GATE EE 2016 Official Paper: Shift 1 (Previous Year Paper)

The sentence is talking about the Municipal Commissioner’s previous profession. There cannot be only one security guard at the University. Thus, the indefinite article should precede the security guard. Since the location where he used to work is specific, definite article ‘the’ needs to precede it. Hence the correct option satisfying this criteria is option 2.

The sentence is speaking about the problem that the Indian cricket team will face in Australia. Thus, the best phrase to replace is ‘put up with’.

The correct answer is praise

Key Points

• The words ‘deride’, ‘mock’, ‘jeer’ are synonyms that are negative words meaning to ridicule or make fun of.
• The other word ‘praise’ is a positive word that means to ‘express admiration’. Hence it is the odd one out.

Option D is generating different pattern and hence it is the odd man out

$\begin{array}{l} \frac{{{\alpha ^n} + {\beta ^n}}}{{{\alpha ^{ - n}} + {\beta ^{ - n}}}} = \frac{{{\alpha ^n} + {\beta ^n}}}{{\frac{1}{{{\alpha ^n}}} + \frac{1}{{{\beta ^n}}}}}\ = \frac{{\left( {{\alpha ^n}{\beta ^n}} \right)\left( {{\alpha ^n} + {\beta ^n}} \right)}}{{{\alpha ^n} + {\beta ^n}}}\ = {\left( {\alpha \beta } \right)^n} = {4^n} \end{array}$

No of members only on Facebook $= x$

No of members only on Whatsapp $= z$

No. of members on both $= y$

No of members on neither $= w$

$\begin{array}{l} x + y + z + w = 150\ x + y = 55\ y + z = 85,;w = 30\ x + y + z = 120\ x = 120-85;\ x = 35 \end{array}$

The passage states how computers have taken over the world today. The last line of the passage helps us infer that though the author understands how the new developments like computers have helped us, he is dubious as to whether these developments are actually beneficial or required.

The first inference does not hold.

The second inference does not hold because the passage clearly states that many people feel that the internet is an unintended invention.

Let us look at the inferences one by one.

Since all hill stations have lakes and Ooty has two lakes, then Ooty must be a hill station.

The second inference does not follow because there can be no fixed rule as to how many lakes must a hill station have.

Thus none of the inferences follow.

Data:

number of rows = n = 2

number of columns = m = 4

Formula:

$number ;of ;rectangles =( \frac{n×(n+1)}{2})×(\frac{m×(m+1)}{2} )$

Calculation:

$number ;of ;rectangles =( \frac{2×(2+1)}{2})×(\frac{4×(4+1)}{2} )$

number of rectangles = 3 × 10 = 30

Putting the values shown is graph, option (C) matches.

Concept:

Bilateral Laplace transform:

\(L\left[ {x\left( t \right)} \right] = x\left( s \right) = ;\mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){e^{ - st}}dt\)

Unilateral Laplace transform:

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

Some important Laplace transforms:

Frequency shifting property:

If X(s) is the Laplace transform of x(t), then

${e^{at}}X\left( s \right) \leftrightarrow X\left( {s - a} \right)$

Calculation:

f(t) = e^2t sin (5t) u(t)

$L\left( {\sin \left( {5t} \right)u\left( t \right)} \right) = \frac{5}{{{s^2} + 25}}$

By using frequency shifting property,

 The Laplace transform of f(t) is

$F\left( s \right) = \frac{5}{{{{\left( {s - 2} \right)}^2} + 25}} = \frac{5}{{{s^2} - 4s + 29}}$

$\frac{{{d^2}y}}{{d{t^2}}} + \frac{{2dy}}{{dt}} + y = 0$

By applying the Laplace transform,

${s^2}Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right) + 2sY\left( s \right) - 2y\left( 0 \right) + Y\left( s \right) = 0$

$\Rightarrow Y\left( s \right) = \frac{{s + 2 + y'\left( 0 \right)}}{{{{\left( {s + 1} \right)}^2}}}$

$= \frac{1}{{s + 1}} + \frac{{1 + y'\left( 0 \right)}}{{{{\left( {s + 1} \right)}^2}}}$

By applying the inverse Laplace transform

\(\begin{array}{*{20}{c}} {y\left( t \right) = {e^{ - t}} + \left( {1 + y'\left( 0 \right)} \right)t{e^{ - t}}} \end{array}\)

y(1) = 3 e^-1

⇒ y(1) = e^-1 + (1 + y’(0)) e^-1 = 3 e^-1

⇒ y’(0) = 1

Now the equation of y(t) becomes

y(t) = (1 + 2t) e^-t

At t = 2,

y(2) = 5 e^-2

Concept:

Cauchy’s Theorem:

If f(z) is an analytic function and f’(z) is continuous at each point within and on a closed curve C, then

$\mathop \oint \limits_C f\left( z \right)dz = 0$

Cauchy’s Integral Formula:

If f(z) is an analytic function within a closed curve and if a is any point within C, then

$f\left( a \right) = \frac{1}{{2\pi i}}\mathop \oint \limits_C \frac{{f\left( z \right)}}{{z - a}}dz$

${f^n}\left( a \right) = \frac{{n!}}{{2\pi i}}\mathop \oint \limits_C \frac{{f\left( z \right)}}{{{{\left( {z - a} \right)}^{n + 1}}}}dz$

Calculation:

$f\left( z \right) = \oint \frac{{2z + 5}}{{\left( {z - \frac{1}{2}} \right)\left( {{z^2} - 4z + 5} \right)}}dz$

the poles are \(z; = ;\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/ \kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} ,2; \pm ;i\)

Since the contour is $\left| z \right| = 1$, hence we will consider only $z = \frac{1}{2}$

The value of integral will be

$2\pi i{\left[ {\left( {z - \frac{1}{2}} \right)\left( {\frac{{2z + 5}}{{\left( {z - \frac{1}{2}} \right)\left( {{z^2} - 4z + 5} \right)}}} \right)} \right]_{z = \frac{1}{2}}}$

$= 2\pi i\left( {\frac{6}{{\frac{1}{4} - 2 + 5}}} \right)$

$= \frac{{48\pi i}}{{13}}$

Concept:

For an LTI system, the sinusoidal input produces a sinusoidal output of the same frequency but different amplitude and phase. The amplitude and phase depend upon the transfer function of the system.

For example:

Let, r(t) = cos(ωt) then

y(t) = a cos(ωt + ϕ)

where a = |H(jω)|, ϕ = ∠ H(jω)

$H(jω)=\frac{Y(jω)}{X(jω)}$

Calculation:

$H(jω)=\frac{jω}{jω+2}$

r(t) = cos (2t)

ω = 2

$H(jω)=\frac{j2}{j2+2}$

$|H(jω)|=\frac{{1}}{{\sqrt 2}}$

∠ H(jω) = 90° - 45°

= +45° 

$y(t)=\frac{1}{\sqrt 2}~cos (2t + 45^{\circ})$

Gain margin (GM): The gain margin of the system defines by how much the system gain can be increased so that the system moves on the edge of stability.

It is determined from the gain at the phase cross over frequency.

$GM = \frac{1}{{{{\left| {G\left( {j\omega } \right)H\left( {j\omega } \right)} \right|}_{\omega = {\omega _{pc}}}}}}$

Phase crossover frequency (ωpc): It is the frequency at which phase angle of G(s) H(s) is -180°.

$\angle G\left( {j\omega } \right)H\left( {j\omega } \right){|_{\omega = {\omega _{pc}}}} = - 180^\circ$

Phase margin (PM): The phase margin of the system defines by how much the phase of the system can increase to make the system unstable.

$PM = 180^\circ + \angle G\left( {j\omega } \right)H\left( {j\omega } \right){|_{\omega = {\omega _{gc}}}} = - 180^\circ$

It is determined from the phase at the gain cross over frequency.

Gain crossover frequency (ωgc): It is the frequency at which the magnitude of G(s) H(s) is unity.

${\left| {G\left( {j\omega } \right)H\left( {j\omega } \right)} \right|_{\omega = {\omega _{gc}}}} = 1$

Calculation:

$\begin{array}{l} G\left( s \right) = \frac{{100}}{{{{\left( {s + 1} \right)}^3}}} = \frac{{100}}{{{s^3} + 1 + 3s\left( {1 + s} \right)}}\ G\left( {j\omega } \right) = \frac{{100\left( {\left( {1 - 3{\omega ^2}} \right) - j\left( {3\omega - {\omega ^3}} \right)} \right)}}{{{{\left( {1 - 3{\omega ^2}} \right)}^2} + {{\left( {3\omega - {\omega ^3}} \right)}^2}}} \end{array}$

at phase cross over frequency, phase = -180°

Hence, the imaginary part should be zero

∴ ${\omega ^2}; = ;3$

$\omega = \sqrt 3$

Concept:

Linear system

The system is said to be linear if it follows Homogenous and Superposition property.

1. The system is homogenous if 

y(t) = α₁x(t) 

y(t) = α₂x(t)

2. The system follows Superposition if

y(t) = α1x(t) + α2x(t)

Time invariant system

if y(t - t₀) = x(t - t0)

and y(t') = x(t - t₀)  

Analysis:

For the given signal check linearity property

y(t) = x(t)cos(t)

1. check homogeneity

y(t) = α1x(t)cos(t)

y(t) = α2x(t)cos(t)

2. check superposition 

y(t) = α1x(t)cos(t) + α2x(t)cos(t)

The system follows both homogeneity and superposition thus the system is Linear.

check the time invariance

y(t - t0) = x(t - t0)cos(t - t0)

and y(t') = x(t - t0)cos(t)

since y(t - t0) ≠ y(t') 

System is not time invariant.

Hence option (3) is correct

Concept:

Shifting property of impulse function

$\rm\displaystyle\int_{-\infty}^{\infty} x(t) \delta (t -a ) dt = x(a)$

Scaling property of impulse function

$\delta (at) = \frac{1}{|a|)} \delta (t)$

Explanation:

Let:

$\rm I =\displaystyle\int_{-\infty}^{\infty}e^{-t} \delta(2t - 2) dt$

$I = \rm\displaystyle\int_{-\infty}^{\infty}e^{-t} \delta[2(t - 1)] dt$

Using scaling property of impulse function in the above equation, we'll get:

$I =\rm\displaystyle\int_{-\infty}^{\infty}e^{-t} \frac{1}{|2|}\delta(t - 1) dt$

Applying Shifting property of impulse function to the above equation, we'll get:

$I =\frac{1}{2}\rm\displaystyle\int_{-\infty}^{\infty}e^{-t} \delta(t - 1) dt$

$I = \frac{1}{2}. \left. e^{-t} \right|_{t = 1}$

$\frac{1}{{2e}}$

Concept:

• Analog-to-Digital Converters (ADCs) transform an analog voltage to a binary number (a series of 1’s and 0’s).
• Then eventually to a digital number (base 10) for reading on a meter, monitor, or chart.
• The ADC resolution depends upon the number of bits used to represent the digit number.
• As the number of bits increases the resolution of an Analog to Digital Converter improves and the quantization error decreases.

Resolution for n – bit A/D converter  will be:

$R= \frac{V_{FS} \ \times \ (2^ib^i)}{{{2^n} - 1}}$

Where 

R = Resolution

VFS is reference voltage 'or' Full-scale voltage

n = number of bits

2ibi gives output voltage value.

Calculation:

Given that,

Resolution = 0.1° C

Final temp = 55° C

Initial temp  = 40° C

calculating the range of temperature

Range of temperature = (Final temp - Initial temp)

= 55° - (-40 °) = 95° C 

No of points required = Range of temp/ resolution

= 95/0.1 = 950

The number of bits (n) for ADC is given by

2ⁿ > No. of points required

n > log₂(No. of points required)

n > 9.891

n = 10

Hence option (2) is correct

Concept:

For a 2 × 1 MUX is shown above, the output function F is expressed as:

F = S̅1 I0 + S1I1

i.e. when S1 = 0, I0 is transmitted to the output.

And when S1 = 1, I1 is transmitted to the output.

Calculation:

Given multiplexer

$F = \overline {A;} \bar B + AB;$

$= \overline {A \oplus B}$

$\nabla \times \vec E = \frac{{ - \partial B}}{{\partial t}}$

A uniformly charged ring is specified in the question, it can be considered as static. A static electric charge produces an electric field.

For absence of rotating field or magnetic field

$\nabla \times \vec E = 0$

For  RA  

${R_A} = \frac{{{R_A}}}{{1 + {R_A}}} + 2$

(RA)² + RA = 2 + 3RA

(RA)² - 2RA - 2  = 0 

On solving we'll get:

R_A = 1 ± √3 ----(1)

${R_B} = \frac{{{R_B} + 2}}{{{R_B} + 3}}$

(RB)2 + 2RB - 2  = 0 

On solving we'll get:

R_B = -1 ± √3   ---(2)  

From the given equation

RA ≠ RB & RA , RB ≠ 0

also RA > RB 

Hence options (D) is correct

Concept:

• From the below diagram we can observe that the maximum torque of an induction motor is independent of rotor resistance
• Slip at which maximum torque occurs depends on rotor resistance and they change on adding the additional resistance to the rotor circuit.

The maximum torque of an induction motor is given by

${T_{max}} = \frac{{kE_{20}^2}}{{2{X_{20}}}}$

∴ Maximum torque is directly proportional to supply voltage & maximum torque is inversely proportional to rotor reactance.

Hence, the maximum torque is dependent on the supply voltage & reactance of the rotor and is independent of the rotor resistance.

Sm the value of slip corresponding to the maximum torque is

${S_m} = \frac{{{R_2}}}{{{X_{20}}}}$

Explanation:

For any voltage and frequency slip corresponding to maximum torque in three phase induction motor is,

${S_m} = \frac{{{R_2}}}{{{X_{20}}}}$

where, X ∝ f

 S_m  ∝ (1/N_s)                         [ NS ∝ f ]

Thus the slip has inverse relation with the synchronous speed,

For PV buses, voltage magnitude was known quantity, if changed to PQ type the voltage magnitude becomes unknown. Also the angle was known quantity before and now it is unknown.

Explanation:

A bus in a power system is a line at which the several components of the power system like generators, loads, and feeders, etc., are connected.

The buses in a power system are associated with four quantities, these quantities are the following:

• The magnitude of the voltage
• Phase angle
• Active power
• Reactive power
  

In the load flow studies, two variable are known, and the other two is to determined.

Depends on the quantity to be specified the buses are classified into three categories as follow:

The table shown below shows the types of buses and the associated known and unknown value.

Generation Bus  or Voltage Control Bus:

• This bus is also called the P-V bus.
• on this bus, the voltage magnitude corresponding to generate voltage and true or active power P corresponding to its rating are specified.
• Voltage magnitude is maintained constant at a specified value by injection of reactive power.
• The reactive power generation Q and phase angle δ of the voltage is to be computed.
  

Load Bus:

• This is also called the P-Q bus
• at this bus, the active and reactive power is injected into the network.
• The magnitude and phase angle of the voltage is to be computed.
• Here the active power P and reactive power Q are specified, and the load bus voltage can be permitted within a tolerable value, i.e., 5 %.
• The phase angle of the voltage, i.e.δ is not very important for the load.
  

Slack, Swing, or Reference Bus:

• Slack bus in a power system absorbs or emits active or reactive power from the power system.
• The slack bus does not carry any load.
• At this bus, the magnitude and phase angle of the voltage is specified.
• The phase angle of the voltage is usually set equal to zero.

• Generally, synchronous machines are designed with over-excitation to improve stability.
• Synchronous motors are designed with leading power factors and synchronous generators are designed with lagging power factors to improve stability.
• An alternator or synchronous generator operating at a lagging power factor means it's overexcited and can supply lagging reactive power (Q) and real power (P). Hence synchronous generator operating at lagging pf is the source of reactive power factor.
• Over excited synchronous machine works at leading power factor and can supply reactive power, hence over-excited synchronous motor is source of reactive power
• The induction generator needs reactive power to function in the absence of rotor dc excitation. Hence it is sink of reactive power.

Concept:

For lap wound, the number of parallel path (A) is equal to the number of poles (P)

Explanation:

Armature current in the each parallel path = I/A ${I_A} = \frac{I}{A}$ 

Where,

A= No. of parallel paths

In this case A=4 for lap connected winding

So, current in armature conductor = 10 A​

${N_S} = \frac{{120f}}{P} = 600 rpm$

 f = 20 H. 

so, T = 50 msec with straight line commutation.

When dc current passes through simulator and goes inside armature, its shape will be that of shown is option (C)

Concept: 

Transformation ratio of transformer is given by K = V2/V1 = E2/E1 = N2/N1.

Where N1 is the number of primary turns

V1 is the primary voltage

N2 is the number of secondary turns

V2 is the secondary voltage

Step-up transformer:

A transformer that increases the voltage from primary to secondary (more secondary winding turns than primary winding turns) is called a step-up transformer.

Therefore, K > 1

Step down transformer:

A transformer that decreases the voltage from primary to secondary (more primary winding turns than secondary winding turns) is called a step-down transformer.

Therefore, K < 1

Explanation:

Transformation ratio (k) = $\frac{N_2}{{{N_1}}} = \frac{1}{{{n}}}$

When Port 2 inductance L is referred to Port 1

Then, inductance will be $= \frac{L}{{{k^2}}} = \frac{L}{{\frac{1}{{{n^2}}}}} = {n^2}L$

Concept:

If λ is an eigenvalue of a matrix A and k  is a scalar then:

1) λm is the eigenvalue of matrix Am (m belongs to N).

2) kλ  is an eigenvalue of matrix kA.

3) λ + k is an eigenvalue of the matrix A + kI.

4) λ - k  is an eigenvalue of matrix A - kI.

Calculation:

For a given matrix, if the eigenvalues are λ₁ and λ₂, then the eigenvalues of Aⁿ will be $λ_1^n ; and ; λ_2 ^n$.

1 and -2 will become 1 and -8;

Although the eigenvectors remain the same.

If $A$ is real

rank of (${A^T};A$) = rank of $A$ = rank of ${A^T}$ = rank of $A{A^T}$

Here starting slope is +20 db/dec

Hence s¹ is at numerator part of T.F.

Now from the straight line equation y = mx + c, we get

0 = 20 log (0.5) + c

∴ c = 6.0205

Now c = 20 log K

6.0205 = 20 log K

K = 2

We can find the value of ω₁ as

12 = 20 log (ω₁) + 6.0205

ω₁ = 1.99 ≈ 2

Now we have to find ω_2,  so that

0 = - 40 log (8) + c

c = 36.1236

∴ 12 = - 40 log (ω₂) + 36.1236

ω₂ = 4

Hence transfer function for given bode plot is

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Concept:

Principle arguments

• It states that if there are “P” poles and “Z” zeroes for a closed, random selected path then the corresponding G(s)H(s) plane encircles the origin with P – Z times.
• Encirclements in s – plane and GH – plane are shown below.

In GH plane Anti clockwise encirclements are taken as positive and clockwise encirclements are taken as negative.

It is applied to the total RH plane by selecting a closed path with r = ∞

Calculation:

Given transfer function,

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

Substitute s = jω in the above equation,

$G\left( j\omega \right)H\left(j\omega \right) = \frac{{j\omega + 3}}{{{(j\omega )^2}\left( {j\omega - 3} \right)}}$

$\angle G\left( {j\omega } \right)H\left( {j\omega } \right) = {\tan ^{ - 1}}\frac{\omega }{3} - 180 - (180 - {\tan ^{ - 1}}\frac{\omega }{4})$

$\angle G\left( {j\omega } \right)H\left( {j\omega } \right) = {\tan ^{ - 1}}\frac{\omega }{3} + {\tan ^{ - 1}}\frac{\omega }{4}$

​$\left| {G\left( {j\omega } \right)H\left( {j\omega } \right)} \right| = \frac{{\sqrt {{\omega ^2} + 9} }}{{{\omega ^2}\sqrt {{\omega ^2} + 9} }}=\frac{1}{\omega ^2}$

In the plot G(s)H(s) encircle -1 + j0 once in clockwise direction.

Given the open-loop transfer function,

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

$CE =1+ \frac{{s + 3}}{{{s^2}\left( {s - 3} \right)}};$

⇒ s3 - 3s2 + s + 3 = 0

By routh hurwitz criterion 

\(\left. {\begin{array}{{20}{c}} {{s^3}}\ {{s^2}}\ {{s^1}}\ {{s^0}} \end{array}} \right|\begin{array}{{20}{c}} 1&{-3}&{}&{}\ 1&3&{}&{}\ {-6}&0&{}&{}\ 3&{}&{}&{} \end{array} \)

There are two sign changes 

Therefore it is unstable with two right half of s-plane poles

Z = 2, P = 1

N = P - Z = 1 - 2 = -1

Once in clockwise direction

${X_1}\left( {j\omega } \right)$ and ${X_2}\left( {j\omega } \right)$ are not conjugate symmetric, hence ${x_1}\left( t \right),;{x_2}\left( t \right)$ are not real

Now, Fourier transform of ${x_1}\left( t \right).;{x_2}\left( t \right)$ will be \(\frac{1}{{2x}}{X_1}\left( {j\omega } \right){X_2}\left( {j\omega } \right)\) and by looking at ${X_1}\left( {j\omega } \right)$ and ${X_2}\left( {j\omega } \right)$ We can say that \({X_1}\left( {j\omega } \right){X_2}\left( {j\omega } \right)\) will be conjugate symmetric and thus ${x_1}\left( t \right).{x_2}\left( t \right)$ will be real.

Concept:

A signal is said to be Eigen function if the output is scalar multiple of input signal and that scalar is referred as Eigen value.

Let input signal is x(t) and it is an Eigen function then

output = K x(t) where K is Eigen value.

Explanation:

The impulse response is real and even, hence $H\left( {j\omega } \right)$ will also be real and even.

Since $H\left( {j\omega } \right)$ is real and even, $H\left( {j{\omega _0}} \right); = ;H\left( { - j{\omega _0}} \right)$

Now $cos\left( t \right)$ is input i.e. $\frac{{{e^{jt}} + {e^{ - jt}}}}{2}$  is input

Output $\frac{{H\left( {j1} \right){e^{jt}} + H\left( { - j1} \right){e^{ - jt}}}}{2} = H\left( {j1} \right)\left[ {\frac{{{e^{jt}} + {e^{ - jt}}}}{2}} \right] = H\left( {j1} \right){\rm{cos}}\left( t \right)$

If $sin\left( t \right)$ is input i.e. $;\frac{{{e^{jt}} - {e^{ - jt}}}}{2j}$ is input

Output will be $\frac{{H\left( {j1} \right){e^{jt}} - H\left( { - j1} \right){e^{ - jt}}}}{2j} = H\left( {j1} \right)\left[ {\frac{{{e^{jt}} - {e^{ - jt}}}}{2j}} \right] = H\left( {j1} \right)\sin \left( t \right)$

So $sin\left( t \right)$ and $cos\left( t \right)$ are eigen signals with same eigen values.

In the given circuit