GATE EE 2014 Official Paper: Shift 3 (Previous Year Paper)

The correct answer is option 4 i.e. Mr. X fell down and was losing consciousness

Let's look at the following points-

• The tense used in the first clause is the present progressive(continuous) tense.
• Whereas the second clause contains simple past tense i.e. "Mr X fell down" and in **past continuous tense i.e. '**Mr. X fell down and was losing consciousness'.

Now, look at the following points-

• When the simple past tense and the past continuous tense are used together in a sentence, it shows that the action referred to by the simple past tense takes place somewhere in the middle of the past continuous action.
• These tenses are often used to show an action interrupting another action.
• For e.g.- I was working when someone knocked at the door.

Explanation

Losing consciousness is an instantaneous action which cannot go on for a period of time. There is a definite 'point' where the person falls unconscious. Hence, we may conclude that in this sentence, it was an effect of Mr. X's falling down and would take the simple past tense. 

​Correct Sentence- While trying to collect an envelope from under the table, Mr. X fell down and lost consciousness.

The correct answer is option 3 i.e. had known, could have

The given sentence is a conditional sentence in the third conditional form. 

Let's look at the following points-

• The given sentence is a conditional sentence, conditional sentences show a possible cause and effect situation
• A conditional sentence has two clauses that depend on each other to make sense, Conditional clause (often referred to as if-clause) is a dependent clause and the main clause is the independent clause
• The tense of these clauses determines the type of conditional sentence

 

Now look at the following point-

• To understand the types of Conditional sentence look at the following description -

 

 

Explanation

• The given sentence is conditional consisting of 'if clause' with a blank and 'Main clause' with a blank preceded by subject and followed by the third form of Verb,
• As the subject and the third form of the verb are given in the 'Main clause' we can easily conclude that the Tense of the 'If clause' will be in 'Past Perfect'
• The third type of conditional sentence has the form'-

• Hence, as per the rule the correct option to be filled in the blank is had known, could have

​

​**Correct sentence-**If she had known how to calibrate the instrument, she could have done the experiment.

The correct answer is option 3, i.e.Rambling

 

Explanation 

Let’s look at the meanings of the given word and the option:

• Coherent: logical and consistent (of an argument, theory, or policy); rational, orderly, clear, methodical
• Rambling:  lengthy and confused or inconsequential (of writing or speech); roundabout, confused, disjointed, incoherent

​​Here we find that both Coherent and Rambling express the opposite meaning

Therefore both the words are Antonyms, and thus Option 3 Rambling is the correct answer

Let's look at the meaning of the other words:

• Sticky: Made of or covered with a substance that stays attached to any surface it touches
• Well-connected: Acquainted with or related to people with prestige or influence
• Friendly: Behaving in a pleasant, kind way towards someone

Series: 2, 5, 10, 17, 26, 37, 50, 64

5 - 2 = 3

10 - 5 = 5

17- 10 = 7

26 - 17 = 9

37 - 26 = 11

50 - 37 = 13

64 - 50 = 14

Sinc the series created by the difference of the consecutive terms is 3, 5, 7, 9, 11, 13 

The last difference is 14 which doesn't fit in the series ∴

Therefore 64 does not belong in the series.

Correct series: 2, 5, 10, 17, 26, 37, 50, 65

For  Q1:

Number of students Answered Correctly  = 21

Number of students Answered Wrongly  = 17

Number of students Answered Wrongly  = 6

∴ number of students = 21 + 17 + 6 = 44

Similary, for Q2 and Q3 sum will be 45

For 1^st question total marks obtained = 2 × 21 = 42

For 2^nd question total marks obtained = 3 × 15 = 45

For 3^rd question total marks obtained = 2 × 23 = 46

Hence total marks obtained by the class are = 42+45+46 = 133

Average marks obtained = $\frac{Total~marks}{ Total ~ students} = \frac{133}{44}$

∴ Average marks obtained = 3.02.

The correct answer is Option 2.

Explanation

Let's look at the key points from the paragraph.

• The dance programme is scheduled for 10:00 am
• Participating students need to come an hour early i.e. at 9:00 am with a parent.
• Other students and parents should come on time i.e. 10:00 am for the programme.

Let's look at the options given.

• Option 1  is wrong because parents and students who are not participating should come at 10:00 am. Only those who are participating (student along with a parent) should come at 9:00.
• Option 2 is correct as it agrees totally with the paragraph and what is mentioned.
• Option 3 is wrong because participating students along with a parent should come at 9:00 am.
• Option 4 is wrong because participating students along with a parent should come at 9:00 am

Hence the only option that agrees with the paragraph is Option 2 and therefore the correct answer.

The correct answer is Option 4 i.e (iv) only

Explanation:

Let's look at the key points from the paragraph.

• Beginning of the 20th century saw several hypotheses being proposed
• These proposed hypotheses brought in a paradigm shift in our understanding of the universe.
• The clinging evidence regarding the shift was provided by the experimental measurements of a star that was directly behind the sun.

Let's look at the options given

• Option 1 is wrong because from the above points we find that statements (i),(ii), given in the question are wrong
• Option 2 is wrong because statement (iii) Stars are important objects in the universe is wrong.
• Option 3 is wrong because the statement (i) given in the question is wrong. Our understanding of the universe changed on the basis of the new hypotheses and not on the basis of the position of the stars.
• Option 4 is correct because the statement (iv) given in the question is correct. Experimental evidence (measurements) was important in confirming the shift.

Hence Option 4 is the correct answer.

Let the original GDP be x.

Therefore, its international value is $\frac{x}{50}$

GDP in Rupees grew at 7%

Therefore, GDP becomes $\frac{107x}{100}$

USD increased from Rs. 50/ USD to Rs. 60/ USD. 

Therefore, its international value is $\frac{\frac{107x}{100}}{60} = \frac{107x}{6000}$

Percentage Increase = $\frac{- \frac{13x}{6000}}{\frac{x}{50}} \times 100 = -\frac{13}{120} \times100 =10.83$

Since the percentage is negative ∴ it is decreased by 10.83 ≈ 11%

The number of female students in 2011 and 2012 is equal.

number of males student in 2011 = number of female students in 2011 = x 

In 2012, Male students ÷ Female students = 1.5

Therefore Male students in 2012 = 1.5 ×  Female students = 1.5 × x

In 2012 and 2011, ratio of Male students to Male students = ( 1.5 × x ) ÷  x = 1.5: 1

Calculation:

(7526)8 − (Y)8 = (4364)8

(7526)8 − (4364)8 = (Y)8

∴ (Y)8 = (8³ × 7 + 82 × 5 + 8¹ × 2 + 80 × 6 ) − (83 × 4 + 82 × 3 + 81 × 6+ 80 × 4)

∴ (Y)8 =  3926 − 2292

∴ (Y)8 =  (1634)₁₀

 

(Y)8 =  (3926)10 = (3142)₈

Concept:

The order of highest ordered non zero minor is called the rank of a matrix.

Example:

If the square matrix is of order n × n then find the determinant of it if the value is non zero then its rank is n if the value comes out to be zero then find the determinant of  (n - 1) order of all sub-matrix if the value comes out to be non zero of any sub-matrix then the rank is n if the value is zero then the same processes continues further and the order at which value of the determinant is non zero is the rank of a matrix, Use this concept only for order  3 × 3 or lower order for more than 3 × 3 order we use the concept of Echelon form matrix.

Calculation:

The determinant of A is, |A| = ps – qr

Determinant of B is, |B| = (p² + q²) (r² + s²) – (pr + qs)²

|B| = (p²s² + q²r² – 2pqrs)

|B| = (ps - qr)²

Now, it is clear that |A|² = |B|

If the rank of A, ρ(A) = 1, that means the determinant of the matrix is zero, and hence the determinant of the matrix also zero. Therefore, the rank of the matrix B is 1

If the rank of A, ρ(A) = 2, that means the determinant of the matrix is non-zero, and hence the determinant of the matrix also non-zero. Therefore, the rank of the matrix B is 2.

Therefore, ρ(B) = ρ(A) = N

 

Echelon form: 

• The no of zeros before non-zero elements in a row are less than such no of zeros in the next row.
• Zero rows ( if any ) must follow non-zero rows.
• The no of non zero's row is called the rank of the matrix when it is in Echelon form.

Concept:

By vector identity, if A̅ is differentiable vector function and f is differential scalar function of position (x, y, z) then

∇⋅(fA̅) = (∇f)⋅A̅ + f(∇⋅A)

Calculation:

Given ∇⋅(fv) = x²y + y²z + z²x,

v = yi + zj + xk

From the property of vector field

∇⋅(fv) = v⋅ (∇⋅f) – f(∇⋅ v)

⇒ v⋅(∇f) = ∇⋅(fv) + f(∇⋅v)

$\nabla \cdot {\rm{v}} = \frac{{\partial y}}{{\partial x}} + \frac{{\partial z}}{{\partial y}} + \frac{{\partial x}}{{\partial z}} = 0$

⇒ f(∇⋅v) = 0

v⋅(∇f) = ∇⋅(fv) - 0

v⋅(∇f) = x²y + y²z + z²x

Concept:

A function is odd, if the function on one side of t-axisx-axis is sign inverted with respect to the other side or graphically, symmetric about the origin.

f(t) = - f(-t)

Symmetry condition of Fourier Transform

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Calculation:

We have the wave form of function f (t) as

From the wave form, f(t) is an odd function

∴ f (t) = - f (- t)

⇒ Fourier transform of the function is imaginary and odd function of ω

Concept:

Balanced star connected system:

Let's consider phase voltages as VR, VY, and VB. and line voltages as VRY, VYB, and VBR.

And for balanced system magnitudes of all the phase voltages equal and 120° phase displacement between them.

Consider the phasor diagram of a star-connected system (positive sequence) for lagging load.

Figure: Phasor diagram of balanced star connected system (positive sequence) for lagging load.

Properties:

1. In the balanced star connection the line current and phase current are equal.
2. Line voltage(VRY) is equal to √3 times of phase voltage VR.
3. From the phasor diagram, we can observe that line voltage(VRY) leads the phase voltage(VR) by 30°.
4. The phase displacement between line voltage and line current is (30°+ ϕ) for lagging loads, 30° for resistive loads, and (30°- ϕ) for leading loads.

 

Calculation:

Line to neutral voltage is nothing but phase voltage of the connection, the relation between line voltage and phase voltage of the balanced star connection is as follows

V_L = √3 V_ph

Given, Phase voltage V_ph = 10∠15°V

VL = √3 × 10∠15°V = 10√3∠15°V

If V_A = 10∠0

Then V_BC = 10√3 ∠-90

Given V_A = 10∠15°

V_BC = 10√3 ∠-90 + 15

VBC = 10√3 ∠-75°

Therefore, The voltage of line B with respect to line C is given by 10√3 ∠-75°

Concept:

By Gauss law, $ϵ \oint \vec{E}.d\vec{s}={{Q}_{encl}}$

Thus, $ϵ \oint \vec{E}.d\vec{s}={{ϵ }_{0}}E\left( 4π {{r}^{2}} \right)$

$ϕ = \oint \vec{E}.d\vec{s}$

where ϕ = flux coming out of the concentric sphere.

r = radius of the sphere

Calculation:

Since electric potential can be defined as:

$V = \frac{Q}{{4πϵ{_0}r}}$

V = 1 V (given)

The total electric flux coming out of the concentric spherical surface = Charge enclosed Q = 4 π ϵ0 r

The s-Domain representation of the given circuit 

$Z\left( s \right) = s + \frac{{\frac{1}{s}\left[ {s + \frac{1}{s}} \right]}}{{\frac{1}{s} + \left( {s + \frac{1}{s}} \right)}}$

$Z\left( s \right) = \frac{{{s^4} + 3{s^2} + 1}}{{{s^3} + 2s}}$

Additional Information

We have, \(x\left( t \right) = \left{ {\begin{array}{*{20}{c}} 1&{\left| t \right|}&{ < 1}\ 0&{\left| t \right|}&{ > 1} \end{array}} \right.\)

Its wave form is given as

Now, the wave form of rectangular function, rect (t) is

$\therefore x\left( t \right) = rect\frac{t}{2}$

The Fourier transform pair for rectangular function is

$rect\left( {\frac{t}{\tau }} \right) \leftrightarrow \tau \sin c\left( {\frac{{\omega \tau }}{{2\pi }}} \right)$

So, $rect\left( {\frac{t}{2}} \right) \leftrightarrow 2\sin c\left( {\frac{\omega }{\pi }} \right)$

Similarly, $rect\left( {\frac{t}{4}} \right) \leftrightarrow 4\sin C\left( {\frac{{2\omega }}{\pi }} \right)$

$\begin{array}{l} = \frac{{4{\rm{sin}}\left( {2\omega } \right)}}{{2\omega }} = \frac{{2{\rm{sin}}\left( {2\omega } \right)}}{\omega }\ rect\left( t \right) \leftrightarrow \sin C\left( {\frac{\omega }{{2\pi }}} \right) = \frac{{\sin \left( {\frac{\omega }{2}} \right)}}{{\frac{\omega }{2}}} \end{array}$

Now, we have the Fourier transform

\(F\left{ {x\left( {2t} \right)} \right} = F\left{ {rect\left( t \right)} \right} = \frac{{sin\left( {\frac{\omega }{2}} \right)}}{{\frac{\omega }{2}}}\)

And \(F\left{ {x\left( {\frac{t}{2}} \right)} \right} = F\left{ {rect\left( {\frac{t}{4}} \right)} \right}\left( {2\omega } \right)\)

$= \frac{{2{\rm{sin}}\left( {2\omega } \right)}}{\omega }$

We get the Fourier transform of y (t) as

\(\begin{array}{l} F\left{ {y\left( t \right)} \right} = F\left{ {x\left( {2t} \right)*x\left( {\frac{t}{2}} \right)} \right}\ = F\left{ {x\left( {2t} \right)} \right}F\left{ {x\left( {\frac{t}{2}} \right)} \right}\ \frac{{sin\left( {\frac{\omega }{2}} \right)}}{{\frac{\omega }{2}}}\frac{{2\sin \left( {2\omega } \right)}}{\omega }\ = \frac{4}{{{\omega ^2}}}sin\left( {\frac{\omega }{2}} \right)\sin \left( {2\omega } \right) \end{array}\)

In synchronous motors, the phenomenon of successive overshoots and undershoots in the motor speed due to sudden changes in the load is called Hunting. After a sudden change of the load in three-phase synchronous machine, the rotor has to search or hunt for its new equilibrium position. It causes hunting which can be damped by rotor copper losses.

Causes of hunting

• Periodic variation of load
• Sudden changes in load
• Faults occurring in the system when supplied by the generator
• Sudden change in the field current
• Cyclic variations of the load torque

 

Effects of hunting

• It can lead to loss of synchronism
• It can cause variations of the supply voltage producing undesirable lamp flicker
• The possibility of a resonance condition increases. If the frequency of the torque component becomes equal to that of the transient oscillations of the synchronous machine, resonance may take place
• Large mechanical stresses may develop in the rotor shaft
• The machine losses increases and the temperature of the machine rises

 

Reduction of Hunting

• Use of damper windings
• Use of flywheels

A 1ϕ IM is having a capacitor and centrifugal switch in series with auxiliary winding. The switch is expected to operate at a speed of 0.7N_s

If the switch operates correctly, then torque-speed characteristic is

But, the switch does not operate due to the malfunctioning.

So there will be no discontinuity in the characteristic curve.

As the capacitor is present permanently, the motor behaves like an unbalanced 2 phase motor.

Thus, we have the modified characteristic curve of the motor as

For Distortion less Transmission line:

r = Resistance per unit length

l = Inductance per unit length

g = Shunt conductance per unit length

c = Capacitance per unit length

$\frac{r}{l} = \frac{g}{c}$

or rc = lg

Additional Information

Phase constant

$\beta = {\rm{\omega }}\sqrt {{\rm{lc}}}$

Attenuation constant

$\alpha = \sqrt {RC}$

$v = \frac{\omega }{\beta } = \frac{1}{{\sqrt {lc} }}$

Purpose of Transposition:

• Transmission lines are transposed to prevent interference with neighbouring telephone lines.
• The transposition arrangement of high voltage lines helps to reduce the system power loss.
• We have developed transposition system for Single circuit tower using same tension tower with reduced deviation angle.
• Transposition arrangement of power line helps to reduce the effect of inductive coupling.
• It is proved more economical Solution, in comparison of the conventional transposition system.

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Important:

Transposition arrangement

The transposition arrangement of the conductor can simply show in the following the figure. The conductor in Position 1, Position 2 and Position 3 changes in a specific arrangement to reduce the effect of capacitance and the electrostatic unbalanced voltages.

Z1 = Z2 = ZS - Zm

Zo = ZS + 2Zm

Where,

Z1 = positive sequence impedance

Z2 = negative sequence impedance

Zm= mutual impedance

ZS= self impedance

Positive and negative sequence impedances are equal.

Concept:

According to Mason’s gain formula, the transfer function is given by

\(TF=\frac{\mathop{\sum }{k-1}^{n}{{M}{k}}{{\Delta }_{k}}}{\Delta }\)

Where,

 n = no of forward paths

Mk = kth forward path gain

Δk = the value of Δ which is not touching the kth forward path

Δ = 1 – (sum of the loop gains) + (sum of the gain product of two non-touching loops) – (sum of the gain product of three non-touching loops)

Application:

Number of forward paths = 2

${P_1} = \frac{{{h_0}}}{{{s^2}}},;{P_2} = \frac{{{h_1}}}{s}$

Number of loops = 2

${L_1} = - \frac{{{a_1}}}{s},{L_2} = - \frac{{{a_0}}}{{{s^2}}}$

${\rm{\Delta }} = 1 + \frac{{{a_1}}}{s} + \frac{{{a_0}}}{{{s^2}}}$

${{\rm{\Delta }}_1} = 1,;{{\rm{\Delta }}_2} = 1 + \frac{{{a_1}}}{s}$

Transfer function $= \frac{{\frac{{{h_0}}}{{{s^2}}}\left( 1 \right) + \frac{{{h_1}}}{s}\left( {1 + \frac{{{a_1}}}{s}} \right)}}{{1 + \frac{{{a_1}}}{s} + \frac{{{a_0}}}{{{s^2}}}}}$

$= \frac{{{h_0} + {h_1}\left( {s + {a_1}} \right)}}{{{s^2} + {a_1}s + {a_0}}}$

$= \frac{{{b_0} - {b_1}{a_1} + {b_1}\left( {s + {a_1}} \right)}}{{{s^2} + {a_1}s + {a_0}}}$

$G\left( s \right) = \frac{{{b_1}s + {b_0}}}{{{s^2} + {a_1}s + {a_0}}}$

Concept:

D(s) = 1 + G(s)H(s)

D(s) gives the roots of characteristic equation i.e. closed-loop poles.

Nyquist stability criteria state that the number of unstable closed-loop poles is equal to the number of unstable open-loop poles plus the number of encirclements of the origin of the Nyquist plot of the complex function D(s).

It can be slightly simplified if instead of plotting the function D(s) = 1 + G(s)H(s), we plot only the function G(s)H(s) around the point and count encirclement of the Nyquist plot of around the point (-1, j0).

From the principal of argument theorem, the number of encirclements about (-1, j0) is

N = P - Z

Where

Where P = Number of open-loop poles on the right half of s plane

Z = Number of closed-loop poles on the right half of s plane

Calculation:

For the given system, we have |G(s)H(s)| < 1

So, we may easily conclude that the Nyquist-plot intersect the negative real axis between 0 and -1 point, i.e. the Nyquist plot does not enclose the point (-1, 0) or in other words number of encirclements is zero.

⇒ N = 0

For a stable closed loop system, we must have Z = 0.

To get Z = 0, from the Nyquist criteria, P must be equal to zero.

Therefore, the system is stable if all poles of G(s)H(s) are in left half of the s-plane.

S₁ → applied to y → vertical plates

S₂ → applied to x → horizontal plates

Observing the waveform, we conclude that

For 0 < t < T/2, 0 < S₁ < 1 and S₂ = 1

And at t = T/2, S₁ = S₂ = 0

For T/2 < t < T, 0 > S₁ > - 1 and S₂ = 1

We can represent this data in tabular form

 

Thus the waveform displayed on the CRO screen is,

From above truth table, state diagram → NAND Gate.

Concept:

• Schmitt trigger is basically bistable multivibrator
• Multivibrator which has both the state stable is called a bistable multivibrator
• Schmitt trigger is to convert any regular or irregular shaped input waveform into a square wave pulse.

Explanation:

As the positive feedback employed to the first operational amplifier,

(i) If v_i > V_sat; then o/p voltage will be V_sat

(ii) If v_i < -V_sat , then o/p voltage will be -V_sat

(iii) If -V_sat < v_i < V_sat , then o/p voltage varies according to the applied input.

For the given op-Amp circuit, we assume that,

-V_sat < v_i < V_sat

Applying KCL at the non-inverting terminal of first op-Amp, we have

$\frac{{{v_i} - 0}}{{{R_1}}} + \frac{{{v_i} - x}}{{{R_2}}} = 0$

${v_i}\left( {\frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}} \right) = \frac{x}{{{R_2}}}$

$x = \left( {\frac{{{R_1} + {R_2}}}{{{R_1}}}} \right){v_i}$

Where x is the voltage at the non-inverting terminal of 2^nd op-Amp.

2^nd op-Amp is a non-inverting amplifier so applying KCL, we get

$\frac{{x - 0}}{R} + \frac{{x - {V_0}}}{R} = 0$

V₀ = 2x

${V_0} = \frac{{2{v_i}}}{{{R_1}}}\left( {{R_1} + {R_2}} \right) = 2{v_i}\left( {1 + \frac{{{R_2}}}{{{R_1}}}} \right)$

Since there is no single output, our assumption is wrong and v_i > +V_sat and v_i < - V_sat.

As it exceeds the supply voltage range. V₀ is +V_sat or - V_sat

For the 8085A microprocessor, given instruction is LHLD 2100_H

This operation load registers L and H with the content in the memory at location 2100_H and 2101_H respectively

i.e. (H) ← M (2101_H) & (L) ← M (2100_H)

where L is lower address data and H is higher address data.

Concept:

Maximum power is transferred tom load impedance when load impedance is complex conjugate Source impedance.

\({{\rm{Z}}{\rm{L}}} = {\rm{Z}}{{\rm{th}}}^{\rm{*}}\)

And the maximum power is given by

\({{\rm{P}}{{\rm{max}}}} = \frac{{{\rm{V}}{\rm{s}}^2}}{{4{{\rm{R}}_{\rm{s}}}}}\)

\({\rm{Where;}}{{\rm{R}}{\rm{s}}} = {\rm{Re}}\left[ {{{\rm{Z}}{\rm{s}}}} \right]\)

Special Case:

Here phase balancing is not possible. So at least magnitude must be equal in order to get maximum power transferred through RL.

\( \Rightarrow {{\rm{R}}{\rm{L}}} = \left| {{{\rm{Z}}{\rm{s}}}} \right|\)

Concept:

• When T_m = T_L then motor runs at constant speed.
• When Tm > T_L then motor will accelerate & speed will increase.
• When Tm < T_L then motor will retard & speed will decrease.

where, T_m = motor torque & T_L = Load Torque

Explanation:

At point 'P'-

If speed slightly increased, then T_L > T_m As a result there will be retardation & it will come back to point P

If speed slightly decreased, then T_L < T_m As a result, there will be acceleration & it will come back to point P

Therefore operation is STABLE at operating point 'P'.

At point Q - 

If speed is slightly increased, then TL < T_m so, there will be acceleration & the speed will further increase.

If speed is slightly decreased, then T_L > T_m then there will be retardation & the speed will decrease further.

Therefore, operation is UNSTABLE at operating point 'Q'.

At point 'R'

If speed slightly increased, then T_L > T_m. there will be Acceleration & motor's speed will increase

Therefore, operation is UNSTABLE at point 'R'.

At point 'S'

If speed slightly increased, then T_L > T_m, then there will be retardation & motor's speed will decrease. As a result motion will back to original speed. therefore, operation is STABLE at point 'S'.

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

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

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

Residue Theorem:

If f(z) is analytic in a closed curve C except at a finite number of singular points within C, then

\(\mathop \smallint \limits_C f\left( z \right)dz = 2\pi i \times \left[ {{\rm{sum;of;residues;at;the;singualr;points;within;C}}} \right]\)

Formula to find residue:

1. If f(z) has a simple pole at z = a, then

$Res;f\left( a \right) = \mathop {\lim }\limits_{z \to a} \left[ {\left( {z - a} \right)f\left( z \right)} \right]$

2. If f(z) has a pole of order n at z = a, then

\(Res;f\left( a \right) = \frac{1}{{\left( {n - 1} \right)!}}{\left{ {\frac{{{d^{n - 1}}}}{{d{z^{n - 1}}}}\left[ {{{\left( {z - a} \right)}^n}f\left( z \right)} \right]} \right}_{z = a}}\)

Application:

Given function is $f\left( z \right) = \frac{{{z^2}}}{{{z^2} - 1}}$

Poles: z = 1, -1

|z – 1| = 1

⇒ |x – 1 + iy| = 1

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

The given region is a circle with the centre at (1, 0) and the radius is 1.

Only pole z = 1, lies within the given region.

Residue at z = 1 is, $\mathop {\lim }\limits_{z \to 1} \frac{{{z^2}}}{{\left( {z + 1} \right)}} = 0.5$

The value of the integral = 2πi × 0.5 = πi

Concept:

The total thickness of the transformer core made using 100 varnish-coated laminations will be:

total thickness = (2 x thickness of laminations) + (2 x thickness of one side insulation) x number of laminations

= (2 x 0.2 mm) + (2 x 0.1 mm) x 100

= 40 mm

The variance of the thickness of each varnish-coated lamination can be calculated by summing the variances of the lamination and one side insulation and multiplying by 2 (since there are two sides):

variance of one varnish coated lamination = 2 x (variance of lamination + variance of one side insulation)

= 2 x (0.02 + 0.01)

= 0.06

The standard deviation of the thickness of each varnish-coated lamination is the square root of its variance:

standard deviation of one varnish coated lamination = sqrt(0.06) = 0.245

To calculate the mean thickness and standard deviation of the core, we can use the properties of the normal distribution. The mean thickness of the core will be equal to the total thickness of the core, and the standard deviation of the core will be equal to the square root of the sum of the variances of the individual laminations:

mean thickness of core = 40 mm

variance of one varnish coated lamination x number of laminations = 0.06 x 100 = 6

standard deviation of core = sqrt(6) = 2.45 mm

Therefore, the mean thickness and standard deviation of the transformer core are 40 mm and 2.45 mm, respectively.

$E = {E_1} + {E_2} = \frac{1}{{4\pi {\varepsilon _0}}}\left[ {\frac{{{Q_1}{R_1}}}{{R_1^3}} + \frac{{{Q_2}{R_2}}}{{R_2^3}}} \right]$

Q₂ = -Q₁

R₁ = (√2, √2, 0) – (0, 0, 2)

R₁ = √2 a_x + √2 a_y – 2 a_z  

R₂ = (√2, √2, 0) – (0, 0, -2)

= √2 a_x + √2 a_y – 2 a_z  

$E = \frac{1}{{4\pi {\varepsilon _0}}}\left[ {\frac{{{Q_1}}}{{16\sqrt 2 }}\left( {\sqrt 2 ;{a_x} + \sqrt 2 ;{a_y} - 2;{a_z}} \right) - \frac{{{Q_1}}}{{16\sqrt 2 }}\left( {\sqrt 2 ;{a_x} + \sqrt 2 ;{a_y} + 2;{a_z}} \right)} \right]$

$= \frac{Q}{{16\sqrt 2 \times 4\pi {\varepsilon _0}}}\left[ { - 4{a_z}} \right] = \frac{{32\sqrt 2 \pi {\varepsilon _0}}}{{16\sqrt 2 \pi {\varepsilon _0}}}\left( { - {a_z}} \right)$

E = -2a_z

Concept:

Circuit diagram of series RLC circuit

For a series RLC circuit, the impedance is given by

$Z=\sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}}$

Power factor $\cos \phi = \frac{R}{{\sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} }}$ 

Where, 

Inductive reatance X_L = ωL = 2πfL

Capacitive reactance X_C = 1 / ωC = 1 / 2πfC

Calculation:

Case 1:

V₁ = 100∠0° V, I₁ = 0.03 ∠31° at ω₁ = 1000 r/sec   

$Z = \frac{{{V_1}}}{{{I_1}}} = \frac{{100\angle 0^\circ }}{{0.03\angle 31^\circ }} = R + j\left( {{X_L} - {X_C}} \right)$

$\phi = 31^\circ = ta{n^{ - 1}}\left( {\frac{{{X_L} - {X_C}}}{R}} \right)$

$⇒ \tan 31^\circ = \left( {\frac{{{X_L} - {X_C}}}{R}} \right)$

$⇒ \tan 31^\circ = \left( {\frac{{{ω _1}L - \frac{1}{{{ω _1}C}}}}{R}} \right)$

$⇒ \left[ {{ω _1}L - \frac{1}{{{ω _1}C}}} \right] = 0.600 \times 50$

Substitute the value of ω1 = 1000 r/sec in the above equation

$⇒ \left[ {{1000 }L - \frac{1}{{{1000}C}}} \right] = 30.04$ → 1

Case 2:

V2 = 100∠0° V, I2 = 2∠0° at ω2 = 2000 r/sec

As the phase difference between them is 0° the circuit will be resistive in nature and capacitive reactance will be equal to inductive reactance 

$R = \frac{{{V_2}}}{{{I_2}}} = \frac{{100}}{2} = 50;{{\Omega }}$

${ω _2}L - \frac{1}{{{ω _2}C}} = 0$

Substitute the value of ω2 = 2000 r/sec in the above equation

${2000}L - \frac{1}{{{2000}C}} = 0$

⇒ ${4\times10^{6}\times}L = \frac{1}{{C}}$

Substitute the above value in equation 1C = 25 μF

$⇒ \left[ {{1000 }L - \frac{4\times10^{6}L}{{{1000}}}} \right] = 30.04$

 

R = 50 Ω ; L = 10 mH ; C = 25 μF

Pole zero plot is redrawn as

As observed from the given pole-zero plot, poles & zeros are located symmetrically.

Therefore, it is all pass filter.

As we know that, from the property of all pass filter is,

• |H(jω)| = constant for all values of ω
• ∠H(jω) = Linearly vary with ω

Hence, the correct option is (4)

Period of sampling train, T_s = 20 ms

$\therefore {f_s} = \frac{1}{{20 \times {{10}^{ - 3}}}} = 50{\rm{Hz}}$

If frequency of x(t) is f_x, then after sampling the signal, the sampled signal has the frequency,

f_s - f_x = 50 - f_x and f_s + f_x = 50 + f_s

 Now, the sampled signal is applied to and ideal low pass filter with cut off frequency.

f_c = 25 Hz

Now, the o/p of filter carried a single frequency component of 20 Hz

∴, only $\left( {{f_s} - {f_x}} \right)$ component passes through the filter, ie.

f_s - f_x < 25

and f_s - f_x = 20

50 - f_x­ = 20

f_x = 50 - 20 = 30 Hz

We have,

              $y\left( t \right) = \frac{{dx\left( t \right)}}{{dt}}$........eq(i)

Since, x(t) is a Differentiable non - constant even function.

so, x(t) = x(-t) and its Fourier transform X(ω) = X(-ω)

Taking Fourier transform of eq(i)

⇒ Y(ω) = jω X(ω)

Y(-ω) = j(-ω)X(-ω)

Y(-ω) = -jω X(ω)

Y(-ω) = -Y(ω) 

So, Y(ω) is odd and imaginary.

Since, x(t) is differentiable, non-constant,  even function. So, its frequency response will be real i.e., X(ω) is real.

Hence, the correct option is (2)

Concept:

At constant V/f ratio, the total core losses of the transformer are,

P_C = P_H f + P_e f²

Where,

f = frequency

P_H = Hysteresis losses constant

P_e = eddy current losses constant

Calculation:

$\frac{{{P_c}}}{f} = {P_H} + {P_e}f$ 

i.e. the plot between $\frac{{{P_c}}}{f}$ and f is a straight line.

From the graph we have value of P_H = 10

& ${P_e} = \frac{{\left( {\frac{{{P_c}}}{f} - {P_H}} \right)}}{f} = \frac{{15 - 10}}{{50}} = \frac{1}{{10}}$ 

Hence, we get P_H × f = 10 × 25 = 250 W

P_e × f₂ $= \frac{1}{{10}} \times {\left( {25} \right)^2} = 62.5;W$

Concept:

The leakage inductance is defined as the ratio of no of turns multiplied by leakage flux divided by current through it.

The magnetizing inductance is the difference of self-inductance of the primary winding and primary leakage inductance

Let N₁ = primary turns,

N₂ = secondary turns

Let a current I₁, flowing through the primary produce a flux ϕ₁, of which part ϕ links with the secondary also while remaining (ϕ₁ – ϕ) is the leakage flux.

Self-inductance of primary ${L_1} = \frac{{{N_1}{\phi _1}}}{{{I_1}}}$

And mutual inductance between primary and secondary, $M = \frac{{{N_2}\phi }}{{{I_1}}}$

The leakage inductance of the primary

${L_{l1}} = \frac{{{N_1}\left( {{\phi _1} - \phi } \right)}}{{{I_1}}} = \frac{{{N_1}{\phi _1}}}{{{I_1}}} - \frac{{{N_1}\phi }}{{{I_1}}}$ 

$= {L_1} - \left( {\frac{{\left( {{N_2}\phi } \right)}}{2}} \right)\left( {\frac{{{N_1}}}{{{N_2}}}} \right)$ 

${L_{l1}} = {L_1} - \left( {\frac{{{N_1}}}{{{N_2}}}} \right)M$

Similarly,

${L_{l2}} = {L_2} - \left( {\frac{{{N_2}}}{{{N_1}}}} \right)M$

Calculation:

L₁ = 45 mH, L₂ = 30 mH, M = 20 mH

L_l1 = 40 – 2(20) = 5 mH

L_l2 = 30 – 0.5 (20) = 20mH

Leakage inductance of secondary referred to primary = 4L_l2 = 80 mH

And the magnetizing inductance L_m (referred to primary) = L₁ – L_l1 = 45 mH – 5 mH = 40 mH

Concept:

According to Mason’s gain formula, the transfer function is given by

\(TF=\frac{\mathop{\sum }{k-1}^{n}{{M}{k}}{{\Delta }_{k}}}{\Delta }\)

Where,

 n = no of forward paths

Mk = kth forward path gain

Δk = the value of Δ which is not touching the kth forward path

Δ = 1 – (sum of the loop gains) + (sum of the gain product of two non-touching loops) – (sum of the gain product of three non-touching loops)

Application:

Number of forward paths = 1

P₁ = G(s)

Number of loops = 3

${L_1} = - \frac{{G\left( s \right)}}{s},;{L_2} = - G\left( s \right),;{L_3} = - sG\left( S \right)$

${\rm{\Delta }} = 1 + \frac{{G\left( s \right)}}{s} + G\left( s \right) + sG\left( s \right)$

$= 1 + G\left( s \right)\left[ {\frac{1}{s} + 1 + s} \right]$

Δ = 1

Transfer function $\frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{G\left( s \right)}}{{1 + G\left( s \right)\left[ {\frac{1}{s} + 1 + s} \right]}}$

It is given that, $\frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{s}{{{s^2} + s + 2}}$

$\Rightarrow \frac{{G\left( s \right)}}{{1 + G\left( s \right)\left[ {\frac{1}{s} + 1 + s} \right]}} = \frac{s}{{{s^2} + s + 2}}$

$\Rightarrow G\left( s \right)\left[ {{s^2} + s + 2} \right] = s + G\left( s \right)\left[ {1 + s + {s^2}} \right]$

⇒ G(s) = s

Concept:

The transfer function from the state space representation is given by:

G(s) = C(SI – A)-1 B + D

KP = position error constant = $\mathop {\lim }\limits_{s \to 0} G\left( s \right)H\left( s \right)$

Kv = velocity error constant = $\mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)$

Ka = acceleration error constant = $\mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)H\left( s \right)$

Steady state error for different inputs is given by

 

From the above table, it is clear that for type – 1 system, a system shows zero steady-state error for step-input, finite steady-state error for Ramp-input and $\infty$ steady-state error for parabolic-input.

Calculation:

From the given state-space representation,

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

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

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

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

$= \frac{1}{{{s^2} + s + 1}}$

Open-loop transfer function $= \frac{{\frac{1}{{{s^2} + s + 1}}}}{{1 - \frac{1}{{{s^2} + s + 1}}}}$

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

It is a type 1 system. The steady-state error for a step input is zero.

Concept:

Bode plot transfer function is represented in standard time constant form as $T\left( s \right) = \dfrac{{k\left( {\dfrac{s}{{{\omega _{{c_1}}}}} + 1} \right) \ldots }}{{\left( {\dfrac{s}{{{\omega _{{c_2}}}}} + 1} \right)\left( {\dfrac{s}{{{\omega _{{c_3}}}}} + 1} \right) \ldots }}$

${\omega _{{c_1}}},{\omega _{{c_2}}},$… are corner frequencies.

In a Bode magnitude plot,

• For a pole at the origin, the initial slope is -20 dB/decade
• For a zero at the origin, the initial slope is 20 dB/decade
• The slope of magnitude plot changes at each corner frequency
• The corner frequency associated with poles causes a slope of -20 dB/decade
• The corner frequency associated with poles causes a slope of -20 dB/decade
• The final slope of Bode magnitude plot = (Z – P) × 20 dB/decade

<br>

Where Z is the number zeros and P is the number of poles

Application:

From the given Bode plot,

The corner frequencies are: 1/3 and 1

There is a slope change of +20 dB/decade at ω = 1/3 and hence it is a zero.

There is a slope change of -20 dB/decade at ω = 1 and hence it is a pole.

At low frequencies, 20 log k = 0 dB

⇒ k = 1

Now, the transfer function is $G\left( s \right) = \frac{{1 + 3s}}{{1 + s}}$

Now it is in the form of $\frac{{1 + asT}}{{1 + sT}}$

By comparing both the transformers, we get a = 3, T = 1

It is a lead compensator.

The maximum phase $= {\sin ^{ - 1}}\left( {\frac{{a - 1}}{{a + 1}}} \right) = {\sin ^{ - 1}}\left( {\frac{{3 - 1}}{{3 + 1}}} \right) = 30^\circ$

The maximum phase occurs at $\omega = \frac{1}{{T\sqrt a }} = \frac{1}{{\sqrt 3 }}$

The gain of the transfer function G(s) is ${G_m} = \frac{{\sqrt {1 + 9{\omega ^2}} }}{{\sqrt {1 + {\omega ^2}} }}$

At $\omega = \frac{1}{{\sqrt 3 }}$, ${G_m} = \frac{{\sqrt {1 + 9{{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}} }}{{\sqrt {1 + {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}} }} = \sqrt 3$

The gain DB is = 20 log √3 = 4.77 dB

Concept:

Moving iron voltmeter measures the RMS value of the measured voltage.

RMS value of a periodic waveform is \( = \sqrt {\frac{1}{T}\mathop \smallint \limits_0^T {{\left[ {V\left( t \right)} \right]}^2}dt} \)

Calculation:

The given periodic waveform is

\(V\left( t \right) = \left{ {\begin{array}{*{20}{c}} {1 + \sin \omega t,;;0 \le \omega t < 6\pi }\ { - 1 + \sin \omega t,;;6\pi \le \omega t < 12\pi } \end{array}} \right.\)

\(\frac{1}{T}\mathop \smallint \limits_0^T {\left[ {V\left( t \right)} \right]^2}dt\)

\( = \frac{1}{{12\pi }}\left[ {\mathop \smallint \limits_0^{6\pi } {{\left[ {1 + \sin \omega t} \right]}^2}dt + \mathop \smallint \limits_{6\pi }^{12\pi } {{\left[ { - 1 + \sin \omega t} \right]}^2}dt} \right]\)

\( = \frac{1}{{12\pi }}\left[ {\mathop \smallint \limits_0^{6\pi } \left[ {1 + 2\sin \omega t + {{\sin }^2}\omega t} \right]dt + \mathop \smallint \limits_{6\pi }^{12\pi } \left[ {1 - 2\sin \omega t + {{\sin }^2}\omega t} \right]dt} \right]\)

\( = \frac{1}{{12\pi }}\left[ {6\pi + 6\pi + \mathop \smallint \limits_0^{6\pi } \left[ {\frac{{1 - \cos 2\omega t}}{2}} \right]dt + \mathop \smallint \limits_{6\pi }^{12\pi } \left[ {\frac{{1 - \cos 2\omega t}}{2}} \right]dt} \right]\)

$= \frac{1}{{12\pi }}\left[ {12\pi + 6\pi } \right] = \frac{3}{2}$

RMS value $= \sqrt {\frac{3}{2}}$

• Mono shot multivibrators are used to generate a single output pulse of a specified width, either ‘HIGH’ or ‘LOW’, when a suitable external trigger signal or pulse T is applied.
• Trigger signal initiates a timing cycle which causes the output of the monostable to change its state at the start of the timing cycle and will remain in this second state.

 

When monoshot multivibrator M₁ is triggered, it produces a pulse of T₁ duration.

Similarly, when monoshot multivibrator M₂ is triggered, it produces a pulse of T₂ duration.

Let, the output Q₂ is initially high, so the output v₀ will be high for a time period T₂

<sub>

</sub>

 

Now for this period, Q̅₂ = 0, so the output of AND gate is low.

For M₁ is positive triggered, M₁ will be OFF for duration T₂.

After T₂ time, Q₂ = 0 and Q̅₂ = 1

Hence the output of AND gate is 1 i.e. M₁ is triggered and the output Q₁ is high for duration T₁.

For M₂ is negative edge triggered, M₂ will be off for the duration T₁ and hence, output v₀ will be as

After T₁ time the output Q₁ becomes low, and M₂ gets ON due to negative edge triggering. Hence the cycle continues as given below.

Concept:

Precision rectifier:

 

A half-wave precision rectifier is implemented using an op-amp and includes the diode in the feedback loop. This effectively cancels the forward voltage drop of the diode (Vd), so very low signals can still be rectified with minimal error.

The characteristics are shown below.

 

The Analysis of the given circuit can be done by considering the following two cases:

Case I: [Vin > 0]

When, Vin > 0, the Op-Amp tries to keep Vout = Vn = Vp = Vin and it does so because the forward-biased diode provides the closed-loop feedback path.

Case II: [Vin < 0]

The diode is reverse biased and the op-amp is working in the open-loop as shown in the above figure.

The input and output waveform are as shown in the figure:

Hence the circuit is a precision half-wave rectifier

Important points:

The other types of the precision rectifier and their characteristics are given below.

  

 

  

Explanation:

Case – I : when V_i > 0

In this case, the diode conducts.

Since, the voltage at negative and positive terminals are same for ideal op – amp. So, for the first op – amp, we have

$V_1^ - = V_1^ + = 0$

∴ the o/p of first op – amp is zero, which is applied to negative terminal of second op – amp.

Thus we get the o/p V_o = 0.

Case  - II : When V_i < 0

For this case, the diode is open and the o/p of first op – amp is

∴ the o/p of first op – amp is zero, which is applied to negative terminal of second op – amp.

Thus we get the o/p V_o = 0.

Case  - II : When V_i < 0

For this case, the diode is open and the o/p of first op – amp is

${V_{{o_1}}} = \frac{{ - R}}{R}{V_i} = - {V_i}$

∴ the o/p of second op – amp is

\(\begin{array}{l} {V_o} = \frac{{ - R}}{R}{V_{01}} = {V_i}\ \therefore {V_o} = \left{ {\begin{array}{{20}{c}} {\begin{array}{{20}{c}} 0&{{V_i} > 0} \end{array}}\ {\begin{array}{*{20}{c}} {{V_i}}&{{V_i} < 0} \end{array}} \end{array}} \right. \end{array}\)

In the given circuit, the 3-bit gray counter converts the given decimal number into gray code.

The output of gray code is connected to the selection lines of the multiplexer.

The output of the multiplexer is pulled high. Therefore, the output is one when the multiplexer is not enabled i.e. when E̅ = 1.

When multiplexer is enabled i.e. E̅ = 0, the multiplexer gives the output according to the selection lines (S₁S₀).

Assuming each secondary winding is identical and has the same resistance R

When the R phase has a maximum voltage, lines a and c will conduct. At this condition, the current will be divided into R and YB winding.

Current through $R = \frac{{{I_0} \times 2R}}{{3R}} = \frac{{2{I_0}}}{3}$

Current through B and Y = $\frac{{{I_0}}}{3}$

And ${I_0} = \frac{{2I}}{3}$

Current through R and Y $= \frac{{{I_0}}}{3}$

Current through B $= \frac{{2{I_0}}}{3}$

The current through Δ-phase winding

Current through primary ${I_p} = k{I_s}$

It is given that load current is sinusoidal, so the current conduction is continuous & hence, it is zero at 0, π, 2π,...........

For 0 to θ : S₁ is OFF, D₃ & D₂ is ON

$V_{Ao} = - \frac{V_{DC}}{2}$

For 0 to π - θ :

S₁ S₄ : ON

$\rm V_{AO} = \frac{V_{DC}}{2}$

For π - θ to θ :

D₂D₃ : ON

$\rm V_{AO} = \frac{-V_{DC}}{2}$

For π to π + θ :

D₁D₄ : ON

$\rm V_{AO} = \frac{V_{DC}}{2}$

For π + θ to 2π - θ :

S₂S₃ : ON

$\rm V_{AO} =- \frac{V_{DC}}{2}$

For 2π - θ to 2π

D₁D₄ : ON

$\rm V_{AO} = \frac{V_{DC}}{2}$

Hence, the load voltage waveform of V_AO is,