GATE EE 2015 Official Paper: Shift 2 (Previous Year Paper)

Given $f\left( x \right) = 1-\left| x \right|$ in $- 1 \le x \le 1$

Since, $\left| x \right| > 0,;for;all;x$

So, $\left| x \right|$can take minimum value of 0 and maximum value of 1. Also on increasing value of $\left| x \right|$, value of $f\left( x \right)$ decreases. At $\left| x \right| = 0$, we obtain

$f\left( x \right) = 1 - \left| x \right|$

$= 1 - 0 = 1$ (function has maximum value)

Again, at $\left| x \right| = 1$, we obtain

$f\left( x \right) = 1 - \left| x \right|$

$= 1 - 1 = 0$ (function has minimum value)

Thus, at $\left| x \right| = 0$ value of $f\left( x \right)$ is maximum and at $\left| x \right| = 1$ value of $f\left( x \right)$ is minimum.

The word ‘personnel’ means 'people who are employed in an organization'.

Out of the given options, this meaning is accurately represented only in the sentence from option 3. The other options are instead referring to the word 'personal'.

Hence, option 3 is the answer.

Here are the meanings of the given words:

• Fabric - cloth produced by weaving or knitting fibres
• Textile - a type of cloth or woven fabric
• Fibre - a thread or filament from which textile is formed
• Apparel - clothing in general
  

We can see that the word that most accurately defines the given set of words is ‘apparel’.

Hence, option 4 is correct.

The given problem requires the total weight of 10 poles, each of same weight.

We check the given statements for evaluation of the total weight.

Statement I:

Given that one fourth of weight of a pole is 5 kg.

Let weight of a pole $= x;kg$

$\begin{array}{l} \frac{x}{4} = 5\ x = 20;kg \end{array}$

Total weight of 10 poles $= 10x = 200;kg$

Statement II:

The total weight of 10 poles is 160 kg more than the total weight of two poles.

Let weight of a pole $= x;kg$

So, total weight of 10 poles $= 10x$

$\begin{array}{l} 10x = 2x + 160\ 8x = 160\ x = 20;kg \end{array}$

Hence, total weight of 10 poles $= 10x = 20 \times 10 = 200;kg$

Thus, the given problem can be solved by using any of the two statements.

'Completely’ and ‘just’ are considered adverbs of emphasis and should be placed in front of the verbs they describe. Even if a verb is in its negative form, for eg. "don't know", the adverb should be placed before the entire form.

The only option that satisfies this criteria in both cases is option C.

Hence, option C is the correct answer.

From the given data, we draw the schematic as

Let coordinate of N = (0,0)

Coordinate of M = (0,4)

Coordinate of O = $\left( { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right)$

Coordinate of P = $\left( { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }} - 2} \right)$

Hence by distance formula, we obtain

$\begin{array}{l} = \sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}} \ = \sqrt {{{\left( {0 - \frac{1}{{\sqrt 2 }}} \right)}^2} + {{\left( {4 + 2 - \frac{1}{{\sqrt 2 }}} \right)}^2}} \ = 5.34;km \end{array}$

Out of all the given sentences, only the sentence in option 1 is entirely correct with respect to grammar and usage.

Here are the errors in the other sentences:

• Option 2: The verb 'were' should actually be 'was' since the subject 'information' is singular.
• Option 3: The word 'real' should instead be 'really' - an adverb to describe the adjective 'useful'.
• Option 4: 'Would not had been' is an ungrammatical tense construction.

Hence, the correct answer is option 1.

We redraw this triangle as

In triangle, the internal bisector of an angle bisect the opposite side in the ratio of the other two sides. So, we have

$\begin{array}{l} \frac{{QS}}{{SR}} = \frac{{PQ}}{{PR}}\ \frac{{QS}}{{SR}} = \frac{r}{q}\ \frac{{QS + SR}}{{SR}} = \frac{{r + q}}{q}\ \frac{{SR}}{{QR}} = \frac{q}{{r + q}}\ SR = \frac{{qp}}{{r + q}} \end{array}$

By using cosine formula in triangle ${\rm{\Delta }}PSR$,

$\begin{array}{l} \cos 60^\circ = \frac{{{{\left( {PR} \right)}^2} + {{\left( {PS} \right)}^2} - {{\left( {SR} \right)}^2}}}{{\left( {PR} \right)\left( {PS} \right)}}\ \frac{1}{2} = \frac{{{{\left( q \right)}^2} + {{\left( {PS} \right)}^2} - {{\left( {\frac{{qp}}{{r + q}}} \right)}^2}}}{{\left( q \right)\left( {PS} \right)}}\ PS = \frac{{qr}}{{r + q}} \end{array}$

Here $P,R,S,T,U$ are in A.P. So, the difference between two consecutive numbers will be equal, i.e.

$d = R - P = S - R = T - S = U - T$

Where $d$ is common difference for the given A.P. Now, we check the given sequences

Sequence I:

$2P,;2R,;2S;,;2T;,;2U$

The differences between two consecutive numbers are obtained as

$\begin{array}{l} D = 2R - 2P = 2\left( {R - P} \right) = 2d\ D = 2S - 2R = 2\left( {S - R} \right) = 2d\ D = 2T - 2S = 2\left( {T - S} \right) = 2d \end{array}$

Since, the differences are same in each case, so it is in A.P.

Sequence II:

$P - 3,;R - 3,;S - 3,;T - 3,;V - 3$

The differences between two consecutive numbers are obtained as

$\begin{array}{l} D = \left( {R - 3} \right) - \left( {P - 3} \right)\ = R - P - 3 + 3 = R - P = d\ D = \left( {S - 3} \right) - \left( {R - 3} \right)\ = S - R - 3 + 3 = S - R = d \end{array}$

Again, the differences are same in each case, so it is in A.P.

Sequence III:

${P^2},;{R^2},;{S^2},;{T^2},;{V^2}$

The differences between two consecutive numbers are obtained as

$\begin{array}{l} D = {R^2} - {P^2} = \left( {R - P} \right)\left( {R + P} \right)\ = d\left( {R + P} \right)\ D = {S^2} - {R^2} = \left( {S - R} \right)\left( {S + R} \right)\ = d\left( {S + R} \right) \end{array}$

In this case, the differences are not same, so it is not in A.P.

Thus, sequences I and II will be in A.P.

Concept:

Let f(z) = u + iv be the analytic function,

Cauchy-Riemann equations are 

vy = ux

vx = - uy

Calculation:

Given $f\left( z \right) = g\left( z \right) + h\left( z \right)$

Let $g\left( z \right) = gu\left( z \right) + j;gv\left( z \right)$

$h\left( z \right) = hu\left( z \right) + j;hv\left( z \right)$

If g and h are differentiable, then it will satisfy C-R equations. So, we have

\(\begin{array}{*{20}{c}} {g{u_x} = g{v_y},}&{g{u_y} = - g{v_x}}\ {h{u_x} = h{v_y},}&{huy = - h{v_x}} \end{array}\)

$f\left( z \right) = \left( {gu\left( z \right) + hu\left( z \right)} \right) + j;(\left( {gv\left( z \right) + hv\left( z \right)} \right)$

By observing the above equations, we get

\(\begin{array}{*{20}{c}} {\left( {g{u_x} + h{u_x}} \right) = \left( {g{v_y} + h{v_y}} \right)}\ {\left( {g{v_y} + h{v_y}} \right) = - \left( {g{v_x} + h{v_x}} \right)} \end{array}\)

Hence, form above two equations f(z) is differentiable (because it satisfies C- R equation).

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 the following matrices.

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

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 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 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 rank of the augmented matrix, then the system is inconsistent, and it has no solution.

The rank of A ≠ Rank of an augmented matrix

Explanation:

If determinant of coefficient matrix is non-zero then there is a unique solution. If determinant is non – zero. So all the Eigen values are non-zero. If Eigen values are non – zero then question are linearly independent.

we have a set of three linear equations with three variables.

i.e., [A]_3×3 [X]_3×1 = [B]_3×1

Statement P:

In this case,

P(A) = P(A : B) = Number of variables

⇒ P(A) = P(A : B) = 3

∴ three independent rows exist in matrix A irrespective of matrix B.

Also, |A| ≠ 0     ---(i)

Hence, P = Q = S

Statement R:

Let λ_1, λ₂ & λ₃ are eigen values of matrix A

using the properties,

λ1 λ2 λ₃ = |A|   ----(ii)

from equation (i) & (ii),

λ1 λ2 λ3 ≠ 0

⇒ λ1 ≠ 0, λ2 ≠ 0 & λ3 ≠ 0

Therefore, from above discussion, it is inferred that,

P = Q = R = S

Stoke’s theorem: The line integral of a vector around closed path L is equal to the integral of curl over the open surface is enclosed by the closed path L.

$\oint \vec A \cdot d\vec l$=$\iint \left( {\vec \nabla \times \vec A} \right).d\vec s$

Gauss's theorem: The total flux coming out of a closed surface is equal to the change enclosed.

$\oint D.ds = Q$

Divergence theorem: The total outward flux of a vector field $\vec F$ through a closed surface is equal to the volume integral of the divergence of $\vec F$

\(\mathop \oint \limits_s \vec F.d\vec s = \mathop \smallint \limits_v \left( {\vec \nabla .\vec F} \right)dv\)

Cauchy's integral theorem: $\oint f\left( z \right).dz = 0$

Concept:

$x\left( t \right)\mathop \leftrightarrow \limits^{LT} X\left( s \right)$

$\frac{{dx\left( t \right)}}{{dt}}\mathop \leftrightarrow \limits^{L.T} sX\left( s \right)$

$sX\left( s \right)\mathop \leftrightarrow \limits^{I.L.T} \frac{{dx\left( t \right)}}{{dt}}$

\(L\left{ {\frac{{x\left( t \right)}}{{t}}} \right} = \frac{}{}\mathop \smallint \limits_s^0 \left{ {x\left( t \right)} \right}ds\ \)

Calculation:

Given that Laplace transform $f\left( t \right) = 2\sqrt {\frac{t}{\pi }}$ is ${s^{ - \frac{3}{2}}}$

Given as $g\left( t \right) = \frac{1}{{\sqrt {\pi t} }}$

\(\begin{array}{l} \Rightarrow g\left( t \right) = \frac{{2\sqrt {\frac{t}{\pi }} }}{{2t}} = \frac{{f\left( t \right)}}{{2t}}\ L\left{ {g\left( t \right)} \right} = L\left{ {\frac{{f\left( t \right)}}{{2t}}} \right} = \frac{1}{2}\mathop \smallint \limits_s^0 \left{ {f\left( t \right)} \right}ds\ = \frac{1}{2}\mathop \smallint \limits_s^\infty {s^{ - \frac{3}{2}}}ds = \frac{1}{2}\left( {\frac{{{s^{\frac{{ - 3}}{2} + 1}}}}{{\frac{{ - 3}}{2} + 1}}} \right)_s^\infty \ = \frac{1}{2}\left( { - 2} \right)\left[ {0 - {s^{ - \frac{1}{2}}}} \right] = {s^{ - \frac{1}{2}}} = \frac{1}{{\sqrt s }} \end{array}\)

• Permanent Magnet Moving Coil (PMMC) is only used for DC measurements.
• Moving Iron (MI) type instruments can be used for both AC & DC measurements. But Moving Iron connected through current Transformer block DC supply. So that Moving Iron connected through current Transformer are only used for AC measurements.
• Rectifier type instruments are used for both AC & DC measurements.
• Induction type instruments are only used for AC measurements.

Concept:

In a two-wattmeter method, for lagging load

The reading of first wattmeter (W1) = VL IL cos (30° + ϕ)

The reading of second wattmeter (W2) = VL IL cos (30° - ϕ)

Total power in the circuit (P) = W1 + W2

Calculation:

Given,

Total power consumed by load W₁ + W₂ = 5 kW

Power factor cos ϕ = 0.707

⇒ ϕ = cos^-1 (0.707) = 45° 

The reading of first wattmeter (W1) = VL IL cos (30° + ϕ)

W1 = VL IL cos (30° + 45°) = 0.2588 V_L I_L

The reading of second wattmeter (W2) = VL IL cos (30° - ϕ)

W2 = VL IL cos (30° - 45°) = 0.966 VL IL

The total power consumed by the load is 

W1 + W2 = 5 kW = (0.2588 + 0.966) VL IL

⇒ VL IL = 4.082 kW

Therfore the reading in each wattmeter is 

W₁ = 0.966 VL IL = 0.966 × (4.082 kW) 

W₁ = 3.94 kW

W2 = 0.2588 VL IL = 0.2588 × (4.082 kW)

W₂ = 1.06 kW

Concept:

• Once the capacitor gets charged to a particular voltage, the capacitor behaves like a voltage source of that particular voltage and the diode responsible for the charging of the capacitor will always be off.
• The voltage across the capacitor will always only depend on the peak to the peak value of the input and not on the shape of the input waveform.
• If the given input waveform is anything other than a square wave, then during the analysis consider it as a square wave between the same two levels (+Vm, -Vm)

Explanation:

During the positive half cycle, the circuit under steady-state is as shown below.

The voltage across R, |V_R| = 2 V_m

During the negative half-cycle, the circuit under steady-state is as shown below.

The voltage across R, |V_R| = 2 V_m

The voltage across resistance R = 2 V_m = 2 × 100 = 200 V

Given y= 0 or x-z plane

x and z are tangential components, y is normal component

y < 0, ϵ₁ = 2

E₁ = 3a_x + 4a_y + 2a_z

Normal component is E_n1 = 4a_y 

y > 0, ϵ2 = 5

ϵ₁ E_n1 = ϵ2 En2

⇒ E_n2 = (2/5) (4a_y) 

${E_2} = 3{a_x} + \frac{2}{5}\left( {4{a_y}} \right) + 2{a_z}$

E₂ = 3a_x + 1.6 a_y + 2 a_z

Concept:

The SOP representation of the circuit is:

F = Σm (minterms)

Minterm: a minterm of n variables is a product of the variables in which each appears exactly once in true or complemented form.

The POS representation of the circuit:

F = ΠM (max terms)

Maxterm: a maxterm of n variables is a sum of the variables in which each appears exactly once in true or complemented form.

Calculation:

Given,

$F=ABC+\bar A\bar B C+A\bar BC+\bar ABC+\bar A\bar B\bar C$

For the function, we form the K-map as:

Hence, the function in the form of minterms is expressed as:

f(A, B, C) = Σm (0,1,3,5,7)

Now, we put 0 in each block of the K-map excluding the blocks corresponding to the terms in the above function.

Grouping the 0’s in K-map, we obtain the max terms as

F = πM(2, 4, 6)

F = πM(2, 4, 6) = $\left( {A + \bar B + C} \right)\left( {\bar A + B + C} \right)\left( {\bar A + \bar B + C} \right)$

In the given figure, (a) is low pass filter and (b) high pass filter. Also, we have

${f_1} < {f_2}$

For $f < {f_1}$, we have the op- omp circuit as

Again, for ${f_1} < f < {f_2}$, the op – amp circuit is

for $f > {f_2}$, the op – omp circuit is

Hence, the output graph is

Thus, the resultant circuit represents a band stop filter.

In Bipolar transistor, saturation mode occurs when

1. Collector – base (CB) is forward biased.
2. Base-emitter (BE) is also forward biased.

Hence, for the saturation condition, bath the junction should be in forward-bias, as shown in the figure below.

Analysis:

In the given line diagram

Line 1 is tripped at time t1 by opening the circuit breakers at the two ends (although there is no fault)

As the line is disconnected the reactance of the system increases

The maximum power transfer increases (as power is inversely proportional to the reactance of the line $P_{max}=\frac{EV}{X}$)

It is also given that mechanical power input to the generator and the voltage magnitude E is constant.

Therefore the new electrical power is to be the same as the old power, so δ value should be increased (given in the line reactance diagram the steady-state value of δ before the transient is positive)

The rotor angle transient will be

Concept:

For a 3 bus power system

For the above figure, the admittance matrix is as shown below.

\({y_{bus}} = \left[ {\begin{array}{*{20}{c}} {{y_{11}} + {y_{12}} + {y_{13}}}&{ - {y_{12}}}&{ - {y_{13}}}\ { - {y_{12}}}&{{y_{21}} + {y_{22}} + {y_{23}}}&{ - {y_{23}}}\ { - {y_{31}}}&{ - {y_{23}}}&{{y_{31}} + {y_{32}} + {y_{33}}} \end{array}} \right]\)

Diagonal elements of the Bus Admittance matrix are known as self-admittances and the off-diagonal elements are known as mutual admittances.

Calculation:

Given bus admittance matrix of the uncompensated line is

\(Y=\left[ {\begin{array}{*{20}{c}} { - j6}&{j3}&{j4}\ {j3}&{ - j7}&{j5}\ {j4}&{j5}&{ - j8} \end{array}} \right]Pu\)

By comparing the above matrix with standard 3 bus matrix

y₁₃ = -j4

y₃₂ = -j5

⇒ y₃₁ + y₃₂ + y₃₃ = -j8

⇒ y_33­ = j

After compensating, ${y_{33}} = \frac{j}{2}$

Y_33(new) = -8.5 j.

Given as zero initial condition:

I(0^-) = I(0⁺) = 0

When the switch is closed the inductor will act as an open circuit

∴ I(0⁺) = 0

After some time inductor will start charging and at t → ∞

I(∞) = V/R

Now The current through the inductor is given by:

Putting all the value in the above equation

Differentiate the equation first time then we get

Differentiate the equation a second time then we get

Hence option (4) is correct.

Concept:

Referred value in Transformer:

In order to simplify the calculation, it is theoretically possible to transfer the voltage, current, and impedance of one winding to the other winding and combined them to a single value for each quantity.

Considered a transformer has turns ration 'a' which is given by,

$a = \frac{{{N_2}}}{{{N_1}}} = \frac{{{V_2}}}{{{V_1}}} = \frac{{{I_1}}}{{{I_2}}}$

Where,

I1 and I2 are primary and secondary current respectively.

V1 and V2 are primary and secondary voltage respectively.

N1 and N2 are numbers of turn in primary and secondary respectively.

For an ideal transformer:

Input Power = Output Power

$\frac{{V_1^2}}{{{Z_1}}} = \frac{{V_2^2}}{{{Z_2}}}$

$\frac{{V_2^2}}{{V_1^2}} = \frac{{{Z_2}}}{{{Z_1}}}$

${\left( {\frac{{{V_2}}}{{{V_1}}}} \right)^2} = \frac{{{Z_2}}}{{{Z_1}}}$

${a^2} = \frac{{{Z_2}}}{{{Z_1}}}$

Equivalent secondary Impedance in terms of primary Impedance:

${Z_1} = \frac{{{Z_2}}}{{{a^2}}}$

Equivalent primary Impedance in terms of secondary Impedance:

${Z_2} = {a^2}{Z_1}$

Calculation:

Given,

Zin is resistive and equal to 2.5 Ω

Angular frequency ω = 5000 rad/s

Inductive reactance X_L = jωL = j 5000 × 1 × 10^-3 = j 5 Ω 

Capative reactance X_C = 1 / jωC = 1 / j(5000 × 10 × 10^-6) = - j 20 Ω 

The input impedance of the transformer from the secondary side of the transformer will be

Z' = XL + R/a² = j 5 + 2.5/a²

Circuit diagram will be as follows  

The input impedance of the transformer from the primary side of the transformer will be

$⇒ {Z_{in}} = \frac{1}{{{b^2}}}\left[ {\frac{{2.5}}{{{a^2}}} + j5} \right] - j20$

$⇒ {Z_{in}} = \left[ {\frac{{2.5}}{{{a^2}\times {b^2}}} +\frac{{j5}}{{{b^2}}}} \right] - j20$

It is given that the input impedance is resistive, therefore there won't be any reactance term in the input impedance

As there is no reactance in the input impedance, make reactance equal to zero in the above equation

$⇒- 20 + \frac{5}{{{b^2}}} = 0$

⇒ b² = 0.25

⇒ b = 0.5

Now the given resistive term of the input impedance is equal to 2.5 Ω 

$⇒ \frac{{2.5}}{{{a^2}{b^2}}} = 2.5$

$⇒ \frac{{2.5}}{{{a^2}\times{0.5^2}}} = 2.5$

⇒ a = 2

Therefore the value of a = 2.0, b = 0.5

Wave wound

For wave wound, number of parallel pat (A₁) = 2

∴ $E_1=\frac{pϕ Nz}{60\ A_1}$

⇒ $E_1∝\frac{1}{A_1}$ {p, ϕ, N & z are constants}

Lap wound

for lap wound,

Number of parallel path = Number of poles

i.e., A₂ = P = 4

∴ $E_2∝\frac{1}{A_2}$

So,

$\frac{E_2}{E_1}=\frac{A_2}{A_1}$

⇒ $E_2=\frac{E_1}{2}$

It is given that armature resistance (R_a) = 0

so, E ∝ V

⇒ $\frac{E_2}{E_1}=\frac{V_1}{V_2}$

⇒ $V_2=\frac{E_2}{E_1}V_1$

⇒ $V_2=\frac{V_1}{2}$

so, power (p) = V I_a

⇒ $I_a=\frac{P}{V}$

⇒ $I_a\propto\frac{1}{V}$

so,

$\frac{I_{a_2}}{I_{a_1}}=\frac{V_1}{V_2}\ {V_2=\frac{V_1}{2}}$

$I_{a_2}=2I_{a_1}$

The power of the machine will remain same as P_lap = P_wave

The machine with lap wound armature coils will have rated voltage.

$V_2=\frac{V_1}{2}=\frac{230}{2}=115$ V

∴ P_lap = P_wave = p = 5 kW

From the Nyquist plot, we have

$\begin{array}{l} {F_{G1}}\left( S \right) = \frac{{{K_1}}}{s}\ {F_{G2}}\left( S \right) = {K_2}s; \end{array}$

So, the product of ${F_{G1}}$ and ${F_{G2}}$ is obtained as

$\begin{array}{l} F\left( s \right) = {F_{G1}}\left( s \right).{F_{G2}}\left( s \right)\ = \frac{{{k_1}}}{s}.{k_2}s = {k_1}{k_2}; \end{array}$

Given events ${H_R}{H_S},;{T_R}{T_S},{H_R}{T_S},{T_R}{H_S}$

If coins are independent, corresponding probabilities will be

\(\begin{array}{*{20}{c}} {\frac{1}{2}.\frac{1}{2},\frac{1}{2}.\frac{1}{2}}&{\frac{1}{2}.\frac{1}{2},}&{\frac{1}{2}.\frac{1}{2}} \end{array}\)

$= \frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4}$ respectively

But given probabilities are 0.28, 0.18, 0.3, 0.24 respectively. We cannot decide whether R is fair or S is fair

⇒ The coin tosses are dependent.

Concept:

The Fourier Transform of a continuous-time signal x(t) is given as:

\(X\left( \omega \right) = \mathop \smallint \limits_{ - \infty}^{\infty} x(t) ~{e^{ - j\omega t}}~dt \)

Analysis:

Given:

x(t) = e^j10t  defined from t = -1 to 1. 

\( X\left( \omega \right) = \mathop \smallint \limits_{ - 1}^1 {e^{j10t}}.{e^{ - j\omega t}}dt = \mathop \smallint \limits_{ - 1}^1 {e^{j\left( {10 - \omega } \right)t}}dt\)

$X(\omega) = \left. {\frac{{{e^{j\left( {10 - \omega } \right)t}}}}{{j\left( {10 - \omega } \right)}}} \right|_{ - 1}^1 = \frac{{2\sin \left( {\omega - 10} \right)}}{{\left( {\omega - 10} \right)}}$

Concept:

• BIBO (Bounded Input Bounded output stable system is a type of system in which for given bounded input output is bounded.
• On applying unit step input to the system then the response which we get is called unit step response.
• Condition for absolutely integrable.

$\rm\displaystyle\int_{-\infty}^{\infty}|x(t)| dt < \infty$

• Bounded signals are those whose value always remains less than a certain value & finite support implies that the duration of the signal is finite.

Explanation:

• Since the system is BIBO stable & the unit step function is bounded, the unit step response will also be bounded.

Note:

$\rm\displaystyle\int_{-\infty}^{\infty}|u(t)| dt = \rm\displaystyle\int_{-\infty}^{\infty}1 dt = \infty$

Therefore unit step response is NOT absolutely integrable.

Concept:

 The first difference of any signal t[n] is t[n] - t[n-1]

Explanation: 

As per the question,

$y\left( n \right)$ is first difference of $x\left( n \right)$ 

So, 

      $y\left( n \right){\rm{;}} = {\rm{;}}x\left( n \right){\rm{;}}-{\rm{;}}x\left( {n - 1} \right)$

⇒ $Y\left( z \right){\rm{;}} = {\rm{;}}x\left( Z \right)\left( {1 - {z^{ - 1}}} \right){\rm{;}} = {\rm{;}}X\left( z \right){\rm{;}}-{\rm{;}}{z^{ - 1}}{\rm{;}}X\left( z \right)$

$\begin{array}{l} Y\left( z \right){\rm{;}} = {\rm{;}}\left[ {2x{\rm{;}} + {\rm{;}}4{\rm{;}}-{\rm{;}}4{z^{ - 1}}{\rm{;}} + {\rm{;}}3{z^{ - 2}}\left] {{\rm{;}}-{\rm{;}}} \right[2{\rm{;}} + {\rm{;}}4{z^{ - 1}}{\rm{;}} - {\rm{;}}4{z^{ - 2}}} \right]{\rm{;}}\ = 2x{\rm{;}} + {\rm{;}}4{\rm{;}}-{\rm{;}}4{z^{ - 1}} + {\rm{;}}3{z^{ - 2}}{\rm{;}}-{\rm{;}}2{\rm{;}}-{\rm{;}}4{z^{ - 1}}{\rm{;}}-{\rm{;}}4{z^{ - 2}}{\rm{;}}-{\rm{;}}3{z^{ - 3}}\ = {\rm{;}}2z{\rm{;}} + {\rm{;}}2{\rm{;}}-{\rm{;}}8{z^{ - 1}}{\rm{;}} + {\rm{;}}7{z^{ - 2}}{\rm{;}}-{\rm{;}}3{z^{ - 3}} \end{array}$

The force F₁ acting will be ${F_1} = \frac{{Q \times \left( { - Q} \right)}}{{4\pi \epsilon{_0}{{\left( {2d} \right)}^2}}}\left( {\frac{{2d{{\hat a}_x}}}{{2d}}} \right)$

The force F2 acting will be ${F_2} = \frac{{Q \times \left( { - Q} \right)}}{{4\pi \epsilon{_0}{{\left( {2d} \right)}^2}}}\left( {\frac{{2d{{\hat a}_y}}}{{2d}}} \right)$

The force F₃ acting will be ${F_3} = \frac{{Q \times Q}}{{4\pi \epsilon{_0}{{\left( {\sqrt 8 ;d} \right)}^2}}}\left( {\frac{{2d\hat a + 2d\hat y}}{{\sqrt 8 d}}} \right)$

The net force on Q due to the remaining charges will be

F = F₁ + F₂ + F₃

$F = \frac{{Q \times \left( { - Q} \right)}}{{4\pi \epsilon{_0}{{\left( {2d} \right)}^2}}}\left( {\frac{{2d{{\hat a}_x}}}{{2d}}} \right)+\frac{{Q \times \left( { - Q} \right)}}{{4\pi \epsilon{_0}{{\left( {2d} \right)}^2}}}\left( {\frac{{2d{{\hat a}_y}}}{{2d}}} \right)+\frac{{Q \times Q}}{{4\pi \epsilon{_0}{{\left( {\sqrt 8 ;d} \right)}^2}}}\left( {\frac{{2d\hat a + 2d\hat y}}{{\sqrt 8 d}}} \right)$

$F= \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q^2}}}{{{{\left( {2d} \right)}^3}}}\left[ { - 2d{a_x} - 2d{a_y} + \frac{1}{{2\sqrt 2 }}\left( {2d{a_x} + 2d{a_y}} \right)} \right]$

$F = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q^2}}}{{{d^2}}};\left[ {\frac{{1 - 2\sqrt 2 }}{{8\sqrt 2 }}{a_x} + \frac{{1 - 2\sqrt 2 }}{{8\sqrt 2 }}{a_y}} \right]$

By comparing the above equation with $\frac{{{Q^2}}}{{4\pi {\varepsilon _0}}}\frac{K}{{{d^2}}}$

$\Rightarrow K = \frac{{1 - 2\sqrt 2 }}{{8\sqrt 2 }}\hat i + \frac{{1 - 2\sqrt 2 }}{{8\sqrt 2 }}\hat j$

Given counter is 

The same clock pulse is given to both the flip-flops, therefore it is synchronous counter.

For every four cycles, the output is repeated 

So, dividing 333 by 4, we get 83.25

So, at 332 (83 × 4 = 332) clock pulse the output will be 0 0 (Q₁Q₀) 

At 333^rd clock pulse, Q₁ Q₀ = 0 1

Given maxterm

F (A,B,C,D) = π (1,5,12,15)

The minterms of the given max terms will be

f (A,B,C,D) = Σm (0,2,3,4,6,7,8,9,10,11,13,14) 

Therefore, the correct inputs to pins 0, 1, 2, 3, 4, 5, 6, 7 in order is D’, 1, D’, 1, 1, 1, D, D’

Concept:

An astable multivibrator is also called a Square wave generator or Free running oscillator.

It generates a square wave output.

The output waveform:

It generates a square wave output with time (T ) = $\ T = 2{R_c}\log \frac{{\left( {1 + \beta } \right)}}{{\left( {1 - \beta } \right)}}$

Calculation:

Astable multivibrator produces square wave.

$\begin{array}{l} \beta = \frac{{{R_2}}}{{{R_1} + {R_2}}} = \frac{2}{4} = 0.5\ T = 2{R_c}\log \frac{{\left( {1 + \beta } \right)}}{{\left( {1 - \beta } \right)}} = 2 \times 1 \times {10^3} \times 0.25 \times {10^{ - 6}} \times \log \left( {\frac{{1 + 0.5}}{{1 - 0.5}}} \right)\ T = 0.55;ms \end{array}$

Square wave of period 0.55 ms.