GATE EE 2013 Official Paper (Previous Year Paper)

The correct answer is 'fall out'.

Key Points

• Quarrel: a heated argument or disagreement, typically about a trivial issue and between people who are usually on good term.
• Fall out: have an argument.
• Thus, from above we can conclude that the correct answer is option 4.

Additional Information

• let's explore other options:
• Make out: manage with some difficulty to see or hear someone or something.
• Call out: an instance of being summoned to deal with an emergency or do repairs.
• Dig out: extract something from the ground by breaking up and moving earth

(Monday + Tuesday + Wednesday) / 3 = 41

(Monday + Tuesday + Wednesday) = 123

Tuesday + Wednesday = 123 - Monday      ----(1)

(Tuesday + Wednesday + Thursday) / 3= 43

(Tuesday + Wednesday + Thursday) = 129      ----(2)    

We know, Thursday = 1.15 Monday

⇒ 123 - Monday + 1.15 Monday = 129

⇒ Monday = 40⁰C

Equation (2) becomes

⇒ 123 - Monday + Thursday = 129  

⇒ 123 - 40 + Thursday = 129

⇒ Thursday = 460C

Hence, the correct answer is 460C.

The correct answer is 'commit'.

Key Points

• Dare is both a main verb and a semi-modal verb. Dare can mean ‘challenge somebody’. With this meaning, it is a main verb and requires an object. Any verb that follows it is in the to-infinitive.
• Example: Some snakes can bite but I dare you to hold this big snake.
• Dare also means ‘to be brave enough or rude enough to do something’. With this meaning it is used as a semi-modal verb followed by an infinitive without to.
• Example: No one dare go there.
• The context of 'Dare' in the given sentence is 'brave enough or rude enough to do something', thus, it will be followed by an infinitive without 'to'.
• Therefore, the correct answer is option 1.

The correct sentence is: 'Dare commit mistakes.'

The correct answer is 'Two and Two make four'.

Key Points

• In the above-given sentences, 'Two' and 'Two' are two different subjects connected with conjunction 'and' which will be treated as plural subject and therefore, will require a plural verb i.e. 'make'.
• Thus, the correct answer is option 4.

Additional Information

• As per the rule, Whenever two subjects are connected with the conjunction 'and' it will be treated as plural subjects.

The correct answer is 'Because a friend in need is a friend indeed.'

Key Points

• A friend in need is a friend indeed: It is a popular proverb which means a person who helps at a difficult time is a person who you can really rely on.
• Let's refer to the lines:
• You can always give me a ring whenever you need.
• Thus, from above we can infer that the correct answer is 'Because a friend in need is a friend indeed' because with the context of the sentence we can say that the person is expressing his desire to be with another person in his difficult time.
• Therefore, the correct answer is option 3.

A leap year has 366 days or 52 weeks and 2 odd days.

The two odd days can be (Saturday,Sunday), (Sunday,Monday), (Monday,Tuesday), (Tuesday,Wednesday), (Wednesday,Thursday), (Thursday,Friday),(Friday,Saturday).

So there are 7 possibilities out of which 2 have a Saturday.

So the probability of 53 Saturday is 2/7. 

Hence, the correct answer is 2/7.

The correct answer is 'Nationalism in India is heterogeneous.'

Key Points

• Let's refer to the lines:
• There were different streams of freedom movements in colonial India carried out by the moderates, liberals, radicals, socialists, and so on.
• Heterogeneous: diverse in character or content.
• From above, we can infer that the 'Nationalism in India is heterogeneous' because it included different factions of India and these different factions brought diverse ideologies with the same intent of serving and supporting our own nation. Our then Prime minister Jawarharlal Nehru has beautifully quoted it as 'Unity in Diversity'.
• Thus, the correct answer is option 4.

p(p−1) < 0, because product of roots is a negative number.

Thus p must be less than 1 and greater than 0

We know that,

Speed = Distance/Time

Now, according to the question, the car has traveled 30 km in three-quarters of an hour.

Total time taken by the car = (15 + 15 + 15) min = 45/60 hour = 3/4 hour.

So, the average speed of the car = Total distance/Total time

\(= \frac{{30}}{{\frac{3}{4}}} = 10 \times 4 = 40\frac{{km}}{h}\)

Hence, the required answer is 40 km/h.

The given series is a sum of Geometric progression and Arithmetic Progression.

The given series is

⇒ 10 + 84 + 734 + _ _ _ _ _

⇒ (9¹ + 1) + (9² + 3) + (9³ + 5) +  _ _ _ _ _

Thus the formula becomes

⇒ Σ (GP + AP)

Where, GP = Geometric Progression 

AP = Arithmetic Progression

⇒ a₁ (rⁿ - 1) / (r - 1) + [(n / 2) × (2 a₂ + (n - 1) d)]

Where, a1 = First term of GP i.e. 9

a₂ = First term of AP i.e. 1

r = ratio in GP of second term to first term i.e. 9

d = difference of two consecutive terms in AP i.e. 2

n = number of terms in AP

Since, ΣGP with r > 1 is [a1 (rn - 1) / (r - 1)]

ΣAP is [(n / 2) × (2 a2 + (n - 1) d)]

⇒ 9 (9ⁿ - 1) / (9 - 1) + [(n / 2) × (2 + (n -1) 2)]

⇒ 9 (9n - 1) / 8 + [(n / 2) × 2n]

⇒ 9(9n - 1)/8 + 2n²/2

⇒ \(\frac{9(9^n-1)}{8}+n^2\)

Hence, the correct answer is \(\frac{9(9^n-1)}{8}+n^2\).

Concept:

Advantage of negative feedback

The open-loop gain of an opamp depends on temperature, frequency, process variation, manufacturer etc… with the negative feedback we desensitise the output from the open-loop gain such that the overall closed-loop gain depends only on resistors, therefore the gain is predictable.

Opamp is amplifying the difference. We can say this as an error amplifier.

In the negative feedback, the error converges to give stability.

Analysis:

V₂ = β V₀

\( = \beta \frac{A}{{1 + A\beta }}{V_1}\)

\({V_2} = \frac{{{V_1}}}{{1 + \frac{1}{{A\beta }}}}\)

If loop gain (Aβ ) is large, ideally ∞

then V₂ = V₁  ( Virtual ground )

NOTE: virtual ground only exists in negative feedback because in positive feedback error will diverge.

Calculation:

Given that the opamp is ideal and is in negative feedback. So we can apply the virtual ground concept.

From the diagram voltage at the negative terminal also zero by applying virtual ground.

The current through a resistor is

\(I = \frac{{5 - 0}}{{1K{\rm{\Omega }}}} = 5mA\)

This current completely flows through the BJT since ideal opamp doesn’t take any current.

For BJT V_C = 0 volts and V_B = 0 volts, I_C = 5 mA

The base-collector junction is reverse biased ( ∵  zero volts ), therefore the collector current can have a value only if the base-emitter junction is forward biased.

Forward bias ⇒ V_BE = 0.7 volts.

V_B – V_E = 0.7

0 – V_E = 0.7

From the given circuit V_out = V_E

So V_out = - 0.7 volts

 Important points:

1. V₁ = V₂ valid only if open loop gain ( A ) is ∞ and in negative feedback
2. V₀ = 0 only when output resistance r₀ = 0 Ω
3. I_op = 0 only when input resistance R_id = ∞

Concept:

The transfer function of a linear time-invariant function is defined to be the ratio of the Laplace transform of the output variable to the Laplace transform of the input variable under the assumption that all initial conditions are zero.

Order: The highest power of the complex variable ‘s’ in the denominator of the transfer function determines the order of the system.

In the Laplace domain, inductor impedance is written as “sL”

and capacitance impedance as:

\(\frac{1}{{sC}}\) .

For a resistor, there is no change.

Calculation:

From the given circuit capacitor is of 100 μ F value and resistor of 10 KΩ  

Capacitive impedance is

 \(\frac{1}{{sC}} = \frac{1}{{s100 \times {{10}^{ - 6}}}}\) 

\( = \frac{1}{{s{{10}^{ - 4}}}}\)

\( = \frac{{{{10}^4}}}{s}\)

Assume current I is flowing in a circuit and the circuit in Laplace domain will be

Output is

\({V_2}\left( s \right) = I\left( s \right)\left( {10 \times {{10}^3} + \frac{{{{10}^4}}}{s}} \right)\)

Applying KVL to the loop we get

 \( - {V_1}\left( s \right) + I\left( s \right)\left( {\frac{{{{10}^4}}}{s}} \right) + I\left( s \right)\left( {{{10}^4} + \frac{{{{10}^4}}}{s}} \right) = 0\)

\({V_1}\left( s \right) = I\left( s \right)\left( {2 \times \frac{{{{10}^4}}}{s} + {{10}^4}} \right)\)

Now the ratio of Laplace transform is:

 \(\frac{{{V_2}\left( s \right)}}{{{V_1}\left( s \right)}} = \frac{{I\left( s \right)\left( {\frac{{s + 1}}{s}} \right){{10}^4}}}{{I\left( s \right)\left( {\frac{{s + 2}}{s}} \right){{10}^4}}} \) 

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

 Important points:

NOTE: poles and zeroes are nothing but the negative of the inverse of time constant

From the given block diagram:

\(\frac{y\left( s \right)}{u\left( s \right)}=\frac{1}{s}\)

For unit step input u(t), the response will be:

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

Applying Inverse Laplace transform, we get:

⇒ y(t) = t u(t)

\(y\left( t \right) = ;u\left( {t - 1} \right)*tu\left( t \right)\)

Taking Laplace Transform,

\(Y\left( s \right) = \frac{{{e^{ - s}}}}{s} \times \frac{1}{{{s^2}}}\)

\(Y\left( s \right) = \frac{{{e^{ - s}}}}{{{s^3}}}\)

\(Y\left( s \right)\mathop \to \limits^{ILT} \frac{{{{\left( {t - 1} \right)}^2}}}{2}u\left( {t - 1} \right)\)

[As \(\frac{1}{{{s^2}}}\mathop \to \limits^{ILT} t;u\left( t \right)\)&  shifting in time leads to phase in frequency]

Concept:

Continuous-time signal: A signal of continuous amplitude and time is known as a continuous-time signal or an analog signal.

Discrete-time signal: Unlike a continuous-time signal, a discrete-time signal is not a function of a continuous argument; however, it may have been obtained by sampling from a continuous-time signal. When a discrete-time signal is obtained by sampling a sequence at uniformly spaced times, it has an associated sampling rate.

Analysis:

We will check each option to find out the incorrect one.

All the poles of the system must lie on the left side of the jω axis.

• The system is stable if and only if all the roots lie on the left side of the S plane or jω axis. Also, we can say all the poles have a negative real part.

Example:

If ROC is Re[s]>-a, the system h(t) =e^-at u(t) is causal.

Its Laplace transform \({\rm{H}}\left( {\rm{s}} \right) = \frac{1}{{{\rm{a}} + {\rm{s}}}}\)

If a<0, i.e., imaginary axis Re[s] =0 can be included in the ROC, the system is stable.

Zeros of the system can lie anywhere on the plane.

When zeros are in the right half of the S plane or outside the unit circle in the Z- plane, it does not cause the system to be unstable.

All the poles of the system must lie within |s|=1

• The system is stable if and only if the ROC of the system function H(z) includes the unit circle |z|=1 and not |s|=1.
• The system is stable in the case of rational system function H(z) if and only if all the poles of H(z) lie inside the unit circle, they must all have a magnitude smaller than unity.

All the roots of the characteristics equation must lie on the left side of the jω axis.

For stability, all the roots of the characteristics equation must lie on the left side of the s-plane, even if one root lies on the right side the system becomes unstable.

Conclusion:

So, we can say options a, b and d are satisfying property of a continuous-time casual and stable system.

Option c is correct for a discrete-time casual and stable LTI system.

Concept:

The systems which are cascaded in the time domain are always convolved but in the time domain and are multiplied in the frequency domain.

Given two impulse responses are h1(t) and h2(t).

These are cascaded, so the resulting system response will be a convolution of these two systems.

Properties:

Commutative property:

 x(t) ∗ h(t) = h(t) ∗ x(t)

Associative property: 

x(t) ∗ [h1(t) ∗ h2(t)] = [x(t) ∗ h1(t)] ∗ h2(t)

Distributive property:

x(t) ∗ [h1(t) + h2(t)] = x(t) ∗ h1(t) + x(t) ∗  h2(t)

Convolution with impulse property:

 x(t) ∗ δ (t) = x(t)

Width property: if the durations(widths) of x(t) and h(t) are finite and given by Wx and Wh,

Then the duration(width) of the x(t) ∗ h(t) is Wx + Wh

Important points:

Convolution of two equal width rectangles will give resultant as a triangular profile.

Convolution of two unequal width rectangles will give resultant as a trapezoidal profile.

For the purely resistive load, maximum average power is transferred when

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Concept:

The complex power across the load is given by

\(S = V × {I^*}\) = P + j Q

S = Apprent power

V = Voltage across the load

I* = Conjugate of current through the load

P = Active Power 

Q = Reactive power

Calculation:

Given that,

Load current outgoing ILO  = 10∠-150° A

Load current incoming ILi = - ILO = - 10∠-150° A = 10∠30° A

Load voltage VL = Source voltage VS = 100∠60° V

Source current IS = - ILO = - 10∠-150° A = 10∠30° A

(Outgoing load current is equal to incoming load current)

Power across load SL = V × ILO*

⇒ \({I_{LO}^*} = 10\angle150^{\circ}; A;\)

SL = VILO*

= (100 ∠60°) (10∠150°) = 1000∠210°

SL = - 866.02 - j 500 = P + j Q 

Since the current was going from load, so the power is going to deliver by the load to the source.

∴ The load will deliver the - 866.02 W of active power and - 500 VAR of reactive power.

or we can also say that load absorb 866.02 W of active power and 500 VAR of reactive power.

Alternate MethodPower across load SL = V × ILi*

⇒ \({I_{Li}^*} = 10\angle-30^{\circ}; A;\)

SL = VILi*

= (100 ∠60°) (10∠-30°) = 1000∠30°

SL = 866.02 + j 500 = P + j Q 

Since in this case we have taken as incoming current, so the power is going to absorb by the load.

From the value of apparent power, we can observe that both active and reactive are positive.

The power is positive across any element means the element absorbing the power, so here the load is absorbing both active and reactive power to the source.

Important Points

Case 1: Q > 0:

Load is inductive in nature.

Absorbed Lagging VAr by inductor

Deliver Leading VAr by inductor

Case 1: Q < 0:

Load is capacitive in nature.

Absorbed Leading VAr by capacitor

Delivered Lagging VAr by source.

Concept:

The efficiency of Transformer \((\eta ) = \dfrac{{x\ S\cos \phi }}{{x\ S\cos \phi \ +\ {P_i} \ +\ {X^2}\ {P_{cu FL}}}}\)

Where,

x = Fraction of load

S = Apparent power in kVA

Pi = Iron losses

PcuFL = Full load copper losses

Maximum efficiency of transformer occurred at a fraction of load at

 \(x = \sqrt {\dfrac{{{P_i}}}{{{P_{cuFL}}}}}\)

A short circuit test is used in a transformer to find copper losses and the open circuit test is used to find core losses.

Calculation:

Given: core losses P_i = 64 W, 

copper losses P_cu at 90% of load = 81 W

P_cu = x² P_cuFL

P_cuFL = P_cu / x²

PcuFL = 81 / 0.9² = 100 W

For maximum Efficiency to Occur,

The Transformer must be operated at load \(x = \sqrt {\dfrac{{{P_i}}}{{{P_{cuFL}}}}}\)

\(x = \sqrt {\dfrac{{{64}}}{{{100}}}}=0.8\)

= 80 % of Rated current

We know that,

∇.B = 0

\(\left( {\frac{\partial }{{\partial x}}{a_x} + \frac{\partial }{{\partial y}}{a_y} + \frac{\partial }{{\partial z}}a_z} \right)\left( {4 x{a_x} + 2{ky}{a_y} + 8{a_z}} \right) = 0\)

4 + 2K = 0

K = -2

Given

\(f\left( x \right) =e^{-x}, 0<x<∞\)

Calculation

P(X > 1) = \(\mathop \smallint \nolimits_1^∞\)e^-x dx

⇒ [-e ^-x ] |₁^∞ 

⇒ -e^-∞ + e^-1

⇒P(X > 1) = 0 + .3678 = .368

Concept:

1. Gradient:

The gradient of a scaler field gives direction and magnitude of the maximum rate of change of that scaler field.

 If A is a scalar field, i.e. a scalar function of position A(x,y,z) in 3 dimensions, then its gradient at any point is defined in Cartesian co-ordinate as: \(grad\space A = \frac{dA}{dx}a_x+ \frac{dA}{dy}a_y+\frac{dA}{dz}a_z\)                                               

It converts a scaler identity into a vector identity.

And usually represented as:   

\(grad \space A= ∇ A\)

Gradient in Cartesion Co-ordinate:

\(∇ A = \frac{dA}{dx}a_x+ \frac{dA}{dy}a_y+\frac{dA}{dz}a_z\)

Gradient in Cylindrical Coordinate:

\(∇ A =\frac{dA}{d\rho}a_\rho+ \frac{1}{\rho}\frac{dA}{d\phi}a_\phi+\frac{dA}{dz}a_z\)

Gradient in Spherical Coordinate:

\(∇ A = \frac{dA}{dr}a_x+ \frac{1}{r}\frac{dA}{d\theta}a_\theta+ \frac{1}{rsin\theta}\frac{dA}{d\phi}a_\phi\)

2. Curl:

The curl of a vector field gives the idea about the circulation per unit area of that vector field.

 If A is a vector field, i.e. a vector function of position A(x,y,z) in 3 dimensions, then its curl at any point is defined in Cartesian co-ordinate as:

\(∇ \times \vec A = \begin{vmatrix} ̂ a_x & ̂ a_y & ̂ a_z \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ A_x & A_y & A_z \end{vmatrix}\)

Curl in Cylindrical Coordinate:

\(∇ \times \vec A = \frac{1}{\rho} \begin{vmatrix} ̂ a_\rho & \rho ̂ a_\phi & ̂ a_z \\ \frac{\partial}{\partial \rho} & \frac{\partial}{\partial \phi} & \frac{\partial}{\partial z} \\ A_\rho & \rho A_\phi & A_z \end{vmatrix}\)

Curl in Spherical Coordinate:

\(∇ \times \vec A = \frac{1}{r^2 \sin \theta} \begin{vmatrix} ̂ a_r & r̂ a_\theta & r \sin \theta ̂ a_\phi \\ \frac{\partial}{\partial r} & \frac{\partial}{\partial \theta} & \frac{\partial}{\partial \phi} \\ A_r & rA_\theta & r \sin \theta A_\phi \end{vmatrix}\)                                                                                    

Calculation:

V = 2x²y + 3y²z + 4z²x

Grad V = ∇V = \(\frac{\partial V}{\partial x} a ̂ x + \frac{\partial V}{\partial y} a ̂ y + \frac{\partial V}{\partial z} a ̂ z\)

=(4xy + 4z²) â_x + (6yz + 2x²)â_y + (3y² + 8zx)â_z

\(\nabla \times \nabla V = \begin{vmatrix} ̂ i & ̂ j & ̂ k \\ \frac{\partial}{\partial z} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ (4xy + 4z^2) & (6yz + 2x^2) & (3y^2 + 8zx) \end{vmatrix}\)

= î(6y - 6y) - ĵ(8z - 8z) + k̂(4x - 4x)

= 0

Important PointsSome definitions involving div, curl, and grad A vector

• field with zero divergences is said to be solenoidal.
• A vector field with zero curls is said to be irrotational.

There are 4 possible combinations for Voltage and Current with which we can sample at the output and mix the feedback to the input.

 Sampling:

• At the output side, we will take a sample of output since we want to check the behavior of output and we don't want to disturb the output when we take the sample.
• That's why when the voltage is sampled, it is in parallel (as the voltage is the same in parallel) and current in series (as the current is the same in series).

Mixing:

• In mixing end we want to affect the signal that is provided to the amplifier since that is the actual fundament of giving the feedback.
• So, the voltage will be in series and current will be in parallel. So that they can change the input and effect the change in output.

The four basic feedback topologies are as shown:

i) Voltage-sampling voltage-mixing (series-shunt) topology.

ii) current-sampling voltage-mixing (series-series) topology.

iii) current-sampling current-mixing (shunt-series) topology.

iv) voltage-sampling current-mixing (shunt-shunt) topology.

PMMC reads the average value,

The circuit looks like a Half-wave rectifier circuit.

Vavg = (Vm / π)

The voltage across the 100 k Ω is the average output voltage ( V0) 

And V0 = Reading of the voltmeter

V0= (100k / (100k + 1k)) x Vavg

V0 = (100 / 101) x (Vm / π)

V0 = (100 / 101) x (14.14 / π)

V0 = 4.46 V

⇒ Voltmeter reading = 4.46 V

Concept:

Purpose of Bode plot

• To draw the frequency response of the open-loop transfer function.
• To find closed-loop system stability
• To find the gain margin, phase margin, gain crossover frequency, phase crossover frequency
• To find the relative stability. The largest Gain margin and Phase margin gives more relative stability.

It consists of both magnitude and phase plot

NOTE: whenever the transfer function consists of poles and zeroes at the origin then magnitude plot starts with a magnitude of the opposite sign of slope at a frequency of 0.1

The optimum value of the Gain margin is 5 dB to 10 dB and

Phase margin is 30° to 40°.

\({M_{start}} = 20logk + {\left. {20log} \right|_{poles;or;zeroes;at;origin}}\)

When k = 1,

 M_0.1 = - slope and M_{w = 1} = 0 dB

Gain margin is defined as:

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

\( - 20\log {\left| {G\left( {j\omega } \right)H\left( {j\omega } \right)} \right|_{\omega = {\omega _{pc}}}}\)  dB

Phase margin is defined as:

 = \({\left. {180^\circ + \angle GH\left( {j\omega } \right)} \right|_{\omega = {\omega _{gc}}}}\) 

Calculation:

From the given plot, we obtain the slope as

Slope = \(\frac{{20log{G_2} - 20log{G_1}}}{{log{w_2} - log{w_1}}}\) 

From the figure:

 20log G₂ = - 8 dB

 20log G₁ = 32 dB

So, the slope:

 = \(\frac{{ - 8 - 32}}{{log10 - log1}}\)

= - 40 dB 

The slope represents that two poles are present.

The transfer function can be given as:

 \(G\left( s \right) = \frac{k}{{{s^2}}}\) 

At ω = 1

\(\left| {G\left( {j\omega } \right)} \right| = \frac{k}{{\left| {{w^2}} \right|}}\)

= k 

In dB \(20log\left| {G\left( {j\omega } \right)} \right| = 20logk\)

\(20logk = 32\)

\(k = {10^{\frac{{32}}{{20}}}}\)

k = 39.8

∴ The transfer function is:

 \(G\left( s \right) = \frac{k}{{{s^2}}} \) 

\(= \frac{{39.8}}{{{s^2}}}\)

We need a response as shown:

OR

This type of truth table corresponds to that of an XOR gate.

Given, the signal

V (t) = 30 sin 100t + 10 cos 300 t + 6 sin (500t+π/4)

So, we have

ω₁ = 100 rads

ω₂ = 300 rads

ω₃ = 500 rads

∴ The respective time periods are

\(\begin{array}{l} {T_1} = \frac{{2\pi }}{{{\omega _1}}} = \frac{{2\pi }}{{100}}sec\ {T_1} = \frac{{2\pi }}{{{\omega _2}}} = \frac{{2\pi }}{{300}}sec\ {T_3} = \frac{{2\pi }}{{500}}sec \end{array}\)

So, the fundamental time period of the signal is

\(LCM\left( {{T_1},{T_2},{T_3}} \right) = \frac{{LCM\left( {2\pi ,2\pi ,2\pi } \right)}}{{HCF\left( {100,\ 300,\ 500} \right)}}\)

as \({T_0} = \frac{{2\pi }}{{100}}\)

∴ The fundamental frequency, \({\omega _0} = \frac{{2\pi }}{{{T_0}}} = 100\ rad/s\)

The maximum frequency of the band-limited signal.

f_m = 5 kHz

According to the Nyquist sampling theorem, the sampling frequency must be greater than the Nyquist  frequency which is given as

f_N = 2 f_m = 2 × 5 = 10 kHz

So the sampling frequency f_s ≥ f_N

f_s ≥ 10 kHz

Only the option (A) does not satisfy the condition.

∴ 5 kHz is not a valid sampling frequency.

Concept: 

For interconversions of delta and star, when all resistances are the same we can directly conclude two points

1. From Delta to Star resistance will be divided by 3
2. From Star to Delta resistance is multiplied by 3

Shortcuts for conversion:

Delta connection is shown below

Star connection of resistors is shown as

To convert Star from Delta

Branch resistance = product of connected resistance/sum of all resistances

For branch resistance PS, R_a and R_b are connected

\({R_A} = \frac{{{R_a} \times {R_b}}}{{{R_a} + {R_b} + {R_c}}}\)

For branch resistance QS, R_a and R_c are connected

\({R_B} = \frac{{{R_a} \times {R_c}}}{{{R_a} + {R_b} + {R_c}}}\)

For branch resistance, SR, R_b and R_c are connected

\({R_C} = \frac{{\left( {{R_c} \times {R_b}} \right)}}{{{R_a} + {R_b} + {R_c}}}\)

To convert Delta from Star

Branch resistance = ∑ connected resistances + \(\frac{{product;of;connected;resistance}}{{resistance;which;is;not;connected}}\)

For branch resistance PQ, R_A and R_B are connected

\({R_a} = {R_A} + {R_B} + \frac{{{R_A} \times {R_B}}}{{{R_C}}}\)

For branch resistance PR, R_B and R_C are connected

\({R_b} = {R_C} + {R_B} + \frac{{{R_C} \times {R_B}}}{{{R_C}}}\)

For branch resistance QR, R_A and R_C are connected

\({R_c} = {R_C} + {R_A} + \frac{{{R_A} \times {R_C}}}{{{R_B}}}\)

Calculation:

Given that all resistances in Delta connections are scaled by ‘K’

From the formula let us calculate R_A which is PS branch resistance,

\({R_{{A_1}}} = \frac{{k{R_a} \times k{R_b}}}{{k{R_a} + k{R_b} + k{R_c}}}\)

\( = \frac{{{k^2}\left( {{R_a} \times {R_b}} \right)}}{{k\left( {{R_a} + {R_b} + {R_c}} \right)}}\)

\( = k\frac{{{R_a} \times {R_c}}}{{{R_a} + {R_b} + {R_c}}}\)

\( = k{R_A}\)

the resistance in the Star network also scales by the same factor as in Delta network.

Swing Equation:

• A power system consists of a number of synchronous machines operating synchronously under all operating conditions.
• The equation describing the relative motion is known as the swing equation, which is a non-linear second order differential equation that describes the swing of the rotor of the synchronous machine.
• The transient stability of the system can be determined by the help of the swing equation given below

\({P_m} - {P_e} = M\frac{{{d^2}\delta }}{{d{t^2}}}\)

Also, \(M = \frac{H}{{180;{f_0}}}\)  for unit quantity

So that swing equation becomes

\({P_m} - {P_e} = \frac{H}{{180{f_o}}}\frac{{{d^2}\delta }}{{d{t^2}}}\)

Where,

Pm = Mechanical power input

Pe = Electrical power output

Pa = Accelerating power

δ = angular displacement of an axis fixed to the rotor with respect to a synchronously rotating axis

M = Angular momentum of the rotor

H = Per unit inertia constant

f0 = Frequency

Note: The swing equation gives the relation between the accelerating power and angular acceleration. It describes the rotor dynamics of the synchronous machines and it helps in stabilizing the system.

Concept of Leakage Flux:

• The Leakage flux in an Induction motor is basically due to the air gap between the stator and rotor which links to either stator winding or rotor winding but not links to both windings.
• The presence of an air gap between the stator and rotor of an induction motor increases the reluctance of the magnetic circuit.
• Consequently, an induction motor draws a large magnetizing current (Im) to produce the required flux in the air gap. That is why the small air gap is preferred for the induction motor which helps to reduce the magnetizing current.
• Hence for any machine, we have to always try for less leakage flux and more linkage flux so that the machine becomes more efficient.

Important Points

If the air gap of an induction motor is increased,

1. The permeability of the magnetic circuit will decrease
2. The magnetizing inductance of the motor will decrease
3. Leakage reactance will increase
4. Leakage flux will increase
5. The magnetizing current will increase
6. This will cause a poorer power factor at all loads

Concept:

Moving iron ammeter:

Moving iron ammeter is used to measure both AC and DC values in the circuit.

The reading of the MI instrument is in RMS value.

Calculation:

The voltage across inductor  = V₂  = j IX_L

The voltage across capacitor = V₁ = -j IX_C

Then the net voltage V = I (jX_L - jX_C)

\(V = {V_2} - {V_1}\)

We know that,

i = e^iπ/2

⇒ -i = e-iπ/2

⇒ √-i = ± [e^(-iπ/2)]^1/2 = ± e^-iπ/4

= ± [cos (π/4) - i sin (π/4)]

= cos (π/4) - i sin (π/4);  - cos (π/4) + i sin (π/4)

= cos (-π/4) + i sin (-π/4);  cos (3π/4) + i sin (3π/4)

Concept:

Line Integral:

A line integral is an integral where the function to be integrated is evaluated along a curve.

The terms path integral, curve integral, and curvilinear integral are also used.

Calculation:

\(=\displaystyle\int \vec F . \vec{dl}\)

\(=\displaystyle\int (y^2 \times \hat a_z - yz \hat a_y - x^2 a_z) \vec{dl}\)

\(=\displaystyle\int (y^2 \times \hat a_z - yz \hat a_y - x^2 \hat a_z) .[dx \hat a_x + dy \hat a_y + dz \hat a_y]\)

\( =\displaystyle\int y^2 dx + \displaystyle\int (-y^2) dy - \displaystyle\int x^2 dz\)

along x - axis

y = 0, z = 0

dz = 0

\( =\displaystyle\int 0 dx + \displaystyle\int 0 dy - \displaystyle\int x^2 dz\)

= 0

Homogenous system of linear equations:

Homogenous equations are the set of equations linear equations that can be written in matrix form as AX = 0.

Where,

A is a matrix of the coefficients and X is the column matrix of the variables.

Theorem:

The system AX = 0 has,

• Unique solution (zero solution or trivial solution), if the rank of A (ρ(A) is equal to number of variables(n).
• Infinitely many solutions non-zero solutions, if ρ(A) < n.
• In this case, if |A| = 0 then the system posses infinitely many non-trivial solutions i.e., non-zero solutions.
• In this case, if |A| ≠ 0 then the system posses only a trivial solution (i.e., zero solution or unique solution.)
• If ρ(A) = r, and number of variables = n, then the number of linearly independent solution of AX = 0 is 'n-r'

Calculation:

Given that,

\( \left[\begin{matrix} 2 && -2 \\ 1 && -1 \end{matrix}\right] \left[\begin{matrix} x_1 \\ x_2 \end{matrix}\right]= \left[ \begin{matrix} 0 \\ 0 \end{matrix} \right]\)

⇒ A = \( \left[\begin{matrix} 2 && -2 \\ 1 && -1 \end{matrix}\right] \)

⇒ |A| = 0

As det A is zero, the rank of the matrix ((ρ (A) =1) < order of the matrix (n = 2).

∴ The given system posses infinitely many non-trivial solutions i.e., non-zero solutions.

Concept:

Strain Gauge:

• A strain gauge is a type of electrical sensor. Its primary use is to measure force or strain.
• The resistance of a strain gauge changes when force is applied and this change will give a different electrical output.
• Strain gauges use this method to measure pressure, force, weight and tension.

Calculation:

\(V_i= 20\times 10^{-3}\times (300+300)=12\space volts\)

After a 1% increment in the resistance of the strain gauge.

\(R_s=300+( \frac{300\times 1}{100}) = 303 \Omega\)

So,

\(V_o=(\frac{300}{300+300}-\frac{300}{300+303})\times 12= 29.85 mv\)

I = I_z + I_Load

\(\frac{{\left( {10 - 5} \right)}}{{100}} = {I_{zmin}} + {I_{Lmax}}\)

50 mA = 10 mA + I_Lmax

\(\begin{array}{l} \therefore {I_{Lmax}} = 40mA = \frac{{{V_z}}}{{{R_{L\left( {min} \right)}}}}\ \therefore {R_{L\left( {min} \right)}} = \frac{5}{{40 \times {{10}^{ - 3}}}} = 125\ {\rm{\Omega }} \end{array}\)

Maximum Power rating of the Zener Diode P_zmax = V_z × I_z(max) 

where Iz(max) is the maximum current through Zener Diode in reverse bias.

Iz(max) = 50 mA (When there is no Load i.e. R_L=∞)

Pzmax = 5 × 50 mA = 250 mWatt

Option (2) is correct

Concept:

If G(s) is the open-loop transfer function then the closed-loop transfer function with feedback H(s) is written as:

 \(CLTF = \frac{{G\left( s \right)}}{{1 + G\left( s \right)H\left( s \right)}}\) 

for negative feedback and

\(CLTF = \frac{{G\left( s \right)}}{{1 - G\left( s \right)H\left( s \right)}}\)

for the positive feedback.

The time constant form of system is represented as:

 \(G\left( s \right) = \frac{{k\left( {1 + s{\tau _1}} \right)\left( {1 + s{\tau _2}} \right) \cdots \cdots }}{{{s^n}\left( {1 + s{\tau _a}} \right)\left( {1 + s{\tau _b}} \right) \cdots \cdots }}\) 

If \(G\left( s \right) = \frac{k}{{s + a}}\) 

and time-domain signal with respect to this is:

 e^-at and corresponding time constant is 1/a

The time constant for the RC circuit is:

 τ = R_eqC_eq. R_eq, and C_eq are found by de-energizing the independent sources.

De energizing the Independent sources

Voltage source: Replace this by a short circuit since its internal resistance is zero.

Current source: Replace this by an open circuit since its internal resistance is ∞

The time constant of RL circuit is:

 \(\tau = \frac{{{L_{eq}}}}{{{R_{eq}}}}\) 

Calculation:

Given the open-loop transfer function is:

\(G\left( s \right) = \frac{{10{k_a}}}{{1 + 10s}}\)

\( = \frac{{{k_a}}}{{s + \frac{1}{{10}}}}\)

By taking the inverse Laplace transform we get

 \(g\left( t \right) = {e^{ - \frac{t}{{10}}}}\) 

Comparing with the standard form

 Ae^-t/τ we get

 τ_ol = 0.1

We obtain the closed-loop transfer function for the given system as:

\(H\left( s \right) = \frac{{10{k_a}}}{{1 + 10s + 10{k_a}}}\)

\(H\left( s \right) = \frac{{{k_a}}}{{s + \left( {{k_a} + \frac{1}{{10}}} \right)}}\)

By taking inverse Laplace transform, we get

\(h\left( t \right) = {k_a}{e^{ - \left( {{k_a} + \frac{1}{{10}}t} \right)}}\)

the time constant of the closed-loop system is obtained as

\({\tau _{cl}} = \frac{1}{{{k_a} + \frac{1}{{10}}}}\)

\({\tau _{cl}} = ; \approx \frac{1}{{{k_a}}}\)

Now given that K_a reduces open loop time constant by a factor of 100

\({\tau _{cl}} = \frac{{{\tau _{ol}}}}{{100}}\)

\(\frac{1}{{{k_a}}} = \frac{{10}}{{100}} = \frac{1}{{10}}\)

K_a = 10

For evaluating the Thevenin voltage seen by the load R_L , we open the circuit across it (also its dependent source) The equivalent circuit is shown below.

As the circuit opens across R_L so

I₂ = 0

j 40 I₂ = 0

The dependent source in Loop 1 is short-circuited therefore:

\({V_{L1}} = \frac{{\left( {j4} \right){V_S}}}{{j4 + 3}}% MathType!End!2!1! \)

V_th = 10 V_L1

_{\( = \frac{{j40}}{{j4 + 3}} \times 100\angle 53.13^\circ % MathType!End!2!1! \)}

= 800 ∠90°

Consider that the voltage across the three capacitors C₁, C₂ and C₃ are V₁, V₂ and V₃ respectively. So, we can write

V₂/V₃ = C₃/C₂

Since, voltage is inversely proportional to capacitance. Now, given that

C₁ = 10 μF ; (V₁)_max = 10 V

C₂ = 5 μF ; (V₂)_max = 5 V

C₃ = 2 μF ; (V₃)_max = 2 V

V₂/V₃ = 2/5

For (V₃)_max = 2

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V₂ < (V₂)_max

Hence this is voltage at C₂ . Therefore,

V₃ = 2 Volt

V₂ = 0.8 Volt

V₁ = V₂ + V₃ = 2.8 Volt

Equivalent capacitance across the terminal is

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Equivalent voltage is (max.value)

V_max = V₁ = 2.8

So, charge stored in the effective capacitance is

Q = C_eq V_max

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Input voltage V_YZ = 1000 sinωt.

During positive cycle ie V_YZ > 0 → \(V_Y\) > \(V_Z\) So all Diodes will be in cutoff region. Therefore, there is no voltage Difference between X and W node. Ie \(V_{WX}\) = 0.

In negative half cycle ie V_YZ < 0. So now all Diodes are short circuited

∴ \(V_{WX}\) = 0

∴ \(V_{WX}\) = 0 for all t

E = V− I_aR_a =150 − 20 × 1 = 130V

I_1a = I_a/2 (Half-rated torque)

⇒V₁ = E + I_1aR_a = 130 + 10 × 1 = 140V

Step down chopper

\(\begin{array}{l} \Rightarrow {V_o} = D{V_s} \Rightarrow D = \frac{{{V_o}}}{{{V_s}}} = \frac{{140}}{{200}} = 0.7 \end{array}\)

Calculation:

Power system having n nodes.

Z₃₃ = j 0.5 PU

V₃ = 1.3∠ -10° PU

A capacitor having a reactance of -j 3.5 PU is now added to the network

\({I_C} = \frac{{{V_3}}}{{{Z_{33}}\left( {new} \right)}} = \frac{{{V_3}}}{{{Z_{33}} - j3.5}}\)

\(\frac{{1.3\angle - 10^\circ }}{{j0.5 - j3.5}} = \frac{{1.3\angle - 10^\circ }}{{-j3}} = \frac{{1.3\angle - 10^\circ }}{{3\angle - 90^\circ }} = 0.433\angle 80^\circ \)

I_C = 0.433∠ 80°

Energy density stored in a dielectric medium obtained as

W_E = (1/2) ε|E|² J/m²

The electric field inside the dielectric will be same to given field in free space only if the field is tangential to the interface

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The total stored energy is

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= ε₀ 100 × 10⁶/mm² × (500 × 500) mm² × (0.4)

= ε₀ 100 × 10⁶ × 0.4 × 25 × 10⁴

= 8.85 × 10^-12 × 10¹³

= 88.5 J

For a matrix A, whose eigen value is \(λ\) and corresponding eigen vector is X, C.E. equation is given by

AX=λ X

Assume, \(A= \begin{pmatrix} a & b\ c & d \end{pmatrix}\)

For eigen value λ = -1⇒

\(\begin{pmatrix} a & b\ c & d \end{pmatrix}\begin{pmatrix} +1 \ -1 \end{pmatrix}=(-1)\begin{pmatrix} +1 \ -1 \end{pmatrix} \)

\(a-b=-1\space\space\space\space\space\space \space...(i)\)

\(c-d=-1\space\space\space\space\space\space \space...(ii)\)

For eigen value λ = -2⇒

\(\begin{pmatrix} a & b\ c & d \end{pmatrix}\begin{pmatrix} +1 \ -2 \end{pmatrix}=(-2)\begin{pmatrix} +1 \ -2 \end{pmatrix} \)

\(a-2b=-2\space\space\space\space\space\space \space...(ii)\)

\(c-2d=4\space\space\space\space\space\space \space...(iv) \)

From (i) - (iii)

b= -1+2 = 1

from (i)

a = 0

From (ii) - (iv)

d = -3

From (ii)

c = -2

So matrix \(A= \begin{pmatrix} 0 & 1\ -2 & -3 \end{pmatrix} \)

Concept:

According to the residue theorem, we have:

\({\displaystyle \operatorname {Res} (f,c)={1 \over 2\pi i}\oint _{\gamma }f(z),dz}\)

\(2\pi i\space {\displaystyle \operatorname {Res} (f,c)={}\oint _{\gamma }f(z),dz}\)

Removable singularities:

If the function f can be continued to a holomorphic function on whole disk 

{\displaystyle |y-c|<R}

, then Res(f, c) = 0. The converse is not generally true.

Simple poles:

if a simple pole c, the residue of f is given by:

\(\operatorname {Res} (f,c)=\lim _{z\to c}(z-c)f(z) \)

Calculation:

\(\displaystyle\int \frac{z^2 - 4}{z^2 + 4} dx\)

around circle → |z - i| = 2

= (x + iy - i) = 2

= [x + i(y - 1)] = 2

= x² + (y - 1)² = 2²

\(= \displaystyle\int \frac{z^2 - 4}{z^2 + 4} dx = \displaystyle\int \frac{z^2 - 4}{(z + 2i)(z - 2i)}\)

z = -2i → outside given circle

z = 2i → inside given circle

\(\displaystyle\int \frac{z^2 - 4}{(z + 2i)(z - 2i)}dz = 2\pi i \left{ \displaystyle\lim_{z \rightarrow 2i} \left[ \frac{(z^2 - 4)}{(z + 2i)(z - 2i)}.(z - 2i) \right] \right}\)

\(= 2\pi i \left{ \frac{-4 - 4}{2i + 2i} \right}\)

\(= 2\pi i \left{ \frac{-8}{4i} \right} = - 4\pi \)

The given circuit is in toggle mode, so frequency will get half.

Output frequency = 0.5 kHz

The transistor – resistor logic circuit has output of \(\bar X\)

The diode is ideal for Y = 1 i.e reverse voltage of diode is = 5V hence Z and Y are open circuit then Z = \(\bar X\)

For Y = 0 i.e reverse voltage of diode is zero hence Z = 0

Hence Z = \(\bar XY\)

\(\begin{array}{l} V_2^ - = V_2^ + = {V_2}\ V_1^ + = V_1^ - = {V_1}\ {V_1} = {V_{out}} \times \frac{1}{2} \end{array}\)

V_out = 2V₁

As the i/p current in o/p – amp is always zero.

∴ there will be no voltage drop across 1kΩ in II op – amp.

i.e. V₂ = 1V

\(\therefore \frac{{{V_1} - {V_2}}}{1} = \frac{{{V_2} - \left( { - 2} \right)}}{1}\)

V₁ – 1 = 1 + 2

V₁ = 4

∴ V_out = 2V₁ = 8 V.

Concept:

Signal flow graph

• It is a graphical representation of a set of linear algebraic equations between input and output.
• The set of linear algebraic equations represents the systems.
• The signal flow graphs are developed to avoid mathematical calculation.

Maon gain formula is used to find the ratio of any two nodes or transfer function.

T F = \(\mathop \sum \limits_{k = 1}^i \frac{{{P_k}{{\rm{\Delta }}_k}}}{{\rm{\Delta }}}\) 

Where P_k = k^th forward path gain

Δ = 1- ∑ individual loop gain + ∑ two non-touching loops gain - ∑ the gain product of three non-touching loops + ∑ gain of four non-touching loops

Shotcut: while writing Δ take the opposite sign for the odd number of non-touching loops snd the same sign for the even the number of non-touching loops.

Δ_K is obtained from Δ by removing the loops touching the K^th forward path.

Calculation:

For the given SFG two forward paths

\({P_{K1}} = 1\left( {{s^{ - 1}}} \right)\left( {{s^{ - 1}}} \right)\left( 1 \right) = {s^{ - 2}}\)

\({P_{k2}} = 1\left( {{s^{ - 1}}} \right)\left( 1 \right)\left( 1 \right) = {s^{ - 1}}\)

Since all loops are touching the paths P_K1 and P_K2 so Δ_K1 = Δ_K2 = 1

We have Δ = 1- ∑ individual loops + ∑ non-touching loops gain

Loops are

\({L_1} = \left( { - 4} \right)\left( 1 \right) = - 4\)

\({L_2} = \left( { - 4} \right)\left( {{s^{ - 1}}} \right) = - 4{s^{ - 1}}\)

\({L_3} = ; - 2\left( {{s^{ - 1}}} \right)\left( {{s^{ - 1}}} \right) = - 2{s^{ - 2}}\)

\({L_4} = - 2\left( {{s^{ - 1}}} \right)\left( 1 \right) = - 2{s^{ - 1}}\)

As all the loops are touching each other we have

Δ = 1 – ( L1 + L2 + L3 + L4)

Δ = 1 – ( - 4 – 4s^-1 – 2s^-2 -2s^-1 )

Δ = 5 + 6s^-1 + 2s^-2

\(T.F = \frac{{{s^{ - 2}} + {s^{ - 1}}}}{{5 + 6{s^{ - 1}} + 2{s^{ - 2}}}}\)

\( = \frac{{s + 1}}{{5{s^2} + 6s + 2}}\)

The convolution of a signal is given by:

y(t) = h(t) * x(t)

Given, h(t) = δ(t-1) + δ(t-3)

and x(t) = u(t) (Step Input)

y(t) = (δ(t-1) +δ(t-3)) * u(t)

y(t) = u(t-1) + u(t-3)

The required value of y(2) is:

y(2) = u(2-1) + u(2-3) = u(1) + u(-1)

Since u(-1) = 0

∴ y(2) = u(1) = 1

The quality factor of the inductances are given by

\(% MathType!Translator!2!1!AMS LaTeX.tdl!AMSLaTeX! % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGXbWdamaaBaaaleaapeGaaGymaaWdaeqaaOWdbiabg2da9maa % laaapaqaa8qacqaHjpWDcaWGmbWdamaaBaaaleaapeGaaGymaaWdae % qaaaGcbaWdbiaadkfapaWaaSbaaSqaa8qacaaIXaaapaqabaaaaOWd % biaacYcacaWGXbWdamaaBaaaleaapeGaaGOmaaWdaeqaaOWdbiabg2 % da9maalaaapaqaa8qacqaHjpWDcaWGmbWdamaaBaaaleaapeGaaGOm % aaWdaeqaaaGcbaWdbiaadkfapaWaaSbaaSqaa8qacaaIYaaapaqaba % aaaaaa!490A! {q_1} = \frac{{\omega {L_1}}}{{{R_1}}},{q_2} = \frac{{\omega {L_2}}}{{{R_2}}}% MathType!End!2!1! \)

So, in series circuit, the effective quality factor is given by

\(% MathType!Translator!2!1!AMS LaTeX.tdl!AMSLaTeX! % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGrbGaeyypa0ZaaSaaa8aabaWdbmaaemaapaqaa8qacaWGybWd % amaaBaaaleaapeGaamita8aadaWgaaadbaWdbiaadwgacaWGXbaapa % qabaaaleqaaaGcpeGaay5bSlaawIa7aaWdaeaapeGaamOua8aadaWg % aaWcbaWdbiaadwgacaWGXbaapaqabaaaaOWdbiabg2da9maalaaapa % qaa8qacqaHjpWDcaWGmbWdamaaBaaaleaapeGaaGymaaWdaeqaaOWd % biabgUcaRiabeM8a3jaadYeapaWaaSbaaSqaa8qacaaIYaaapaqaba % aakeaapeGaamOua8aadaWgaaWcbaWdbiaaigdaa8aabeaak8qacqGH % RaWkcaWGsbWdamaaBaaaleaapeGaaGOmaaWdaeqaaaaaaaa!518D! Q = \frac{{\left| {{X_{{L_{eq}}}}} \right|}}{{{R_{eq}}}} = \frac{{\omega {L_1} + \omega {L_2}}}{{{R_1} + {R_2}}}% MathType!End!2!1! \)

\(% MathType!Translator!2!1!AMS LaTeX.tdl!AMSLaTeX! % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacqGH9aqpdaWcaaWdaeaapeWaaSaaa8aabaWdbiabeM8a3jaadYea % paWaaSbaaSqaa8qacaaIXaaapaqabaaakeaapeGaamOua8aadaWgaa % WcbaWdbiaaigdaa8aabeaak8qacaWGsbWdamaaBaaaleaapeGaaGOm % aaWdaeqaaaaak8qacqGHRaWkdaWcaaWdaeaapeGaeqyYdCNaamita8 % aadaWgaaWcbaWdbiaaikdaa8aabeaaaOqaa8qacaWGsbWdamaaBaaa % leaapeGaaGymaaWdaeqaaOWdbiaadkfapaWaaSbaaSqaa8qacaaIYa % aapaqabaaaaaGcbaWdbmaalaaapaqaa8qacaaIXaaapaqaa8qacaWG % sbWdamaaBaaaleaapeGaaGOmaaWdaeqaaaaak8qacqGHRaWkdaWcaa % WdaeaapeGaaGymaaWdaeaapeGaamOua8aadaWgaaWcbaWdbiaaigda % a8aabeaaaaaaaaaa!4F28! = \frac{{\frac{{\omega {L_1}}}{{{R_1}{R_2}}} + \frac{{\omega {L_2}}}{{{R_1}{R_2}}}}}{{\frac{1}{{{R_2}}} + \frac{1}{{{R_1}}}}}% MathType!End!2!1! \)

\(% MathType!Translator!2!1!AMS LaTeX.tdl!AMSLaTeX! % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacqGH9aqpdaWcaaWdaeaapeGaamyCa8aadaWgaaWcbaWdbiaaigda % a8aabeaak8qacaWGsbWdamaaBaaaleaapeGaaGymaaWdaeqaaOWdbi % abgUcaRiaadghapaWaaSbaaSqaa8qacaaIYaaapaqabaGcpeGaamOu % a8aadaWgaaWcbaWdbiaaikdaa8aabeaaaOqaa8qacaWGsbWdamaaBa % aaleaapeGaaGymaaWdaeqaaOWdbiabgUcaRiaadkfapaWaaSbaaSqa % a8qacaaIYaaapaqabaaaaaaa!455B! = \frac{{{q_1}{R_1} + {q_2}{R_2}}}{{{R_1} + {R_2}}}% MathType!End!2!1! \)

\(\rm \frac{{{V_2}}}{{{V_1}}} = \frac{{{N_2}}}{{{N_1}}}\)

∴ \(\rm {V_2} = 1.25 \times {V_1}\)

if V₁= 100V 

then V₂ = 125V

Also,

V_yz= 0.8 x 125 = 100V

V_yz / V_wx1 = 100/100

case 2

I=0

Since for ideal transformer it takes zero magnetising current to set up the rated flux 

∴ \(\rm \frac{{{V_{{WX}_2}}}}{{{V_{{YZ}_2}}}} = \frac{{80}}{{100}}\)

\(\begin{array}{l} \omega = \frac{1}{{\sqrt {LC} }}\ = \frac{{2\pi }}{T}\ \Rightarrow T = 200\mu s. \end{array}\)

Circuit conducts only for the positive half cycle so conducting angle will be \(\frac{T}{2} = 100\mu s.\)

Given,

Supply  frequency (f) = 50 Hz

Poles  = 4

Synchronous speed (N_s) \( = \frac{{120 \times f}}{p}\)

\( = \frac{{120 \times 50}}{4} = 1500\) rpm

Slip (s) \( = \frac{{{N_s} - {N_r}}}{{{N_s}}} = \frac{{1500 - 1440}}{{1500}} = 0.04;\)

Now the electrical frequency of the induced negative sequence current in the rotor is obtained as

f₂ = (2 – s)f

'f' is the stator frequency (50 Hz)

f₂ = (2 – 0.04) × 50

= 98 Hz.

y = f(x) = 5x2 + 10x in the internal x = (1, 2)

Since, the function y is continuous in the interval (1, 2), as well as is differentiable at each point so, from Lagrange mean value theorem there exist at least a point where:

\(f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\)

Here, we have

a = 1, b = 2

So, for x = a = 1, we obtain:

y = f(a) = f(1) = 5(1)2 + 10(1) = 15

and for x = b = 2

y = f(b) = f(2) = 5(2)2 + 10(2) = 40

Therefore:

\(f'\left( c \right) = \frac{{40 - 15}}{{2 - 1}} = 25\)

Concept:

Newton Raphson Method:

\(\begin{array}{l} {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\ \end{array}\)

Calculation:

Given:

f(x) = x3 + 2x - 1 = 0

f'(x) = 3x2 + 2 

\(\begin{array}{l} {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\ \end{array}\)

\(\begin{array}{l} {x_{1}} = {x_0} - \frac{{f\left( {{x_0}} \right)}}{{f'\left( {{x_0}} \right)}}\ \end{array}\)

Initial guess = x0 = 1.2

f(1.2) = (1.2)3 + 2 × 1.2 -1 = 3.128

f'(2) = (3 ×1.2²) + 2  = 6.32

\(\begin{array}{l} {x_{1}} = {1.2} - \frac{{f\left( {{1.2}} \right)}}{{f'\left( {{1.2}} \right)}}\ \end{array}\)

\(\therefore;\begin{array}{l} {x_{1}} = {1.2} - \frac{{\left( {{3.128}} \right)}}{{\left( {{6.32}} \right)}}\ \end{array}=.705\)

Concept:

Volt - sec balance:

V_S  ∝T + Vs (1 - D) – V₀ (1 - D)T = 0

\({{\rm{V}}_0} = \frac{{{V_s}}}{{1 - ; \propto }}\)

Ampere – sec balance:

• I_o ∝T + (I_L – I₀) (1 - ∝) T = 0
• I_o ∝ T + I_L (1 - ∝) T – I₀ (1 - ∝) T = 0

\({{\rm{I}}_L} = \frac{{{I_0}}}{{1 - \propto }}\)    

Ripple current:

V_L(ON) = V_s

\(\begin{array}{l} L\frac{{\Delta {\rm{I}}}}{{ \propto T}} = {V_s}\ \Delta {\rm{I}} = \frac{{ \propto {V_s}}}{{FL}}\ \Rightarrow {I_{Lmax}} = {I_L} + \frac{{\Delta {{\rm{I}}_L}}}{2}\ = \frac{{{{\rm{I}}_0}}}{{1 - \propto }} + \frac{{ \propto {V_s}}}{{2FL}} \end{array}\)

Calculation:

The circuit is a boost converter

\(\begin{array}{l} {V_o} = \frac{{{V_{dc}}}}{{1 - d}} = \frac{{12}}{{1 - 0.4}} = 20V\ {I_o} = \frac{{{V_o}}}{R} = 1A\ {I_s} = \frac{{{I_o}}}{{1 - d}} = \frac{1}{{0.6}} = \frac{5}{3}A \end{array}\)

As the current from source of 12 V is the same as that pass through inductor. So, the peak to peak current ripple will be equal to peak to peak inductor current.

Now, the peak to peak inductor current can be obtained as

\({I_{L\left( {Peak\ to\ peak} \right)}} = \frac{{{V_S}}}{L}D{T_S}\)

Where, V_S → source voltage = 12 V

L → inductance = 100 μH = 10^-4 + 1

D → Duty ratio = 0.4

T_S → switching time period of  

and fs → switching frequency = 250 kHz.

∴ we get

\({I_{L\left( {Peak\ to\ Peak} \right)}} = \frac{{12}}{{{{10}^{ - 4}}}} \times 0.4 \times \frac{1}{{250 \times {{10}^3}}}\)

= 0.192 A

This is the peak to peak source current ripple.

Concept:

Consider an LTI system:

\(X'=AX+BU \)

\(Y=CX+DX\)

Controllability: 

The necessary and sufficient condition for controllability is

\(Q_c=[B \space\space AB\space \space A^2B.....] \)

so, the given matrix is controllable if,

\(Q_c≠0; \)

so, the given matrix is uncontrolled,

\(Q_c=0; \)

Observability: 

The necessary and sufficient condition for observability is

\(Q_b=[C' \space\space A'C'\space \space A'^2C'.....] \)

so, the given matrix is observability if,

\(Q_b≠0; \)

so, the given matrix is unobservable,

\(Q_b=0; \)

Solution:

Compare the given system with standard equations:

\(X'=AX+BU \) &

\(Y=CX+DX\)

\(A=\begin{pmatrix} -2 & 0\ 0 & -1 \end{pmatrix}\) , \(B=\begin{pmatrix} 1\ 1 \end{pmatrix}\) ,  \(C=\begin{pmatrix} 1 &1\ \end{pmatrix}\), \(D=0\)

Compare given Question:

\(Q_c = \begin{vmatrix} \begin{bmatrix} \begin{matrix} 1 \\ 1 \end{matrix} \begin{bmatrix} -2 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} \end{bmatrix} \end{vmatrix} \)

\(= \begin{vmatrix} \begin{bmatrix} 1 & -2 \\ 1 & -1 \end{bmatrix} \end{vmatrix}\)

\(= \begin{bmatrix} 1 & -2 \\ 1 & -1 \end{bmatrix} \)

= -1 + 2 = 1 ≠ 0 →  controlable

\(Q_b = \begin{vmatrix} \begin{bmatrix} \begin{matrix} 1 \\ 0 \end{matrix} \begin{bmatrix} -2 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} \end{bmatrix} \end{vmatrix} \)

\(= \begin{vmatrix} 1 & -2 \\ 0 & 0 \end{vmatrix} \)

= 0 ⇒ not observable

Given state model is

\(\left[ \begin{matrix} x_1 \\ x_2 \end{matrix} \right] = \left[ \begin{matrix} -2 && 0 \\ 0 && -1 \end{matrix} \right]\left[ \begin{matrix} x_1 \\ x_2 \end{matrix} \right]+\left[ \begin{matrix} 1 \\ 1 \end{matrix} \right]u \)    ------- (1)

\(y = \left[ \begin{matrix} 1 &&0 \end{matrix} \right]\left[ \begin{matrix} x_1 \\ x_2 \end{matrix} \right]\)       ------- (2)

From equation (1),

(dx₁/dt) = 2 x₁ + u

Taking Laplace transform, we get

s x₁(s) - x₁(0) = 2 x₁(s) + (1/s)

⇒ x₁ (s) = 1/(s × (s+2))

From equation (2),

y = x₁

Taking Laplace transform we get,

⇒ y(s) = x₁(s)

⇒ y(s) = 1/(s × (s+2))

⇒ y(s) = \(\frac{1}{2}[\frac{1}{s}-\frac{1}{s+2}]\)

Taking inverse Laplace transform, we get

y(t) = \(\frac{1}{2}-\frac{1}{2}e^{-2t}\)

Concept:

Power Flow in Transmission Line:

• Active Power flows from higher load angle bus to lower load angle bus.
• Reactive Power flows from a higher voltage bus to a lower voltage bus.

When the line is lossless (i.e. resistive part absent), the active power at both the buses will be equal and is given by:

P12  = \( {{V_1}{V_2}} \over X_{12}\) × sin(θ1 - θ2)

where, P12 = Power transferred between Bus 1 and Bus 2

V₁ = Voltage at Bus 1

V₂ = Voltage at Bus 2

θ1 = Load Angle at Bus 1

θ₂ = Load Angle at Bus 2

X₁₂ = Transmission Line Reactance between Bus 1 and Bus 2

Calculation:

In the figure, Bus 3 is absorbing power, and Bus 1 and 2 are delivering power.

Assuming absorbed power as negative and delivering power as positive.

P2 = 0.1 pu and P₃ = -0.2 pu

Conservation of Power states that- "The sum of total generated and delivered power in a system is always equal to zero."

P₁ + P₂ + P₃ = 0

P1 + 0.1 - 0.2 = 0

P1 = 0.1 pu

By power flow equation:

P₁₂  = \( {{V_1}{V_2}} \over X_{12}\) × sin(θ₁ - θ₂)

  0.1 = \( {{1}\times{1}} \over {1}\) × sin(θ1 - θ2)

sin(θ1 - θ2) = 0.1

θ1 - θ2 = sin^-1(0.1)

θ1 - θ2 = 0

θ1 = θ2 = 0

P23  = \( {{V_2}{V_3}} \over X_{23}\) × sin(θ2 - θ3)

 0.1= \( {{1}\times{1}} \over {1}\) × sin(0 - θ3)

 0 - θ₃= sin-1(0.1)

θ3= -0.1

Consider the voltage phase angles at buses 2 and 3 be Q₂ and Q₃ since, all the three buses have the equal voltage magnitude. Which is 1 pu, so, it is a D.C. load flow. The injections at Bus 2 and 3 are respectively P₂ = 0.1 Pu

P₃ = -0.2 Pu

P₁ + P₂ + P₃ = 0

P₁ – P₂ – P₃ = -0.1 + 0.2 = 0.1 Pu

Now, the apparent power delivered to base is,

\(\left| S \right| = \frac{{{V^2}}}{R} = \frac{{{{\left( {100 \times {{10}^3}} \right)}^2}}}{{100}} = 100 \times {10^6}VA\)

The real power delivered by slack bus

P = P₁ |S| = (0.1) (100 × 10⁶)

= 10 × 10⁶ watt = 10 MW

Case I: When Q₁ and Q₂ ON

In this case, the positive terminal of V₀ will be at a higher voltage. i.e V₀ > 0 and so i₀ > 0 (i.e it will be positive).

Case II: When Q₃, Q₄ ON, and Q₁, Q₂ OFF

In this case, the negative terminal of applied voltage V₀ will be at a higher potential i.e., V₀ < 0, and since, inductor opposes the change in current so, although the polarity of Voltage is inversed, current remains the same in the inductor i.e., I₀>0.

In this condition since, IGBT's can't conduct reverse currents, therefore, current will flow through D_3, D₄ until I_D becomes zero.

• ZCS can eliminate the switching losses at the turnoff and reduce the switching losses at turn-on.
• Zero Voltage Switching (ZVS) switches when the voltage is zero and is different from Zero Current Switching (ZCS) which switches when the voltage and current are both zero; referred to as the “zero-crossing” in a sinewave.
• ZVS is easier to implement into relay-based devices and whilst it works well for capacitive loads (switch mode power supplies). Zero voltage switching is not suitable for inductive loads including transformers and motors.