GATE EE 2021 Official Paper (Previous Year Paper)

Given data from the chart,

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Fail in the first year will give an exam in the second year.

Students will pass on 2^nd attempt.

It means fail in 1^st year students will become pass in 2^nd year.

The number of students who took the exam for the first time in the year 2^nd = 50 + 5 = 55

The number of students who took the exam for the first time in the year 3^rd = 45 + 3 = 48

Concept:

The mean or expectation of a continuous random variable is given by

\(E\left( x \right) = \mathop \smallint \limits_{ - \infty }^\infty xf\left( x \right)dx\)

Where f(x) is the probability density function (PDF).

Application:

Given probability density function of X 

f(x) = 0.01 for 0 ≤ X ≤ 100

The mean or expectation of a continuous random variable is given by

\(E\left( x \right) = \mathop \smallint \limits_{ 0}^{100} x\left( 0.01 \right)dx=50\)

Additional Information

 Properties of a valid probability density function (PDF):

1) ${f_X}\left( x \right) \ge 0,;\forall ;x;\epsilon;R$

∴ PDF will be bounded between 0 and 1.

2) \(\mathop \smallint \limits_{ - \infty }^\infty {f_X}\left( x \right)dx = {F_X}\left( \infty \right) = 1\)

3) \({F_X}\left( \infty \right) = \begin{array}{*{20}{c}} {{\rm{lim}}}\ {x \to \infty } \end{array}{F_X}\left( x \right)\)

∴ PDF will always be a monotonically increasing function as the probability is always greater than or equal to 0.

The correct answer is 'who'.

Key Points

We use who to refer to the subject of a sentence.

• In the above sentence, 'The people' is the subject of the sentence.
• Thus, we will use the pronoun 'who' as we are referring to the subject 'The people'.

Additional Information

Pronouns are words that take the place of a noun. 

• Relative pronouns are used at the beginning of an adjective clause (a dependent clause that modifies a noun).
• The relative pronouns are "that," "which," "who," "whom," and "whose."
• "Whom" is an object pronoun like "him," "her" and "us."  We use "whom" to ask which person receives an action.
• "Whose" is a possessive pronoun like "his," "her" and "our." We use "whose" to find out which person something belongs to.
• We use "Which" is used for animals in general or things.

Concept:

(aⁿ - bⁿ) is divisible by (a+b), only when 'n' is even.

Application: 

$\dfrac {{11^{52}-1}} {{11^{13}+1}}=\dfrac {({{11^{13})}^4-1^4}}{{11^{13}+1}}$

⇒ $\dfrac {{11^{52}-1}}{{11^{13}+1}}=\dfrac {{[{{{(11^{13})}^2}-1^2}][{{{(11^{13})}^2}+1^2}]}}{{{{11^{13}}}+1}}$

⇒ $\dfrac {{11^{52}-1}}{{11^{13}+1}}=\dfrac {{[{{{11^{13}}}-1}][{{{11^{13}}}+1}][{{{(11^{13})}^2}+1^2}]}}{{{{11^{13}}}+1}}$

⇒ $\dfrac {{11^{52}-1}}{{11^{13}+1}}=[11^{13}-1][{({11^{13})}^2}+1]$

∴ (1152 - 1) is exactly divisible by (1113 + 1)

The correct answer is ​To do well in an exam, adequate sleep must be part of the preparation.

Key Points

• Look at the lines: Research has consistently shown that sleep deprivation greatly reduces the ability to recall the material learnt. Hence, cutting down on sleep to study longer hours can be counterproductive.
• Upon the perusal of the above lines, it can be concluded that adequate sleep is needed to do well in the exam as it increases the ability to recall the material learnt.
• Hence, option 1 is the most appropriate answer choice.

Additional Information

• Overlook: to fail to notice or consider something
• Deprivation:  a situation in which you do not have things or conditions that are usually considered necessary for a pleasant life
• Counterproductive: having an effect that is the opposite of what you intend or desire

The arrangement can be written as,

Arrangement 1: Cars S and V also should be parked next to each

Arrangement 2: P and Q cannot be parked next to each other.

Arrangement 3: Q and S must be parked next to each other.

Arrangement 4: R is parked to the immediate right of V.

Arrangement 5: T is parked to the left of U.

From Arrangement 1 and Arrangement 4 the arrangement will be,

From Arrangement 3,

From Arrangement 2 and Arrangement 5 the final arrangement will be,

Conclusion:

V is the only car parked in between S and R: Right

Car P is parked at the extreme end: Right

Q and R are not parked together: Right

There are two cars parked in between Q and V: Wrong

The correct answer is Water.

Key Points

• An Oasis is a small fertile or green (wet) area that springs up out of the desert. Desert is covered with sand.
• An Island is a small land (dry) area that springs up out of the ocean. Ocean is filled with water.

Thus, Oasis is to sand as island is to water. This maintains a similar logical relation.

Length of the side of larger square L₁ = 10 cm

The first small square is cutting midway of the larger square, so  $L_2={10\over \sqrt 2}; cm$

The second small square is cutting midway of the first small square, so $L_3={{10\over (\sqrt 2)^2}}; cm$

Similarly, 5 square are made, so the length of the smallest square $L_6={{10\over (\sqrt 2)^5}}; cm$

Area of the smallest square = (L₆)² = $[{{10\over (\sqrt 2)^5}}]^2; (cm)^2 = 3.125;{cm}^2$

The area of the smallest square (shaded) = 3.125 cm²

Area of largest square = 100 cm²

The first small square is dividing in half, so its area = ${100\over 2};{cm}^2$

Similarly, 5 squares are made, area of smallest square = ${100\over 2^5};{cm}^2 = 3.125;{cm}^2$

Concept:

The formula for the interior angle (θ_i) of a 'n' side regular polygon:

 $θ _i={{(n-2)\over n}\times180}$

Calculation:

Given:

n = 10

$\theta _i={{(10-2)\over 10}\times180} = 144$

∴ The interior angle = 144°

The pattern which will appear when the transparent sheet is folded along the dotted line is,

Let, X, be the number of tosses,

The probability that a coin toss "head" is p, where 0 < p < 1

The coin is tossed repeatedly until the first "head" appears.

\(E\left[ X \right] = \mathop \sum \limits_{i = 1}^n {X_i}p\left( {{X_i}} \right) = p + 2p\left( {1 - p} \right) + 3p{\left( {1 - p} \right)^2} + \ldots \)

$E\left[ X \right] = p\left[ {1 + 2\left( {1 - p} \right) + 3{{\left( {1 - p} \right)}^2} + \ldots } \right]$

Let,

(1 - p) = a

$E\left[ X \right] = p\left[ {1 + 2a + 3{a^2} + \ldots } \right]$

Using Binomial expression,

$\left( {1 + 2a + 3{a^2} + \ldots } \right) = {\left( {1 - a} \right)^{ - 2}}$

$E\left[ X \right] = p{\left( {1 - \left( {1 - p} \right)} \right)^{ - 2}}$

$E\left[ X \right] = p{p^{ - 2}}$

$E\left[ X \right] = \frac{1}{p}$

​Inductance can be measured by Maxwell bridge

Important Points

Concept:

Consider a function y = f(x) on a defined interval of x.

The function attains extreme values (the value can be maximum or minimum or both).

For maxima:

• Local maxima: A point is the local maxima of a function if there is some other point where the maximum value is greater than the local maxima but that point doesn’t exist nearby local maxima.
• Global maxima: It is the point where there is no other point has in the domain for which function has more value than global maxima.

 

For minima:

• Local minima: A point is the local minima of a function if there is some other point where the minimum value is less than the local minima but that point doesn’t exist nearby local minima.
• Global minima: It is the point where there is no other point has in the domain for which function has less value than global minima.

 

Stationary Points: Points where the derivative of the function is zero i.e., f’(x) = 0. The points can be:

• Inflection point
• Local maxima
• Local minima

 

Second derivative test: Let the function has a stationary point x = a

• If ${\left( {\frac{{{d^2}f}}{{d{x^2}}}} \right)_{x = a}} < 0$  then x = a, is a point of maxima.
• If ${\left( {\frac{{{d^2}f}}{{d{x^2}}}} \right)_{x = a}} > 0$  then x = a, is a point of minima.

 

Application:

Given f(x) is a real -valued function such that f'(x0) = 0 for some x0 ∈ (0, 1)

Also given f"(x) > 0 for all x ∈ (0, 1)

So, Then f(x) has exactly one local minimum in (0, 1), called the point of minima.

Number of buses in the system = N

Number of PQ buses = N_L

Number of slack buses = 1

Number of PV buses = N - 1 - N_L

Newton Raphson method for load flow study in polar form

\(\left[ {\begin{array}{{20}{c}} {Δ P}\ {Δ Q} \end{array}} \right] = J\left[ {\begin{array}{{20}{c}} {Δ δ }\ {Δ V} \end{array}} \right]\)

Where \(J = \left[ {\begin{array}{*{20}{c}} H&S\ M&R \end{array}} \right]\)

The submatrix M relates between [ΔQ] and [Δδ]

Number of elements in ΔQ vector = Number of known Q 

Number of elements in ΔQ vector = Number of PQ buses = N_L

Number of elements in Δδ vector = Number of unknown δ = N - 1

Size of matrix M = NL × (N - 1)

Size of other sub-matrix:

H = (N - 1) × (N - 1)

S = (N - 1) ×  N_L

R = NL × NL

Given:

Forward path gain = G

Feedback path gain = H

Input signal = R(s)

Output signal = C(s)

Error signal = E(s)

C(s) = G × E(s)

Error signal = Input signal - Feedback signal

E(s) = R(s) - H × C(s)

E(s) = R(s) - H × G × E(s)

E(s) + H × G × E(s) = R(s) 

E(s) [1 + GH] = R(s)

${E(s)\over R(s)} = \dfrac{1}{ (1+GH)}$

Concept:

Economic distribution of the loads between the units of the power plant:

To determine the economic distribution of a load amongst the different units of a plant, the variable operating costs of each unit must be expressed in terms of its power output.

The fuel cost is the main cost and it must be expressed in Rs./ hour

The fuel cost of a unit can be expressed as

Fi = a Pi2 + b Pi + c Rs / hour

The incremental cost is expressed as

$IC_i = \frac{dF_i}{dP_i} = 2aP_i + b$ in Rs / MWh or Rs / kWh.

For the economic operation of the system, the incremental cost of each generator must be equal.

Calculation:

Total delivering power is

P1 + P2 = 260 MW .....(1)

The incremental cost of Generator-1 is 

$IC_1=\frac{dF_1}{dP_1}=40+0.2P_1$

The incremental cost of Generator-2 is 

$IC_2=\frac{dF_2}{dP_2}=32+0.4P_2$

For the optimum cost, the incremental cost of each generator must be equal.

IC1 = IC2 

⇒ 40 + 0.2 P1 = 32 + 0.4 P2

⇒ 0.2 P₁ - 0.4 P₂ = - 8 .....(2)

By solving equations (1) and (2), we get

P1 = 160 MW

P2 = 100 MW

Hunting:

Oscillations of the rotor about its new equilibrium position, due to sudden application or removal of load is called swinging or hunting in the synchronous machine.

Causes for Hunting:

• Sudden change in load.
• Sudden change in the supply system or in the field system.
• Load containing harmonic torque.

​

Methods for eliminating hunting:

1. By designing the machine with a suitable synchronizing power coefficient.
2. By using the flywheel.
3. By using damper winding.

​

Damper winding or bars:

• Damper winding is made with low resistance copper, aluminum, or brass.
• They are inserted in the slots made under the pole shoes.
• With respect to damper winding, the rotor behaves like a squirrel cage rotor of an induction motor.

​

Functions of damper winding:

1. In alternator to eliminate hunting and to suppress the negative sequence field.
2. In synchronous motor to eliminate hunting and for starting purpose.

 

Skewing:

• ​In the squirrel cage induction motor, the rotor slots in lamination or rotor core are not made parallel to the rotor shaft. A slight angle is maintained between the rotor slots and the rotor shaft due to some operational advantages. This is called rotor skewing.
• As the rotor slot of the induction motor skewed through some angle so that the bars lie under alternate harmonic poles of the same polarity.​

 

​The functions of skewed rotor slots in induction motor:

1. The skewed rotor slot increases the length of the copper bar thereby increases the resistance of the rotor bars hence starting torque of the machine can be improved and also the starting current is drawn by the machine can be reduced.
2. It makes the air gap flux distribution uniform thereby reduces harmonic torque produced by the machine.
3. As harmonic torque is reduced, the cogging or magnetic locking phenomenon due to harmonic torque can also be reduced. Here the magnetic locking tendency occurs when rotor teeth remain directly under stator teeth thus they might be magnetically locked.
4. And we can also eliminate a particular harmonic by selecting a proper skew angle.
5. The humming noise can be reduced.

 

Compensating windings:

• The cross-magnetizing effect of armature reaction may cause trouble in dc machines subjected to large fluctuations in load.
• In order to neutralize the cross magnetizing effect of the armature reaction, a compensating winding is used. The compensating windings consist of a series of coils embedded in slots in the pole faces. These coils are connected in series with the armature.
• The series-connected compensating windings produce a magnetic field, which varies directly with armature current. Because the compensating windings are wound to produce a field that opposes the magnetic field of the armature, they tend to cancel the cross magnetizing effect of the armature magnetic field.

Concept:

The z-transform of a unit impulse function or Kronecker delta δ [n] ↔ 1

The time-shifting affects the z-transform as:

x[n - n0] = z -n0 X(z)

Application:

Given:

h1[n] = δ[n - 1] + δ[n + 1] 

h2[n] = δ[n] + δ[n - 1] 

If h₁[n]and h₂[n] are cascaded connected then h[n] = h₁[n] * h₂[n]

Where '*' denotes convolution.

h[n] = h1[n] * h2[n]

Taking z-transform both side

H[z] = H₁[z] ⋅ H₂[z]

H[z] = (z^-1 + z) ⋅ (1 + z^-1) = (z^-1 + z-2 + z + 1 )

Taking inverse z-transform both side

h[n] = δ[n-1] + δ[n-2] + δ[n+1] + δ[n]

∴ Impulse response of the cascaded system is δ[n - 2] + δ[n - 1] + δ[n] + δ[n + 1]

Concept:

The current through the inductor in transient is given by:

$i\left( t \right) = i(\infty ) + \left( {i\left( 0 \right) - i\left( ∞ \right)} \right){e^{ - \frac{{tR}}{L}}}$ ....(1)

i(∞) = Steady-state/final value of the inductor

i(0) = Initial stored current

Also, an inductor does not allow a sudden change in voltage, i.e.

i(0-) = i(0+)

τ = Time Constant

The time constant in the RL series circuit is given by:

$τ=\dfrac{L}{R}$

Calculation:

For t < 0:

Initially, for t < 0, the switch is closed for a long time.

The 4 Ω resistor and 30 V source are bypassed and the inductor is short-circuited.

Current flowing in the circuit will be:

$i_L(0^-)=\dfrac{10}{1}=10~A$

For t > 0:

After an infinite time when the switch is closed, the inductor is short-circuited and the current will be:

$i_L(\infty)=\dfrac{10+30}{4+1}=8~A$

τ = L/R

$\rm \tau = \frac{{\frac{1}{2}}}{5} = \frac{1}{{10}}$

From equation (1)

i_L(t) = 8 + [10 - 8] e^-10t

i_L(t) = 8 + 2e^-10t

Concept:

Intermediate Value Theorem:

It states that if 'f' is a continuous function whose domain contains the interval [a, b], then it takes any given value between f(a) and f(b) at some point within the interval.

Considerd an interval of I = [a, b], and a continuous function f(x) then,

If 'u' is the number between f(a) and, f(b), then

$\min \left[ {f\left( a \right),;f\left( b \right)} \right] < u < \max \left[ {f\left( a \right),f\left( b \right)} \right]$

$c \in \left( {a,b} \right);and,;f\left( c \right) = u$

Calculation:

Given polynomial,

p(x) = x4 - 4x3 + 2

Using intermediate value theorem,

p(0) = 0 - 0 + 2 = 2

p(1) = 1 - 4 + 2 = -1

Since,

p(0) > 0 .... (1)

p(1) < 0 .... (2)

From equation (1) and (2), there exist a value 'x' in between 0 and 1 where,

p(x) = 0

Hence, in the open interval (0, 1), the polynomial p(x) has one real root.

Concept:

Thevenin’s Theorem:

Thevenin’s theorem states that a linear-bilateral two-terminal circuit can be replaced by an equivalent circuit consisting of a voltage source Vth in series with a resistor Rth.

Where V_th is the open-circuit voltage at the terminals and R_th is the input or equivalent resistance at the terminals when the independent sources are turned off.

 

Different case of Thevenin's theorem:

Case 1: Circuit with only dependent source,

In this case, R_th is finding by the voltage test method

And, V_th = 0

Case 2: Circuit with both dependent source and independent source,

In this case, Rth is finding by the voltage test method

And, Vth is finding by the same method above discuss.

Calculation:

Given circuit,

Step 1: Finding Vth across terminal ab,

V_th = V_10Ω + 3i₁

5 A = i₁ + 0

∴ i₁ = 5 A

V_th = 10(5) + 3(5) = 65 volt

Step 2: Finding R_th across terminal ab,

To finding R_th we have to open-circuited the current source by its internal resistance and applied test voltage across the terminal ab,

${R_{th}} = \frac{{{V_T}}}{{{I_T}}}$

I_T = i₁

V_T = 13i₁ + 2i₁ = 15I_T

${R_{th}} = \frac{{{V_T}}}{{{I_T}}} = \frac{{15}}{1} = 15\Omega$

Concept:

Bus Admittance Matrix (Y_Bus):

Consider a small power system network consisting of two generating stations, three transmission lines, one load, and a static capacitor connected to load bus 3.

Assumed that the network is symmetrical and operating under the balance conditions.

The node voltage equation of the system can be written as,

${I_1} = \left( {{y_{12}} + {y_{31}}} \right){V_1} - {y_{12}}{V_2} - {y_{31}}{V_3}$

${I_2} = - {y_{12}}{V_1} + \left( {{y_{12}} + {y_{23}}} \right){V_2} - {y_{23}}{V_3}$

$- {I_3} = - {y_{31}}{V_1} - {y_{23}}{V_2} + \left( {{y_{31}} + {y_{23}} + {y_{30}}} \right){V_3}$

Where,

${y_{12}} = \frac{1}{{{z_{12}}}},;{y_{23}} = \frac{1}{{{z_{23}}}},;{y_{31}} = \frac{1}{{{z_{31}}}}$

The above equation can be written in form of a matrix,

\(\left[ {\begin{array}{{20}{c}} {{I_1}}\ {{I_2}}\ { - {I_3}} \end{array}} \right] = \left[ {\begin{array}{{20}{c}} {{y_{12}} + {y_{31}}}&{ - {y_{12}}}&{ - {y_{31}}}\ { - {y_{12}}}&{{y_{12}} + {y_{23}}}&{ - {y_{23}}}\ { - {y_{31}}}&{ - {y_{23}}}&{{y_{31}} + {y_{23}} + {y_{30}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_1}}\ {{V_2}}\ {{V_3}} \end{array}} \right]\)

\(\left[ {\begin{array}{{20}{c}} {{I_1}}\ {{I_2}}\ { - {I_3}} \end{array}} \right] = \left[ {\begin{array}{{20}{c}} {{Y_{11}}}&{{Y_{12}}}&{{Y_{13}}}\ {{Y_{21}}}&{{Y_{22}}}&{{Y_{23}}}\ {{Y_{31}}}&{{Y_{32}}}&{{Y_{33}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_1}}\ {{V_2}}\ {{V_3}} \end{array}} \right]\)

Where,

Y₁₁ = y₁₂ + y₃₁, Y₂₂ = y₂₁ + y₂₃, Y₃₃ = y₃₀ + y₃₁ + y₃₂

Y₁₂ = Y₂₁ = - y₁₂, Y₂₃ = Y₃₂ = -y₂₃, Y₃₁ = Y₁₃ = -y₃₁

The element Y₁₁, Y₂₂, Y₃₃ is termed as self admittances.

The elements Y₁₂, Y₁₃, Y₂₁, Y₂₃, Y₃₁, Y₃₂ termed as mutual admittance.

Calculation:

Given circuit can be drawn as,

Applying KCL at node V,

I₁ + I₂ + I₃ + I₄ = 0

$\frac{{{E_1} - V}}{{j1}} + \frac{{{E_2} - V}}{{j1}} + \frac{{{E_3} - V}}{{j1}} + \frac{{ - V}}{{j1}} = 0$

4V = E₁ + E₂ + E₃

${I_1} = \frac{{{E_1} - V}}{{j1}} = \frac{{{E_1} - \left( {\frac{{{E_1} + {E_2} + {E_3}}}{4}} \right)}}{{j1}}$

${I_1} = j\frac{{ - 3}}{4}{E_1} + j\frac{1}{4}{E_2} + j\frac{1}{4}{E_3}$

${I_2} = \frac{{{E_2} - V}}{{j1}} = \frac{{{E_2} - \left( {\frac{{{E_1} + {E_2} + {E_3}}}{4}} \right)}}{{j1}}$

${I_2} = j\frac{1}{4}{E_1} + j\frac{{ - 3}}{4}{E_2} + j\frac{1}{4}{E_3}$

${I_3} = \frac{{{E_3} - V}}{{j1}} = \frac{{{E_3} - \left( {\frac{{{E_1} + {E_2} + {E_3}}}{4}} \right)}}{{j1}}$

${I_1} = j\frac{1}{4}{E_1} + j\frac{1}{4}{E_2} + j\frac{{ - 3}}{4}{E_3}$

The equation of current can be written in form of the matrix,

\(\left[ {\begin{array}{{20}{c}} {{I_1}}\ {{I_2}}\ {{I_3}} \end{array}} \right] = j\left[ {\begin{array}{{20}{c}} {\frac{{ - 3}}{4}}&{\frac{1}{4}}&{\frac{1}{4}}\ {\frac{1}{4}}&{\frac{{ - 3}}{4}}&{\frac{1}{4}}\ {\frac{1}{4}}&{\frac{1}{4}}&{\frac{{ - 3}}{4}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{E_1}}\ {{E_2}}\ {{E_3}} \end{array}} \right]\)

\({Y_{bus}} = j\left[ {\begin{array}{*{20}{c}} {\frac{{ - 3}}{4}}&{\frac{1}{4}}&{\frac{1}{4}}\ {\frac{1}{4}}&{\frac{{ - 3}}{4}}&{\frac{1}{4}}\ {\frac{1}{4}}&{\frac{1}{4}}&{\frac{{ - 3}}{4}} \end{array}} \right]\)

\(Y_{bus}=\left[ \begin{array}{20 {c*}} {-\frac 3 4 j}&{\frac 1 4 j}&{\frac 1 4 j}\\ {\frac 1 4 j}&{-\frac 3 4 j}&{\frac 1 4j}\\ {\frac 1 4 j}&{\frac 1 4 j}&{-\frac 3 4 j} \end{array} \right]\)

Concept:

Linearity: Necessary and sufficient condition to prove the linearity of the system is that the linear system follows the laws of superposition i.e. the response of the system is the sum of the responses obtained from each input considered separately.

y{ax1[t] + bx2[t]} = a y{x1[t]} + b y{x2[t]}

Conditions to check whether the system is linear or not.

• The output should be zero for zero input.
• There should not be any non-linear operator present in the system.

 

Time-Invariance: If the input to a time-invariant system is shifted in time, its output remains the same signal, but is shifted equally in time.

If the output for an input x(t) is y(t), then a time shift of t0 in the input gives the t0 shift in the output.

x(t) → y(t), then x(t – t0) → y(t – t0)

Application:

Given y(t) = max (0, x(t))

\(y(t)= \left{ {\begin{array}{*{20}{c}} {0,;;;;x(t) < 0}\ {x(t),x(t) > 0} \end{array}} \right.\)

Linearity checking:

For input x₁(t) = -1 , Output y₁(t) = 0

For input x₂(t) = 1 , Output y₂(t) = 1

y(x1(t) + x2(t)) = y(-1 + 1) = y(0) = 0

y1(t) + y2(t) = 0 + 1 = 1

⇒ y(x1(t) + x2(t)) ≠ y1(t) + y2(t)

Hence the system is non linear.

Time-invariance checking:

Let the output is delayed by t_o

\(y(t-t_o)= \left{ {\begin{array}{*{20}{c}} {0,;;;;;;;;;;;x(t-t_o) < 0}\ {x(t-t_o),x(t-t_o) > 0} \end{array}} \right.\)

Let the output of the system be g(t) for a delayed input x(t - t_o)

\(y_1(t)= \left{ {\begin{array}{{20}{c}} {0,;;;;g(t) < 0}\ {g(t),g(t) > 0} \end{array}} \right.\;\= \left{ {\begin{array}{{20}{c}} {0,;;;;;;;;;;;x(t-t_o) < 0}\ {x(t-t_o),x(t-t_o) > 0} \end{array}} \right.\;\=y(t-t_o)\)

Hence, the given system is time-invariant

Concept:

Matrix:

• If A is any square matrix of order n, we can form the matrix [A – λI], where I is the nth order unit matrix.
• The determinant of this matrix equated to zero i.e. |A – λI| = 0 is called the characteristic equation of A.
• The roots of the characteristic equation are called Eigenvalues or latent roots or characteristic roots of matrix A.

 

Properties of Eigenvalues:

• The sum of Eigenvalues of a matrix A is equal to the trace of that matrix A
• The product of Eigenvalues of a matrix A is equal to the determinant of that matrix A

 

Calculation:

Given:

p2 + q2 = 1

Let \(A = \left[ {\begin{array}{*{20}{c}} p&{ q}\ { q}&-p \end{array}} \right]\)

|A – λI| = 0

\( \Rightarrow \left| {\begin{array}{*{20}{c}} {p - λ }&{ q}\ { q}&{-p - λ } \end{array}} \right| = 0\)

⇒ (p - λ)(- p - λ) - q² = 0

⇒ - p² - pλ + pλ + λ2 - q² = 0

⇒ λ2 - 1 = 0

⇒ λ = ± 1

 

Let the Eigenvalues be a, b

Determinant of matrix = -p² - q² = - 1 

Product of Eigenvalues = a × b = -1 

Checking from the given option

⇒ a = 1, b = - 1.

Concept:

Residue Theorem: 

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

∫_cf(z) dz = 2πj × [sum of residues at the singular points within C]

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 (-1 - j), (3 - j), (3 + j) and (-1 + j) are the vertices of a rectangle C in the complex plane

f(z) from the given data is,

$f(z)=\dfrac{1}{z^2 (z-4)}$

Poleas of f(z) is

z = 0 of order n = 2, lies in side the closed curve.

z = 4 of order n = 1, lies outside the closed curve.

∴ $\displaystyle\oint_C \dfrac{dz}{z^2 (z-4)}=2π j;Res;f\left( 0 \right)$ 

⇒ \(Res;f\left( 0 \right) = \frac{1}{{\left( {2 - 1} \right)!}}{\left{ {\frac{{{d^{2 - 1}}}}{{d{z^{2 - 1}}}}\left[ {{{\left( {z } \right)}^2}\dfrac{1}{z^2 (z-4)}} \right]} \right}_{z = 0}}\)

$Res;f(0)=\dfrac{d}{dz}\left[\dfrac{1}{z-4}\right]_{z=0}=-\dfrac{1}{16}$

$\displaystyle\oint_C \dfrac{dz}{z^2 (z-4)}=2π j\left(\dfrac{-1}{16}\right)=\dfrac{-jπ }{8}$

Additional Information

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

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

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

Concept:

Causal Signals:

• Causal Signals are signals that are zero for all negative time.
• Causality in a system determines whether a system relies on future information of a signal x [n+1].
• Talking about “causality” in signals, it means whether they are zero to the left of t = 0 or zero to the right of t = 0.
• A causal signal is zero for t < 0.
• A system is said to be causal if its output depends upon present and past inputs, and does not depend upon future input.

 

Calculation:

Given a casual signal,

X (Z) = z2 (z - a)-2

$X(Z) = \frac{{{z^2}}}{{{{\left( {a - z} \right)}^2}}}$

From standard Z-Transform,

$n{\left( a \right)^n}u\left( n \right) \leftrightarrow \frac{{az}}{{{{\left( {z - a} \right)}^2}}}$

$n{\left( a \right)^{n - 1}}u\left( n \right) \leftrightarrow \frac{z}{{{{\left( {z - a} \right)}^2}}}$

From the time-shifting property,

$x\left( {n - {n_0}} \right) \leftrightarrow {z^{ - {n_0}}}X\left( z \right)$

$\left( {n + 1} \right){\left( a \right)^{n + 1 - 1}}u\left( {n + 1} \right) \leftrightarrow z\left[ {\frac{z}{{{{\left( {z - a} \right)}^2}}}} \right]$

$\left( {n + 1} \right){\left( a \right)^n}u\left( {n + 1} \right) \leftrightarrow \frac{{{z^2}}}{{{{\left( {z - a} \right)}^2}}}$

Since,

$X\left( z \right) = \frac{{{z^2}}}{{{{\left( {z - a} \right)}^2}}}$

Therefore,

$x\left( n \right) = \left( {n + 1} \right){\left( a \right)^n}u\left( {n + 1} \right)$

The given signal is causal, therefore,

$x\left( n \right) = \left( {n + 1} \right){\left( a \right)^n}u\left( n \right)$

Important Points

Z-Transform basic functions:

Concept: 

Power Transmission Capability of Synchronous Generator:

Considered a synchronous generator connected to the infinite bus through a transmission line of reactance Xl.

Assumed that the resistance and capacitance of the system are zero.

Where,

V = V∠0 = voltage at infinite bus

E = E∠δ = voltage behind direct axis synchronous reactance of the generator

Xd = synchronous or transient reactance of machine

Xl = line reactance

Complex power delivered by the generator to the system is,

\(S = V{I^} = V{\left[ {\frac{{E\angle {δ _;} - V}}{{j\left( {{X_d} + {X_l}} \right)}}} \right]^}\)

Let, X = Xd + Xl

Active power transferred to the system,

$P = \frac{{EV}}{X}sinδ$

The maximum steady-state power transfer occurs when,

δ = 90°

Reactive power transferred to the system,

$Q = \left( {\frac{{EV}}{X}cosδ - \frac{{{V^2}}}{X}} \right)$

Calculation:

Given,

X_s1 = X_s2 = j1.5 p.u.

X_m = j0.5 p.u.

Equivalent reactance (X) will be written as,

$X = \frac{{{X_{s1}}{X_{s2}} - X_m^2}}{{{X_{s1}} + {X_{s2}} - 2{X_m}}}$

$\frac{{{L_1}{L_2} - {M^2}}}{{{L_1} + {L_2} - 2M}}$

From the above concept, 

The maximum steady-state power transfer occurs when δ = 90°

${P_m} = \frac{{\left| E \right|\left| V \right|}}{X} = \frac{{\left| E \right|\left| V \right|}}{{j1}} = \left| E \right|\left| V \right|$

Important Points

The equivalent inductance of parallel aiding connection is

Leq = $\frac{{{L_1}{L_2} - {M^2}}}{{{L_1} + {L_2} - 2M}}$

The equivalent inductance of parallel opposing connection is

Leq = $\frac{{{L_1}{L_2} - {M^2}}}{{{L_1} + {L_2} + 2M}}$

Concept:

Excitation Table / Transition Table:

D Flip-Flop:

The single input is called the 'DATA' input. If this data input is held HIGH, the flip flop would be 'SET' and when it is LOW the flip flop would change and become 'RESET'.

 

Application:

Given:

000 → 001 → 011 → 010 → 110 → 111 → 101 → 100 → 000 ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅

Table for the given condition will be

 

D₁ will have high logic at (1,3,2,6) ⇒ D₁ = ∑m (1,3,2,6)

∴ D₁ = Q̅₂ Q₀ + Q₁ Q̅₀

Concept:

Separation of Iron loss in Transformer:

In transformer iron loss (P_i) has two components namely Hysteresis loss (P_h) and Eddy current loss (P_e).

P_i = P_h + P_e

${P_i} = {k_h}fB_m^n + {k_e}{f^2}B_m^2$  …. (1)

Where,

K_h and k_e are loss coefficient constant and their value depends on the type of material

B_m is the maximum value of flux density

f is supply frequency

${P_h} \propto f$

${P_h} = af$

${P_e} \propto {f^2}$

${P_e} = b{f^2}$

Where a and b are constants.

${P_i} = af + b{f^2}$

$\frac{{{P_i}}}{f} = a + bf$

Note: For the separation of these two losses, the no-load test is performed on the transformer.

Calculation:

Given,

V₁ = 440 V

f1 = 50 Hz

V2 = 220 V

f₂ = 25 Hz

$\frac{{{V_1}}}{{{f_1}}} = \frac{{{V_2}}}{{{f_2}}} = 8.8$

Hence, (V / f) ratio is constant,

P₁ = 2500 W

P₂ = 850 W

From the above concept,

P_i = af + bf²

2500 = 50a + 2500b .... (1)

850 = 25a + 625b .... (2)

From equation (1) and (2),

a = 18 and b = 0.64

Iron loss at 50 Hz,

Hysteresis loss (P_h) = af = 18 × 50 = 900 W

Eddy current loss (P_e) = bf² = 0.64 × 2500 = 1600 W

Concept:

Even Signal:

A signal f(t) is called even signal if, 𝑓(𝑡) = 𝑓(−𝑡)

It is given by,

\({f_e}\left( t \right) = 2\mathop \smallint \limits_0^{\frac{T}{2}} {f_e}\left( t \right)dt \)

Differentiation and Integration Properties of Fourier transform:

If,

$x\left( t \right)\mathop \leftrightarrow \limits^{F.T} X\left( \omega \right)$

Then differentiation property states that,

$\frac{{dx\left( t \right)}}{{dt}}\mathop \leftrightarrow \limits^{F.T} j\omega X\left( \omega \right)$

$\frac{{{d^n}x\left( t \right)}}{{d{t^n}}}\mathop \leftrightarrow \limits^{F.T} {\left( {j\omega } \right)^n}X\left( \omega \right)$

Then integration property states that,

$\smallint x\left( t \right)dt\mathop \leftrightarrow \limits^{F.T} \frac{1}{{j\omega }}X\left( \omega \right)$

$\smallint \smallint \smallint \ldots \smallint x\left( t \right)dt\mathop \leftrightarrow \limits^{F.T} \frac{1}{{{{\left( {j\omega } \right)}^n}}}X\left( \omega \right)$

Calculation:

Given function is even,

$\dfrac{dF(ω)}{dω} = -ω F(ω)$ .... (1)

From differentiation property,

$tf\left( t \right) = \frac{{jdF\left( \omega \right)}}{{d\omega }}$

Applying IFT to the above equation,

$- jtf\left( t \right) = \frac{{jdf\left( t \right)}}{{df}}$

$\frac{{df\left( t \right)}}{{dt}} = - tf\left( t \right)$ .... (2)

From equation (1) and (2) it is clear that the f (t) is the change of Gaussian function, it can be written as,

$f\left( t \right) = \frac{1}{{\sqrt {2\pi } }}{e^{\frac{{ - {t^2}}}{2}}}$ .... (3)

Note:

$\left[ {{e^{ - a{t^2}}} \leftrightarrow \sqrt {\frac{\pi }{a}} {e^{\frac{{ - {\omega ^2}}}{{4a}}}}} \right]$

$\left[ {{e^{ - \frac{{{t^2}}}{2}}} \leftrightarrow \sqrt {2\pi .} {e^{\frac{{ - {\omega ^2}}}{2}}}} \right]$

From equation (3),

$\frac{1}{\sqrt{{2\pi }}}{e^{\frac{{ - {t^2}}}{2}}} \leftrightarrow {e^{;\frac{{ - {\omega ^2}}}{2}}}$

$f\left( t \right) = \frac{1}{\sqrt{{2\pi }}}{e^{\frac{{ - {t^2}}}{2}}}$

$f\left( 0 \right) = \frac{1}{\sqrt{{2\pi }}}$

$f\left( 0 \right) = 0.3989$

$\therefore {\bf{f}}\left( 0 \right) < 1$

Concept:

Mutual Inductance:

When two coils are placed close to each other, a change in current in the first coil produces a change in magnetic flux, which cuts not only the coil itself but also the second coil as well. The change in the flux induces a voltage in the second coil, this voltage is called induced voltage and the two coils are said to have a mutual inductance.

Consider a pair of coupled inductors with self-inductance L1 and L2, magnetically coupled through coupling coefficient k.

Input and output voltage expressions are given as

V1 = jωL1I1 + jωMI2   ...(1)

V2 = jωL2I2 + jωMI1    ...(2)

Where,

ω = 2πf

$M = K\sqrt {{L_1}{L_2}}$

M = Mutual inductance

L1 = Inductance of coil one

L2 = Inductance of coil two

Calculation:

The given circuit in the Laplace domain is 

Apply KVL in the input loop,

V₁(s) = (4 + 6s)I₁(s) + sI₂(s)  ...(1)

Apply KVL in the output loop

I₂(s)[5 +4s] + sI₁(s) = 0

$I_2(s)=\dfrac{{-sI_1(s)}}{{4s + 5}}$

Substitute I₂(s) in equation(1)

$V_1(s)=(4+6s)I_1(s)-\dfrac{{s^2I_1(s)}}{{4s + 5}}$

$\dfrac{{V_1(s)}}{{I_1(s)}}=\dfrac{{(6s+4)(4s+5)-s^2}}{{4s+5}}$

$Z_{in}(s)=\dfrac{{24s^2+16s+30s+20-s^2}}{{4s+5}}$

$Z_{in}(s)=\dfrac{{23s^2+46s+20}}{{4s+5}}$

Concept:

The general form of a cubic equation is ax³ + bx² + cx + d = 0.

Where a, b, c, and d are constants and a ≠ 0.

Let the roots be p, q, and r

• The sum of the roots (p + q + r) = - b/a
• The product of the roots (pqr) = - d/a
• The sum of the product of any two roots (pq + qr + rp) = c/a

 

Calculation:

Given p(z) = z3 + (1 + j) z2 + (2 + j) z + 3

Sum of the roots (p + q + r) = - (1 + j)

Product of the roots (pqr) = - 3

Sum of the roots is complex, so all the roots cannot be real.

Concept:

The magnetic field forms a closed loop, i.e. the amount of field leaving a point equals the amount entering. i.e. Magnetic monopoles do not exist.

Since the divergence of a field gives the net outflow of a field and is calculated as $∇ .\vec F$

So, ∇ ⋅ B = 0

B = Magnetic flux density

Since B is related to H (magnetic field intensity) as:

B = μH

so, ∇ ⋅ H = 0

Application:

Option: 1

B = 10x x̂  - 30z ŷ + 20y ẑ  

$∇ ⋅ B = \frac{{\partial {B_x}}}{{\partial x}} + \frac{{\partial {B_y}}}{{\partial y}} + \frac{{\partial {B_z}}}{{\partial z}} = \frac{{\partial 10x}}{{\partial x}} + \frac{{\partial ( - 30z)}}{{\partial y}} + \frac{{\partial 20y}}{{\partial z}} = 10$

∇ ⋅ B ≠ 0, so function does not represent magnetic field.

Option: 2

B = 10y x̂ + 20x ŷ - 10z ẑ 

$\nabla ⋅ B = \frac{{\partial {B_x}}}{{\partial x}} + \frac{{\partial {B_y}}}{{\partial y}} + \frac{{\partial {B_z}}}{{\partial z}} = \frac{{\partial 10y}}{{\partial x}} + \frac{{\partial 20x}}{{\partial y}} + \frac{{\partial ( - 10z)}}{{\partial z}} = - 10$

∇ ⋅ B ≠ 0, so function does not represent magnetic field.

Option: 3

B = 10x x̂ + 20y ŷ - 30z ẑ 

$\nabla ⋅ B = \frac{{\partial {B_x}}}{{\partial x}} + \frac{{\partial {B_y}}}{{\partial y}} + \frac{{\partial {B_z}}}{{\partial z}} = \frac{{\partial 10x}}{{\partial x}} + \frac{{\partial 20y}}{{\partial y}} + \frac{{\partial ( - 30z)}}{{\partial z}} = 10 + 20 - 30 = 0$

∇ ⋅ B = 0, so function represent magnetic field.

Option: 4

B = 10z x̂ + 20y ŷ - 30x ẑ 

$\nabla ⋅ B = \frac{{\partial {B_x}}}{{\partial x}} + \frac{{\partial {B_y}}}{{\partial y}} + \frac{{\partial {B_z}}}{{\partial z}} = \frac{{\partial 10z}}{{\partial x}} + \frac{{\partial 20y}}{{\partial y}} + \frac{{\partial ( - 30x)}}{{\partial z}} = 20$

∇ ⋅ B ≠ 0, so function does not represent magnetic field.

Concept:

Fortescue’s Theorem:

A unbalance set of ‘n’ phasors may be resolved into (n - 1) balance n-phase system of different phase sequence and one zero phase sequence system.

A zero-phase sequence system is one in which all phasors are of equal magnitude and angle.

Considered three phasors are represented by a, b, c in such a way that their phase sequence is (a b c).

The positive phase sequence will be (a b c) and the negative phase sequence will be (a c b).

Assumed that subscript 0, 1, 2 refer to zero sequences, positive sequence, negative sequence respectively.

Current I_a, I_b, I_c represented an unbalance set of current phasor as shown,

Each of the original unbalance phasor is the sum of its component and it can be written as,

I_a = I_a0 + I_a1 + I_a2

I_b = I_b0 + I_b1 + I_b2

I_c = I_c0 + I_c1 + I_c2

For a balance position phase sequence (a b c) we can write the following relation,

I_a0 = I_b0 = I_c0

I_b1 = α² I_a1

I_c1 = α I_a1

I_b2 = α I_a2

I_c2 =α² I_a2

From the above equation I_a, I_b, I_c can be written in terms of phase sequence component,

I_a = I_a0 + I_a1 + I_a2

I_b = I_b0 +  α² I_a1 + α I_a2

I_c = I_a0 + α I_a1 + α² I_a2

The above equation can be written in form of Matrix as shown,

\(\left[ {\begin{array}{{20}{c}} {{I_a}}\ {{I_b}}\ {{I_c}} \end{array}} \right] = \left[ {\begin{array}{{20}{c}} 1&1&1\ 1&{{\alpha ^2}}&\alpha \ 1&\alpha &{{\alpha ^2}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{I_{a0}}}\ {{I_{a1}}}\ {{I_{a2}}} \end{array}} \right]\)

Calculation:

Given,

IB0 = 0.1 ∠0° p.u

IA0 = IB0 = IC0 = 0.1 ∠0° p.u

IA = 1.1 ∠0° p.u.

IC = (1 ∠120° + 0.1) p.u.

From the above concept,

I_A = I_A0 + I_A1 + I_A2

I_A - I_A0 = I_A1 + I_A2 = 1.1 ∠0° + 0.1 ∠0° p.u = 1 ∠0° p.u

I_A1 + I_A2 = 1 ∠0° p.u .... (1)

I_C = I_C0 + I_C1 + I_C2 = I_A0 + α I_A1 +α² I_A2

1 ∠120° + 0.1 = 0.1 ∠0° + α I_A1 +α² I_A2

α I_A1 +α² I_A2 = 1 ∠120° .... (2)

Adding equation (1) and (2),

(I_A1 + I_A2) + (α I_A1 +α² I_A2) = 1 ∠0° + 1 ∠120°

I_A1(α + 1) + I_A2(α² + 1)  = 1 ∠0° + 1 ∠120° (Since, 1 + α + α2 = 0 ⇒ 1 + α = -α²)

IA1(-α²) + IA2(-α)  = 1 ∠0° + 1 ∠120°

α2 I_A1 + α I_A2 = 1 ∠-120° .... (3)

From above concept,

I_B = I_A0 + α² I_A1 + α I_A2

I_B = 0.1 ∠0° + 1 ∠-120°