 100\)  = 30.43%

The given excerpt mentions that Indian history which was written by British historians was 'extremely well documented and researched, but not always impartial'. The word 'impartial' means 'treating all rivals or disputants equally; unbiased'. This meaning is accurately expressed in option 3, 'was well documented and researched but was sometimes biased'. 

Option 1 is incorrect as it says that the Indian history which was written by British historians was 'always biased'.

Option 2 is incorrect as it says that the Indian history which was written by British historians was 'not well documented and researched and was always biased'.

Option 4 is incorrect as it says that the Indian history which was written by British historians was 'not well documented and researched and was sometimes biased'.

Hence option 3 is the correct answer. The complete sentence is: Indian history written by British historians was well documented and researched but was sometimes biased.

210° = 7 hrs

Total hours worked by P: 7 hrs

Total break time:

\(= \frac{{15}}{{60}} \times 2 = \frac{1}{2}hrs\)

Net-work time \(= 7 - \frac{1}{2} \)

\(= 6.5;hrs = \frac{{13}}{2}hrs\)

For Q

240° = 8 hrs

Works hours for Q = 8 hrs

\(Break;time = \frac{{20}}{{60}}hr = \frac{1}{3}hr\)

Net work time \(= 8-\frac{{1}}{3} = \frac{{23}}{3}hr\)

Total work of P and Q combined will be:

\( \left( {\frac{{13}}{2} + \frac{{23}}{3}} \right)\)

\(= \frac{{85}}{6}hours\)

Payment for 1 hr = 200

For \(\frac{{85}}{6}hr = \frac{{85}}{6} \times 200\) 

= 2833.33

Money left = (3000 – 2833.33) $

= 166.67 $

At present

P, Q sit together

R, T sit together

Only option ‘c’ satisfies the given conditions.

(i) ln z

At, z = 0

The function is not analytic

(ii) e^1/z

Using the expansion formula for e^x

\({e^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \ldots \)

\({e^{1/z}} = 1 + \left( {\frac{1}{z}} \right) + \frac{1}{{2!}}\left( {\frac{1}{{{z^2}}}} \right) + \frac{1}{{3!}}\left( {\frac{1}{{{z^3}}}} \right) + \ldots \)

At, z = 0, e^1/z is not analytic

(iii) \(\frac{1}{{1 - z}}\)

At, z = 1, the function is not analytic

(iv) cos z

expansion of cos z is:

\(\cos z = 1 - \frac{{{z^2}}}{{2!}} + \frac{{{z^4}}}{{4!}} - \frac{{{z^6};}}{{6!}} + \ldots \)

The function cos z is analytic over the entire z.

\(\frac{{dy}}{{dx}} = - {\left( {\frac{x}{y}} \right)^n}\)

At, n = -1

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

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

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

\(\frac{{dy}}{y} = - \frac{{dx}}{x}\)

Integrating

\(\smallint \frac{{dy}}{y} = - \smallint \frac{{dx}}{x}\)

⇒ ln y = -ln x + ln c

Ln (xy) = ln c

xy = c

at, n = +1

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

Ydy = -xdx

Integrating

\(\frac{{{y^2}}}{2} = \frac{{ - {x^2}}}{2} + c\)

x² + y² = c

Given:

\(P\left( z \right) = H\left( z \right)H\left( {\frac{1}{z}} \right)\)

\(P\left( {\frac{1}{z}} \right) = H\left( {\frac{1}{z}} \right)H\left( z \right)\)

Thus, P(z) = P(z^-1)

In such a condition, if z = z₀ is a zero then:

\(z_0^*,z = \frac{1}{{{z_0}}}\) and \(z = \frac{1}{{z_0^*}}\) are also zeroes.

Given, \({z_0} = \frac{{1 + j}}{2}\) is zero

\(z = z_0^* = \frac{{1 - j}}{2}\) is also zero

\(z = \frac{1}{{{z_0}}} = \frac{{2\left( {1 - j} \right)}}{{\left( {1 + j} \right)\left( {1 - j} \right)}} = \frac{{2 - 2j}}{2} = 1 - j\) is also a zero

\(z = \frac{1}{{z_0^*}} = \frac{2}{{\left( {1 - j} \right)\left( {1 + j} \right)}} = \frac{{2\left( {1 + j} \right)}}{2} = 1 + j\) is also a zero

The corresponding pole-zero plot will be:

Reciprocity Theorem:

In any passive linear bilateral network, if a single voltage source ‘V’ in the branch AB produces the current response ‘I’ in the branch CD, then the removal of voltage source from the branch AB and its insertion in the branch CD will produce same current ‘I’ in the branch AB.

In the reciprocity theorem, if the position of excitation and responses are interchanged, then their ratio remains the same.

\(\frac{{{V_1}}}{{{I_1}}} = \frac{{{V_2}}}{{{I_2}}}\)

[Applicable for linear bilateral network]

Application:

\(\Rightarrow \frac{{{V_1}}}{{{I_1}}} = \frac{{{V_2}}}{{{I_2}}}\)

\(\Rightarrow \frac{5}{1} = \frac{5}{{{I_2}}}\)

I₂ = 1 A

Concept:

The output response of a system is equal to the sum of natural response and forced response.

Forced response: The response generated due to the pole of the input function is called the forced response.

Natural response: The response generated due to the pole of system function is called the natural response.

Calculation:

The output y(s) is given as

\(y\left( s \right) = \frac{{G\left( s \right)}}{s} = \frac{{3 - s}}{{s\left( {s + 1} \right)\left( {s + 3} \right)}}\)

Converting into partial fractions

\(y\left( s \right) = \frac{{3 - s}}{{s\left( {s + 1} \right)\left( {s + 3} \right)}} = \frac{A}{s} + \frac{B}{{\left( {s + 1} \right)}} + \frac{C}{{\left( {s + 3} \right)}}\)

Multiply whole equation LHS and RHS by s and put s = 0

A = 1

Multiply whole equation by (s + 1) and put s = -1

B = -2

Multiply whose equation by (s + 3) and put s = -3

C = 1

\(y\left( s \right) = \frac{1}{s} - \frac{2}{{s + 1}} + \frac{1}{{s + 3}}\)

Taking the ILT, we get:

\(y\left( t \right) = \mathop {\underbrace {u(t)}_ \downarrow }\limits_{\begin{array}{*{20}{c}} {Forced;response}\ \end{array}} - \mathop {\underbrace {2{e^{ - t}}u\left( t \right) + {e^{ - 3t}}u\left( t \right)}_ \downarrow }\limits_{\begin{array}{*{20}{c}} {Transient;response}\ \end{array}}\)

Concept:

• The slope of the bode plot decreases by 20 dB/decade at the pole
• The slope of the bode plot increases by 20 dB/decade at zero

Application:

 

Total poles: 6 = N_P

Total zeros: 3 = N_Z

• A linear code is an error correcting code.
• Any linear combination of codewords is also a code word.
• A Hamming code (7, 4) = (n, k) can correct any single bit error. 4-bit data is encoded with 7-bit code-word
• Hamming distance is the distance between the code-words.
• The minimum distance tells us how many errors can be corrected.

 

 

0010 = (0001) ⊕ → EXOR (0011)

To get the code word, Exor the two code words.

(0000111) ⊕ (1100110)

= 1100001

Concept:

1. Fermi level is straight/aligned at equilibrium

2. Fermi level is closer valence band for p-type semiconductor and moves into valence band for high doping

3. Fermi level is closer to conduction band for n-type material and moves into the conduction band for high doping.

Applications:

Since fermi level is not aligned in option (2) and (3). They are eliminated

For p++ region fermi-level will lie inside the valance band. Hence option (4) is eliminated.

Correction option is (1) which satisfies all properties.

Points to note to solve this question.

1. Emitter and collector are n-type so the transistor will be n-p-n

∴ Options (3) and (4) are eliminated

2. The diode is connected between base and collector

Base is n-type

Collector is n⁺⁺ type

Hence collector will be of n-type and base will be p-type.

1. Flat band voltage is the voltage where there is a no-charge present in the oxide or oxide semiconductor interface.

2. As the gate voltage V_a increase, more electrons are attracted to the oxide-semiconductor interface. A-n type channel is formed.

The inversion takes place at R

3. For negative gate voltage, more holes are accumulated near the gate.

Accumulations take place at P.

Concept:

Gauss law states that:

\(\oint D.ds = Charge;Enclosed\)

Analysis:

Charge enclosed = line charge density × height of cylinder

= Q × H = QH

\(\vec D = \epsilon\vec E\)

\(\oint \vec E.d\vec s = \frac{1}{\epsilon}\left[ {QH} \right]\)

This is the electric flux through the full cylinder.

For quarter-cylinder, flux is obtained by dividing full flux by 4

\(= \frac{1}{{4\epsilon}}\left[ {QH} \right]\)

Various Maxwell laws are shown in the table

Concept:

The behavior of the CMOS inverter for static conditions of operation is described by the voltage transfer characteristics (VTC) and for dynamic operation, the condition is described by the time response during switching conditions.

V_OH = V_DD

\({V_{IL}} = \frac{{2{V_0} - \left| {{V_{Top}}} \right| - {V_{DD}} + {K_r}{V_{Ton}}}}{{1 + {K_r}}}\)

\({V_0} = {V_{in}} - {V_{Top}} + \sqrt {{{\left( {{V_{in}} - {V_{DD}} - {V_{Top}}} \right)}^2} + {k_r}{{\left( {{V_{in}} - {V_{top}}} \right)}^2}} \)

\({k_r} = \frac{{{\beta _n}}}{{{\beta _p}}} = \frac{{{\mu _n}{c_{OX}}{{\left( {\frac{W}{L}} \right)}_n}}}{{{\mu _p}{c_{OX}}{{\left( {\frac{W}{L}} \right)}_p}}}\)

\({V_{IH}} = \frac{{{V_{DD}} + {V_{Top}} + {k_r}\left( {2{V_0} + {V_{Top}}} \right)}}{{1 + {k_r}}}\)

\({V_0} = {V_{in}} - {V_{Ton}} + \sqrt {{{\left( {{V_{in}} - {V_{Ton}}} \right)}^2} + \frac{1}{{{k_r}}}{{\left[ {{V_{in}} - {V_{DD}} - {V_{Top}}} \right]}^2}} \)

V_OL = 0

Observations:

NM_L = V_IL - V_OL

NM_H = V_OH - V_IH

∴ W_p ↑ → NM_L ↑

W_p ↑ → NM_H ↓

(1) Concept

NAND gate truth table

 

When one of the input is zero then the output = 1

When one of the input = 1, output is the complement of other

(2) NOR-GATE

 

When one of the input = 1, output = 0

When one of the input = 0, output is the complement of other input.

(3) NMOS is  ON when V_G is high and PMOS is ON when V_G is low.

Application:

Case I:

When EN = 0

Output of NAND = 1

PMOS = OFF

Output of NOR = 0

NMOS = OFF

Output = Hi – z

Case 2:

When EN = 1

Output of NAND = D̅

If D = 1 D̅ = 0

If D = 0 D̅ = 1 → PMOS OFF

Output of NOR = D̅

So for EN = 1

Circuit is

The output F = (D̅)' = D

(Basic inverter operation)

Case A:

A = 0, B = 0

NMOS 1 → OFF

NMOS 2 → OFF

PMOS 3 → ON

PMOS 4 → ON

Output = 1

Case B and Case C

A = 0, B = 1 or

A = 1, B = 0

In both cases either of NMOS is ON and one of the PMOS is OFF.

Case D:

A = 1, B = 1

Both NMOS are ON

NMOS will pass logic high from input A and B

F = 1

Truth table

 

This is an EXNOR logic.

\(\frac{1}{{2\pi j}}\oint {\left( {z + \frac{1}{z}} \right)^2}dz\)

\(\Rightarrow \frac{1}{{2\pi j}}\oint \left( {{z^2} + \frac{1}{{{z^2}}} + 2} \right)dz\)

\(\Rightarrow \frac{1}{{2\pi j}}\left[ {\oint {z^2}dz + \oint \frac{1}{{{z^2}}}dz + \oint 2dz} \right]\)

The first and third term are analytic. Hence integral = 0

For integral

\(\oint \frac{1}{{{z^2}}}dz\)

\(\begin{array}{*{20}{c}} {{\rm{lim}}}\ {z \to {z_0}} \end{array}\frac{1}{{\left( {n - 1} \right)!}}\frac{{{d^{n - 1}}}}{{d{z^{n - 1}}}};{\left( {z - {z_0}} \right)^n}f\left( z \right)\)

\(\begin{array}{*{20}{c}} {{\rm{lim}}}\ {z \to {z_0}} \end{array}\frac{1}{{\left( {2 - 1} \right)!}}\frac{d}{{dz}}{\left( {z - 0} \right)^2}\frac{1}{{{{\left( z \right)}^2}}}\)

\(\begin{array}{*{20}{c}} {{\rm{lim}}}\ {z \to {z_0}} \end{array}\frac{1}{{1!}}\frac{d}{{dz}}\left( 1 \right)\)

= 0

Hence, \(\mathop \smallint \limits_{2\pi i}^{} {\left( {z + \frac{1}{z}} \right)^2}dz = 0\)

|z| = 1

\(A = \left[ {\begin{array}{*{20}{c}} 2&2&3&3\ 0&1&1&1\ {\begin{array}{*{20}{c}} 0\ 0 \end{array}}&{\begin{array}{*{20}{c}} 0\ 0 \end{array}}&{\begin{array}{*{20}{c}} 3\ 0 \end{array}}&{\begin{array}{*{20}{c}} 3\ 2 \end{array}} \end{array}} \right]\)

The given matrix is a upper triangular matrix

For upper triangular matrix, eigen values are the diagonal elements

Hence,

λ₁ = 2 = λ₄

λ₂ = 1

λ₃ = 3

Hence, there are three distinct eigen values.

Concept:

E[x + y] = E[x] + E[y]

E [Ax] = A E[x]

Where E [x] represents the expected or mean value of the random variable 'x'

Application:

E[2x + y] = 0

2E[x] + E[y] = 0    ----(1)

E[x + 2y] = 33

E[x] + 2E[y] = 33    ----(2)

Adding (1) and (2)

3E[x] + 3E[y] = 33

E[X] + E[Y] = 11

\(Given:;\mathop \smallint \limits_{y = 0}^π \mathop \smallint \limits_{x = y}^π \frac{{\sin x}}{x}dxdy\)

Changing the order and take vertical strip, we get:

\(I = \mathop \smallint \limits_{x = 0}^π \mathop \smallint \limits_{y = 0}^{y = x} \frac{{\sin x}}{x}dxdy\)

\(\mathop \smallint \limits_{x = 0}^π \frac{{\sin x}}{x}dx\mathop \smallint \limits_{y = 0}^{y = x} dy\)

\(\Rightarrow \mathop \smallint \limits_{x = 0}^π \frac{{\sin x}}{x}dx\left[ y \right]_0^x\)

\(\Rightarrow \mathop \smallint \limits_{x = 0}^π \frac{{\sin x}}{x}dx\left[ {x - 0} \right]\)

\(\Rightarrow I = \mathop \smallint \limits_{x = 0}^π \sin xdx\)

\(I = (-cos x)^\pi_{0}\)

I = (1 – (-1))

I = 2

Concept: 1. \(PDF = \frac{{d\left( {CDF} \right)}}{{dz}}\) 

                2. \(P\left( {\frac{A}{B}} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}\)                 3. CDF gives the probability i.e. Fx(x) = P (X ≤ x)

Calculation:

Method – I:

mean = 1

\(CDF = {F_z}\left( x \right) = \left{ {\begin{array}{*{20}{c}} {1 - {e^{ - x}}}&{x \ge 0}\ 0&{x < 0} \end{array}} \right.\)

\(PDF = \frac{{d\left( {CDF} \right)}}{{dz}}\)

\({f_z}\left( x \right) = \frac{d}{{dz}}\left( {1 - {e^{ - x}}} \right) = {e^{ - x}}\)

\({f_z}\left( x \right) = \left{ {\begin{array}{*{20}{c}} {{e^{ - x}}}&{x \ge 0}\ 0&{x < 0} \end{array}} \right.\)

From concept of Probability:

\(P\left( {\frac{A}{B}} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}\)

\(P\left( {\frac{{z > 2}}{{z > 1}}} \right) = \frac{{P\left( {z > 2} \right) \cap P\left( {z > 1} \right)}}{{P\left( {z > 1} \right)}}\)

\(= \frac{{P\left( {z > 2} \right)}}{{P\left( {z > 1} \right)}}\)

Area under Pdf gives probability

\(P\left( {z > 2} \right) = \mathop \smallint \limits_2^\infty {e^{ - x}}dx\)

\(\Rightarrow \left[ { - {e^{ - x}}} \right]_2^\infty\)

\(\Rightarrow - \left[ {{e^{ - \infty }} - {e^{ - 2}}} \right]\)

= e^-2

\(P\left( {z > 1} \right) = \mathop \smallint \limits_1^\infty {e^{ - x}}dx\)

= e^-1 (Simply replace ‘2’ by ‘1’ in above result)

\(P\left( {\frac{{z > 2}}{{z > 1}}} \right) = \frac{{{e^{ - 2}}}}{{{e^{ - 1}}}} = \frac{1}{e} = 0.367\)

Method – 2:

\(P\left( {\frac{{z > 2}}{{z > 1}}} \right) = \frac{{P\left( {z > 2} \right)}}{{P\left( {z > 1} \right)}}\)

CDF gives the probability

i.e. F_x(x) = P (X ≤ x)

P(z > 2) = 1 – P(z ≤ 2)

= 1 – CDF(z = 2)

= 1 – (1 – e^-2)

= e^-2

P(z > 1) = 1 – P(z ≤ 1)

= 1 – CDF (z = 1)

= 1 – (1 – e^-1)

= e^-1

\(P\left( {\frac{{z > 2}}{{z > 1}}} \right) = \frac{{{e^{ - 2}}}}{{{e^{ - 1}}}} = \frac{1}{e}\)

= 0.367

Hence pdf and then Integral need not be done in this method.

Concept:

A signal x(t) is a said to be periodic with a period T if x(t ± T) = x(t)

The time period for a signal in the form of A cos (ωt + ϕ) & A sin(ωt + ϕ)  is \(T = \frac{{2\pi }}{\omega}\)

For the sum of the two continuous-time signals with period T1 and T2 to be periodic, there must exist non-zero integers a, b such that:

a × T1 = b × T2   = c × T₃

If T1 / T2 = b / a , T2 / T3 = c / b  is rational, then only the signal is periodic and the period is given by LCM (T1, T2, T₃ ) otherwise it is aperiodic or not periodic.

Calculation:

\(f\left( t \right) = 1 + 2\cos \left( {\pi t} \right) + 3\sin \left( {\frac{{2\pi }}{3}t} \right) + 4\cos \left( {\frac{{\pi t}}{2} + \frac{\pi }{4}} \right)\)

The DC term ‘1’ will give the offset

\({T_1} = \frac{{2\pi }}{\pi } = 2\)

\({T_2} = \frac{{2\pi }}{{\frac{{2\pi }}{3}}} = 3\)

\({T_3} = \frac{{2\pi }}{{\frac{\pi }{2}}} = 4\)

Time Period: T = LCM (T₁, T₂, T₃)

= LCM (2, 3, 4)

= 12

Concept: The instantaneous angular frequency will be obtained by differentiating the instantaneous phase as: \({\omega _i} = \frac{{d\theta }}{{dt}} = 2\pi fc + k\frac{{dm\left( t \right)}}{{dt}}\) if θ (t) = 2π fct + k m(t)

Calculation:  Given Φ(t) = cos (2π f_ct + km(t))

f_c = 50 kHz

K = 10 π

\(\left| {\frac{{fi\left( {min} \right)}}{{fi\left( {max} \right)}}} \right| = ?\)

 

Phase of the modulated signal is given by:

θ (t) = 2π f_ct + k m(t)

The instantaneous angular frequency will be obtained by differentiating the instantaneous phase as:

\({\omega _i} = \frac{{d\theta }}{{dt}} = 2\pi fc + k\frac{{dm\left( t \right)}}{{dt}}\)

\(f_i = {f_c} + \frac{{10\pi }}{{2\pi }}\frac{d}{{dt}}m\left( t \right)\)

\(f_i = {f_c} + 5\frac{d}{{dt}}m\left( t \right)\)

\(\frac{d}{{dt}}m\left( t \right)\) is the slope of the message signal:

maximum positive slope (between 0 ≤ t ≤ 1) is 2

Maximum negative slope (between +1 < t < 3) is:

\(= \frac{{ - 2}}{2} = - 1\)

f_{i max} = 50 + 5(2) = 60 kHz

f_{i min} = 50 – 5(1) = 45 kHz

\(\frac{{f_{min};;}}{{fi_{max};}} = \frac{{45}}{{60}} = \frac{3}{4} = 0.75\)

Concept:

The radiation resistance of dipole is given by:

\({R_{rad}} = 80{\pi ^2}{\left( {\frac{{dl}}{λ }} \right)^2}\)

Application:

As \({R_{rad}} = 80{\pi ^2}{\left( {\frac{{dl}}{λ }} \right)^2}\) Taking log

log(R_rad) = log(80π²) + 2 log dl – 2 log λ

Differentiating the above, we get:

\(\frac{{d{R_{rad}}}}{R} = \frac{{2dl}}{l}\) (As λ = constant, its derivative will be 0

Multiply both the sides by 100, the above equation becomes:

\(\frac{{d{R_{rad}}}}{R} \times 100 = 2\left( {\frac{{dl}}{l} \times 100} \right)\)

% R_rad = 2 × 1%

= 2%

In the first half cycle 0 < t <T/2

During the positive half of Input cycle, diode is off

\({V_{out}} = {V_s}\left( {\frac{{{R_2}}}{{{R_1} + {R_2}}}} \right) = \frac{{{V_s}}}{2}\)

\(= \frac{{ + 8}}{2} = + 4V\)

In the second half cycle T/2 < t < T

During the negative half of the input cycle, diode is on

Equivalent circuit:

V_out = V_s

output voltage waveform is:

\(Average = \frac{1}{T}\mathop \smallint \limits_0^{\frac{T}{2}} 4dt + \frac{1}{T}\mathop \smallint \limits_{\frac{T}{2}}^T \left( { - 10dt} \right)\)

\(= \frac{1}{T}4\left( {\frac{T}{2}} \right) - 10\left( {T - \frac{T}{2}} \right)\)

\(\Rightarrow \frac{1}{T} \times \frac{T}{2}\left[ {4 - 10} \right]\)

\(= \frac{{ - 6}}{2}\)

= -3 V: = -3 V

let Initial states be Q₁ = Q₂ = 0

\({D_1} = \begin{array}{*{20}{c}} {\overline {{Q_1}} }&{\overline {{Q_2}} } \end{array}\)

D₂ = Q₁

Hence after 3 clock pulses, the output repeats

It is mod – 3 counter

out frequency \(= \frac{{12}}{3} = 4\) kHz.

Given:

f(-1) = 0

|f’(x)| ≤ 2

-2 ≤ f ’(x) ≤ 2

Using Lagrange mean value theorem:

\(f'\left( x \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\)

Since value of f (-1) = 0 is given, we take the interval [-1, 2], i.e.

-2 ≤ f’(x) ≤ 2

\(- 2 \le \frac{{f\left( 2 \right) - f\left( { - 1} \right)}}{{2 - \left( { - 1} \right)}} \le 2\)

\(- 2 \le \frac{{f\left( 2 \right)}}{3} \le 2\)

-6 ≤ f(2) ≤ 6

We observe that the option 2 satisfies this condition.

Concept:

Green’s theorem states that:

\(\mathop \smallint \limits_c^; Pdx + Qdy = \int!!!\int \left( {\frac{{\partial Q}}{{\partial x}} - \frac{{\partial P}}{{\partial y}}} \right)dxdy\)

Application:

Given: \(\mathop \smallint \limits_c^; \left( {xdy - ydx} \right)\), i.e.

P = -y

Q = x

\(\frac{{\partial Q}}{{\partial x}} = 1;;,;;\frac{{\partial P}}{{\partial Y}} = - 1\)

\(= \int!!!\int \left( {1 - \left( { - 1} \right)} \right)dxdy\)

\(\Rightarrow 2\int!!!\int dxdy\)

= 2 × Area enclosed

= 2 [Area of rectangle + Area of semicircle]

\(= 2\left[ {3 \times 2 + \frac{1}{2}\pi {{\left( 1 \right)}^2}} \right]\)

⇒ 12 + π

Six point DFT of x(n) is given by:

\(X\left( k \right) = \mathop \sum \limits_{n = 0}^5 x\left( n \right)W_6^{kn}\)

For X (1), K = 1

\(X\left( 1 \right) = \mathop \sum \limits_{n = 0}^5 x\left( n \right)W_6^n\)

\(X\left( 1 \right) = x\left( 0 \right)W_6^0 + x\left( 1 \right)W_6^1 + x\left( 2 \right)W_6^2 + x\left( 3 \right)W_6^3 + x\left( 4 \right)W_6^4 + x\left( 5 \right)W_6^5\)

\(X\left( 1 \right) = x\left( 0 \right).1 + \left[ {x\left( 1 \right) - x\left( 4 \right)} \right]W_6^1 + \left[ {x\left( 2 \right) - x\left( 5 \right)} \right]W_6^2 + x\left( 3 \right)W_6^3\)

\(\because W_6^4 = W_6^3 - W_6^1 = - W_6^1\) , \(W_6^5 = - W_6^2]\) 

From the signal flow graph:

X(1) = a₁ [x(0) – x(3)] + a₂ [x(1) – x(4)] + a₃ [x(2) – x(5)]

On comparing the coefficient, we can write:

a₁ = 1

\({a_2} = W_6^1\)

\({a_3} = W_6^2\)

h(n) = [1, a, b]

\(x\left( n \right) = {C_1}{e^{ - j\frac{\pi.n }{{{2}}}}} + {C_2}{e^{j\frac{\pi.n }{{{2}}}}}\)

y(n) = 0

In frequency domain

H(e^jω) = 1 + ae^-jω + be^{-j 2ω}

Input signal is given by:

\(x\left( n \right) = {C_1}{e^{ - j\frac{\pi.n }{{{2}}}}} + {C_2}{e^{j\frac{\pi.n }{{{2}}}}}\)

For exponential input to the system output \(= \left| {H\left( {{e^{j{\rm{\omega }}}}} \right)} \right|x\left( n \right){e^{\left( {j{\rm{\omega }} + {\rm{\Phi }}} \right)}}\) 

Since these are two frequencies magnitude of |H(e^jω)| has to be evaluated separately.

\(at;{\rm{\omega }} = \frac{{ - {\rm{\pi }}}}{2}\)

\(H\left( {{e^{ - j\frac{\pi }{2}}}} \right) = 1 + a{e^{ - j\left( { - \frac{\pi }{2}} \right)}} + b{e^{ - j2\left( { - \frac{\pi }{2}} \right)}}\)

\( = 1 + a{e^{j\frac{\pi }{2}}} + b{e^{j\pi }}\)

e^jπ/2 = j

e^jπ = -1

= 1 + a(j) + b(-1)

= (1 - b) + ja

\(\left| {H\left( {{e^{ - j\frac{\pi }{2}}}} \right)} \right| = \left| {\left( {1 - b} \right) + ja} \right|\)

\(= \sqrt {{{\left( {1 - b} \right)}^2} + {a^2}}\)

at ω = π/2

\(H\left( {{e^{j\frac{\pi }{2}}}} \right) = 1 + a{e^{ - j\left( {\frac{\pi }{2}} \right)}} + b{e^{ - j2\left( {\pi /2} \right)}}\)

= (1 - b) – ja

\(\left| {H\left( {{e^{j\frac{\pi }{2}}}} \right)} \right| = \sqrt {{{\left( {1 - b} \right)}^2} + {a^2}} \)

\(\left| {H\left( {{e^{j\frac{\pi }{2}}}} \right)} \right| = \left| {H\left( {{e^{ - j\frac{\pi }{2}}}} \right)} \right|\)

\(= \sqrt {{{\left( {1 - b} \right)}^2} + {a^2}} \)

Expression of y(n)

\(y\left( n \right) = {\left[ {{{\left( {1 - b} \right)}^2} + {a^2}} \right]^{1/2}}\left[ {{C_1}{e^{ - j;\left( {\frac{{n\pi }}{2} + \phi } \right)}} + {C_2}{e^{j;\left( {\frac{\pi }{2}n + {\phi _2}} \right)}}} \right]\)

For y(n) = 0

\(\left| {H\left( {{e^{j\frac{\pi }{2}}}} \right)} \right| = 0 = \left| {H\left( {{e^{j\frac{\pi }{2}}}} \right)} \right|\)

\(\sqrt {{{\left( {1 - b} \right)}^2} + {a^2}}\)

from the options

b = 1

a = 0

Concept:

Star to delta transformation

\({Z_A} = {Z_1} + {Z_2} + \frac{{{Z_1}{Z_2}}}{{{Z_3}}}\)

\( = \frac{1}{{sC}} + \frac{1}{{sC}} + \frac{1}{{sC}}\)

\(Z_A= \frac{3}{{sC}}\)

\(Z_A=\frac{1}{{s\left( {\frac{C}{3}} \right)}}\)

Equivalent capacitance here is C/3 in Δ connection.

Equivalent circuit is then redrawn as:

Let \(R||\frac{1}{{\frac{sC}{3}}} = {Z_1}\) 

\({Z_1} = \frac{{R \times \frac{{\frac{1}{{sC}}}}{3}}}{{R + \frac{{\frac{1}{{sC}}}}{3}}} = \frac{R}{{1 + sR\frac{C}{3}}}\)

Z_eq = Z₁ || 2Z₁

\(= \frac{2}{3}{Z_1}\)

\(= \frac{2}{3}\left( {\frac{R}{{1 + sR\frac{C}{3}}}} \right)\)

s = jω

ω = 1000

\(= \frac{2}{3}\left( {\frac{{1000}}{{1 + \frac{{j\left( {\left( {1000} \right)\left( {1000} \right){{10}^{ - 6}}} \right)}}{3}}}} \right)\)

\(= \frac{2}{3} \times \frac{{{{10}^3}}}{{\left( {1 + \frac{j}{3}} \right)}}\)

\(= \frac{2}{3} \times \frac{{3 \times {{10}^3}}}{{\left( {3 + j} \right)}} = \frac{{2 \times {{10}^3}}}{{j + 3}}\)

\(i = \frac{V}{{{Z_{eq}}}} = \frac{{2\sin 1000t}}{{\frac{{2 \times {{10}^3}}}{{3 + j}}}}\)

\(= \left[ {\frac{{\left( {3 + j} \right)\sin 1000t}}{{1000}}} \right]A\)

(3 + j) sin 1000 t mA

i(t) = [3 sin 1000 t + j sin 1000 t] mA

j = 1 ∠ 90

i(t) = 3 sin 1000 t + 1 ∠ 90 sin (1000 t)

= 3 sin 1000 t + sin (1000 t + 90)

i(t) = [3 sin 1000 t + cos 1000 t] mA

\(G(s) = \frac{{C\left(s \right)}}{{R\left( s \right)}}= \frac{1}{{s^2} + 2s + 1}\)

\(C\left( s \right) = \frac{1}{{\left( {{s^2} + 2s + 1} \right)}} \times \frac{1}{s}\)

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

\(C\left( s \right) = \frac{A}{s} + \frac{B}{{s + 1}} + \frac{C}{{{{\left( {s + 1} \right)}^2}}}\)

A = 1

B = -1

C = -1

\(C\left( s \right) = \frac{1}{s} - \frac{1}{{s + 1}} - \frac{1}{{{{\left( {s + 1} \right)}^2}}}\)

Applying inverse laplace transform:

C(t) = 1 – e^-t – te^-t

at steady state i.e. at t → ∞

\(\mathop {{\rm{lt}}}\limits_{t \to \infty } C\left( t \right) = 1 - {e^{ - \infty }} - t{e^{ - \infty }}\)

C(t) = 1

94% of steady state \( = \frac{{94}}{{100}} \times 1\) 

= 0.94.

0.94 = 1 – e^-t – te^-t

Substitute all options

Let us substitute options

option (a) t= 5.25

1 – e^-5.25 – 5.25 e^-5.25

\(\Rightarrow 1 - \left( {\frac{{1 + 5.25}}{{{e^{5.25}}}}} \right)\)

= 0.967

option (b)

t = 4.50

\(1 - \frac{{\left( {1 + 4.50} \right)}}{{({e^{4.50}})}}\)

1 – 0.938

≈ 0.94

Solving the loop:

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

Solving the second loop, we get:

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

\(H(s)= \frac{{{s^2} + 1}}{{{s^3} + 2{s^2} + s + 1}}\)

For the A – matrix

\(H\left( s \right) = \frac{1}{{{s^3} +3{s^2} + 2s + 1{} }}\)

Write these terms in this order with a negative sign, i.e.

\(A = \left[ {\begin{array}{*{20}{c}} 0&1&0\ 0&0&1\ { - 1}&{ - 2}&{ - 3} \end{array}} \right]\)

For the output:

\(H\left( s \right) = \frac{{Y\left( s \right)}}{{U\left( s \right)}} = \frac{{Y\left( s \right)}}{{{X_1}\left( s \right)}}.\frac{{{X_1}\left( s \right)}}{{U\left( s \right)}}\)

\(H(s)= 1\times\frac{1}{{{s^3} + 3{s^2} + 2s + 1}}\)

\(\frac{{Y\left( s \right)}}{{{X_1}\left( s \right)}} = 1\)

Y = X₁(s)

Y = x₁

\(y = \left[ {\begin{array}{*{20}{c}} 1&0&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\ {{x_2}}\ {{x_3}} \end{array}} \right]\)

= CX

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

Concept: In constellation diagram d_min is the minimum distance between two adjacent points and Probability of error is given by \({P_e} = Q\left( {\sqrt {\frac{{d_{min}^2}}{{2No}}} } \right)\)

Analysis: \(P\left( 0 \right) = P\left( 1 \right) = \frac{1}{2}\)

Case 1:

p(t) = Φ₁(t)

q(t) = -Φ₁(t)

The system is binary PSK

Probability of error is given by:

\({P_e} = Q\left( {\sqrt {\frac{{d_{min}^2}}{{2No}}} } \right)\)

d_min = 1 – (-1) = 2

\({P_e} = Q\left( {\sqrt {\frac{4}{{2No}}} ;} \right) = Q\left( {\sqrt {\frac{2}{{No}}} ;} \right) \ldots \left( 1 \right)\)

Case 2: When p(t) = Φ₁(t)

\(q\left( t \right) = \sqrt E {\phi _2}\left( t \right)\)

Since Φ₁(t) and Φ₂(t) are orthonormal, the signalling scheme is BFSK.

Constellation Diagram:

(d_min)² = (√E)² + (1)²

d_min² = E + 1

\({P_e} = Q\left( {\sqrt {\frac{{d_{min}^2}}{{2No}}} } \right)\)

\({P_e} = Q\sqrt {\frac{{E + 1}}{{2No}}} \ldots \left( 2 \right)\)

Comparing (1) and (2)

\(\frac{2}{{No}} = \frac{{E + 1}}{{2No}}\)

4 = E + 1

E = 3

Responsivity (R) is defined as the ratio of the photon current to the optical power, i.e.

\(R = \frac{{{I_p}}}{{{P_{opt}}}}\)

If the electron generation rate is r_e and photon incident rate is r_p, then Responsivity will be:

\(R= \frac{{{I_p}}}{{{P_{opt}}}} = e\frac{{{r_e}}}{{h\nu {r_p}}}\)

\(R= \frac{e}{{h\nu }}\left( η \right)\)

where η is the quantum efficiency defined as:

\(η = \frac{{Electron;generation;rate}}{{Incident;photon;rate}} \)

\(R = \frac{{η e}}{{h\nu }}\left[ {\because h\nu = \frac{{1.24\left( {eV} \right)}}{{\lambda ;\mu m}}} \right]\)

\(R = \frac{{n\lambda }}{{1.24}}\frac{A}{W}\)

Concept:

The magnetic field at a distance ‘R’ from a current-carrying conductor carrying current I is given by:

\(B = \frac{{{\mu _0}I}}{{2\pi R}}\)

Application:

Since the two wires are carrying current in the opposite direction, the magnetic field due to wire ‘1’ will be:

\({B_1} = \frac{{{\mu _0}I}}{{2\pi r}}\)

The direction, according the right-hand rule, will be downwards into the plane of the paper/screen.

Magnetic field due to wire ‘2’

Similarly \({B_2} = \frac{{{\mu _0}\left( {2I} \right)}}{{2\pi \left( {3r} \right)}}\)

\(B_2= \frac{{2{\mu _0}I}}{{2\pi 3r}}\) 

The direction for this is also into the plane of the paper.

Adding the two we get:

B = B₁ + B₂

\(= \frac{{{U_0}I}}{{2\pi r}} + \frac{{2{\mu _0}I}}{{2\pi 3r}}\)

\(= \frac{{{u_0}I}}{{2\pi r}}\left[ {1 + \frac{2}{3}} \right]\)

\(B = \frac{{5{u_0}I}}{{6\pi r}}\)

Concept:

Phase velocity:

The rate at which the phase of the wave propagates in space

\({V_p} = \frac{\omega }{\beta }\)

Group Velocity:

The velocity with which overall envelope of the wave travels

\({V_g} = \frac{{d\omega }}{{d\beta }} = \frac{{d\omega }}{{dk}}\)

\(K\left( \omega \right) = \frac{1}{c}\sqrt {\omega - \omega _0^2} \ldots \left( 1 \right)\)

V_g = 2 × 10⁸ m/s (green)

\({V_g} = \frac{{d\omega }}{{dk}}\)

\(\frac{{dk}}{{d\omega }} = \frac{1}{{{V_g}}}\)

differentiating (1)

\(\frac{{dk}}{{d\omega }} = \frac{1}{{{V_g}}} = \frac{{1 \times 2\omega }}{{2C\sqrt {{\omega ^2} - \omega _0^2} }} = \frac{1}{{{V_g}}}\)

\(\Rightarrow \frac{\omega }{{3 \times {{10}^8}\sqrt {{\omega ^2} - \omega _0^2} }} = \frac{1}{{2 \times {{10}^8}}}\)

\(\sqrt {{\omega ^2} - \omega _0^2} = \frac{{2\omega }}{3}\)

\({V_p} = \frac{\omega }{K} = \frac{{\omega c}}{{\sqrt {{\omega ^2} - \omega _0^2} }}\)

\(= \frac{{\omega c}}{{\left( {\frac{{2\omega }}{3}} \right)}}\)

\({V_p} = \frac{{3c}}{2} = \frac{{3 \times {{10}^8} \times 3}}{2} = 4.5 \times {10^8}m/s\)

Shortcut:

Group velocity × Phase velocity = c²

2 × 10⁸ × phase velocity = (3 × 10⁸)²

\({V_p} = \frac{{9 \times {{10}^{16}}}}{{2 \times {{10}^8}}}\)

= 4.5 × 10⁸ m/s

For the zenor diode to be on minimum voltage across it should be 20V.

\({V_z} = {V_{RL}} = {V_i}\left( {\frac{{{R_L}}}{{{R_1} + {R_L}}}} \right)\)

\({V_z} = {V_i}\left( {\frac{{1000}}{{1000 + 200}}} \right) \ge 20\)

\(Vi \ge \frac{{20 \times 1200}}{{1000}}\)

Vi ≥ 24 V

Only option (B) satisfies this condition.

This objective approach solves time to calculate the upper bound on V_i

Case 1:

When Q = 0

A = 0

NAND: 0 NAND Q̅ = 1

on next Clk pulse: Q will become 1

Case 2:

When Q = 0

A = 1

NAND output: Q NAND Q’

0 NAND 1 = 1

Q_n+1 = 1

only option (c) satisfies both the conditions (for objective purpose question can be stopped here)

Case 3:

Q = 1

A = 0

NAND output = 1 NAND Q’

= 1 NAND 0

= 1

Q_n+1 = 1

Case 4:

When

Q = 1

A = 1

NAND output: 1 NAND Q

= 1 NAND 1

= 0

Combining the above four cases:

Concept:

For the pass transistor logic:

V_D = V_G – V_T

Application:

V_G2  = V_D1 = V_G1 - V_T

= 3 – 0.6

= 2.4

V_D2 = V_G2 – V_T = V_{out 1}

= 2.4 - 0.6

= 1.8 = V_{out 1}

Case 2:

V_D = V_G1 – V_T

= 3 – 0.6 = 2.4

V_D2 = V_G2 – V_T

= 3 – 0.6 = 2.4

V_D3 = V_G3 – V_T = V_{Out 2}

= 3 – 0.6 = 2.4

Applying KCL, we get:

\(C\frac{{dv\left( t \right)}}{{dt}} + \frac{{v\left( t \right) - v\left( s \right)}}{R} = 0\)

\(\frac{{dv\left( t \right)}}{{dt}} + \frac{{v\left( t \right) - v\left( s \right)}}{{RC}} = 0\)

C = 1

\(R = {R_o}\left( {1 - \frac{t}{T}} \right)\)

\(\frac{{dv\left( t \right)}}{{dt}} + \frac{{v\left( t \right) - 1}}{{\left( {1 - \frac{t}{T}} \right)}} = 0\)

T = 3R_oC = 3(1) (1) = 3 sec

\(\frac{{dv\left( t \right)}}{{dt}} + \frac{{v\left( t \right) - 1}}{{\left( {1 - \frac{t}{3}} \right)}} = 0\)

\(\frac{{dv\left( t \right)}}{{v\left( t \right) - 1}} = \frac{{ - dt}}{{1 - \frac{t}{3}}}\)

\(\Rightarrow \frac{{dv\left( t \right)}}{{v\left( t \right) - 1}} = \frac{{ + 3dt}}{{t - 3}}\)

Integrating both the sides, we get:

ln(v(t) - 1) = 3 ln (t - 3) + ln C

where C = constant

ln(v(t) - 1) = ln(t - 3)³ + ln C

\(\ln \left( {\frac{{v\left( t \right) - 1}}{{{{\left( {t - 3} \right)}^3}}}} \right) = \ln C\)

Taking anti-log both sides

v(t) – 1 = C(t - 3)³

v(t) = 1 + C(t - 3)³

\(i\left( t \right) = C\frac{{dv\left( t \right)}}{{dt}} = \frac{{dv\left( t \right)}}{{dt}}\)

\(= \frac{d}{{dt}}\left[ {1 + C{{\left( {t - 3} \right)}^3}} \right]\)

i(t) = 3C (t - 3)²

at t = 0, i = 1

i(0) = 3C (0 - 3)² = 1

\(C = \frac{1}{{27}}\)

\(at;t = \frac{T}{2} = \frac{{3{R_o}C}}{2} = \frac{3}{2} = 1.5\)

i(t = 1.5) = 3C (1.5 - 3)²

= 3 C (1.5)²

\(3 \times \frac{1}{{27}} \times \frac{{15}}{{10}} \times \frac{{15}}{{10}}\)

\(\frac{1}{4} = 0.25A\)

The OLTF of unity feedback system is:

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

Characteristic equation = 1 + GH = 0

= s³ + 3s² + 2S + K = 0

For two symmetric poles, 1 row of routh – array should be zero

 

\(\frac{{6 - K}}{3} = 0\)

K = 6

The given equation is homogeneous ordinary differential equation.

\({x^2}\frac{{{d^2}y}}{{d{x^2}}} - 3x\frac{{dy}}{{dx}} + 3y = 0 \ldots \left( 1 \right)\)

x = e^z    

z = ln x

Equation (1) can be written as:

D(D - 1)y – 3Dy + 3y = 0

D² – 4D + 3 = 0

D² – 3D – D + 3 = 0

D(D - 3) – 1 (D - 3) = 0

(D - 1)(D - 3) = 0

D = 1, 3

general solution:

y = C₁e^z + C₂ e^3z

z = ln x

y = C₁ x + C₂ x³      ----(1)

To find C₁ and C₂ the initial condition

y(1) = 1

y(2) = 14

substituting in (1)

C₁ + C₂ = 1

2C₁ + 8C₂ = 14

Solving above equations give

C₁ = -1 and C₂ = 2

y = -x + 2x³

y(1.5) = -1.5 + 2(1.5)³

y(1.5) = 5.25

Given:

h[n] = [-1, -2, -3, 4, 3, 2, 1]

g(n) is derived from h(n) that means its sample values will be one of the sample values of h(n).

To find: 10 g[-1] + g[1]

 g[n] must have sample at 1 and -1.

The probable g[n] filter is

y[n] = {a, b, c}.

\(E\left( {h,g} \right) = \mathop \smallint \limits_{ - \pi }^{ + \pi } {\left| {H\left( {{e^{j\omega }}} \right) - G\left( {{e^{j\omega }}} \right)} \right|^2}d\omega\)

From Parseval's theorem energy in the time domain is equal to the energy in the frequency domain.

Energy of \(\left[ {h\left( n \right) - g\left( n \right)} \right] = \frac{1}{{2\pi }}\mathop \smallint \limits_{ - \pi }^\pi {\left| {H\left( {{e^{j\omega }}} \right) - G\left( {{e^{j\omega }}} \right)} \right|^2}d\omega\)

\(2\pi \mathop \sum \limits_{n = - \infty }^{ + \infty } {\left| {h\left( n \right) - g\left( n \right)} \right|^2} = \mathop \smallint \limits_{ - \pi }^{ + \pi } {\left| {H\left( {{e^{j\omega }}} \right) - G\left( {{e^{j\omega }}} \right)} \right|^2}d\omega\)

To minimise energy of h(n) – g(n) minimise h(n) – g(n)

h(n) – g(n) = { -1, -2, -3 – a, 4 – b, 3 – c, 2, 1]

E[h, g] = 2π ∑ |h (n) – g(n)|²

= 2π ∑ [h(n) – g(n)]²

E[h, g] = 2π [(-1)² + (-2)² + (-3 - a)² + (4 - b)² + (3 - c)² + 2 + 1²]

= 2π [10 + (-3 - a)² + (4 - b)² + (3 - c)²]

To have minimum energy E(h, g)

2π [10 + (-3 - a)² + (4 - b)² + (3 - c)²]

To have minimum energy

(-3 - a)² = 0 → a = -3

(4 - b)² = 0 → b = 4

(3 - c)² = 0 → c = 3

y(n) = [a, b, c] = [-3, 4, 3]

10g (-1) + g(1) = 10a + c

= [10(-3)] + 3

= -30 + 3

= -27

Concept: Average power = Area bounded between the Power Spectral Density - frequency curve.

Calculation:

Given:

y(t) = h(t) * x(t) + z(t)

Power in y(t) = Power [h(t) * x(t)] + Power [z(t)]

H(f) will determine the band width of x(t).

h(t) * x(t) is bandlimited to – 5 Hz to + 5 Hz

\(Power\left[ {h\left( t \right)*x\left( t \right)} \right] = \mathop \smallint \limits_{ - \infty }^{ + \infty } {\left| {H\left( f \right)} \right|^2}5 \times \left( f \right)\)

\(= \mathop \smallint \limits_{ - 5}^{ + 5} {\left( {0.5} \right)^2}5df\)

\(\Rightarrow \mathop \smallint \limits_{ - 5}^{ + 5} \left( {0.25} \right)\left( {0.5} \right)df\)

= 10(1.25) = 12.5 W

power [z (t)] = Area under power spectral density

\({P_z}\left( t \right) = \left( {\frac{1}{2}} \right)\left( {10} \right)\left( 1 \right) = 5W\)

Total power = (5 + 12.5) W

= 17.5 W

Concept:  

1. In case of Bandpass sampling

Nyquist criteria \(= \frac{{2{f_H}}}{k}\)

2. In case of Baseband and for Band limited signals Nyquist criteria F_s = 2 F_m

Calculation: The bandlimited signal is represented by:

After modulation, f(t) becomes:

f(t) = A(1 + m(t) cos 2π f_ct

= A cos 2π f_ct + A m(t) cos 2π f_ct

The frequency spectrum is drawn as:

In case of Bandpass sampling

Nyquist criteria \(= \frac{{2{f_H}}}{k}\)

where \(k = \left[ {\frac{{{f_H}}}{{BW}}} \right]\)

BW = 615 – 585 = 30

\(k = \left[ {\frac{{615}}{{30}}} \right]\)

= 20

Nyquist criteria \(= \frac{{2 \times 615}}{{20}} = 61.5~kHz\)

Sampling frequency = 1.2 × 61.5 = 73.8 kHz

Number of Quantization levels = 256

2ⁿ = 256

n = 8

Bit rate = nf_s

= 8 × 73.8

= 590.4 kbps

= 0.5904 Mbps

Note: The answer given in the official gate key is as per the consideration that the AM modulated signal is a bandpass signal. In that case

Bit rate = 8 × 1.2 × 2 × 615 ≈ 11.81 Mbps

Noise is distributed uniformly

PDF of noise

When – 1 is transmitted

PDF of Received signa;

Let x = threshold voltage

\({P_{{e_1}}} = \left( {0.2} \right)\left( {\frac{1}{4}} \right)\left( {1 - x} \right)\)

When +1 is transmitted

Received pdf

\({P_{{e_2}}} = \left( {0.8} \right)\left( {\frac{1}{4}} \right)\left( {1 + x} \right)\)

Total probability of error

\({P_e} = {P_{{e_1}}} + {P_{{e_2}}}\)

\(= \left( {0.2} \right)\left( {\frac{1}{4}} \right)\left( {1 - x} \right) + \left( {0.8} \right)\left( {\frac{1}{4}} \right)\left( {1 + x} \right)\)

\(\Rightarrow \frac{{\left( {0.2} \right)\left( {1 - x} \right) + \left( {0.8} \right)\left( {1 + x} \right)}}{4}\)

\({P_e} = \frac{{1 + 0.6x}}{4}\)

-1 ≤ x ≤ 1

Minimum probability is obtained when x = -1

\({P_e} = \frac{{1 + \left( {0.6} \right)\left( { - 1} \right)}}{4}\)

\(\Rightarrow \frac{{0.4}}{4} = 0.1\)

Incident power = 20 mW = P_i

Reflected power (P_r) from SiO₂ is one-fourth, i.e.

\(P_r= \frac{{{P_i}}}{4}\)

Power left will be:

\({P_i} - \frac{{{P_i}}}{4} = \frac{{3P_i}}{4}\)

From this \(\frac{{3{P_i}}}{4}\) power, the power reflected from SiO2 – Ge Interface is:

\(= \frac{1}{3}\left( {\frac{{3{P_i}}}{4}} \right)\)

\(= \frac{{{P_i}}}{4}\)    ----(1)

Now, the power absorbed in the Germanium layer is:

\(= \frac{1}{3}\left( {\frac{{3{P_i}}}{4}} \right) = \frac{{{p_i}}}{4}\)     ----(2)

Power Transmitted through the other side of the sample is:

\(= \frac{1}{3}\left( {\frac{{3{P_i}}}{4}} \right) = \frac{{{P_i}}}{4}\)      ----(3)

The power entering Germanium Sample = Power absorbed + Power Transmitted, i.e.

\(= \frac{{{P_i}}}{4} + \frac{{{P_i}}}{4} = \frac{{2{P_i}}}{4} = \frac{{{P_i}}}{2}\)

\({P_o} = P_i'{e^{ - \alpha t}}\)

\(\frac{{{P_i}}}{4} = \frac{{{P_i}}}{2}{e^{ - \alpha t}}\)

α = absorption coefficient

α t = ln 2

\(t = \frac{{\ln 2}}{\alpha } = \frac{{\ln 2}}{{3 \times {{10}^4}}}cm\)

t = 0.231 μm

The diode current equation is given as \(I = {I_o}\left( {{e^{\left( {\frac{V}{{\eta {V_T}}}} \right)}} - 1} \right)\)

given η = 1 (Ideality factor)

\({V_T} = \frac{T}{{11600}} = \frac{{300}}{{11600}} = 0.0258V\)

= 25.8 mV

Given: I = -0.75 I_o

\(- 0.75{I_o} = {I_o}\left( {{e^{\frac{V}{{{V_T}}}}} - 1} \right)\)

\(1 - 0.75 = {e^{V/{V_T}}}\)

\(0.25 = {e^{V/{V_T}}}\)

V = V_T ln(0.25)

V = -0.0357 Volts

V = -35.7 mV

|V| = 35.7 mV

L = 1 μm

W = 10 μm

N_A = 5 × 10¹⁶ cm^-3

μ_n = 800 cm²/v-sec

C_ox = 3.45 × 10^-7 F/cm²

V_T = 0.7 V

V_G = 5 V

I_Dsat = ?

The saturation current in MOSFET is given by:

\({I_{Dsat}} = \frac{{{\mu _n}{C_{ox}}W}}{{2L}}{\left( {{V_{GS}} - {V_T}} \right)^2}\)

\(= \frac{{800 \times 3.45 \times {{10}^{ - 7}} \times 10}}{{2 \times 1}}{\left( {5 - 0.7} \right)^2}\)

= 13.8 × 10^-4 × (4.3)²

I_Dsat = 25.516 mA

Concept:

The cutoff frequency of TE_mn mode is given by:

\({f_c} = \frac{c}{2}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}}\)

Application:

for TE₁₀ mode

\({f_{{c_{10}}}} = \frac{c}{2}\left( {\frac{1}{a}} \right)\)        ----(1)

for TE11 mode

\({f_{{c_{11}}}} = \frac{c}{2}\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}}\)        ----(2)

\(\frac{{eqn\left( 1 \right)}}{{eqn\left( 2 \right)}} = \frac{1}{2}\)

\(\frac{1}{2} = \frac{{\frac{c}{2}\left( {\frac{1}{a}} \right)}}{{\frac{c}{2}\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}} }}\)

\(= \frac{1}{2} = \frac{{\frac{c}{{2a}}}}{{\frac{c}{2}\sqrt {\frac{{{a^2} + {b^2}}}{{ab}}} }} = \frac{b}{{\sqrt {{b^2} + {a^2}} }}\)

\(\frac{1}{2} = \frac{1}{{\sqrt {1 + {{\left( {\frac{a}{b}} \right)}^2}} }}\)

\({\left( {\frac{a}{b}} \right)^2} = {\left( {\frac{w}{h}} \right)^2} = 4 - 1 = 3\)

\(\frac{w}{h} = \sqrt 3 = 1.732\)

The trick in the question is:

Time period T = 4 msec

So after half the time period i.e T = 2 msec the diode will be off, and voltage at end of 3 msec will equal the voltage at the end of 2 milli-seconds.

V(t) = V(∞) + (V(0) – V(∞))e ^-t/RC

= 10 + (0 - 10)e ^-t/RC

RC = 500 × 10 × 10^-6

= 5 msec

at t = 2 msec 

V(t) =10 (1 - e ^-2/5)

V(t) = 3.3 V 

So, the voltage across the capacitor at t = 3 ms is equal to the voltage at t = 2 ms, i.e. 3.3 V

The threshold  voltage of CMOS inverter is given by:

\({V_{th}} = \frac{{{V_{T,n}} + \sqrt {\frac{{{K_p}}}{{{K_n}}}} \left( {{V_{DD}} + {V_{T,P}}} \right)}}{{1 + \sqrt {\frac{{{K_p}}}{{{K_n}}}} }}\)

given threshold voltage \({V_{th}} = \frac{{{V_{DD}}}}{2}\)

\(\frac{{{V_{DD}}}}{2} = \frac{{{V_{T,n}} + \sqrt {\frac{{{K_p}}}{{{K_n}}}} \left( {{V_{DD}} + {V_{T,p}}} \right)}}{{1 + \sqrt {\frac{{{K_p}}}{{{K_n}}}} }}\)

\(\frac{{{V_{DD}}}}{2} + \frac{{{V_{DD}}}}{2}\sqrt {\frac{{{K_p}}}{{{K_n}}}} = {V_{T,n}} + \sqrt {\frac{{{K_p}}}{{{K_n}}}} {V_{DD}} + \sqrt {\frac{{{K_p}}}{{{K_n}}}} {V_{T,p}}\)

\(\frac{{{V_{DD}}}}{2} - {V_{T,n}} = \left( {{V_{DD}} + {V_{T,p}} - \frac{{{V_{DD}}}}{2}} \right)\sqrt {\frac{{{K_p}}}{{{K_n}}}} ;\)

\(\left( {\frac{{{V_{DD}}}}{2} - {V_{T,n}}} \right) = \left( {\frac{{{V_{DD}}}}{2} + {V_{T,p}}} \right)\sqrt {\frac{{{K_p}}}{{{K_n}}}} \)

\(\frac{{{K_n}}}{{{K_p}}} = {\left( {\frac{{\frac{{{V_{DD}}}}{2} + {V_{T,p}}}}{{\frac{{{V_{DD}}}}{2} - {V_{T,n}}}}} \right)^2}\)

\(\frac{{{\mu _n}{C_{ox}}{{\left( {\frac{W}{L}} \right)}_n}}}{{{\mu _p}{C_{ox}}{{\left( {\frac{W}{L}} \right)}_p}}} = {\left( {\frac{{1.5 + \left( { - 0.9} \right)}}{{1.5 - 0.7}}} \right)^2}\)

\(\frac{{100{{\left( {\frac{W}{L}} \right)}_n}}}{{40{{\left( {\frac{W}{L}} \right)}_p}}} = {\left( {\frac{{0.6}}{{0.8}}} \right)^2}\)

\(\frac{{{{\left( {\frac{W}{L}} \right)}_n}}}{{{{\left( {\frac{W}{L}} \right)}_p}}} = \frac{4}{{10}} \times \frac{{36}}{{64}} = 0.225\)

Given: |V_tp| = V_tn = 1 V

r_ds = 6 MΩ

For NMOS:

μ_nC_ox = 60 μA/V²

\({\left( {\frac{W}{L}} \right)_{nmos}} = 5\)

For PMOS:

μ_pC_ox = 30 μA/V²

\({\left( {\frac{w}{L}} \right)_{pmos}} = 10\)

Current through PMOS M1 and PMOS M2 will be same because they are connected in series.

Both the transistors will be in saturation because the drain is connected to the gate.

\({V_{SG\left( {M1} \right)}} = {V_{SG\left( {M2} \right)}} = \frac{{{V_{DD}}}}{2}\)

\({V_{SG\left( {M1} \right)}} = \frac{4}{2} = 2V\)

For MOS M3 and M1, the source voltages are the same and the gates are shorted, which gives:

V_SG(M3) = V_SG(M1) = 2 Volts

\({I_2} = \frac{1}{2}\mu_p{C_{ox}}{\left( {\frac{W}{L}} \right)_p}{\left[ {{V_{SG\left( {M3} \right)}} - \left| {{V_{tp}}} \right|} \right]^2}\)

\(I_2= \frac{1}{2} \times 30 \times {10^{ - 3}} \times 10{\left( {2 - 1} \right)^2}\)

I₂ = 0.15 mA

I₂ = 150 uA

The same current I₂ will pass through M4

The transconductance of M4 can be given as:

\({g_{m4}} = \frac{{\partial {I_{DS}}}}{{\partial {V_{GS}}}} = {K_N}'\left( {\frac{W}{L}} \right)\left( {{V_{GS}} - {V_{tn}}} \right) \ldots 1)\)

where K_n’ = u_nC_ox = 60 μA/V²

For NMOS (M4)

\({I_{DS4}} = \frac{1}{2}{u_n}{C_{ox}}\left( {\frac{W}{L}} \right){\left( {{V_{gs}} - {V_{tn}}} \right)^2}\)

\({V_{GS}} - {V_{tn}} = \sqrt {\frac{{2{I_{DS4}}}}{{{K_N}'\left( {\frac{W}{L}} \right)}}} \)

Substituting this in (1), we get:

\({g_{m4}} = \sqrt {2{I_{DS4}}} ;\sqrt {{K_N}'\left( {\frac{W}{L}} \right)} \)

I_DS4 = I₂ = 150 uA

\({g_{m4}} = \sqrt {2 \times 150} \times \sqrt {60 \times 5}\)

g_m4 = 300 uA/V

Drawing the small-signal model of m4.

Note: MOS – 3 is replaced by its output drain resistance r_d3 = 6mΩ

\(\frac{{{V_o}}}{{{V_{in}}}} = \frac{{\left( { - {g_m}{V_{gs}}} \right)({r_{d4}}||r{d_3})}}{{{V_{gs}}}}\)

= -g_m(rd₄ || rd₃)

= -300 × 10^-6 (6 × 10⁶ || 6 × 10⁶)

= -300 × 10^-6(3 × 10⁶)

= -900

magnitude wise

\(\left| {\frac{{{V_o}}}{{{V_{in}}}}} \right| = 900\)

The given circuit shows various stages of current mirror circuit.

From the first stage

\(\frac{{{I_1}}}{{1mA}} = \frac{{{{\left( {\frac{W}{L}} \right)}_2}}}{{{{\left( {\frac{W}{L}} \right)}_1}}}\)

\({I_1} = \frac{3}{2}\left( {1mA} \right)\)

= 1.5mA

Since V₁ = 0 transistor 3 is off

Hence no current will flow from transistor 3 and 4

I₁ = I₃ = 1mA

Now transistor 5 and 7 are matched

\({I_3}{\left( {\frac{W}{L}} \right)_7} = {I_{out}}{\left( {\frac{W}{L}} \right)_5}\)

(1.5)(40) = I_out (10)

\({I_{out}} = \left( {\frac{{40}}{{10}}} \right)\left( {1.5} \right)\)

I_out = 6mA