GATE EE 2016 Official Paper: Shift 2 (Previous Year Paper)

The context of the sentence requires a phrasal verb that will mean to ‘endure’. The phrasal verb from the options with that meaning is ‘bear with’. ‘Bore with’ is in the past tense and cannot be used because the blank is preceded by ‘to’. Thus the base form of the verb needs to be used. ‘Bare’ means to disclose or uncover and thus does not fit the context.

The correct spelling is ‘manageable’

In the group of given numbers, all are prime numbers but 33 is not.

There are two facts. R2D2 is a robot who can repair aeroplanes and no other robot can do this work. The other functions of R2D2 is not known and thus inference 1 does not follow.

We do not know whether R2D2 can repair anything else other than aeroplanes. Thus option 3 does not follow.

There can be other robots and thus option 4 is eliminated.

Only option 2 is the logical inference that follows.

$\begin{array}{l} \left| {9y - 6} \right| = 3\ 9y - 6 = \pm 3 \end{array}$

Case 1:

$9y - 6 = 3 \Rightarrow y - 1$

Case 2:

$9y - 6 = - 3 \Rightarrow y = \frac{1}{3}$

Substitute $y = 1$

${y^2} - \frac{{4y}}{3} = - \frac{1}{3}$

Substitute $y = \frac{1}{3}$

${y^2} - \frac{{4y}}{3} = - \frac{1}{3}$

Let us examine the options one by one:

1. Various engineering disciplines have not been talked about in the paragraph and thus this inference does not follow. We cannot make a general statement with respect to various disciplines when only mechanical engineering has been talked about.
2. The poll taken was for students and the gender of the students participating in the poll has not been mentioned. Thus this inference does not hold.
3. This inference hold because the research study found that the women with a high degree in mechanical engineering were successful in their profession. Thus it can be a suitable profession for them.
4. We cannot infer anything about the number of women pursuing mechanical engineering because no information has been provided regarding the same.

Thus option 3 is the answer.

Let us examine the options one by one.

1. No inference can be drawn about the degrees of the students because no such information has been provided in the passage.
2. Since the institutes are being set up to cater to the technological development needs, it can be inferred they have a national objective. Thus this options follows
3. Though SITs re being established in collaboration with IITs it is not always true that they cannot be set up independently. Thus this option does not hold.
4. The passage states that the SITs are being set up keeping in mind the technological developments. Thus this inference definitely follows.

Hence options 2 and 4 are logical inferences.

n = No. of attempts = 10

x = free throws successfully = 6

$p = 60\% = 0.6$

The probability that he will successfully make exactly 6 free throws in 10 attempts

$\begin{array}{l} = {n_{{c_x}}}{p^x}{q^{n - x}}\ = {10_{{c_6}}}{\left( {0.6} \right)^6} \times {\left( {0.4} \right)^4}\ = 210 \times 0.046656 \times 0.0256\ = 0.2508 \end{array}$

Concept:

3 variable K-maps:

• For a 3-variable Boolean function, there is a possibility of 8 output minterms.
• The general representation of all the minterms using 3-variables is shown below.

Calculation:

Given k-map is as shown

It can be grouped as follows

Therefore in SOP form (sum of products) output is Y = A + C̅

Concept:

Capacitive reactance(X_c) 

$X_c = \frac{1}{ω C}$

Where C = capacitance in farad, ω = angular frequency in rad/sec

For finding the type of filter find its low frequency and high-frequency response :

At Low-Frequency open circuit, the Capacitors.

At a High-frequency short circuit, the capacitors

For ideal op-amp according to the virtual short concept 

V- = V+

Calculation:

At low frequencies i.e. at ${\rm{ω }} = {\rm{;}}0{\rm{;}} \Rightarrow {{\rm{X}}_{\rm{c}}} = \frac{1}{{{\rm{ω c}}}} = \infty$

∴ The capacitor acts like an open circuit

Due to the virtual short concept V- = V+ = 0 (Virtual Ground)

Apply KCL at node 'a' and we get,

\(\therefore {{\rm{V}}_{{\rm{out}}}} = \frac{{ - {{\rm{R}}_2}}}{{{{\rm{R}}1}}}{{\rm{V}}{{\rm{in}}}}\)

At high frequencies i.e. at ${\rm{ω }} = \infty \Rightarrow {{\rm{X}}_{\rm{c}}} = \frac{1}{{{\rm{ω c}}}} = 0$

I.e. capacitor acts like a short circuit

Input directly connected to the output so

$\therefore {\rm{;}}{{\rm{V}}_{{\rm{out}}}}{\rm{;}} = {\rm{;}}0$    (Due to virtual short concept)

So the given circuit passes only low frequencies hence it acts as a low pass filter.

Concept:

Laplace transform is an important tool for converting differential equations in the time domain to algebraic equations in the frequency domain.

This analysis involves three important processes:

1) The transformation from the time domain to the frequency domain.

2) Manipulate the algebraic equations to form a solution.

3) The inverse transformation from the frequency to the time domain.

Mathematically, the Laplace transform of a function f(t) is defined as:

\(L[f(t)]=F(s)=\mathop \smallint \limits_0^\infty f\left( t \right){e^{ - st}}dt\)

s = σ + jω is the complex frequency 

Applications of the Laplace Transform:

1. Solve differential equations (both ordinary and partial)

2. Application in RLC circuit analysis.

Calculation:

Given $\frac{{\rm{d}}}{{{\rm{dt}}}}{\rm{y}}\left( {\rm{t}} \right) + \frac{1}{6}{\rm{y}}\left( {\rm{t}} \right) = 3{\rm{x}}\left( {\rm{t}} \right)$

Apply Laplace transform we get

${\rm{sY}}\left( {\rm{s}} \right) + \frac{1}{6}{\rm{Y}}\left( {\rm{s}} \right) = 3{\rm{X}}\left( {\rm{s}} \right)$ But ${\rm{X}}\left( {\rm{s}} \right) = \frac{3}{{\left( {{\rm{s}} + \frac{1}{3}} \right)}}$ 

$\begin{array}{l} \therefore {\rm{Y}}\left( {\rm{s}} \right) = \frac{{3{\rm{X}}\left( {\rm{s}} \right)}}{{\left( {{\rm{s}} + \frac{1}{6}} \right)}} = \frac{9}{{\left( {{\rm{s}} + \frac{1}{6}} \right)\left( {{\rm{s}} + \frac{1}{3}} \right)}}\ {\rm{Y}}\left( {\rm{s}} \right) = \frac{9}{{\left( {{\rm{s}} + \frac{1}{6}} \right)\left( {\frac{{ - 1}}{6} + \frac{1}{3}} \right)}} + \frac{9}{{\left( {{\rm{s}} + \frac{1}{3}} \right)\left( {\frac{{ - 1}}{3} + \frac{1}{6}} \right)}}\ {\rm{Y}}\left( {\rm{s}} \right) = \frac{{54}}{{{\rm{s}} + \frac{1}{6}}} - \frac{{54}}{{\left( {{\rm{s}} + \frac{1}{3}} \right)}} \end{array}$

Apply Inverse Laplace transform we get

${\rm{y}}\left( {\rm{t}} \right) = 54\left[ {{{\rm{e}}^{ - \frac{1}{6}{\rm{t}}}} - {{\rm{e}}^{ - \frac{1}{3}{\rm{t}}}}} \right]{\rm{u}}\left( {\rm{t}} \right)$

Concept:

Function of a Complex Variable: ω = f (z) = u (x, y) + i v (x, y) Where x, y ϵ R  and  z = x + iy

u (x, y), v (x, y) are real valued functions. 

Analytic function: A function f (z) is said to be Analytic in a region R of z-plane if it is differentiable at every point of R. 

Necessary Condition for function f (z) to be Analytic: 

1. $\frac{\partial u}{\partial x},\ \frac{\partial u}{\partial y}, \ \frac{\partial v}{\partial x}, \ \frac{\partial v}{\partial y}$  exist 2. f (z) to satisfy the Cauchy-Riemann (C-R) equation.

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

Calculation: 

$\begin{array}{l} {\rm{z;}} = {\rm{;x;}} + {\rm{;iy;}} \Rightarrow {\rm{;}}{{\rm{z}}^{\rm{*}}}{\rm{;}} = {\rm{;x;}}-{\rm{;iy}}\ \therefore {\rm{;f}}\left( {\rm{z}} \right){\rm{;}} = {\rm{;}}\left( {{\rm{x;}} + {\rm{;iy}}} \right){\rm{;}} + {\rm{;}}\left( {{\rm{x;}} - {\rm{;iy}}} \right){\rm{;}} = {\rm{;}}2{\rm{x}} \end{array}$

$\therefore ;f\left( z \right)$ is finite everywhere and it is continuous and differentiable.

But it doesn't satisfy the CR equations.

So we can say that $f\left( z \right)$ is continuous but not analytic.

Concept:

CAYLEY-HAMILTON THEOREM:

Statement: Every square matrix satisfies its own characteristic equation.

The Cayley–Hamilton theorem states that substituting the matrix A for x in polynomial, p(x) = det(xI_n – A), results in the zero matrices, such as:

p(A) = 0

It states that a ‘n x n’ matrix A is demolished by its characteristic polynomial det(tI – A), which is monic polynomial of degree n.

Uses of Cayley-Hamilton theorem:

(1) To calculate the positive integral powers of A

(2) To calculate the inverse of a square matrix A

Calculation:

${P^3} = P$

From Cauley Hamilton theorem ${\lambda ^3} = \lambda$ 

$\begin{array}{l} \Rightarrow \lambda ;\left( {{\lambda ^2}-1} \right) = 0\ \lambda = 0,; + 1,; - 1 \end{array}$

Concept:

 Laplace transform of the first two derivatives.

L{y^''} = s²Y(s) - sy(0) - y'(0)

L{y'} = sY(s) - y(0)

Calculation:

Given ${\rm{y''}}\left({\rm{t}} \right){\rm{;}} + {\rm{;}}2{\rm{y'}}\left({\rm{t}} \right){\rm{;}} + {\rm{;y}}\left( {\rm{t}} \right){\rm{;}} = {\rm{;}}0$

Apply Laplace transform we get

$\left[ {{{\rm{s}}^2}{\rm{Y}}\left( {\rm{s}} \right){\rm{;}}-{\rm{;sy}}\left( 0 \right){\rm{;}}-{\rm{;y'}}\left( 0 \right)\left] {{\rm{;}} + {\rm{;}}2} \right[{\rm{sy}}\left( {\rm{s}} \right){\rm{;}}-{\rm{;y}}\left( 0 \right)} \right]{\rm{;}} + {\rm{;Y}}\left( {\rm{s}} \right){\rm{;}} = {\rm{;}}0$

$\begin{array}{l} \left( {{{\rm{s}}^2}{\rm{;}} + {\rm{;}}2{\rm{s;}} + {\rm{;}}1} \right){\rm{;Y}}\left( {\rm{s}} \right){\rm{;}}-{\rm{;}}1{\rm{;}} = {\rm{;}}0\ {\rm{Y}}\left( {\rm{s}} \right) = \frac{1}{{{{\left( {{\rm{s}} + 1} \right)}^2}}} \Rightarrow {\rm{y}}\left( {\rm{t}} \right) = {\rm{t}} \cdot {{\rm{e}}^{ - {\rm{t}}}}{\rm{u}}\left( {\rm{t}} \right) \end{array}$

Concept:

When two points (x₁, y₁. z₁) and (x₁, y₁. z₂) are mentioned find the relation in terms of the third variable in terms of x,y, and z:

$\dfrac{{{{x}} - x_1}}{{x_2 - x_1}} = \dfrac{{{{y}} - y_1}}{{y_2 - y_1}} = \dfrac{{{{z}} - z_1}}{{z_2 - z_1}} = {{t}}$

Put the value of z,y, and z and use the end-points of one variable.

Calculation:

Given:

${\rm{I}} = \smallint \left( {2{\rm{x}}{{\rm{y}}^2}{\rm{dx}} + 2{{\rm{x}}^2}{\rm{ydy}} + {\rm{dz}}} \right)$, A (0, 0, 0) and B(1, 1, 1).

Equation of line i.e. path

$\frac{{{\rm{x}} - 0}}{{1 - 0}} = \frac{{{\rm{y}} - 0}}{{1 - 0}} = \frac{{{\rm{z}} - 0}}{{1 - 0}} = {\rm{t}}$

$\therefore {\rm{;x;}} = {\rm{;y;}} = {\rm{;z;}} = {\rm{;t;and;t;}}:{\rm{;}}0{\rm{;}} \to {\rm{;}}1$

\( \therefore {\rm{I}} = \mathop \smallint \limits_0^1 \left( {2{{\rm{t}}^3}{\rm{dt}} + 2{{\rm{t}}^3}{\rm{dt}} + {\rm{dt}}} \right) \)

$= 4 \cdot \left[ {\frac{{{{\rm{t}}^4}}}{4}} \right]_0^1 + \left[ {\rm{t}} \right]_0^1 = 1 + 1 = 2$

$\therefore \smallint \left( {2{\rm{x}}{{\rm{y}}^2}{\rm{dx}} + 2{{\rm{x}}^2}{\rm{ydy}} + {\rm{dz}}} \right) = 2$

Concept

\(\begin{array}{l} {\rm{X}}\left( {\rm{t}} \right) = {a_0} + \mathop \sum \limits_{n = 1}^\infty {a_n}cosn{\omega _0}t + {b_n}sinn{\omega _0}t\ Where;{a_0} = dc;value;of;signal = \mathop \smallint \limits_0^{{T_0}} x\left( t \right)dt\ {a_n} = \frac{2}{{{T_0}}}\mathop \smallint \limits_0^{{T_0}} ;x\left( t \right).cos;n{\omega _0}t;\ {b_n} = \frac{2}{{{T_0}}}\mathop \smallint \limits_0^{{T_0}} ;x\left( t \right).sin;n{\omega 0}t = - {b{ - n}} \to odd;function;w.r.t;n;;; \end{array}\)

Solution:

Given that $f\left( { - x} \right) = - f\left( x \right)$ 

So, cosine term and DC term will be zero. It has only sine terms.

\(\therefore {\rm{f}}\left( {\rm{x}} \right) = \mathop \sum \limits_{{\rm{k}} = 1}^\infty {{\rm{b}}{\rm{k}}} \cdot \sin \left( {{{\rm{k}}{\rm{x}}}} \right)\)

It is a series RL circuit and

\({{\rm{V}}{\rm{S}}} = \sqrt {{\rm{V}}{\rm{R}}^2 + {\rm{V}}_{\rm{C}}^2}\)

$\therefore 100 = \sqrt {{{\left( {60} \right)}^2} + {{\left( {80} \right)}^2}}$

Given the network is:

In the given network,

R₂, R₄, and R₅ are directional overcurrent relays. R₁, R₃ and R₆ are non-directional overcurrent relays.

For the fault on line 2 i.e. L₂, R3 and R4 must be operated. If R₄ is not operated then R₁ and R₆ will operate.

Therefore, back up for R₄ are R1 and R6

The relay R₃ is directly connected to relay R₄, so that relay R₃ can’t provide remote backup.

Number of buses $\left( n \right); = 100$

Generator busses $\left( m \right) = 10$

Order of Jacobian matrix $= \left( {2n - 1 - m} \right) \times \left( {2n-1-m} \right)$

$\begin{array}{l} = \left( {200-1-10} \right) \times \left( {200-1-10} \right)\ = 189 \times 189 \end{array}$

Surge impedance $= \sqrt {\frac{L}{C}}$

$= \sqrt {\frac{{1.6 \times {{10}^{ - 3}}}}{{10 \times {{10}^{ - 9}}}}} = 400{\rm{\Omega }}$

Surge impedance loading $= \frac{{{V^2}}}{{{Z_c}}} = {\left( {\frac{{400 \times {{10}^3}}}{{400}}} \right)^2}$

$= 400;MW$

Given Load $= 300;MW$

Load is lesser than surge impedance loading.

So in order to maintain rated voltage at receiving end shunt inductance is required.

Concept:

Capacitors in Series: When capacitors are connected in series, the equivalent capacitance (C_eq) is less than the smallest individual capacitance in the series combination.

The voltage across each capacitor can be different, but the charge (Q) on each capacitor is the same.

Equivalent Capacitance: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \cdots + \frac{1}{C_n} \$

Voltage Distribution: The total voltage (V_total) across the series combination is the sum of the voltages across each capacitor:

$V_{total} = V_1 + V_2 + V_3 + \cdots + V_n \$

Here, V_i is the voltage across capacitor (C_i).

Since the same charge (Q) is on each capacitor:

$Q = C_1 V_1 = C_2 V_2 = C_3 V_3 = \cdots = C_n V_n \$

Capacitors in Parallel: When capacitors are connected in parallel, the total (equivalent) capacitance (C_{eq}) is the sum of the individual capacitances. The voltage (V) across each capacitor is the same, but the charge on each capacitor can be different.

Equivalent Capacitance: $C_{eq} = C_1 + C_2 + C_3 + \cdots + C_n \$

Voltage Distribution: Each capacitor has the same voltage applied to it: $V_{total} = V_1 = V_2 = V_3 = \cdots = V_n \$

Each capacitor stores a charge (Q_i) given by:  $Q_i = C_i V$ 

The total stored charge (Q_total) is the sum of the individual charges:  

$Q_{total} = Q_1 + Q_2 + Q_3 + \cdots + Q_n \$

Equivalent capacitance: $\frac{1}{C_{eq}} = \sum \frac{1}{C_i}\$

Explanation:

From the given figure, it is clear that two capacitors are connected in parallel. So, the voltage across them is same.

d is also same for both capacitors.

Hence, the electric field is also the same by the following formula.

${\rm{E}} = \frac{{{{\rm{ }}}{{\rm{ }}_{\rm{}}}{\rm{V}}}}{{\rm{d}}}$

So, electrical filed in the region B is 4 kV/cm

Single-phase capacitor run induction motor

• The direction of rotation of a single-phase capacitor run induction motor is reversed by changing the direction of the rotating magnetic field produced by the main and starter winding or auxiliary winding.
• This can be accomplished by reversing the polarity of the starter or auxiliary winding.
• Basically, it is done by interchanging the connections on either end of the starter or auxiliary winding.
• Sometimes this reversal is only done by the auxiliary winding & sometimes reversal is accomplished by winding, switch, and capacitor. While the order of the switch and the capacitor does not affect the reversal, as long as they are in series connection.

Important Points

Advantages of capacitor run induction motor:

• A Centrifugal switch is not required.
• It has high efficiency
• It has a high power factor because of a permanently connected capacitor.
• It has a higher pull-out torque

Limitations:

• Electrolytic capacitors cannot be used for continuous running. Therefore, paper-spaced oil-filled capacitors are used.
• It has relatively low starting torque compared to the capacitor start induction motor.

Concept:

Kirchoff’s second law:

• This law is also known as loop rule or voltage law (KVL) and according to it “the algebraic sum of the changes in potential in the complete traversal of a mesh (closed-loop) is zero”, i.e. Σ V = 0.
• This law represents “conservation of energy” as if the sum of potential changes around a closed loop is not zero, unlimited energy could be gained by repeatedly carrying a charge around a loop.

For any load, current enter through positive polarity and exit from negative polarity and for source it will be reversed.

Calculation:

Apply KVL in the loop as shown below,

The current flow through 2 Ω is 5 A as shown in fig and it will give a drop of 10 V.

From the figure we get,

• V_out + 10 + 10 = 0

Vout = 20 V

Concept:

Nodal Analysis:

Nodal analysis is a method of analyzing networks with the help of KCL equations.

For a network of N nodes, the number of simultaneous equations to be solved to get the unknowns

= Number of KCL equations

= N - 1

Mesh Analysis:

Mesh analysis is a method of analyzing networks with the help of KVL equations.

For a network having N nodes and B branches, the number of simultaneous equations to be solved to get the unknowns

= Number of KVL equations

= number of independent loop equations

= B - N + 1

Where, B = no of the branch, N = No of node

Calculation:

Number nodes n = 5

Branches b = 7

Number of KCL equations = 4

Number of KVL equations = 3

As shown in the figure 

The direction of electric field is always outward  to corresponding surface

Electric field at point D is zero, because it is at the center of electrode.

Electric field at point C have subtractive effect due to second electrode ( i.e. E = E₂ - E' )

Electric field at point A have addetive effect due to second electrode ( i.e. E = E1 + E' ) 

So that at point A, the electric field is maximum.

Concept:

De Morgan’s law states that:

$\overline {\left( {{A_1}.{A_2} \ldots {A_n}} \right)} = \left( {\overline {{A_1}} + \overline {{A_2}} + \ldots + \overline {{A_n}} } \right)$

Also,

$\overline {\left( {{A_1} + {A_2}+ \ldots + {A_n}} \right)} = \left( {\overline {{A_1}} .\overline {{A_2}} ...\overline {{A_n}} } \right)$

Analysis:

Given

\({\rm{F}} = \overline {\left( {{\rm{a}} + {\rm{̅ b}} + {\rm{c}} + {\rm{̅ d}}} \right) + \left( {{\rm{b}} + {\rm{̅ c}}} \right)}\)

Applying the De-Morgans theorem in the above function F

\(\begin{array}{l} = \overline {\left( {{\rm{a}} + {\rm{̅ b}} + {\rm{c}} + {\rm{̅ d}}} \right)} \cdot \overline {\left( {{\rm{b}} + {\rm{̅ c}}} \right)} \ \end{array}\)

= a̅.b.c̅.d.b̅. c 

As, b.b̅  = c.c̅ = 0

∴ F = a̅.b.c̅.d.b̅. c  = 0

Hence option (4) is correct

Concept: 

Virtual ground -

The concept of the virtual ground is stated as if anyone of the i/p terminals is grounded physically the other i/p terminal will also be at ground potential even though, it is not grounded physically.

• One key feature of an Op-Amp is the differential input, and when put together in a circuit, this can form a virtual ground.
• The virtual ground concept is helpful for the analysis of Op Amps. This concept makes Op-Amp circuit analysis much easier.

Calculation:

Voltage at Non-inverting terminal, ${{\rm{v}}_ + } = \frac{{{{\rm{v}}_1}\left( 4 \right)}}{{4 + 1}} + \frac{{{{\rm{v}}_2}\left( 1 \right)}}{{1 + 4}}$

${{\rm{v}}_ + } = \frac{{\left( {4{{\rm{v}}_1} + {{\rm{v}}_2}} \right)}}{5}$

Voltage at inverting ${{\rm{v}}_ - } = {{\rm{v}}_ + } = \left( {\frac{{4{{\rm{v}}_1} + {{\rm{v}}_2}}}{5}} \right)$

Apply nodal at inverting terminal we get

\(\begin{array}{l} \left( {\frac{{4{{\rm{v}}_1} + {{\rm{v}}_2}}}{5}} \right) - {{\rm{v}}_3} + \frac{{\left[ {\left( {\frac{{4{{\rm{v}}_1} + {{\rm{v}}2}}}{5}} \right) - {{\rm{v}}{{\rm{out}}}}} \right]}}{9} = 0\ \left( {\frac{{4{{\rm{v}}1} + {{\rm{v}}2}}}{5}} \right)\left[ {1 + \frac{1}{9}} \right] = {{\rm{v}}3} + \frac{{{{\rm{v}}{{\rm{out}}}}}}{9}\ 8{V_1} + 2{V_2}-9{V_3} = {V{out}}\ \therefore ;{V{out}} = 8{V_1} + 2{V_2}-9{V_3} \end{array}\)

Concept:

• Lower limit of y(t) is sum of lower limit of x(t) & h(t).
• Upper limit of y(t) is sum of upper limit of x(t) & h(t).

Explanation:

Fourier transform of x1(t) = x1(ω)

Maximum frequency components of

x1(ω) = B, rad/sec

Fourier transform of x2(t) = x2(ω)

maximum frequency component of x2(ω) = B2 rad/sec

∴ x(t) = x1(t). x2(t)

Using multiplication property in Fourier transform,

$F{x(t)}=x(ω)=\frac{1}{2\pi}{x_1(ω)*x_2(ω)}$

So, maximum frequency component of x(ω) = B1 + B2 rad/sec

Impulse response  $h(t)=e^{-2|t|}$

Taking Fourier transform of h(t) is,

$H(ω)=\frac{4}{4+ω^2}$

Maximum frequency component of H(ω) = ∞

∴ y(t) = x(t) * h(t)

Fourier Transform, Y(ω) = x(ω) H(ω)

$=\frac{1}{2\pi}{x_1(ω) *x_2(ω)}.H(ω)$

Highest frequency component of

Y(ω) = min {x(ω), H(ω)}

Y(ω) = min {B1 + B2, ∞}

Y(ω) = B1 + B2

Minimum sampling rate = Nyquist rate

= 2 × maximum frequency component of Y(ω)

= 2 (B1 + B2) rad/sec

Concept:

Duality Property

If $\rm x(t)\overset{F.T.}{\leftrightarrow}x(\omega)$

then $\rm x(t)\overset{F.T.}{\leftrightarrow}2\pi x(-\omega)$

Fourier time of signal x(t) is defined as,

$x(\omega)=\displaystyle\int^{\infty}_{-\infty}x(t).e^{-j\omega t}dt$

putting t = 0;

$x(0)=\displaystyle\int^{\infty}_{-\infty}x(t) dt$

Explanation:

We know that \(\mathop \smallint \limits_{ - \infty }^\infty {\rm{x}}\left( {\rm{t}} \right) \cdot {\rm{dt}} = {\rm{X}}\left( 0 \right)\)

Also we know

Where $X\left( 0 \right) = 1$

\(\therefore 2\mathop \smallint \limits_{ - \infty }^\infty \left( {\frac{{{\rm{sin}}2{\rm{\pi t}}}}{{{\rm{\pi t}}}}} \right) \cdot {\rm{dt}} = 2{\rm{X}}\left( 0 \right) = 2\left( 1 \right) = 2\)

​

Given \({\rm{P}} = \left[ {\begin{array}{*{20}{c}} 3&1\ 1&3 \end{array}} \right]\)

\(\left[ {\begin{array}{{20}{c}} {\rm{a}}\ {\rm{b}} \end{array}} \right] = {\rm{P}}\left[ {\begin{array}{{20}{c}} {\rm{x}}\ {\rm{y}} \end{array}} \right] = \left[ {\begin{array}{{20}{c}} 3&1\ 1&3 \end{array}} \right]\left[ {\begin{array}{{20}{c}} {\rm{x}}\ {\rm{y}} \end{array}} \right]\)

∴ a = 3x + y and b = x + 3y

Given ${a^2} + {b^2} = 1$

$\begin{array}{l} \therefore ;{\left( {3x + y} \right)^2} + {\left( {x + 3y} \right)^2} = 1\ 10{x^2} + 10{y^2} + 12xy-1 = 0 \end{array}$

${x^2} + {y^2} + 1.2xy-0.1 = 0$

$\Rightarrow \frac{{{{\left( {x + y} \right)}^2}}}{{1/8}} + \frac{{{{\left( {x - y} \right)}^2}}}{{1/2}} = 1$

It represents an ellipse a < b

Major axis is x + y = 0 and minor axis is x – y = 0

∴ So the given equation is the tilted version of ellipse with angle $45^\circ$ with minor axis along \(\left( {\begin{array}{*{20}{c}} 1\ 1 \end{array}} \right)\)

Concept:

A probability density function is a function of continuous random variable, whose integral across an interval gives the probability that the value of the variable lies within the same interval.

For a given probability density function f(x) to be valid, the integration of it over the complete range must equal 1.

Mathematically, it must satisfy:

\(\mathop \smallint \limits_{ - \infty }^\infty f\left( x \right) = 1\)

Calculation:

Given ${{\rm{f}}_{\rm{X}}}\left( {\rm{x}} \right) = \frac{3}{2}{{\rm{e}}^{ - 3{\rm{x}}}}{\rm{u}}\left( {\rm{x}} \right) + {\rm{a}} \cdot {{\rm{e}}^{4{\rm{x}}}}{\rm{u}}\left( { - {\rm{x}}} \right)$ 

We know \(\mathop \smallint \limits_{ - \infty }^\infty {{\rm{f}}_{\rm{x}}}\left( {\rm{x}} \right) \cdot {\rm{dx}} = 1\) 

\(\begin{array}{l} \mathop \smallint \limits_{ - \infty }^0 {\rm{a}} \cdot {{\rm{e}}^{4{\rm{x}}}} + \mathop \smallint \limits_0^\infty \frac{3}{2} \cdot {{\rm{e}}^{ - 3{\rm{x}}}} = 1\ {\rm{a}} \cdot \left[ {\frac{{{{\rm{e}}^{4{\rm{x}}}}}}{4}} \right]{ - \infty }^0 + \frac{3}{2} \cdot \left( {\frac{{{{\rm{e}}^{ - 3{\rm{x}}}}}}{{ - 3}}} \right)0^\infty = 1\ \frac{{\rm{a}}}{4}\left[ {1 - 0} \right] - \frac{1}{2}\left[ {0 - 1} \right] = 1\ \frac{{\rm{a}}}{4} = \frac{1}{2} \Rightarrow {\rm{a}} = 2\ {\rm{P}}\left( {{\rm{X}} \le 0} \right) = \mathop \smallint \limits{ - \infty }^0 {{\rm{f}}{\rm{X}}}\left( {\rm{x}} \right) \cdot {\rm{dx}}\ = \mathop \smallint \limits_{ - \infty }^0 2 \cdot {{\rm{e}}^{4{\rm{x}}}} \cdot {\rm{dx}} = 2 \cdot \left[ {\frac{{{{\rm{e}}^{4{\rm{x}}}}}}{4}} \right]_\infty ^0 = \frac{1}{2}\left[ {1 - 0} \right] = \frac{1}{2} \end{array}\)

Given ${\rm{G}}\left( {\rm{s}} \right) = \frac{{{\rm{K}}\left( {{\rm{s}} + 1} \right)}}{{{\rm{s}}\left( {1 + {\rm{sT}}} \right)\left( {1 + 2{\rm{s}}} \right)}}$

Characteristic equation 1 + G(s) H(s) = 0

As it is a unity feedback control system H(s) = 1

$1+G\left( s \right)H\left( s \right) = 1+\frac{{K\left( {s + 1} \right)}}{{s\left( {1 + Ts} \right)\left( {1 + 2s} \right)}}=0$

⇒ s + 2s² + s²T + 2s³T + Ks + K = 0

⇒ (2T)s³ + (2 + T)s² + (1 + K)s + K = 0

From R-H stability criteria 

\(\begin{array}{{20}{c}} {{s^3}}\ {{s^2}}\ {{s^1}}\ {{s^0}} \end{array}\left| {\begin{array}{{20}{c}} 2T&2+T\ 1+K&{K}\ \frac{{(2+T) \left( {1 + K} \right) - K(2T)}}{1+K }&0\ K&{} \end{array}} \right.\)

For closed-loop system to be stable 

2T > 0, 1 + K > 0, $\frac{{(2+T) \left( {1 + K} \right) - K(2T)}}{1+K }>0$, K > 0

$0 < K < \frac{{T + 2}}{{T - 2}};$

Alternate Method

Consider ${\rm{a}}{{\rm{s}}^3}{\rm{;}} + {\rm{;b}}{{\rm{s}}^2}{\rm{;}} + {\rm{;cs;}} + {\rm{;d;}} = {\rm{;}}0$ from the R – H criteria the system is stable when

$\begin{array}{l} {\rm{bc;}} > {\rm{;ad;;and;a}},{\rm{;b}},{\rm{;c}},{\rm{;d;}} > {\rm{;}}0\ \therefore {\rm{;}}\left( {2{\rm{;}} + {\rm{;T}}} \right){\rm{;}}\left( {1{\rm{;}} + {\rm{;K}}} \right){\rm{;}} > {\rm{;}}2{\rm{TK}}\ 2{\rm{;}} + {\rm{;}}2{\rm{K;}} + {\rm{;T;}} + {\rm{;TK;}}-{\rm{;}}2{\rm{TK;}} > {\rm{;}}0\ \left( {2{\rm{;}} + {\rm{;T}}} \right){\rm{;}} + {\rm{;K}}\left( {2{\rm{;}} - {\rm{;T}}} \right){\rm{;}} > {\rm{;}}0\ {\rm{K}}\left( {2{\rm{;}} - {\rm{;T}}} \right){\rm{;}} > {\rm{;}} - \left( {2{\rm{;}} + {\rm{;T}}} \right)\ {\rm{K}} > \frac{{ - \left( {2 + {\rm{T}}} \right)}}{{\left( {2 - {\rm{T}}} \right)}} ⇒ {\rm{K}} < \frac{{\left( {{\rm{T}} + 2} \right)}}{{\left( {{\rm{T}} - 2} \right)}} \end{array}$

∴ Condition for stability: $0 < {\rm{K}} < \left( {\frac{{{\rm{T}} + 2}}{{{\rm{T}} - 2}}} \right)$

Given ${\rm{\dot x}} = {\rm{Ax}}$

$\begin{array}{l} {\rm{sX}}\left( {\rm{s}} \right){\rm{;}}-{\rm{;x}}\left( 0 \right){\rm{;}} = {\rm{;A;X}}\left( {\rm{s}} \right)\ {\rm{X}}\left( {\rm{s}} \right){\rm{;}}\left[ {{\rm{sI;}} - {\rm{;A}}} \right]{\rm{;\alpha ;}} \Rightarrow {\rm{;X}}\left( {\rm{s}} \right){\rm{;}} = {\rm{;}}{\left( {{\rm{sI;}} - {\rm{;A}}} \right)^{ - 1}}{\rm{;}} \cdot {\rm{;\alpha ;}}\ {\rm{x}}\left( {\rm{t}} \right){\rm{;}} = {\rm{;}}{{\rm{e}}^{{\rm{At}}}}{\rm{;\alpha ;}} \end{array}$

The Eigen value corresponding to $\alpha$ is ${\lambda _1}$

$\therefore {\rm{x}}\left( {\rm{t}} \right) = {{\rm{e}}^{{{\rm{\lambda }}_1}{\rm{t}}}} \cdot {\rm{\alpha }}$

Standard 2^nd order system is \({\rm{T}}\left( {\rm{s}} \right) = \frac{{{\rm{K\omega }}{\rm{n}}^2}}{{{{\rm{s}}^2} + 2{\rm{\xi }}{{\rm{\omega }}{\rm{n}}}{\rm{s}} + {\rm{\omega }}_{\rm{n}}^2}}\) 

Given ${\rm{\xi ;}} = {\rm{;}}0.5{\rm{;and;}}{{\rm{\omega }}_{\rm{n}}}{\rm{;}} = {\rm{;}}10$

Steady state value ${\left. {{\rm{T}}\left( {\rm{s}} \right)} \right|_{{\rm{s}} = 0}} = {\rm{K}} = 1.02$

$\therefore {\rm{T}}\left( {\rm{s}} \right) = \frac{{\left( {1.02} \right)\left( {100} \right)}}{{{{\rm{s}}^2} + 10{\rm{s}} + 100}} = \frac{{102}}{{{{\rm{s}}^2} + 10{\rm{s}} + 100}}$

∇ - side phase voltage $= 110;kV$

Y – side phase voltage $= \frac{{11}}{{\sqrt 3 }}kV$

$\frac{{\nabla - side;phase;current}}{{Y - side;phase;current}} = \frac{{11}}{{\sqrt 3 \left( {110} \right)}} = \frac{1}{{10\sqrt 3 }}$