GATE EE 2012 Official Paper (Previous Year Paper)

The correct answer is 'should be given'; i.e. the error lies in this part of the sentence.

Key Points

• The given sentence is in active voice which represents a form or set of forms of a verb in which the subject is typically the person or thing performing the action and which can take a direct object.
• Thus, the usage of 'be given' is incorrect which is used with a passive voice which is used when we want to emphasize the action (the verb) and the object of a sentence rather than subject.
• Thus, 'should be given' needs to be replaced with 'should give'.

Therefore, the correct sentence is: 'I requested that he should give the driving test today instead of tomorrow.'

Given:

(1.001)1259 = 3.52 and (1.001)2062 = 7.85

Concept used:

X(a + b) = X(a) × X(b)

Calculation:

⇒ (1.001)3321 = (1.001)1259 + 2062 

= (1.001)1259 × (1.001)2062

= 3.52 × 7.85

= 27.64

Hence, the correct answer is "27.64".

The correct answer is 'should have taken'.

Key Points

• Mixed Conditional Sentence: Mixed conditionals are conditional sentences that mix two different times in one sentence or we can say that the if-clause is not the same as the time in the result.
• Structure for Mixed conditional (for present condition, past result):
• If/when + past simple, would have + verb infinitive.
• Example: 
• If Sam spoke Russian, he would have translated the letter for you.
• (But Sam doesn't speak Russian and that is why he didn't translate the letter.)
• Note: Would can be replaced with should/might in the sentence.

Therefore, the correct sentence is: If the tired soldier wanted to lie down, he should have taken the mattress out on the balcony. 

Additional Information

• Conditional sentence:- As the name suggests, these sentences express conditions.
• In conditional sentences, there are two clauses of the sentences. One is a conditional clause and another one is main clause.
• Other Conditional sentences are-

The correct answer is 'nonchalance'.

Key Points

• Let's explore the options:
• beggary: a state of extreme poverty.
• Example: They have no benefits to stand between them and beggary.
• nomenclature: the devising or choosing of names for things, especially in a science or other discipline.
• Example: The Linnean system of zoological nomenclature
• jealousy: the state or feeling of being jealous.
• Example: He broke his brother's new bike in a fit of jealousy.
• nonchalance: the trait of remaining calm and seeming not to care.
• Example: He leaned back in his chair with apparent nonchalance.
• Thus, from above we can refer that the correct answer is option 4.

Therefore, the correct sentence is: 'Given the seriousness of the situation that he had to face, his nonchalance was impressive.'

The correct answer is 'Freedom'.

Key Points

• Latitude: scope for freedom of action or thought.
• Example: Journalists have considerable latitude in criticizing public figures.
• Freedom: the power or right to act, speak, or think as one wants without hindrance or restraint.
• Example: We do have some freedom of choice.
• Thus, the correct answer is option 2.

Additional Information

• Let's explore options:
• Eligibility: the state of having the right to do or obtain something through satisfaction of the appropriate conditions.
• Coercion: the practice of persuading someone to do something by using force or threats.
• Meticulousness: showing great attention to detail; very careful and precise.

A meeting occurs if the person arrives between 1:00 PM and 1:45 PM and the second person arrives in the next 15 minutes or if both the persons arrive between 1:45 and 2:00.

Case 1:

45/60 are favorable cases and hence the probability of first-person arriving between 1:00 and 1:45 is 3/4.

Probability of second person arriving in the next 15 min = 15/60 = 1/4

So, the probability of one person arriving between 1:00 and 1:45 and meeting the other = 3/4 × 1/4 × 2 = 3/8 (2 for choosing the first arriving friend)

Case 2: 

Both friends must arrive between 1:45 and 2:00

Probability = 1/4 × 1/4 = 1/16

So, probability of a meet = 3/8 + 1/16 = 7/16

Hence, the correct answer is 7/16.

The correct answer is 'Thorough regimentation was the main reason for the efficiency of the Roman legions even in adverse circumstances.'

Key Points

• Let's refer to the following lines of the passage:
• In the legions, military law prevailed and discipline was brutal. Discipline on the battlefield kept units obedient, intact and fighting, even when the odds and conditions were against them.
• Thus, from above we can infer that the correct answer is option 1.

Additional Information

• Regimentation: organize according to a strict system or pattern.

Let's consider number of 10 rupee notes as 'x'

And number of 20 rupees notes as 'y'

Then total number of notes = x + y = 14 -------(1)

Total money = 10x + 20y = 230  ---------(2)

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

x = 5 and y = 9

Explanation:

Divide 8 bags into three parts.

2,2 and 4 respectively.

If we compare the 2,2 bags on pans of a balance.

We can identify which side is the lighter bag placed.

And then we will need only one more weighing for identifying the faulty bag.

So only two weighings are required.

Alternative solution:

In the case of a weighing balance (i.e. beam balance) the following model can be observed.

1 - 3 → 1 weighing required

4 - 9 → 2 weighings required

10 - 27 → 3 weighings required

28 - 81 → 4 weighings required

And the process will go so on.

As there are eight objects. The minimum number of weighings required will be only 2.

Total budget = 10500 Rs

Expenditure  other than savings = 9000

The approximate percentage of the monthly budget NOT spent on savings is = (9000/ 10500) × 100

= 85.71 % ≈ 86%

P(max(X,Y) < .5) = P(X <.5, Y < 0.5)

Since X and Y are independent random variable, therefore

P(X <.5, Y < 0.5) = P(X < .5) P(Y < .5)

X and Y are uniformly distributed in [-1,1]

PDF of X and Y are

Thus,

P(X > 2) P(Y > 2)

Let A = xx

Taking logarithm both the sides.

Log A = x log x = i log (0 + x)

As |x| = 1 and arg (x) = π/2,

Log A = x(log1 + xπ/2) = x(0 + x π/2)

Log A = -π/2

Hence, A = e- π/2

\(f\left( z \right); = ;\frac{1}{{z + 1}} - \frac{2}{{z + 3}}; = ;\frac{{ - z + 1}}{{\left( {z + 1} \right)\left( {z +3} \right)}}\)

The poles of f(z) are - 1 and -3 and -3 is outside of the circle |z + 1| = 1.

By Cauchy’s formula,

\(\begin{array}{l} \smallint \frac{{f\left( z \right)}}{{z - a}}dz; = ;2\pi if\left( a \right)\ \smallint \frac{{ - z + 1}}{{\left( {z + 3} \right)\left( {z +1} \right)}}dz; = ;\smallint \frac{{\frac{{ - z +1}}{{\left( {z +3} \right)}}}}{{\left( {z + 1} \right)}}dz \end{array}\)

By Cauchy’s formula,

\(\begin{array}{l} \Rightarrow ;2\pi if\left( { - 1} \right); = ;2\pi i\left( {\frac{{-(-1) + 1}}{{ - 1 + 3}}} \right); = ;2\pi i\ \Rightarrow ;\frac{1}{{2\pi i}}\smallint f\left( z \right)dz; = ;1 \end{array}\)

Applying nodal analysis at top node

\(\frac{{{V_1} + 1\angle 0^\circ }}{1} + \frac{{{V_1} + 1\angle 0^\circ }}{{j1}} = 1\angle 0^\circ\)

V₁ = (j1 + 1) + j1 + 1 ∠ 0° = j1

\(\begin{array}{l} {V_1} = \frac{{ - 1}}{{1 + j1}}\ {I_1} = {V_1} = \frac{{{V_1} + 1\angle 0^\circ }}{{j1}} = \frac{{\frac{{ - 1}}{{j + 1}} + 1}}{{j1}} = \frac{j}{{\left( {1 + j} \right)j}} = \frac{1}{{\left( {1 + j} \right)}}A \end{array}\)

Assume a voltage source V between 1 and 2 ,

then \(\frac{V}{I} = impendance\) at 1 and 2

\({i_b} = \frac{{0 - V}}{{10\ K{\rm{\Omega }}}}\)         ___________(1)

Apply Nodal analysis then

\(- {i_b} + \frac{V}{{100}} - 99{i_b} - I = 0\)

V = 10⁴ i_b + 100 I        ___________(2)

From (1) and (2) we get

\(V = {10^4}\left( {\frac{{ - V}}{{{{10}^4}}}} \right) + 100\ I\)

2V = 100 I

Impendence \(= \frac{V}{I} = 50\ {\rm{\Omega }}\)

∴ Option (1) is correct

\(;\sin \omega t \to \boxed{G\left( s \right)} \to Y\left( s \right) = \left| {G\left( s \right)} \right| \cdot \sin \left( {\omega t + \angle G\left( s \right)} \right)\)

Output will be zero when

|G(s)| = 0

Put s = jω

\(\left| {\frac{{\left( { - {\omega ^2} + 9} \right)\left( {j\omega + 2} \right)}}{{\left( {j\omega + 1} \right)\left( {j\omega + 3} \right)\left( {j\omega + 4} \right)}}} \right| = 0\)

at         ω = 3   , |G(jω)| = 0

Prime implicant is a minterm, which are obtained by combining maximum possible adjacent cell in k-map

\(F\left( {x,y,z} \right) = \bar x\bar y + xy\)

\(\begin{array}{l} x\left[ n \right] = {\left( {\frac{1}{3}} \right)^{\left| n \right|}} - {\left( {\frac{1}{2}} \right)^n}u\left[ n \right]\ = {\left( {\frac{1}{3}} \right)^n}u\left[ n \right] + {\left( {\frac{1}{3}} \right)^{n - 1}}u\left[ { - n - 1} \right] - {\left( {\frac{1}{2}} \right)^n}u\left[ n \right] \end{array}\)

Taking Z-transform

\(\begin{array}{l} X\left[ z \right] = \mathop \sum \limits_{n = - \infty }^\infty {\left( {\frac{1}{3}} \right)^n}{z^{ - n}}u\left[ n \right] + \mathop \sum \limits_{n = - \infty }^\infty {\left( {\frac{1}{3}} \right)^{ - n}}{z^{ - n}}u\left[ { - n - 1} \right] - \mathop \sum \limits_{n = - \infty }^\infty {\left( {\frac{1}{2}} \right)^n}{z^{ - n}}u\left[ n \right]\ = \mathop \sum \limits_{n = 0}^\infty {\left( {\frac{1}{3}} \right)^n}{z^{ - n}} + \mathop \sum \limits_{n = - \infty }^{ - 1} {\left( {\frac{1}{3}} \right)^{ - n}}{z^{ - n}} - \mathop \sum \limits_{n = 0}^\infty {\left( {\frac{1}{2}} \right)^n}{z^{ - n}}\ = \underbrace {\mathop \sum \limits_{n = 0}^\infty {{\left( {\frac{1}{{3Z}}} \right)}^n}}_I + \underbrace {\mathop \sum \limits_{m = 1}^\infty {{\left( {\frac{Z}{3}} \right)}^m}}_{II} - \underbrace {\mathop \sum \limits_{n = 0}^\infty {{\left( {\frac{1}{{2Z}}} \right)}^n}}_{III} \end{array}\)

Series I converges if \(\left| {\frac{1}{{3Z}}} \right| < 1or\left| Z \right| > \frac{1}{3}\)

Series II converges if \(\left| {\frac{1}{3}z} \right| < 1\ or\ \left| Z \right| < 3\)

Series III converges if \(\left| {\frac{1}{{2Z}}} \right| < 1\ or\ \left| Z \right| > \frac{1}{2}\)

Region of convergence of X(Z) will be intersection of above three

So, ROC  \(\frac{1}{2}< |Z| < 3\).

Concept:

Bus Admittance Matrix:

• In a power system, Bus Admittance Matrix represents the nodal admittances of the various buses.
• Admittance matrix is used to analyze the data that is needed in the load or a power flow study of the buses.
• It explains the admittance and the topology of the network.

In a y-bus matrix,

The diagonal element Yii is the sum of all the admittances of the elements connected to the ith bus.

Yii = Σ yik , k = 1, 2, …n and k ≠ i

The off-diagonal element Yij is equal to the minus of the admittance of the element connected between buses i and j.

Yij = -yij

Application:

From the above concept,

_Y22 = Y₂₁ + Y₂₂ + Y₂₃

_Y21 = -y₂₁, y₂₃ = -y₂₃

From the given Y bus matrix \(Y = j\left[ {\begin{array}{*{20}{c}} { - 13}&{10}&5\ {10}&{ - 18}&{10}\ 5&{10}&{ - 13} \end{array}} \right]\)

Y₂₂ = -18, Y₂₁ = 10, Y₂₃ = 10

Y₂₁ = -y₂₁

Y₂₃ = -y₂₃

Y₂₂ = y₂₁ + y₂₂ + y₂₃

-18j = (-10j) + y₂₂ + (-10j) ⇒ y₂₂ = 2j

similarly, y₁₁ = y33 = 2j

Here 2j is shunt capacitance at each bus divides into half in π model at

the buses located

Therefore,the system looks like

y₁₁ = y₀₁ + y₀₁ = 2 y₀₁

2j = 2 y01

y01 = j

similarly, y₀₂ = j

The magnitude of the shunt susceptance of the line connecting bus

1 and 2 is y₀₁ + y₀₂

j + j = 2j Ω

The slip of an induction motor is given by,

\(s = \frac{{{N_s} - {N_r}}}{{{N_s}}} \times 100\)

It directly depends on rotor speed and synchronous speed. It is indirectly depends on shaft torque but it is independent of core loss component.

Given that,

\({i_1}\left( t \right) = {I_m}\sin \left( {\omega t - {ϕ _1}} \right),{i_2}\left( t \right) = {I_m}\cos \left( {\omega t - {ϕ _2}} \right)\)

If these currents are balanced,

The angle between i₁(t) and i₂(t) should be 90°

90° = ϕ₂ + 90 - ϕ₁

∴ ϕ₁ = ϕ₂

Concept:

The DC value or the average value of a waveform f(t) is given by the Area under the curve for one complete time period; i.e.,

\({f_{DC}} = \frac{1}{T}\mathop \smallint \nolimits_o^T f\left( t \right)dt\)

RMS value 'or' the effective value of an alternating quantity is calculated as:

\({V_{rms}} = \sqrt{\frac{1}{T}\mathop \smallint \limits_0^T {v^2}\left( t \right)dt} \)

Calculation:

Average value,

\(\begin{array}{l} = \frac{1}{{20}}\left[ {\mathop \smallint \limits_0^{10} tdt - \mathop \smallint \limits_{10}^{12} 5dt + \mathop \smallint \limits_{12}^{20} 5dt} \right]\ = {\rm{ }}4\ {\rm{ }}V \end{array}\)

Heaviside bridge:

• Heaviside bridge is used to measure the mutual inductance.
• The bridge which measures the unknown mutual inductance regarding mutual inductance such type of bridge is known as the Campbell bridge.
• Mutual inductance is the phenomenon in which the variation of current in one coil induces the current in the nearer coil.
• The bridge also used for measuring the frequency by adjusting the mutual inductance until the null point is not obtained.

Important Points

Given differential equation is \(t\frac{{dx}}{{dt}} + x; = ;{t}\)

\(\Rightarrow ;\frac{{dx}}{{dt}} + \frac{1}{t}x; = ;1\)

This is a linear differential equation in t.

\(I.F.; = ;{e^{\smallint \frac{1}{t}dt}}; = ;{e^{\log t}}; = ;t\)

The solution is,

\(\begin{array}{l} x.(IF); = ;\smallint 1.(IF)dt + c\Rightarrow x.t; = ;\smallint tdt + c\ \Rightarrow ;xt; = ;\frac{{{t^2}}}{2} + c \end{array}\)

Given x(1) = .5 ⇒ c = 0

The solution is \(x; = ;\frac{{{t}}}{2}\)

Concept

Multiplication of function in one domain corresponds to the differentiation in another domain

If f(t) having the Laplace transform F(s) then t ⋅ f(t) will have the transform as

  t f(t) ↔\( - \frac{{dF\left( s \right)}}{{ds}}\)

Calculation:

Given function f(t) is . And  g(t) = t ⋅ f(t)

Laplace transform of g(t) is:

\(L[g(t)] = - \frac{d}{{ds}}\left( {\frac{1}{{{s^2}s + 1}}} \right)\)

\( = - \frac{{\frac{{d\left( 1 \right)}}{{ds}}\left( {{s^2} + s + 1} \right) - 1 \times \frac{{d\left( {{S^2} + S + 1} \right)}}{{ds}}}}{{{{\left( {{S^2} + S + 1} \right)}^2}}}\)

\( = - \left( {\frac{{0 - 1\left( {2s + 1} \right)}}{{{{\left( {{s^2} + s + 1} \right)}^2}}}} \right)\)

\( = \frac{{2s + 1}}{{{{\left( {{s^2} + s + 1} \right)}^2}}}\)

In phasor form:

Z = 4 – j3

Z = 5 ∠-36.86° Ω

I = 5 ∠100° A

Average power delivered will be:

\({P_{avg}} = \frac{1}{2}{\left| I \right|^2}Zcos\theta % MathType!End!2!1! \)

= (½) × 25 × 5 cos 36.86°

= 50 W

Alternate Approach:

The power delivered to the load will simply be the power dissipated by the resistor R, i.e.

P_delivered = I_RMS² R

With I_m = 5, I_RMS will be:

\(I_{RMS} = \frac{5}{\sqrt 2}\)

\(P_{delivered} = (\frac{5}{\sqrt 2})^2\times 4\)

\(P_{delivered} =\frac{25}{2}\times 4 = 50~W\)

The S – domain equivalent circuit is

\( I\left( S \right) = \frac{{\frac{{{V_c}\left( 0 \right)}}{S}}}{{\frac{1}{{{C_1}S}} + \frac{1}{{{C_2}S}}}} = \frac{{{V_c}\left( 0 \right)}}{{\left( {\frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}} \right)}}% MathType!End!2!1! \)

V_c(0) = 12 V

\(I\left( S \right) = \left( {\frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}} \right)\left( {12} \right)% MathType!End!2!1! \)

I(S) = 12 C_eq

Taking Inverse Laplace transform for the current in the time domain.

i(t) = 12 C_eq δ(t) 

The above equation shows that it's an impulse.

Let's consider v >.7 and the diode is forward biased. 

By applying KVL, we get

10 - I × 1k - v = 0

10 -[(v-.7)/(500)] × 1000 - v = 0

⇒ v = 3.8 V 

And v > .7 and our assumption is correct,

⇒ i =[(v-.7)/(500)] = [3.8 - 0.7]/500 = 6.2 mA

The only possible combinations are

A = 01 and B = 0 0

A = 10 and B = 00, 01

A = 11 and B = 00, 01, 10

So there are only 6 combinations

Tips and Tricks:

\(\frac{2^{2n} - 2^n}{2}\)

where n = 2 bit

The circuit is a SR flip-flop with input A = S and B = R. In SR flip-flop there is no race around condition for any combination of input.

Note :

11 is a not allowed state because the output Q and Q'  will be 1. 

For race around Q should toggle.

It occurs in JK flip flop when J=K=1 and there is unequal propagation delay.

For economic load dispatch,

\(\begin{array}{l} {L_1}\frac{{d{F_1}}}{{d{P_{G1}}}} = {L_2}\frac{{d{F_2}}}{{d{P_{G2}}}}\ \frac{1}{{1 - \frac{{d{P_L}}}{{d{P_{G1}}}}}}\frac{{d{F_1}}}{{d{P_1}}} = \frac{1}{{1 - \frac{{d{P_L}}}{{d{P_{G2}}}}}}\frac{{d{F_2}}}{{d{P_2}}} \end{array}\)

From the given data

\(\frac{{d{P_L}}}{{d{P_1}}} = 2\left( {0.5} \right){P_F} = {P_{G1}}\) and

\(\begin{array}{l} \frac{{d{P_L}}}{{d{P_{G2}}}} = 0\ \frac{1}{{\left( {1 - {P_{G1}}} \right)}} × 10000 = \frac{1}{{\left( {1 - 0} \right)}} × 12500\ \left( {1 - {P_{G1}}} \right) = \frac{{100}}{{125}} \end{array}\)

1-P_G1 = 0.8 ⇒ P_G1 = 0.2 pu

⇒  PG1 = 0.2 × 100 = 20 MW

But \({P_L} = 0.5P_1^2\) 

= 0.5 (0.2)²

P_L = 0.02 Pu = .02 × 100 = 2 MW

P_L is 10% P_G1

P_G1 + P_G2 = P_D + P_L

⇒ 20 + PG2  = 40 + 2 

⇒ P_G2 = 22 MW

Concept:

In a single line to ground fault, all the sequence components of fault currents are equal.

I_a1 = I_a2 = I_a0

In a double line to ground fault, the sum of all sequence components of fault currents is zero.

I_a1 + I_a2 + I_a0 = 0

Where, Ia1 = Positive sequence component of the current

Ia2 = Negative sequence component of the current

Ia0 = Zero sequence component of the current

Calculation:

Given

I_positive = j 1.5 p.u. 

I_negative = - j 0.5 p.u. 

I_zero = - j 1 p.u.

Ia1 = j 1.5, Ia2 = -j 0.5, Ia0 = -j 1.0

⇒ Ia1 + Ia2 + Ia0 = j 1.5 – j 0.5 – j 1.0 = 0

Therefore, the fault is double line to ground fault.

Answer:-

Fraction of conduction time freewheeling diode conducts

= α /π

Concept:- 

Circuit of single phase half controlled bridge rectifier is supplying on R-L load and Freewheeling diode.

• The circuit consists of a thyristor T, voltage source V_s , a freewheeling diode across the RL load, an inductive load L and resistive load R.

• During the positive half cycle of input voltage, the thyristor T is forward biased but it does not conduct until a gate signal is applied to it.

• When the gate pulse is given to thyristor T, it gets turned on and begins to conduct.

• When the thyristor is on, the input voltage is applied to the load but due to the inductor  present in the load, the current through the load build up slowly.

• During the negative half cycle, the thyristor T gets reverse biased. At this instant i.e. at \(\omega t = ;\pi \), the load current shift its path from the thyristor to the freewheeling diode.

• When the current is shifted from thyristor to freewheeling diode, the thyristor turns off.( Fig.2)

• The current through the inductor slowly decays to zero through the loop R – Freewheeling diode – L.

• So here the thyristor will not conduct in the negative half cycle and turns off at \(\alpha \)

• So the load reverse voltage only during the positive half cycle.

• The average value of output voltage can be varied by varying the firing angle

• The waveform of voltage and current.

The Advantages of using Freewheeling Diode:-

• It prevents the output voltage from becoming negative.
• As the energy stored in the L is transferred to load R through freewheeling diode, the system efficiency is improved.
• The load current waveform is more smooth.
• The power factor is improved.
• The performance of converter is improved with a freewheeling diode.

Latching Current: It is the minimum anode current required to maintain the Thyristor in the ON state immediately after a Thyristor has been turned on and the gate signal has been removed.

Holding Current: It is the minimum anode current to maintain the Thyristor in the on-state.

• Latching current is always greater than holding current.
• Latching current and holding current terms have their significance in making a Thyristor ON and OFF respectively.
• Thyristor goes to on state when Thyristor current is more than latching current.
• Once Thyristor gets ON then there is no meaning of latching current.
• For 20A Thyristor we have to use BT152 and the ratio of latching current to holding current in BT152  is in the range of 2.

Given that,

v(t) = E1sin(ωt) + E3sin(3ωt)

i(t) = I1sin(ωt - Φ1) + I3sin(3ωt - Φ3) + I5sin(5ωt)

As the 5^th harmonic is not part of voltage expression, in the instantaneous power expression also we will have fundamental and third harmonic only.

p(t) = v(t) . i(t)

As wattmeter measures average power, it is given by

P = \(\frac{1}{2\pi}\mathop \smallint \limits_{0}^{2\pi} v(t).i(t) d \omega t\)     ------(1)

We know that,

\(\frac{1}{2\pi}\mathop \smallint \limits_{0}^{2\pi} A sin(\theta+\alpha).Bsin(\theta+\beta) d \theta=\frac{1}{2}ABcos(\alpha-\beta)\) -------(2)

By the use of the above result, if solve equation (1), we get

P = \(\frac{1}{2}[E_1I_1cos\phi_1+E_3I_3cos\phi_3]\)

Concept:

• Cayley - Hamilton theorem states that every square matrix A satisfies its own characteristic equation. i.e. p(A) = 0.
• The characteristic polynomial of a matrix A is defined as: p(λ) = |λI - A|.
• Identity Matrix:

An identity matrix is a matrix in which the diagonal elements are 1 and all the other elements are 0.

A 3×3 identity matrix is I = \(\rm \begin{bmatrix} 1 & 0 & 0 \0 & 1 & 0 \0 & 0 & 1 \end{bmatrix}\).

Matrix Multiplication by an Identity matrix results in the same matrix.

Calculation:

Let us find the characteristic polynomial p(λ) of the given matrix A = \(\rm \begin{bmatrix} -5 & -3\ 2 & 0 \end{bmatrix}\).

p(λ) = |λI - A|

⇒ p(λ) = \(\rm \begin{vmatrix} \lambda+5 & 3\ -2 & \lambda \end{vmatrix}\)

⇒ p(λ) = (λ )(λ +5) + (2)(3)

⇒ p(λ) = λ2 + 5λ + 6 

According to the Cayley - Hamilton theorem, p(A) = 0.

⇒ A2 + 5A + 6I = 0  --------(1)

⇒ A² = - 5A - 6I ---------(2)

By multiplying equation (1) with A, we get

A³ + 5A² + 6A = 0

⇒ A³ = -5 A² -6A = -5 (- 5A - 6I ) -6A = 19 A + 30I

Additional Information

Matrix Multiplication:

• Multiplication is only possible when the number of columns of the first matrix is equal to the number of rows of the second matrix.
• A m×n matrix multiplied by a n×p matrix results in a m×p matrix.
• Matrices are multiplied by multiplying each element of a row of the first m×n matrix with the corresponding elements of all the columns of the second n×p matrix to obtain the first row of the product matrix with p columns, and so on for all the m rows of the first matrix.

\(f\left( x \right) = {x^3} - 9{x^2} + 24x +5;\)

\({f^{'\left( x \right)}} = 3{x^2} - 18x + 24\)

\({f^{''\left( x \right)}} = 6x - 18\)

To find maxima or minima

\({f^{'\left( x \right)}} = 0\)

\(3{x^2} - 18x + 24 = 0\)

\({x^2} - 6x + 8 = 0\)

x = 4 or x = 2

If x = 4

f'’(4) = 6

f'’(4) > 0

∴ minima may fall at 4 in the given domain

If x = 2

f'’(2) = -6

f'’(2) < 0

∴ maxima may fall at 2 in the given domain

Check the boundary condition along with x = 4 and x = 2

∴ The maximum value of given function is 41

V_A – V_B = 6 V

So current in the branch will be

I_AB = 6/2 = 3 A

We can see, that the circuit is a one part circuit looking from terminal BD as shown below

For a one port network current entering one terminal, equals the current leaving the second terminal. Thus the outgoing current from A to B will be equal to the incoming current from D to C as shown.

i.e. I_DC = I_AB = 3 A

The total current in the resistor 1 Ω will be

I₁ = 2 + I_DC

= 2 + 3 = 5 A

V_CD = 1 × (-I₁)

= -5 V

The given circuit is a voltage – shunt or shunt – shunt negative feedback circuit.

R_F = 100 kΩ

R_S = 10 kΩ

V_o = - I_S R_F

\(\begin{array}{l} = - \left( {\frac{{{V_S}}}{{{R_S}}}} \right){R_F}\ {A_V} = \frac{{{V_o}}}{{{V_S}}} = voltage\ gain = \frac{{ - {R_t}}}{{{R_S}}} \end{array}\)

A_V = -10

|A_V| ≈ 10.

Concept:

Output expression (Y) for 2 × 1 MUX is given as

Y = A̅ x0 + A x1          ---(1)

The next state (Qn + 1) of D flip flop is:

Qn + 1 = D           ---(2)

Analysis:

From given circuit:

D = Y, x1 = Q, x0 = Q̅

We know that

Y = A̅ x0 + A x1

From given data,

D = A̅ Q̅ + AQ

From eqn (2) θn+1 = A̅ Q̅ + A Q

So, the state transition diagram will be:

Convolution sum is defined as

\(y\left[ n \right] = h\left[ n \right]*g\left[ n \right] = \mathop \sum \limits_{k = - \infty }^\infty h\left[ k \right]g\left[ {n - k} \right]\)

For causal sequence,

y[n] = h[0] g[n] + h[1] g [n-1] + h[2] g[n-2] +….

For n = 0,

y[0] = h[0] g[0] + h[1] g[-1] +…..

= h [0] g[0]            (g[-1] = g[-2] =……..0)

y[0] = h[0] g[0]

For. n =1,  y [1] = h [0] g [1] + h [1] g [0] + h [2] g [-1] + ……

y [1] = h [0] g [1] + h [1] g [0] +....

\(\frac{1}{2} = g\left[ 1 \right] + \frac{1}{{2}}g\left[ 0 \right]\)

Now, \(g\left[ 0 \right] = \frac{{y\left[ 0 \right]}}{{h\left[ 0 \right]}} = \frac{1}{1} = 1\)

g [1] = 0.5 - 0.5 = 0.

Active high pass filter:

The simplest HPF using an operational amplifier can be achieved by placing a capacitor in series with one of the resistors in the inverting amplifier circuit as shown.

Apply KCL at the inverting terminal node, we get

\(\frac{{{V_i} - 0}}{{{R_1} + \frac{1}{{jω C}}}} = \frac{{0 - {V_0}}}{{{R_2}}}\)

\( ⇒ \frac{{{V_0}}}{{{V_i}}} = \frac{{ - {R_2}}}{{{R_1} + \frac{1}{{jω C}}}}\)

Case 1:

If the frequency ω = 0 rad/sec

 \( \Rightarrow \frac{{{V_0}}}{{{V_i}}} = \frac{{ - {R_2}}}{{{R_1} + \frac{1}{0}}} = \frac{{ - {R_2}}}{{{R_1} + ∞ }} = 0\)

So, for the frequency ω = 0, the output V0 =0

Case 2:

If the frequency ω = ∞ rad/sec

\( \Rightarrow \frac{{{V_0}}}{{{V_i}}} = \frac{{ - {R_2}}}{{{R_1} + \frac{1}{∞ }}} = \frac{{ - {R_2}}}{{{R_1} + 0}} = \frac{{ - {R_2}}}{{{R_1}}}\)

So, for frequency ω = ∞ , the output \({V_0} = \frac{{ - {R_2}}}{{{R_1}}}{V_i}\)

By, observing case 1 and case 2 we can say that the above circuit is allowing high frequencies but the lower frequencies are not allowed. Hence the above circuit is a high pass filter (HPF)

And the cutoff frequency or the breakpoint of the filter can be calculated very easily by working out of the frequency at which the reactance of the capacitor equals the resistance resistor to which the capacitor is connected in series.

So, Xc = R₁

\( \Rightarrow \frac{1}{{{ω _c};C}} = {R_1}\)

\( \Rightarrow {ω _c} = \frac{1}{{{R_1}C}}\)

Where ωc is the cutoff frequency in rad/sec.

Hence the given circuit is a High pass filter with a cutoff frequency \(\frac{1}{{{R_1}C}}\).

Real power \({P_r} = \frac{{\left| {{V_s}} \right|\left| {{V_r}} \right|}}{{\left| x \right|}}\sin \delta\) 

\(\Rightarrow 1 = \frac{{1.0 \times 1.0}}{{0.5}}\sin \delta \Rightarrow \delta = 30^\circ\)

Reactive power \({\phi _r} = \frac{{\left| {{V_S}} \right|\left| {{V_r}} \right|}}{{\left| x \right|}}\cos \delta - \frac{{{V^2}}}{{\left| x \right|}}\)

\( = \frac{{\left( {1.0} \right)\left( {1.0} \right)}}{{0.5}}\cos 30 - \frac{{{{\left( {1.0} \right)}^2}}}{{0.5}}\)

= -0.268

Q_C + Q_r = 0 ⇒ Q_r = 0.268

P_S = P_e1 = 0.5

Before fault \({P_{m1}} = \frac{{{E_V}}}{X} = \frac{{1.5 \times 1.0}}{{1.5}} = 1.0\) 

During fault P_m2 = 0,

After the fault P_m3 = 1.0

\(\begin{array}{l} \delta = {\sin ^{ - 1}}\left( {\frac{{{P_S}}}{{{P_{m1}}}}} \right) = {\sin ^{ - 1}}\left( {\frac{{0.5}}{{1.0}}} \right) = 30^\circ \ {\delta _o} = \frac{{30 \times \pi }}{{180}} = 0.52\ rad\ {\delta _{max}} = 180 - {\sin ^{ - 1}}\left( {\frac{{{P_S}}}{{{P_{m3}}}}} \right)\ = 180 - {\sin ^{ - 1}}\left( {\frac{{0.5}}{1}} \right) = 150^\circ \ {\delta _{\max \left( {radians} \right)}} = \frac{{150 \times \pi }}{{180}} = 2.618rad\ {\delta _C} = {\cos ^{ - 1}}\left[ {\frac{{{P_S}\left( {{\delta _{max}} - {\delta _o}} \right) + {P_{m3}}\cos {\delta _{max}}}}{{{P_{m3}}}}} \right]\ = {\rm{ }}79.45^\circ \end{array}\)

Solution:-

Input voltage, Vin = 24V

Operating frequency, f = 100 kHz

Duty ratio, D = 50%

Inductor peak-to-peak current ripple, ΔIL = 1.6 A

Average inductor current, ILavg = 5 A

The peak inductor current ILpeak is calculated as:

ILpeak = ILavg + (ΔIL / 2) = 5 A + (1.6 A / 2) = 5 A + 0.8 A = 5.8 A

Since the peak current in the switch S is the same as the peak inductor current during the on-state, the peak current in S is:

I_Speak = 5.8 A

\(\begin{array}{l} {E_1} = 220 - 65.8 \times 0.25 = 203.55\ V\ {E_2} = 220 - 51 \times 0.25 = 207.25\ V\ \frac{{{N_2}}}{{{N_1}}} = \frac{{{E_2}}}{{{E_1}}} \times \frac{{{\phi _1}}}{{{\phi _2}}}\ \frac{{1600}}{{1000}} = \frac{{207.25}}{{203.55}} \times \frac{{{\phi _1}}}{{{\phi _2}}}\ \frac{{{\phi _1}}}{{{\phi _2}}} = 1.571\ \frac{{{\phi _1} - {\phi _2}}}{{{\phi _1}}} = 0.3636 \end{array}\)

Hence there is a 36.36% decrease

Concept:

For a geometric progression given as:

a, ar2, ar3, …, ∞

Where,

a > 0 and 0 < r < 1

The sum is given by:

\(Sum=\frac{a}{1-r}\)

Calculation:

The probability of appearing a head is 1/2. If the number of required tosses is odd, we have the following sequence of events.

H, TTH, TTTTH, …..

The required Probability P will be:

\(P = \frac{1}{2} + {\left( {\frac{1}{2}} \right)^3} + {\left( {\frac{1}{2}} \right)^5} + \ldots \)

\( = \frac{{\frac{1}{2}}}{{1 - \frac{1}{4}}} = \frac{2}{3}\)

\(\left| {\vec A} \right| = k.{r^n},{r^2} = {x^2} + {y^2} + {z^2}\)

\( \Rightarrow \vec A = k{r^n}.{\hat a_r}\) ........ (Given radial outward) 

In spherical coordinates

\(\vec \nabla .{\rm{\vec A}} = \frac{1}{{{{\rm{r}}^2}}}\frac{\partial }{{\partial {\rm{r}}}}\left( {{{\rm{r}}^2}{{\rm{A}}_{\rm{r}}}} \right) + 0 + 0 = \frac{1}{{{{\rm{r}}^2}}}{\rm{;}}\frac{\partial }{{\partial {\rm{r}}}}\left( {{{\rm{r}}^2}.{\rm{k}}{{\rm{r}}^{\rm{n}}}} \right)\)

\(\vec \nabla .{\rm{\vec A}} = \frac{{\rm{k}}}{{{{\rm{r}}^2}}}\frac{\partial }{{\partial {\rm{r}}}}\left( {{r^{\left( {n + 2} \right)}}} \right)=\frac{{\rm{k}}}{{{{\rm{r}}^2}}}(n+2){{}}\left( {{r^{\left( {n + 1} \right)}}} \right)\)

⇒ For  \(\vec \nabla .{\rm{\vec A}} = 0,~n+2 =0\)

\(n = - 2\)

\(\frac{{{d^2}y}}{{d{t^2}}} + \frac{{2dy}}{{dt}} + y = \delta \left( t \right)\)

Taking Laplace transform with initial conditions,

\(\begin{array}{l} \left[ {{s^2}y\left( s \right) - sy\left( 0 \right) - \frac{{dy}}{{dt}}{|_{t = 0}}} \right] + 2\left[ {sy\left( s \right) - y\left( 0 \right)} \right] + y\left( s \right) = 1\ \Rightarrow \left( {{s^2}y\left( s \right) + 2s - 0} \right) + 2\left[ {sy\left( s \right) + 2} \right] + y\left( s \right) = 1\ y\left( s \right)\left( {{s^2} + 2s + 1} \right) = 1 - 2s - 4\ y\left( s \right) = \frac{{ - 2s - 3}}{{{s^2} + 2s + 1}} \end{array}\)

We know that, If \(y\left( t \right)\mathop \leftrightarrow \limits^Z y\left( s \right)\)

Then, \(\frac{{dy}}{{dt}}\mathop \leftrightarrow \limits^Z sy\left( s \right) - y\left( 0 \right)\)

So, \(sy\left( s \right) - y\left( 0 \right) = \frac{{s + 2}}{{{{\left( {s + 1} \right)}^2}}} = \frac{1}{{s + 1}} + \frac{1}{{{{\left( {s + 1} \right)}^2}}}\)

Taking inverse Laplace,

\(\frac{{dy}}{{dt}} = {e^{ - t}}u\left( t \right) + t{e^{ - t}}u\left( t \right)\)

At \(t = {0^ + },\frac{{dy}}{{dt}}{|_{t = {0^ + }}} = {e^0} + 0 = 1\)

We obtain Thevenin equivalent of circuit B

Thevenin impedance:

Thevenin Voltage: V_th = 3∠0° V

Now, circuit becomes as

Current in the circuit,

I₁ = (10-3)/(2 + R)

Power transfer from circuit A to B

P = (I₁)²R + 3 I₁

= (42 + 70 R)/(2 + R)²

dP/dR = 0

⇒ (2 + R) [(2 + R) 70 – (42 + 70 R) 2] = 0

⇒ R = 0.8 Ω

\({Q_c} = \left[ {\begin{array}{*{20}{c}} B&{AB}&{{A^2}B} \end{array}} \right]\)

\({Q_c} = \left[ {\begin{array}{*{20}{c}} 0&0&{{a_1}{a_2}}\ 0&{{a_2}}&0\ 1&0&0 \end{array}} \right]\)

System is controllable if |Q_C| ≠ 0

a₁a₂ (0 – a₂) ≠ 0

\(- {a_1} \cdot a_2^2 \ne 0\)

Hence condition for controllability is

a₁ ≠ 0, a₂ ≠ 0, a₃ = 0

\(\begin{array}{l} H\left( {j\omega } \right) = \frac{{\left( {2\cos \omega } \right)\left( {\sin 2\omega } \right)}}{\omega }\ = \frac{{\sin 3\omega }}{\omega } + \frac{{sin\omega }}{\omega } \end{array}\)

We know that the inverse Fourier transform of sinc function is a rectangular function.

\(A~rect(\frac{t}{\tau}) \leftrightarrow A \tau ~Sa(\frac{\omega \tau}{2})\)

Now,

\(Sa(\frac{\omega \tau}{2}) = \frac{sin~(\frac{\omega \tau}{2})}{\frac{\omega \tau}{2}}\)

\(\frac{sin ~3\omega}{\omega} = 3 \frac{sin~(3\omega)}{3\omega}\)

By comparing, we get

\(\tau/2 = 3, A\tau = 3 \)

hence, A = 0.5

2) 

\(\frac{sin ~\omega}{\omega} = 1\times \frac{sin~(\omega \times 1)}{\omega \times 1}\)

By comparing, we get

\(\tau/2 = 1, A\tau = 1\)

hence, A = 0.5

So, inverse Fourier transform of H (jω)

h(t) = h_1­(t) + h₂(t)

h(0) = h₁(0) + h_2­­(0)

\(\frac{1}{2}+\frac{1}{2} = 1\)

Characteristic equation is       s³ + as² + (k + 2) s + (k + 1) = 0

Routh array

System oscillates if \(\frac{{a\left( {k + 2} \right) - \left( {k + 1} \right)}}{a} = 0\)

ak + 2a = k + 1

\(a = \frac{{k + 1}}{{k + 2}}\)

System oscillates at ω = 2 rad/sec

as² + k + 1 = 0

\(\frac{{k + 1}}{{k + 2}}{s^2} + k + 1 = 0\)

s² = – (k + 2)

\(s = \pm \sqrt {k + 2} ;j\)

Comparing \(\sqrt {k + 2} = 2\)

k = 2

\(a = \frac{{k + 1}}{{k + 2}} = \frac{{2 + 1}}{{2 + 2}} = 0.75\)

1. Time Invariant and Time Variant System:

If in a system, a delay in the input leads to same delay in the output, then such system is known as time invariant system. Otherwise It will be time variant system.

2. Stable and Unstable system:

The system is sold to be stable only when the bounded output for the input.

For a bounded input if output is unbounded then system is said to be UNSTABLE.

Condition:

\(\mathop \smallint \nolimits_{ - \infty }^\infty \left| {h\left( t \right)} \right|dt < \infty \) or finite.

Where h(t) is impulse response of system.

Solution:

Given, \(y\left( t \right) = \mathop \smallint \nolimits_{ - \infty }^t x\left( \tau \right)\cos \left( {3\tau} \right)d\tau\) 

Check for time-invariance:

Since y’(t) ≠ y (t – t0) so it is time-variant

Stability:

Suppose input x(t) = cos 3t

\(y\left( t \right) = \mathop \smallint \nolimits_{ - \infty }^t {\cos ^2}3\tau dt = \mathop \smallint \nolimits_{ - \infty }^t \left( {\frac{{1 + \cos 6\tau }}{2}} \right)d\tau ;\)

\(y\left( t \right) = [\left. {\frac{1}{2}t\tau } \right|_{ - \infty }^t + \frac{1}{2}\sin \left. {\frac{{6\tau }}{6}} \right|_{ - \infty }^t\)

Since y(t) → ∞, it is unbounded of P so this is an unstable system.

TRICK:

→ check for stability: try to find at least one input value for which output will be unbounded.

e.g. \(y\left( t \right) = \frac{{{e^x}\left( {t - 5} \right)}}{{t - 5}}\) at t = 5 y(t) → ∞ - unbounded.

→ check for variance: If any coefficient if input is function of time (t) then it will be time variant.

e.g. tx(t), etx(t), sin t x(t)

Multiplier setting = 20 kΩ

Meter resistance = x

Total resistance = 20 kΩ + x = R₁

Voltmeter reading (V₁) = 440 V

\(\begin{array}{l} {V_1}\alpha {I_1}x\ {I_1}\alpha \frac{1}{{{R_1}}} \end{array}\)

For multiplier setting = 80 kΩ

Meter resistance = x

Total resistance = 80 KΩ + x = R₂

Voltmeter reading (V₂) = 352 V

\(\begin{array}{l} {V_2}\alpha {I_2}x\ {I_2}\alpha \frac{1}{{{R_2}}}\ \frac{{{V_2}}}{{{V_1}}} = \frac{{{I_2}}}{{{I_1}}} = \frac{{{R_1}}}{{{R_2}}}\ \frac{{352}}{{440}} = \frac{{20 + x}}{{80 + x}} \Rightarrow x = 220\ k{\rm{\Omega }} \end{array}\)

For multiplier setting = 40 KΩ

R₃ = 260 kΩ

\(\frac{{{V_3}}}{{{V_1}}} = \frac{{{I_3}}}{{{I_1}}} = \frac{{{R_1}}}{{{R_3}}} \Rightarrow \frac{{{V_3}}}{{440}} = \frac{{240}}{{260}} \Rightarrow {V_3} = 406\ V\)

\(\begin{array}{l} {I_{se}} = \frac{V}{{{X_{se}}}}\ \alpha\ \frac{v}{f}\ \therefore {I_{se}}\ nw = 50 \times \frac{{236}}{{230}} \times \frac{{50}}{{57}}\ = 45A \end{array}\)

Core loss α core volume

∴ New loss \({\left( {\sqrt 2 } \right)^3} \times 55 = 155.6W\)

Core loss \({P_i} = V{I_c}\)

\(\begin{array}{l} {I_{{c_1}}} = \frac{{55}}{{1000}} = 0.055\ A\ {I_{{c_2}}} = {\left( {\sqrt 2 } \right)^3}{I_{{c_1}}} \end{array}\)

From the data for 2^nd transformer

\(\begin{array}{l} {I_{{c_2}}} = 0.155\ A\ \therefore {I_{{m_1}}} = 0.496\left[ {\therefore \sqrt {{{\left( {0.5} \right)}^2} - {{\left( {0.055} \right)}^2}} = 0.496} \right] \end{array}\)

Reluctance \({R_2} = \frac{{{R_1}}}{{\sqrt 2 }}\)

\({\emptyset _{{m_1}}} = \frac{{4000}}{{4.44{N_1}f}}\)             \({\emptyset _{{m_2}}} = \frac{{2000}}{{4.44\ {N_1}f}}\)

\(\begin{array}{l} {\emptyset _{{m_2}}} = 2{\emptyset _{{m_1}}}\ {\emptyset _{{m_1}}} = \frac{{{N_1}{I_m}_2}}{{{R_1}}}\ {\emptyset _{{m_2}}} = \frac{{{N_1}{I_m}_2}}{{\frac{{{R_1}}}{{\surd 2}}}}\ \therefore {I_{{m_2}}} = \surd 2{I_{{m_1}}}\ = \sqrt 2 \times 0.496 \end{array}\)

= 0.702 A

\(\begin{array}{l} \therefore {I_0} = \sqrt {I_{{m_1}}^2 + I_{{c_2}}^2} \ = \sqrt {{{\left( {0.7025} \right)}^2} - {{\left( {0.155} \right)}^2}} = 0.718\ A \end{array}\)

Concept:

Calculation:

A three-phase voltage source inverter supplying equivalent star load.

Equivalent star load resistance is,

\({R_{ph}} =20;{\rm{\Omega }}\ {V_{ph}} = \frac{{\sqrt 2 }}{3}{V_{dc}} = \frac{{\sqrt 2 }}{3} \times 300 = 200\sqrt 2 V = 141.42;V\)

Concept:

Calculation:

A three-phase voltage source inverter supplying equivalent star load.

Equivalent star load resistance is,

\(\begin{array}{l} {R_{ph}} =20;{\rm{\Omega }}\ {V_{ph}} = \frac{{\sqrt 2 }}{3}{V_{dc}} = \frac{{\sqrt 2 }}{3} \times 300 = 200\sqrt 2 V = 141.42;V\ P = 3\frac{{{V_p}{h^2}}}{R} = 3 \times \frac{{{{\left( {141.42 } \right)}^2}}}{{20}} = 3kW \end{array}\)

Now, when 6 V connected at port A let thevenin voltage seen at port B is V_{th,6 V} . Here R_L = 1 Ω and I_L = 7/3 A

\(% MathType!Translator!2!1!LaTeX.tdl!LaTeX 2.09 and later! % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGwbWdamaaBaaaleaapeGaamiDaiaadIgacaGGSaGaaGOnaiaa % dAfaa8aabeaak8qacqGH9aqpcaWGsbWdamaaBaaaleaapeGaamiDai % aadIgaa8aabeaak8qacqGHxdaTdaWcaaWdaeaapeGaaG4naaWdaeaa % peGaaG4maaaacqGHRaWkcaaIXaGaey41aq7aaSaaa8aabaWdbiaaiE % daa8aabaWdbiaaiodaaaaaaa!4931! {V_{th,6V}} = {R_{th}} \times \frac{7}{3} + 1 \times \frac{7}{3}% MathType!End!2!1! \)

\(% MathType!Translator!2!1!LaTeX.tdl!LaTeX 2.09 and later! % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacqGH9aqpcaaIYaGaey41aq7aaSaaa8aabaWdbiaaiEdaa8aabaWd % biaaiodaaaGaey4kaSYaaSaaa8aabaWdbiaaiEdaa8aabaWdbiaaio % daaaGaeyypa0JaaG4naiaadAfaaaa!410C! = 2 \times \frac{7}{3} + \frac{7}{3} = 7V% MathType!End!2!1! \)

This is a linear network, so V_th­ at port B can be written as

V_th = V₁ α + β

Where V₁ is the input applied at port A.

We have V₁ = 10 V, V_{th,10 V} = 9 V

9 = 10 α + β

when V₁ = 6 V, V_{th,6 V} = 9 V

7 = 6 α + β

⇒ α = 0.5, β = 4

Thus, with any voltage V₁ applied at port A, thevenin voltage or open circuit voltage at port B will be

\(% MathType!Translator!2!1!LaTeX.tdl!LaTeX 2.09 and later! % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGwbWdamaaBaaaleaapeGaamiDaiaadIgacaGGSaGaamOva8aa % daWgaaadbaWdbiaaigdaa8aabeaaaSqabaGcpeGaeyypa0JaaGimai % aac6cacaaI1aGaaiiOaiaadAfapaWaaSbaaSqaa8qacaaIXaaapaqa % baGcpeGaey4kaSIaaGinaaaa!43E8! {V_{th,{V_1}}} = 0.5;{V_1} + 4% MathType!End!2!1! \)

V₁ = 8 V

V_{th,8 V} = (0.5 × 8) + 4 = 8 = open circuit voltage

When 10 V is connected at port A the network is

Now, we obtain Thevenin equivalent for the circuit seen at load terminal, let thevenin voltage is V_{th ,10 V} with 10 V applied at port A and thevenin resistance is R_th.

\(% MathType!Translator!2!1!LaTeX.tdl!LaTeX 2.09 and later! % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGjbWdamaaBaaaleaapeGaamitaaWdaeqaaOWdbiabg2da9maa % laaapaqaa8qacaWGwbWdamaaBaaaleaapeGaamiDaiaadIgacaGGSa % GaaGymaiaaicdacaWGwbaapaqabaaakeaapeGaamOua8aadaWgaaWc % baWdbiaadshacaWGObaapaqabaGcpeGaey4kaSIaamOua8aadaWgaa % WcbaWdbiaadYeaa8aabeaaaaaaaa!45A9! {I_L} = \frac{{{V_{th,10V}}}}{{{R_{th}} + {R_L}}}% MathType!End!2!1! \)

For R_L = 1 Ω, I_L = 3 A

\(% MathType!Translator!2!1!LaTeX.tdl!LaTeX 2.09 and later! % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaaIZaGaeyypa0ZaaSaaa8aabaWdbiaadAfapaWaaSbaaSqaa8qa % caWG0bGaamiAaiaacYcacaaIXaGaaGimaiaadAfaa8aabeaaaOqaa8 % qacaWGsbWdamaaBaaaleaapeGaamiDaiaadIgaa8aabeaak8qacqGH % RaWkcaaIXaaaaaaa!430C! 3 = \frac{{{V_{th,10V}}}}{{{R_{th}} + 1}}% MathType!End!2!1! \)

For R_L = 2.5 Ω, I_L = 2 A

\(% MathType!Translator!2!1!LaTeX.tdl!LaTeX 2.09 and later! % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaaIYaGaeyypa0ZaaSaaa8aabaWdbiaadAfapaWaaSbaaSqaa8qa % caWG0bGaamiAaiaacYcacaaIXaGaaGimaiaadAfaa8aabeaaaOqaa8 % qacaWGsbWdamaaBaaaleaapeGaamiDaiaadIgaa8aabeaak8qacqGH % RaWkcaaIYaGaaiOlaiaaiwdaaaaaaa!447D! 2 = \frac{{{V_{th,10V}}}}{{{R_{th}} + 2.5}}% MathType!End!2!1! \)

\(% MathType!Translator!2!1!LaTeX.tdl!LaTeX 2.09 and later! % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacqGHshI3daWcaaWdaeaapeGaaG4maaWdaeaapeGaaGOmaaaacqGH % 9aqpdaWcaaWdaeaapeGaamOua8aadaWgaaWcbaWdbiaadshacaWGOb % aapaqabaGcpeGaey4kaSIaaGOmaiaac6cacaaI1aaapaqaa8qacaWG % sbWdamaaBaaaleaapeGaamiDaiaadIgaa8aabeaak8qacqGHRaWkca % aIXaaaaaaa!469D! \Rightarrow \frac{3}{2} = \frac{{{R_{th}} + 2.5}}{{{R_{th}} + 1}}% MathType!End!2!1! \)

⇒ R_th = 2 Ω

V_{th,10 V} = 3 (2 + 1) = 9 V

Note than it is a nonreciprocal two port network thevenin voltage seen at port B depends on the voltage connected at port A. therefore we took subscript V_{th,10 V} . This is thevenin voltage only when 10 V source is connected at input port A. If the voltage connected to port A is different, then thevenin voltage will be different. However, Thevenin’s resistance remains same. Now, the circuit is

For R_L = 7 Ω 

\(% MathType!Translator!2!1!LaTeX.tdl!LaTeX 2.09 and later! % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGjbWdamaaBaaaleaapeGaamitaaWdaeqaaOWdbiabg2da9maa % laaapaqaa8qacaWGwbWdamaaBaaaleaapeGaamiDaiaadIgacaGGSa % GaaGymaiaaicdacaWGwbaapaqabaaakeaapeGaaGOmaiabgUcaRiaa % dkfapaWaaSbaaSqaa8qacaWGmbaapaqabaaaaOWdbiabg2da9maala % aapaqaa8qacaaI5aaapaqaa8qacaaIYaGaey4kaSIaaG4naaaacqGH % 9aqpcaaIXaGaaiiOaiaadgeaaaa!4B6F! {I_L} = \frac{{{V_{th,10V}}}}{{2 + {R_L}}} = \frac{9}{{2 + 7}} = 1;A% MathType!End!2!1! \)

By taking V₁, V₂, V₃ all are phasor voltages.

V₁ = V₂ + V₃

⇒ V₁ = V₂ ∠0° + V₃ ∠θ

V₁ = V₂ + V₃ cos θ + j V₃ sin θ

V1 = (V₂ + V₃ cos θ) + j V₃ sin θ

\(% MathType!Translator!2!1!LaTeX.tdl!LaTeX 2.09 and later! % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaabdaWdaeaapeGaamOva8aadaWgaaWcbaWdbiaaigdaa8aabeaa % aOWdbiaawEa7caGLiWoacqGH9aqpdaGcaaWdaeaapeWaaeWaa8aaba % WdbiaadAfapaWaaSbaaSqaa8qacaaIYaaapaqabaGcpeGaey4kaSIa % amOva8aadaWgaaWcbaWdbiaaiodaa8aabeaak8qacaWGJbGaam4Bai % aadohacqaH4oqCaiaawIcacaGLPaaapaWaaWbaaSqabeaapeGaaGOm % aaaakiabgUcaRmaabmaapaqaa8qacaWGwbWdamaaBaaaleaapeGaaG % 4maaWdaeqaaOWdbiaadohacaWGPbGaamOBaiabeI7aXbGaayjkaiaa % wMcaa8aadaahaaWcbeqaa8qacaaIYaaaaaqabaaaaa!5301! \left| {{V_1}} \right| = \sqrt {{{\left( {{V_2} + {V_3}cos\theta } \right)}^2} + {{\left( {{V_3}sin\theta } \right)}^2}} % MathType!End!2!1! \)

\(% MathType!Translator!2!1!LaTeX.tdl!LaTeX 2.09 and later! % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaaIYaGaaGOmaiaaicdacqGH9aqpdaGcaaWdaeaapeWaaeWaa8aa % baWdbiaaigdacaaIYaGaaGOmaiabgUcaRiaaigdacaaIZaGaaGOnai % aadogacaWGVbGaam4CaiabeI7aXbGaayjkaiaawMcaa8aadaahaaWc % beqaa8qacaaIYaaaaOGaey4kaSYaaeWaa8aabaWdbiaaigdacaaIZa % GaaGOnaiaadohacaWGPbGaamOBaiabeI7aXbGaayjkaiaawMcaa8aa % daahaaWcbeqaa8qacaaIYaaaaaqabaaaaa!5068! 220 = \sqrt {{{\left( {122 + 136cos\theta } \right)}^2} + {{\left( {136sin\theta } \right)}^2}} % MathType!End!2!1! \)

cosθ = 0.45

By taking V₁, V₂, V₃ all are phasor voltages.

V₁ = V₂ + V₃

⇒ V₁ = V₂ ∠0° + V₃ ∠θ

V₁ = V₂ + V₃ cos θ + j V₃ sin θ

V₁ = (V₂ + V₃ cos θ) + j V₃ sin θ

\(\left| {{V_1}} \right| = \sqrt {{{\left( {{V_2} + {V_3}\cos \theta } \right)}^2} + {{\left( {{V_3}\sin \theta } \right)}^2}} \)

\(220 = \sqrt {{{\left( {122 + 136\cos \theta } \right)}^2} + {{\left( {136\sin \theta } \right)}^2}} \)

⇒ cosθ = 0.45

V_RL = V₃ cos θ = 136 × 0.45 = 61.2 V

Power absorbed in R_L

\({P_L} = \frac{{V_{RL}^2}}{{{R_L}}} = \frac{{{{\left( {61.2} \right)}^2}}}{5} = 750;W\)

Concept:

The general expression for a lead compensator is;

\(G\left( s \right)=\frac{\left( \alpha \right)\left( 1+Ts \right)}{\left( 1+\alpha Ts \right)}\)

And for a lead compensator α < 1

Analysis:

Given lead compensator expression is;

\({{G}_{c}}\left( s \right)=\frac{k\left( s+a \right)}{\left( s+b \right)}\)

On comparing it with the standard expression we get;

\(T=\frac{1}{a}~,~~\alpha T=\frac{1}{b}\)

\(\Rightarrow \frac{\alpha }{a}=\frac{1}{b}\Rightarrow \alpha =\frac{a}{b}\)

⇒ a < b

Only two options are valid as per the above condition, options 1 and 3

But we can't go with option 3, because for lead compensator zero is near to the origin compared to pole.

Hence, a = 1, b = 2

Maximum phase at frequency

 \(\rm {\omega _m} = \frac{1}{{T\sqrt \alpha }}\)   \(\rm \left( {{\rm{for}}\ {G_c}\left( s \right) = \frac{{1 + sT}}{{1 + \alpha sT}};\alpha < 1} \right)\)

\(\rm {G_c}\left( s \right) = \frac{{s + 1}}{{s + 2}} = \frac{{1 + s}}{{1 + 0.5s}}\)

Comparing \(\rm T = 1, αT = 0.5\)

\(\rm α = 0.5\)

\(\rm {\omega _m} = \frac{1}{{1\sqrt {0.5} }} = \sqrt 2\ {\rm{rad}}/{\rm{sec}}\)