GATE EE 2017 Official Paper: Shift 2 (Previous Year Paper)

Let us look at the meanings of the given words:

Aversion: a strong dislike or disinclination.

Luminous: giving off light; bright or shining.

Plunder: steal goods from (a place or person), typically using force and in a time of war or civil disorder.

Yielding: (of a substance or object) giving way under pressure; not hard or rigid.

Hence option 4 is correct.

The word bright is used in order to express the meaning to the required degree or extent and is placed after an adjective, adverb, or verb).

Hence option 2 is correct. The other options are grammatically incorrect.

There are five buildings called V, W, X, Y and Z in a row (not necessarily in that order).

V is to the West of W.

Z is to the East of X and the West of V.

W is to the West of Y.

The building in the middle is V.

A test has twenty questions worth 100 marks in total.

There are two types of questions. Multiple choice questions are worth 3 marks each and essay questions are worth 11 marks each.

Let the number of multiple choice questions are x and the number of essay questions are y.

x + y = 20

and 3x + 11y = 100

By solving the above two equations, we get

x = 15, y = 5

So, the number of multiple choice questions = 15

Explanation:

There are 3 red socks, 4 green socks and 3 blue socks.

We need to select 2 socks.

The probability that both the socks of red colour:

\(\frac{{{3_{{C_2}}}}}{{{{10}_{{C_2}}}}} = \frac{3}{{45}}\)

The probability that both the socks of green colour:

\(\frac{{{4_{{C_2}}}}}{{{{10}_{{C_2}}}}} = \frac{6}{{45}}\)

The probability that both the socks of red colour:

\(\frac{{{3_{{C_2}}}}}{{{{10}_{{C_2}}}}} = \frac{3}{{45}}\)

The probability that both are of the same colour:

\(= \frac{3}{{45}} + \frac{6}{{45}} + \frac{3}{{45}} = \frac{{12}}{{45}} = \frac{4}{{15}}\)

Let us look at the meanings of the given words:

Culture: the arts and other manifestations of human intellectual achievement regarded collectively.

Seemingly: so as to give the impression of having a certain quality; apparently; according to the facts as one knows them; as far as one knows.

Urgent: requiring immediate action or attention.

Political: relating to the government or public affairs of a country.

Hence option 2 is correct.

There are three boxes labelled as apples, oranges and both apples and oranges.

All the boxes are incorrectly labelled.

Case 1: Let we open the box labelled with apples.

If the fruit inspected is orange, we can’t say it is the box with oranges or the box with both apples and oranges.

So, this case is not valid.

Case 2: Let we open the box labelled with oranges.

If the fruit inspected is apple, we can’t say it is the box with apples or the box with both apples and oranges.

So, this case is also not valid.

Case 3: Let we open the box labelled with both apples and oranges.

If the fruit inspected is apple, we can conclude that this box contains apples.

Now, the second box is labelled with orange, but it is incorrectly labelled. So, it can’t be orange and hence this box contains both apples and oranges.

Box 1 contains oranges.

So, we need to choose the third box which is labelled as both apples and oranges in order to determine the contents of all boxes.

Calculation:

X is a 30-digit number starting with the digit 4 followed by the digit 7.

Let X = 47 × 10²⁸

Now, X³ = 47³ × 10⁸⁴ = 103823 × 10⁸⁴

So, X have 90 digits.

Let f(x) = e^x + 0.5 x² – 2 = 0

The given range = [-5, 5]

 

From the graph, we can see that the value of f(x) is zero is occurring twice.

So, the number of roots = 2

One root is in between -2 and -1 and the other is in between 0 to 1.

The possibility of a thunderstorm is given by how fast air pressure rises or drops over a region.

The pressure difference between for the given regions is shown in the below table:

 

The air difference is maximum for region R. So, Region R is most likely to have a thunderstorm.

Concept:

In probability theory, the probability measure of an event is made if another event has already occurred is referred to as conditional probability.

For calculating conditional probability, the probability of the preceding event and the probability of succeeding event is multiplied.

Conditional probability is given by

\(P\left( {{E_1}/{E_2}} \right) = \frac{{P\left( {{E_1} \cap {E_2}} \right)}}{{P\left( {{E_2}} \right)}}\)

\(P\left( {{E_2}/{E_1}} \right) = \frac{{P\left( {{E_1} \cap {E_2}} \right)}}{{P\left( {{E_1}} \right)}}\)

Where E1 and E2 are the events.

Calculation:

Given:

Urn contains 5 red balls, 5 black balls.

One ball is picked at random.

Case (i): The first ball is red ball

Probability to get a red ball in the second draw is

\({P_1} = \frac{5}{{10}} \times \frac{4}{9} = \frac{2}{9}\)

Case (ii): The first ball is black ball

Probability to get a red ball in the second draw is

\({P_2} = \frac{5}{{10}} \times \frac{5}{9} = \frac{5}{{18}}\)

Required probability (P) \(= {P_1} + {P_2} = \frac{2}{9} + \frac{5}{{18}} = \frac{1}{2}\)

For a perfect electric conductor, no charge can be distributed inside the surface of conductor.

If one million electrons are added to the sphere, these electrons will be distributed uniformly over the outer surface of the sphere.

Concept:

Divergence of any vector is zero if the total outward flow is equal to the total inward flow.

\(\nabla .\vec A = 0\)

Curl of any vector is zero if there is no rotation in space.

\(\nabla \times \vec A = 0\)

Application:

Vector \(\vec X\)has no rotation in the space and the outward flow is not equal to inward flow.

\(\nabla .\vec X \ne 0,;\nabla \times \vec X = 0\)

The vector\(\vec Y\) has circular rotation and the flow is same in both inward and outward direction.

\(\nabla .\vec Y = 0,;\nabla \times \vec Y \ne 0\)

Vector \(\vec Z\)has circular rotation and the outward flow is not equal to inward flow.

\(\nabla .\vec Z \ne 0,;\nabla \times \vec Z \ne 0\)

Here, option 3 is correct.

Transfer function of P is,

\({H_p}\left( z \right) = \frac{{\left( {z - 1} \right)\left( {z + 1} \right)}}{{{z^2}}} = \frac{{{z^2} - 1}}{{{z^2}}}\) 

At low frequency, i.e. at z = 1, H_p(z) = 0

At high frequency, i.e. at z = -1, H_p(z) = 0

So, it is a band pass filter.

Transfer function of Q is,

\({H_Q}\left( z \right) = \frac{{\left( {z - 1} \right)\left( {z + 1} \right)}}{{\left( {z - j0.5} \right)\left( {z + j0.5} \right)}} = \frac{{{z^2} - 1}}{{{z^2} + 0.25}}\) 

At low frequency, i.e. at z = 1, H_Q(z) = 0

At high frequency, i.e. at z = -1, H_Q(z) = 0

So, it is a bandpass filter.

Transfer function of R is,

\({H_R}\left( z \right) = \frac{{\left( {z - 1} \right)\left( {z + 1} \right)}}{{\left( {z - j} \right)\left( {z + j} \right)}} = \frac{{{z^2} - 1}}{{{z^2} + 1}}\) 

At low frequency, i.e. at z = 1, H_R(z) = 0

At high frequency, i.e. at z = -1, H_R(z) = 0

So, it is a bandpass filter.

Synchronous motor at leading power factor acts as an over excited machine. At this condition, E_f is greater than V_t.

Synchronous motor at lagging power factor acts as a under excited machine. At this condition, E_f is less than V_t.

Closed loop transfer function is,

\(CLTF = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{Ks + b}}{{{s^2} + as + b}}\) 

Open loop transfer function will be,

\(OLTF = \frac{{CLTF}}{{1 - CLTF}} = \frac{{\frac{{Ks + b}}{{{s^2} + as + b}}}}{{1 - \frac{{Ks + b}}{{{s^2} + as + b}}}}\) 

\(= \frac{{Ks + b}}{{{s^2} + as + b - Ks - b}}\) 

\(= \frac{{Ks + b}}{{{s^2} + as - Ks}} = \frac{{Ks + b}}{{s\left( {s + a - K} \right)}}\)

Steady state error for ramp input is given by

\({e_{ss}} = \frac{1}{{{K_v}}}\) 

Velocity error coefficient,

\({K_v} = \begin{array}{*{20}{c}}
{It}\
{s \to 0}
\end{array}s.G\left( s \right)\) 

\(= \begin{array}{*{20}{c}}
{It}\
{s \to 0}
\end{array}s.\frac{{Ks + b}}{{s\left( {s + a - K} \right)}}\) 

\(= \frac{b}{{a - K}}\) 

Steady state error, \({e_{ss}} = \frac{{a - K}}{b}\)

\(C\left( s \right) = \frac{{\left( {1 + \frac{s}{{0.1}}} \right)\left( {1 + \frac{s}{{100}}} \right)}}{{\left( {1 + s} \right)\left( {1 + \frac{s}{{10}}} \right)}}\)

\(= \frac{{\left( {s + 0.1} \right)\left( {s + 100} \right)}}{{\left( {s + 1} \right)\left( {s + 10} \right)}}\)

Pole zero diagram of the above transfer function is

It represents a standard lead lag compensator. For a lead lag compensator, maximum lead occurs at initial frequency.

So, maximum lead occurs at 0.1 < ω < 1.

Nominal resistance values = R₁ and R₂

Additive uncertainties = ΔR₁ and ΔR₂

Standard deviation of the error in the equivalent resistance R is

\(\sigma = \pm \sqrt {{{\left( {\frac{{\partial R}}{{\partial {R_1}}}} \right)}^2}\sigma _1^2 + {{\left( {\frac{{\partial R}}{{\partial {R_2}}}} \right)}^2}\sigma _2^2} \) 

\(= \pm \sqrt {{{\left( {\frac{{\partial R}}{{\partial {R_1}}}} \right)}^2}{{\left( {{\rm{\Delta }}{R_1}} \right)}^2} + {{\left( {\frac{{\partial R}}{{\partial {R_2}}}} \right)}^2}{{\left( {{\rm{\Delta }}{R_2}} \right)}^2}} \)

Given that,

Horizontal tangencies (n_H) = 3

Vertical tangencies (n_V) = 2

Horizontal frequency (f_H) = 3 kHz

n_Hf_H = n_Vf_V

\(\Rightarrow {f_V} = \frac{{{n_H}{f_H}}}{{{n_V}}} = \frac{{3 \times 3 \times {{10}^3}}}{2} = 4.5;kHz\)

\(Z = \overline {{\overline{P ̅ Q}} .Q.\overline {QR} }\)

\(= P ̅ Q + ̅ Q + QR\)

= Q̅ (1 + P) + QR

= Q̅ + QR 

= Q̅(1 + R) + QR

= Q̅ + Q̅R + QR

= Q̅ + R (Q + Q̅)

= Q̅ + R

Explanation:

Single phase full bridge rectifier

At firing angle 'α' for highly inductive load,

observed from above waveform,

\(\rm T_0 = \frac{T_{in}}{2}\)

⇒ f₀ = 2 f_in

∴ Ripple frequency at D.C side = 2f_in = 2 × 50 = 100 Hz

The small inductance given in the circuit acts as a source inductance.

The effect of source inductance is as shown in figure below.

μ is an overlap angle.

Both diodes D₁ and D₂ conducts for (180 + μ) as shown in the figure.

Concept:

Calculation:

The waveform of output voltage in a three-phase voltage source inverter operating in 180° conduction mode is shown below.

Fourier series expansion of line to neutral voltage is

\({V_{Rn}} = \mathop \sum \limits_{n = 6K \pm 1}^\infty \frac{{2{V_{dc}}}}{{n\pi }}\sin n\omega t\)

For the fundamental voltage, K = 0

\({V_{R1}} = \frac{{2{V_{dc}}}}{\pi }\sin \omega t\)

Peak value \(= {V_{R1}} = \frac{{2{V_{dc}}}}{\pi }\)

Concept:

In a three-phase induction motor,

• The stator is stationary, and the rotor is a rotating part
• Both the stator field and rotor field rotate with synchronous speed
• The rotor rotates with a speed less than synchronous speed
• The relative speed between the rotor field and a stator is synchronous speed
• The relative speed between the rotor field and stator field is zero
• The relative speed between stator filed and stator is synchronous speed

 

The speed of rotor field flux is synchronous speed (Ns) and the speed of the stator is zero as the stator doesn't rotate.

The speed of rotor field flux with respect to the stator = Ns - 0 = Ns

Calculation:

Given that,

Number of poles (P) = 4

Frequency (f) = 50 Hz

Synchronous speed is given by

\({N_s} = \frac{{120f}}{P} = \frac{{120 \times 50}}{4} = 1500;rpm\)

\({\omega _s} = \frac{{2\pi {N_s}}}{{60}}\)

\(\Rightarrow {\omega _s} = \frac{{2\pi \times 1500}}{{60}} = 157;rad/sec\)

The speed of the rotor flux is nothing but the synchronous speed.

The speed of the rotor flux = 157 rad/sec

\(V_1^1 = \frac{{{V_2}}}{\alpha }\)

I₁ = α I₂

\({V_1} = V_1' + {I_1}z = \frac{{{V_2}}}{\alpha } + \left( {\alpha z} \right){I_2}\)

T - parameter matrix,

\(\left[ {\begin{array}{*{20}{c}} {{V_1}}\ {{I_1}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{1}{\alpha }}&{\alpha z}\ 0&\alpha \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_2}}\ {{I_2}} \end{array}} \right]\)

Since A ≠ D, the network is asymmetric and hence Y_bus and Z_bus will also be asymmetric.

Concept:

• Whenever current flow in the load is continuous, freewheeling diode comes into Conduction
• The freewheeling diode comes into action only for inductive loads (L, RL, RLE etc...) in controlled or uncontrolled rectifiers.

 

Explanation:

As observed from the given circuit, the load is purely resistive, so the conduction does not go beyond 180° because there is no stored energy in the resistor, Due to this, there is no flow of current in the freewheeling diode 'D3' Here current in load is discontinuous. Therefore, the RMS value of current through diode 'D3' is OA.

Let X be the time at which vehicle arrives the traffic junction

Let Y be the waiting time

If vehicle arrives between X = 0 and X = 2, the waiting time will be zero.

If vehicle arrives at X ≥ 2, the waiting time will be (5 - X).

 

\(Y = \left{ {\begin{array}{*{20}{c}}
{0,}&{X < 2}\
{5 - X,}&{X \ge 2}
\end{array}} \right.\) 

Expected waiting time,

\(= \smallint xf\left( x \right)dx\) 

\(= \mathop \smallint \limits_2^5 \left( {5 - x} \right)\left( {\frac{1}{5}} \right)dx\) 

\(= \mathop \smallint \limits_2^5 \left( {1 - \frac{x}{5}} \right)dx\) 

\(= \left[ {x - \frac{{{x^2}}}{{10}}} \right]_2^5\) 

\(= 3 - \frac{1}{{10}}\left( {25 - 4} \right)\) 

= 0.9

Concept:

In Partial Differentiation, all variables are considered as a constant except the independent derivative variable i.e If f(x,y,z) is a function, then its partial derivative with respect to x is calculated by keeping y and z as constant.

Calculation:

f(x, y, z) = (x² + y² – 2z²)(y² + z²)

\(\frac{{\partial f}}{{\partial x}} = \left( {2x} \right)\left( {{y^2} + {z^2}} \right)\)

At the point, x = 2, y = 1 and z = 3 is

\(\frac{{\partial f}}{{\partial x}} = 2\left( 2 \right)\left( {{1^2} + {3^2}} \right) = 40\)

2x² + y² = 34

x + 2y = 11

⇒ x = 11 – 2y

⇒ 2(11 – 2y)² + y² = 34

⇒ 2(121 + 4y² – 44y) + y² = 34

⇒ 242 + 8y² – 88y + y² = 34

⇒ 9y² – 88y + 208 = 0

⇒ y = 4, 5.77

As y is an integer, y = 4

⇒ x = 11 – 2(4) = 3

Value of x + y = 7.

y² – 2y + 1 = x

\(\sqrt x + y = 5\)

\(\Rightarrow \sqrt x = 5 - y\)

⇒ x = (5 - y)²

⇒ y² – 2y + 1 = (5 – y)²

⇒ y² – 2y + 1 = 25 + y² – 10y

⇒ 8y = 24 ⇒ y = 3

x = (5 - y)² = (5 - 3)² = 4

x + √y = 4 + √3 = 5.732

Concept:

The Z-parameter for a T-equivalent two-port network is,

\(\left[ Z \right] = \left[ {\begin{array}{*{20}{c}} {{Z_{11}}}&{{Z_{12}}}\ {{Z_{21}}}&{{Z_{22}}} \end{array}} \right]\)

Where

Z11 is an open circuit impedance

Z12 = Z21 = Transfer impedance

Z22 = Open circuit output impedance

Calculation:

By using delta to star conversion, the given circuit can be reduced.

\({R_1} = \frac{{4 \times 2}}{{4 + 2 + 2}} = \frac{8}{8} = 1;{\rm{\Omega }}\)

\({R_2} = \frac{{2 \times 2}}{{4 + 2 + 2}} = \frac{4}{8} = \frac{1}{2};{\rm{\Omega }}\)

\({R_3} = \frac{{4 \times 2}}{{4 + 2 + 2}} = \frac{8}{8} = 1;{\rm{\Omega }}\)

The modified circuit is,

Now, by comparing the above circuit with T-equivalent network of Z-parameter matrix,

\(\left[ Z \right] = \left[ {\begin{array}{*{20}{c}} 4&3\ 3&{3.5} \end{array}} \right]\)

Transfer impedance = Z21 = 3 Ω

Alternate Method

Transfer impedance \({z_{21}} = {\left. {\frac{{{V_2}}}{{{I_1}}}} \right|_{{I_2} = 0}}\)

V₁ = Z₁₁ I₁ + Z₁₂ I₂

V₂ = Z₂₁ I₁ + I₂₂ I₂

At I₂ = 0,

By applying KVL in the loop 1,

(I₁ – I₃) (2) + (I₁ – I₃)(2) – 4(I₃) = 0

⇒ 4I₁ – 8I₃ = 0 ⇒ I₁ = 2I₃

By applying KVL in the loop 2,

-V₂ + 2(I₁ – I₃) + 2I₁ = 0

⇒ V₂ = 4I₁ – 2I₃ = 4(2I₃) – 2I₃ = 6I₃

⇒ V₂ = 6(I₁/2)

\(\Rightarrow {Z_{21}} = \frac{{{V_2}}}{{{I_1}}} = 3\)

Given circuit diagram:

The above circuit can be redrawn as, (the above circuit forms a balanced bridge condition)

The equivalent resistance of all the 5 Ω resistors is 5 Ω 

Current in the circuit will be 

i(t) = I e^-4τ

Time constant τ = RC = 5 secs

\(I = \frac{V}{R} = \frac{{10}}{5} = 2A\)

\(i\left( t \right) = 2{e^{ - \frac{t}{5}}} = 2{e^{ - 0.2t}}\)

Energy transferred from the DC source,

\(E = \mathop \smallint \limits_0^\infty 10 \times 2{e^{ - 0.2t}}dt\)

\(= 20\mathop \smallint \limits_0^\infty {e^{ - 0.2t}}\)

\(= 20\left[ {\frac{{{e^{ - 0.2t}}}}{{ - 0.2}}} \right]_0^\infty = 20\left[ {0 + \frac{1}{{0.2}}} \right] = 100;J\)

Concept:

RMS (Root mean square) value:

• RMS value is based on the heating effect of wave-forms.
• The value at which the heat dissipated in the AC circuit is the same as the heat dissipated in the DC circuit is called the RMS value provided, both the AC and DC circuits have equal values of resistance and are operated at the same time.

RMS value 'or' the effective value is calculated as:

\(= \frac{1}{T}\mathop \smallint \limits_0^T {\left( {f\left( t \right)} \right)^2}dt\)

Where T is the time period.

Calculation:

In the given waveform 

\(f\left( t \right) = \left{ {\begin{array}{*{20}{c}} { - 0.3 < t < 0.7,;f\left( t \right) = 4}\ {0.7 < t < 2.7,f\left( t \right) = - 2}\ {2.7 < t < 3.7,f\left( t \right) = 0} \end{array}} \right.\)

The time period of the given waveform = 3.7 - (- 0.3) = 4

Mean square value \(= \frac{1}{4}\mathop \smallint \limits_{ - 0.3}^{3.7} {\left( {f\left( t \right)} \right)^2}dt\)

\(= \frac{1}{4}\left[ {\mathop \smallint \limits_{ - 0.3}^{0.7} {{\left( 4 \right)}^2}dt + \mathop \smallint \limits_{0.7}^{2.7} {{\left( { - 2} \right)}^2}dt + \mathop \smallint \limits_{2.7}^{3.7} {{\left( 0 \right)}^2}dt} \right]\)

\( = \frac{1}{4}\left[ {16\left( {0.7 + 0.3} \right) + 4\left( {2.7 - 0.7} \right)} \right]\)

\(= \frac{1}{4}\left[ {16 + 8} \right] = 6\)

Additional Information

• Average or mean value of alternating current is that value of steady current which sends the same amount of charge through the circuit in a certain interval of time as is sent by alternating current through the same circuit in the same interval of time
• The ratio of the maximum value (peak value) to the RMS value is known as the peak factor or crest factor.
• \(Peak;factor = \frac{{maximum;value}}{{rms;value}}\)
• The ratio of the RMS value to the average value is known as the form factor.
• \(Form;factor = \frac{{rms;value}}{{average;value}}\)

IMPORTANT EVALUATIONS:

Characteristic impedance of the transmission line is,

\({Z_c} = \sqrt {\frac{Z}{Y}}\) 

Given nominal - π circuit is,

Z = 100 ∠80° Ω

X = 3300 Ω

\(\frac{Y}{2} = \frac{1}{X} = \frac{1}{{3300}}\) 

\(\Rightarrow Y = \frac{2}{{3300}}mho\) 

\({Z_c} = \sqrt {\frac{Z}{Y}} = \sqrt {\frac{{100}}{{\frac{2}{{3300}}}}} = \sqrt {\frac{{3300 \times 100}}{2}} = 406.2{\rm{\Omega }}\)

Size of Jacobian matrix = (2n – m – 2) × (2n – m - 2)

Where n = number of buses

m = number of pv buses

Given that,

Size of Jacobian matrix = 100 × 100

Number of PV buses (m) = 20

⇒ (2n – m - 2) = 100

⇒ 2n – 20 – 2 = 100

⇒ n = 61

Concept:

A function f(x) is said to be continuous at a point x = a, in its domain if,

 \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{a}} \right)\) exists or its graph is a single unbroken curve.

In other words,

A function f(x) is continuous at x = a if,

Left limit = Right limit = Function value = Real and finite

Analysis:

\(f\left( x \right) = \left{ {\begin{array}{*{20}{c}} {1 - x,}&{x \le 0}\ {{x^2},}&{x > 0} \end{array}} \right.\)

\(g\left( x \right) = \left{ {\begin{array}{*{20}{c}} { - x,}&{x \le 1}\ {x + 1,}&{x \ge 1} \end{array}} \right.\)

For x ≤ 0, f(x) = 1 – x

g (x) = -x

(fog)(x) = f(g(x)) = f(-x) = 1 + x

y = 1 + x is a continuous function.

Hence, the number of discontinuities = 0

Concept:

For a given complex function with poles, the complex integral \(I = \mathop \oint \limits_C f\left( z \right)dz\) is given by

Residue theorem as;

\(I = \mathop \oint \limits_C f\left( z \right)dz = 2\pi ;i \times \left{ {Sum;of;residue;of;poles;inside;or;on;C} \right}\)

Calculation:

\(f\left( z \right) = \oint \frac{{{z^3} - 2z + 3}}{{z - 2}}dz\)

Contour: |z|= 3

Simple pole, z = 2 and it is lies inside the contour.

Residue of f(z) at z = 2 is,

\(\mathop {{\rm{lt}}}\limits_{z \to 2} \left( {z - 2} \right)\frac{{\left( {{z^3} - 2z + 3} \right)}}{{\left( {z - 2} \right)}} = {2^3} - 2\left( 2 \right) + 3 = 7\)

f(z) = 2πi(7) = 14πi

Concept:

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

The roots of the characteristic equation are called Eigenvalues or latent roots or characteristic roots of matrix A.

Properties of Eigenvalues:

• The sum of Eigenvalues of a matrix A is equal to the trace of that matrix A
• The product of Eigenvalues of a matrix A is equal to the determinant of that matrix A
• If λ is an eigenvalue of a matrix A, then λn will be an eigenvalue of a matrix An.
• If λ is an eigenvalue of a matrix A, then kλ will be an eigenvalue of a matrix kA where k is a scalar.

 

Calculation:

\(A = \left[ {\begin{array}{*{20}{c}} 0&1&0\ 0&0&1\ 0&{ - 3}&{ - 4} \end{array}} \right]\)

|A - λI| = 0

\(A - \lambda I = \left[ {\begin{array}{*{20}{c}} 0&1&0\ 0&0&1\ 0&{ - 3}&{ - 4} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} \lambda &0&0\ 0&\lambda &0\ 0&0&\lambda \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - \lambda }&1&0\ 0&{ - \lambda }&1\ 0&{ - 3}&{ - 4 - \lambda } \end{array}} \right]\)

|A - λI| = 0

\(\Rightarrow \left| {\begin{array}{*{20}{c}} { - \lambda }&1&0\ 0&{ - \lambda }&1\ 0&{ - 3}&{ - 4 - \lambda } \end{array}} \right| = 0\)

⇒ (-λ) ((-λ)(-4 - λ) + 3) = 0

⇒ (-λ)(4λ + λ² + 3) = 0

⇒ λ = 0, λ² + 4λ + 3 = 0

⇒ λ = 0, -1, -3

Concept:

Power absorbed (D) = √3 VL IL cos ϕ

Where, V_L = line to line voltage , IL = line current

ϕ = power factor angle of the load

\(Impedance\left( z \right) = \frac{{{V_{ph}}}}{{{I_{ph}}}} \)

Calculation:

Given that,

Line to line voltage (V_L) = 208 V

Power absorbed by the load (P) = 432 W

Power factor (cos ϕ) = 0.6 leading

Power absorbed (D) = √3 × 208 × IL × 0.6 = 432

⇒ I_L = 2 A

\(Impedance\left( z \right) = \frac{{{V_{ph}}}}{{{I_{ph}}}} = \frac{{\frac{{208}}{{\sqrt 3 }}}}{2} = 60;{\rm{\Omega }}\)

Power factor angle (ϕ) = cos^-1 (0.6) = 53.1°

As the power factor is leading, ϕ = -53.1°

⇒ Z = 60 ∠-53.1° Ω

Concept:

Maximum power transfer theorem for AC circuits:

The maximum power transfer theorem states that the maximum power flow through an AC circuit will occur when the load impedance is equal to the complex conjugate of the source impedance as viewed from its output terminals.

When the load impedance matches the Thevenin equivalent resistance of the given circuit, maximum power is transferred to it. i.e.

If RL = Rth, Maximum power is transferred to the Load.

ZL = ZS* 

Calculation:

Given circuit diagram:

Inductive reactance = jωL

Capacitive reactance \(= \frac{1}{{j\omega C}}\)

Load impedance (Z_L) will be:

\(\left( {{Z_L}} \right) = j\omega L + \frac{{\left( 1 \right)\left( {\frac{1}{{j\omega C}}} \right)}}{{1 + \frac{1}{{j\omega C}}}}\)

\(= j\omega L + \frac{1}{{1 + j\omega C}}\)

\(= j\omega L + \frac{{1 - j\omega C}}{{1 + {\omega ^2}{C^2}}}\)

\(= j\omega L - \frac{{j\omega C}}{{1 + {\omega ^2}{C^2}}} + \frac{1}{{1 + {\omega ^2}{C^2}}}\)

For maximum power transfer,

Z_s = Z_L* 

\(\Rightarrow 0.5 = \frac{1}{{1 + {\omega ^2}{C^2}}} + j\left( {\omega L - \frac{{\omega C}}{{1 + {\omega ^2}{C^2}}}} \right)\)

As there is no reatance term present in the source, therefore active component of load impedance should be equal to active component of source for maximum power transfer 

\( \Rightarrow \frac{1}{{1 + {\omega ^2}{C^2}}} = 0.5\)

\(\Rightarrow \frac{1}{{1 + {{\left( {100} \right)}^2}{C^2}}} = 0.5\)

⇒ 1 + (100)² C² = 2

⇒ (100)² C² = 1

\(\Rightarrow C = \frac{1}{{100}} = 10;mF\)

Concept:

Duality property,

A rect \(\left(\frac{t}{\tau}\right)\rm\overset{F.T.}{\leftrightarrow}A\tau \sin C(f\tau)=A\tau \sin C \left(\frac{\omega \tau}{2\pi}\right)\)

\(A\tau \sin C \left(\frac{t \tau}{2\pi}\right)\rm\overset{F.T.}{\leftrightarrow}2\pi A\ rect\left(\frac{\omega}{\tau}\right)\)

Explanation:

 

z(t) = x(t) cos (1000 πt)

\({\rm{Z}}\left( {\rm{\omega }} \right) = \frac{1}{2}\left[ {{\rm{x}}\left( {{\rm{\omega }} - 1000{\rm{;\pi }}} \right) + {\rm{x}}\left( {{\rm{\omega }} + 1000{\rm{\pi }}} \right)} \right]\)

Spectrum of impulse response is,

\(\frac{{\sin {\rm{at}}}}{{{\rm{\pi t}}}} \leftrightarrow \left{ {\begin{array}{*{20}{c}} {1,}&{\left| {\rm{\omega }} \right| < {\rm{a}}}\ {0,}&{\left| {\rm{\omega }} \right| > {\rm{a}}} \end{array}} \right.\)

Spectrum of output is,

Maximum frequency of signal y(t), f_m = 750 Hz

Sampling rate, f_s = 2f_m = 2 × 750 = 1500 Hz

NOTE:

\(\frac{\sin xt}{\pi t}\rm \overset{F.T.}{\leftrightarrow}\left{\begin{matrix}1;&|\omega|<x\\ 0;&|\omega|>x\end{matrix}\right.\)

Concept:

Series connected system:

h(n) = h1(n) + h2(n)

Cascade system:

h(n) = h1(n) * h2(n)

Explanation:

Using Tabular method,

h(n) = h1(n) * h2(n)

where, \(h_1(n)=\left{\begin{matrix}1,&-1\\ {\uparrow}\end{matrix}\right}\)

\(h_2(n)=\left{\begin{matrix}1,&1\\ {\uparrow}\end{matrix}\right}\)

\(h(n)=\left{\begin{matrix}1,&0&-1\\ {\uparrow}\end{matrix}\right}\)

consider, x(n) = {a, a', a'', a'''}

y(n) = {a, a', a'' - a, a''' - a', -a'', - a'''}

Now, comparing above expression with given

\(y(n)=\left{\begin{matrix}1,&2&1&-1&-2&-1\\ {\uparrow}\end{matrix}\right}\)

a = 1, a' = 2, a'' - a = 1, a''' - a' = -1, -a'' = -2, -a''' = -1

we get, a = 1, a' = 2, a'' = 2, a''' = 1

Hence,

\(x(n)=\left{\begin{matrix}1,&2&2&1\\ {\uparrow}\end{matrix}\right}\)

Voltage (V) = 220 V

Torque (τ) = 70 N-m

Speed (N) = 900 rpm

Armature resistance (R_a) = 0.02 Ω

Power (P) = 10 kW

⇒ E_b I_a = 10 × 10³

⇒ (V - I_a R_a) I_a = 10,000

⇒ [220 – I_a(0.02)] I_a = 10,000

⇒ I_a = 45.75 A

P = τω

\(\Rightarrow 10 \times {10^3} = {\rm{\tau }}\left( {\frac{{2{\rm{\pi N}}}}{{60}}} \right)\) 

\(\Rightarrow {\rm{\tau }} = \frac{{10 \times {{10}^3} \times 60}}{{2{\rm{\pi }} \times 900}} = 106.1{\rm{;N}} - {\rm{m}}\) 

Torque α ϕ I_a

For separately excited motor

T α I_a

T₁ = 106.1 N-m, I_a1 = 45.75 A

T₂ = 70 N-m

\({{\rm{I}}_{{{\rm{a}}_2}}} = \frac{{{{\rm{T}}_2} \times {{\rm{I}}_{{{\rm{a}}_1}}}}}{{{{\rm{T}}_1}}} = \frac{{70 \times 45.75}}{{106.1}} = 30.18{\rm{;A}}\)

s² + 6Ks + 2s + 5 = 0

\(\Rightarrow {\rm{G}}\left( {\rm{s}} \right){\rm{H}}\left( {\rm{s}} \right) = \frac{{6{\rm{Ks}}}}{{{{\rm{s}}^2} + 2{\rm{s}} + 5}}\)

The point at which root locus enter real axis is break in point

\(\frac{{{\rm{dK}}}}{{{\rm{ds}}}} = 0\)

\({\rm{K}} = - \frac{{\left( {{{\rm{s}}^2} + 2{\rm{s}} + 5} \right)}}{{6{\rm{s}}}}\)

\(\frac{{{\rm{dK}}}}{{{\rm{ds}}}} = \frac{{ - 6{\rm{s}}\left( {2{\rm{s}} + 2} \right) + \left( {{{\rm{s}}^2} + 2{\rm{s}} + 5} \right)\left( 6 \right)}}{{{{\left( {6{\rm{s}}} \right)}^2}}} = 0\)

⇒ -12s² – 12s + 6s² + 12s + 30 = 0

\(\Rightarrow {\rm{s}} = \pm \sqrt 5\)

Break-in point always exist on root locus plot,

\(\Rightarrow {\rm{s}} = - \sqrt 5\)

s³ + 3s² + 2s + K = 0

\(\left. {\begin{array}{*{20}{c}} {{{\rm{s}}^3}}\ {{{\rm{s}}^2}}\ {{{\rm{s}}^1}}\ {{{\rm{s}}^0}} \end{array}} \right|\begin{array}{*{20}{c}} 1\ 3\ {\frac{{6 - {\rm{K}}}}{3}}\ {\rm{K}} \end{array}\begin{array}{*{20}{c}} 2\ {\rm{K}}\ {0{\rm{;}}}\ {} \end{array}\)

To have all the roots are in the left half of the complex plane, there should be no sign changes in the first column of RH table.

⇒ 6 – K > 0 and K > 0

⇒ 6 > K and K > 0

⇒ 0 < K < 6

Peak over-shoot \( = {{\rm{e}}^{ - \frac{{{\rm{\pi \xi }}}}{{\sqrt {1 - {{\rm{\xi }}^2}} }}}}\) 

Peak over-shoot is maximum for minimum damping ratio (ξ).

Standard second order transfer function is

\(= \frac{{{\rm{\omega }}_{\rm{n}}^2}}{{{{\rm{s}}^2} + 2{\rm{\xi }}{{\rm{\omega }}_{\rm{n}}}{\rm{s}} + {\rm{\omega }}_{\rm{n}}^2}}\) 

1) \(\frac{{100}}{{{{\rm{s}}^2} + 10{\rm{s}} + 100}}\)

\({\rm{\omega }}_{\rm{n}}^2 = 100 \Rightarrow {{\rm{\omega }}_{\rm{n}}} = 10\) 

\(2{\rm{\xi }}{{\rm{\omega }}_{\rm{n}}} = 10 \Rightarrow {\rm{\xi }} = \frac{{10}}{{2 \times 10}} = 0.5\) 

2) \(\frac{{100}}{{{{\rm{s}}^2} + 15{\rm{s}} + 100}}\)

\({\rm{\omega }}_{\rm{n}}^2 = 100 \Rightarrow {{\rm{\omega }}_{\rm{n}}} = 10\) 

\(2{\rm{\xi }}{{\rm{\omega }}_{\rm{n}}} = 15 \Rightarrow {\rm{\xi }} = \frac{{15}}{{2 \times 10}} = 0.75\) 

3) \(\frac{{100}}{{{{\rm{s}}^2} + 5{\rm{s}} + 100}}\)

\({\rm{\omega }}_{\rm{n}}^2 = 100 \Rightarrow {{\rm{\omega }}_{\rm{n}}} = 10\) 

\(2{\rm{\xi }}{{\rm{\omega }}_{\rm{n}}} = 5 \Rightarrow {\rm{\xi }} = \frac{5}{{2 \times 10}} = 0.25\)

3) \(\frac{{100}}{{{{\rm{s}}^2} + 20{\rm{s}} + 100}}\)

\({\rm{\omega }}_{\rm{n}}^2 = 100 \Rightarrow {{\rm{\omega }}_{\rm{n}}} = 10\) 

\(2{\rm{\xi }}{{\rm{\omega }}_{\rm{n}}} = 20 \Rightarrow {\rm{\xi }} = \frac{{20}}{{2 \times 10}} = 1\) 

Option (3) has minimum damping ratio and it will have maximum peak overshoot.

Applying KVL in the input loop,

V_CC­ - I_E(4R) - I_BR_B - V_BE - I_ER = 0

⇒ 10 – (1 + β) I_B(4R) - I_BR_B – 0.7 – (1 + β) I_BR = 0

⇒ 10 – 120 I_BR - I_BR_B – 0.7 – 30 I_BR = 0

\( \Rightarrow {{\rm{I}}_{\rm{B}}} = \frac{{9.3}}{{150{\rm{;R}} + {{\rm{R}}_{\rm{B}}}}}\)     ----(1)

Applying KVL in output loop,

V_CC­ - I_E(4R) - V_CE - I_ER = 0

⇒ 10 – 120 I_BR – 5 – 30 I_BR = 0

\(\Rightarrow {{\rm{I}}_{\rm{B}}} = \frac{5}{{150{\rm{;R}}}} = \frac{1}{{30{\rm{;R}}}}\)     ----(2)

From equation (1) & (2)

\(\frac{{9.3}}{{150{\rm{;R}} + {{\rm{R}}_{\rm{B}}}}} = \frac{1}{{30{\rm{;R}}}}\)

⇒ 279R = 150 R + R_B

⇒ R_B = 129 R

\( \Rightarrow \frac{{{{\rm{R}}_{\rm{B}}}}}{{\rm{R}}} = 129\)

Concept:

Virtual ground -

The concept of the virtual ground is stated as if anyone of the i/p terminals is grounded physically the other i/p terminal will also be at ground potential even though, it is not grounded physically.

• One key feature of an Op-Amp is the differential input, and when put together in a circuit, this can form a virtual ground.
• The virtual ground concept is helpful for the analysis of Op Amps. This concept makes Op-Amp circuit analysis much easier.

Explanation:

By applying nodal analysis at inverting node

\(\frac{{{{\rm{V}}_{\rm{a}}}}}{{\rm{R}}} + \frac{{{{\rm{V}}_{\rm{a}}} - {{\rm{V}}_1}}}{{\rm{R}}} = 0\)

⇒ V₁ = 2V_a       ----(1)

By applying nodal analysis at non-inverting node

\(\frac{{{{\rm{V}}_{\rm{a}}}}}{{2{\rm{R}}}} + \frac{{{{\rm{V}}_{\rm{a}}} - {{\rm{V}}_{\rm{s}}}}}{{2{\rm{R}}}} = 0\)

⇒ V_s = 2V_a       ----(2)

Form eq. 1) and 2)

⇒ V_s = V₁ = 2V_a

By applying nodal analysis at V₁,

\(\frac{{{{\rm{V}}_1} - {{\rm{V}}_0}}}{{\rm{R}}} + \frac{{{{\rm{V}}_1} - {{\rm{V}}_{\rm{a}}}}}{{\rm{R}}} + \frac{{{{\rm{V}}_1}}}{{\rm{R}}} = 0\)

⇒ 3V₁ – V_a = V₀

\(\Rightarrow {{\rm{V}}_{\rm{o}}} = 3{{\rm{V}}_{\rm{s}}} - \frac{{{{\rm{V}}_{\rm{s}}}}}{2} = \frac{{5{{\rm{V}}_{\rm{s}}}}}{2} = 2.5{{\rm{V}}_{\rm{s}}}\)

Concept:

From sinusoidal PWM control technique:

The peak value of fundamental voltage is given as,

Vm1 = MVdc    ---(1)

Explanation:

Given that,

RL load: R = 40 Ω, \(L = \frac{0.3}{π} \) H

f = 50 Hz

P = 1.44 kW

Modulation index (M) = 0.6

At 50 Hz,

XL = 2πfL = 100 × \(\frac{0.3}{\pi} \) = 30 Ω

Z = R + jXL = 40 + j30

|Z| = 50 Ω

∵ P = 1.44 × 103

⇒ \(\rm I_{rms}^2 R = 1440\)

\(⇒ \rm I_{rms}^2 R = \frac{1440}{40}\)

⇒ Irms = 6 A

we know that,

\(\rm Z = \frac{V_{rms}}{I_{rms}}\)

⇒ Vrms = Irms × Z

⇒ Vrms = 6 × 50

⇒ Vrms = 300 V

so, peak value Vm = 300√2

From equation (1)

300√2 = 0.6 VDC

⇒ VDC = \(\rm \frac{300√ 2}{0.6}\)

∴ VDC = 500√2 V

Concept:

Random variables:

Random variable assigns a real number to each possible outcome.

Let X be a discreet random variable, then

Expectation E(x) = Σxp(x)

The variance of X = R =E[X2]- (E[X])2 

E(x2) = Var(x) + {E(x)}2

\(Standard;Deviation;σ=\sqrt{Variance}\)

Calculation:

A person decided to toss a fair coin repeatedly until he gets head.

Maximum number of tosses = 3

The event stops if he gets a head (or) at 3 tosses.

The possible outcome and their probabilities are

i) Gets head at first toss. (H)

Probability = ½

ii) Gets head at second toss (T H)

Probability = ½ × ½ = ¼ 

iii) Gets head at third toss (T T H)

Probability = ½ × ½ × ½ = 1/8

iv) Not getting head after three tosses (T T T)

Probability = ½ × ½ × ½ = 1/8

Y denote the number of heads

 

Expectation E(y) = Σy p(y)

\(= \left( {1 \times \frac{1}{2}} \right) + \left( {1 + \frac{1}{4}} \right) + \left( {1 + \frac{1}{8}} \right) + \left( {0 \times \frac{1}{8}} \right) = \frac{7}{8}\)

E(y²) = Σy² P(y)

\(= \left( {{1^2} \times \frac{1}{2}} \right) + \left( {{1^2} + \frac{1}{4}} \right) + \left( {{1^2} + \frac{1}{8}} \right) + \left( {{0^2} \times \frac{1}{8}} \right) = \frac{7}{8}\)

Variance of y.

\({\rm{Var}}\left{ {\rm{y}} \right} = {\rm{E}}\left[ {{{\rm{y}}^2}} \right] - {\left[ {{\rm{E}}\left[ {\rm{y}} \right]} \right]^2}\)

\(= \frac{7}{8} - {\left( {\frac{7}{8}} \right)^2}\)

\(= \frac{7}{{64}}\)

Thevenin’s voltage is the open circuit voltage across the terminals a and b.

By using source transformation,

By using source transformation once again,

By applying KVL

30 + 15 I + 5I + 8 = 0

⇒ I = -1.9 A

By applying KVL

30 + 15I + V_ab = 0

⇒ V_ab = -15I – 30 = -15(-1.9) - 30

= -1.5 V

Voltage (V) = 120 V

shunt field resistance (R_sh) = 240 Ω

Shunt field current \(\left( {{{\rm{I}}_{{\rm{sh}}}}} \right) = \frac{{120}}{{240}} = 0.5{\rm{;A}}\)

Armature resistance (R_a) = 0.8 Ω

No load:

Load current (I_LO) = 2 A

Armature current (I_ao) = I_Lo – I_sh

= 2 – 0.5 = 1.5 A

Let No load speed is N_o

Back emf (E_bo) = V – I_ao R_a = 120 – (1.5) (0.8) = 118.8 V

Full load:

Load current (I_Lf) = 7 A

Armature current (I_af) = I_Lf – I_sh

= 7 – 0.5 = 6.5 A

Back emf (E_bf) – V - I_af R_a

= 120 – (6.5) (0.8)

= 114.8 V

Full load speed (N_f) = 1200 rpm

In a DC shunt motor

E_b ∝ N

\(\Rightarrow {{\rm{N}}_{\rm{o}}} = \frac{{{{\rm{E}}_{{\rm{bo}}}}}}{{{{\rm{E}}_{{\rm{bf}}}}}} \times {{\rm{N}}_{\rm{f}}} = \frac{{118.8}}{{114.8}} \times 1200 = 1241.8{\rm{;rpm}}\)

Concept:

The current is given as for star connected motor,

\(\rm I=\frac{V_{phase}}{\sqrt{\left(R_1+\frac{R_2}{s}\right)^2+(X_1+X_2)^2}}\)

Calculation:

Given that,

R1 = 0.3 Ω, R2 = 0.3 Ω, X1 = 0.41 Ω, X2 = 0.41 Ω

Frequency (f) = 60 Hz

For star connected induction motor,

\(\rm I=\frac{V_{phase}}{\sqrt{\left(R_1+\frac{R_2}{s}\right)^2+(X_1+X_2)^2}}\)

At starting {s = 1}

\(\rm I_{st}=\frac{V_{phase}}{\sqrt{\left(R_1+R_2\right)^2+(X_1+X_2)^2}}\)

When supply is charged to 80 V (line), 20 Hz, reactance offered by stator & rotor will be changed because of change in frequency {∴ x ∝ f}

\(x^{'}_1=\frac{f_2}{f_1}x_1=\frac{20}{60}\times0.41=0.13666 \ \Omega\)

\(x^{'}_2=\frac{f_2}{f_1}x_2=\frac{20}{60}\times0.41=0.13666 \ \Omega\)

\(V^{'}_{ph}=\frac{80}{\sqrt{3}}\) at 20 Hz

At 20 Hz, equivalent circuit can be drawn as,

\(\rm I_{st}=\frac{80/\sqrt{3}}{\sqrt{(0.3+0.3)^2+(0.1366+0.1366)^2}}\)

I_st = 70.05 A

kVA rating = 25 kVA

Rated voltage = 400 V

Rated current \(\left( {{{\rm{I}}_{\rm{L}}}} \right) = \frac{{25 \times {{10}^3}}}{{400 \times \sqrt 3 }} = 36.08{\rm{;A}}\)

In delta connection \({{\rm{I}}_{{\rm{ph}}}} = \frac{{{{\rm{I}}_{\rm{L}}}}}{{\sqrt 3 }} = 20.83{\rm{;A}}\)

Open circuit voltage (V_oc) = 360 V

\({{\rm{X}}_{\rm{s}}} = \frac{{{{\rm{V}}_{{\rm{oc}}}}}}{{{{\rm{I}}_{{\rm{ph}}}}}} = \frac{{360}}{{20.83}} = 17.28{\rm{\Omega }}/{\rm{phase}}\)

Power factor = 0.8 leading

cosϕ = 0.8 and sinϕ = 0.6

\({\rm{E}} = \sqrt {{{\left( {{\rm{Vcos}}\phi + {{\rm{I}}_{\rm{a}}}{{\rm{R}}_{\rm{a}}}} \right)}^2} + {{\left( {{\rm{V;sin;}}\phi - {{\rm{I}}_{\rm{a}}}{{\rm{X}}_{\rm{s}}}} \right)}^2}} \)

\(= \sqrt {{{\left( {400 \times 0.8 + 0} \right)}^2} + {{\left( {400 \times 0.6 - 20.83 \times 17.28} \right)}^2}} = 341.74{\rm{;V}}\)

Voltage regulation \(= \frac{{{\rm{E}} - {\rm{V}}}}{{\rm{V}}} \times 100\)

\(= \frac{{341.74 - 400}}{{400}} \times 100 = - 14.56{\rm{;% }}\)

Concept:

In a three-phase system,

• The current passing through any two lines of supply is called as line current
• The current passing through a supply line is called as phase current
• The voltage across any single phase is called as phase voltage
• The voltage difference between any two phases is called as line voltage

Important Point:

In a star-connected three-phase system,

VL = √3 × Vph

And IL = Iph

\({I_{ph}} = \frac{{{V_L}}}{{\surd 3Z}}\)

In a delta connected three-phase system,

VL = Vph

IL = √3 × Iph

\({I_{ph}} = ;\frac{{{V_L}}}{Z}\)

Where,

VL is line voltage

Vph is phase voltage

IL is line current

Iph is phase current

Explanation:

Primary side rating: 3.3 kV(Y), 25 kVA

Phase current in primary side \(= \frac{{25}}{{\sqrt 3 \times 3.3}} = 4.37{\rm{;A}}\)

Phase turns ratio = 5 : 1

Phase current in secondary side = 5 × 4.374 = 21.87 A

Line current in secondary side \(= 21.87 \times \sqrt 3 = 37.88{\rm{;A}}\)

Form the given space representation,

\(\begin{array}{*{20}{c}}
{A = \left[ {\begin{array}{*{20}{c}}
0&1\
0&{ - 2}
\end{array}} \right]}&{B = \left[ {\begin{array}{*{20}{c}}
0\
1
\end{array}} \right]}&{C = \left[ {\begin{array}{*{20}{c}}
1&0
\end{array}} \right]}
\end{array}\)

State transition matrix

\(\phi \left( s \right) = {\left[ {SI - A} \right]^{ - 1}}\)

\(SI - A = \left[ {\begin{array}{*{20}{c}}
s&0\
0&s
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
0&1\
0&{ - 2}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
s&{ - 1}\
0&2
\end{array}} \right];;;\)

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

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

By taking inverse Laplace transform

\({e^{At}} = \phi \left( t \right) = \left[ {\begin{array}{*{20}{c}}
{u\left( t \right)}&{\frac{1}{2}\left( {1 - {e^{ - 2t}}} \right)u\left( t \right)}\
0&{{e^{ - 2t}}u\left( t \right)}
\end{array}} \right]\)

\({e^{At}}x\left( 0 \right) = \left[ {\begin{array}{*{20}{c}}
{u\left( t \right)}&{\frac{1}{2}\left( {1 - {e^{ - 2t}}} \right)u\left( t \right)}\
0&{{e^{ - 2t}}u\left( t \right)}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
0\
1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{{\rm{u}}\left( {\rm{t}} \right)}\
0
\end{array}} \right]\)

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

\(= \left[ {\begin{array}{*{20}{c}}
{\frac{1}{{2{s^2}}} - \frac{1}{{4s}} + \frac{1}{{4\left( {s + 2} \right)}}}\
{\frac{1}{{2s}} - \frac{1}{{2\left( {s + 2} \right)}}}
\end{array}} \right]\)

By taking inverse Laplace transform

\(x\left( t \right) = \left[ {\phi \left( t \right)} \right]\left[ {x\left( 0 \right)} \right] + {L^{ - 1}}\left{ {\phi \left( s \right)Bu\left( s \right)} \right}\)

\(= \left[ {\begin{array}{*{20}{c}}
{u\left( t \right)}\
0
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
{\left( {\frac{1}{2}t + \frac{3}{4} + \frac{1}{4}{e^{ - 2t}}} \right)u\left( t \right)}\
{\left( {\frac{1}{2} - \frac{1}{2}{e^{ - 2t}}} \right)u\left( t \right)}
\end{array}} \right]\)

y (t) = [1 0] x(t)

\(= \left[ {\begin{array}{*{20}{c}}
1&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{\left( {\frac{1}{2}t + \frac{3}{4} + \frac{1}{4}{e^{ - 2t}}} \right)u\left( t \right)}\
{\left( {\frac{1}{2} - \frac{1}{2}{e^{ - 2t}}} \right)u\left( t \right)}
\end{array}} \right]\)

\(= \frac{1}{2}t + \frac{1}{4}{e^{ - 2t}} + \frac{3}{4}\)

At t = 1 & C, y(t) \(= \frac{1}{2} + \frac{1}{4}\left( {{e^{ - 2}}} \right) + \frac{3}{4} = 1.283\)

For frequency mode of operation

\(f = \frac{n}{t}\)

f is frequency of input signal

n is number of cycles

t is gating time

n = 100 × 10⁶

t = 1 sec

f = 10⁸ cycles/sec

In period made period P of input signal is the inverse of its frequency

\(P = \frac{1}{f} = \frac{1}{{{{10}^8}}} = 10;ns\)

On 10 ½ digit display,

Frequency is displayed as 10000000.00 Hz

Period is displayed as 10.00000000 ns

Concept:

 

 

P = V_srI_sr cos ϕ

P = V_srI_sr1 cos ϕ₁

\(\rm P = I_{or}^2 R = \frac{V_{or}^2}{R}\)

\(\rm Input \ power \ factor = \frac{ power \ delivered \ to \ load}{VA \ rating \ of \ transformer}\)

where,

           cos ϕ = input supply power factor

           cos ϕ₁ = input fundamental power factor

               I_sr = supply rms current

               I_sr1 = supply fundamental rms current

               I_or = output rms current

               V_or = output rms voltage

                V_sr = supply rms voltage

Calculation:

output voltage waveform of half wave rectifier with resistive load is shown below,

\(\rm V_{0(rms)}= \frac{V_m}{2}\)

\(∵ IPF = \frac{P_0}{V_{s(rms)}I_{s(rms)}} = \frac{I^2_{0(rms)}R}{V_{s(rms)}I_{s(rms)}}\)

\(\rm \Rightarrow IPF = \frac{I^2_{0(rms)}R}{V_{s(rms)}I_{s(rms)}}\)      {∵ I_s(rms) = I_0(rms)}

\(\rm \Rightarrow IPF = \frac{I_{0(rms)}R}{V_{s(rms)}} = \frac{V_{0(rms)}}{V_{s(rms)}}\)

\(\Rightarrow \rm IPF = \frac{V_m/2}{V_m/\sqrt 2}= \frac{1}{\sqrt 2}\)

∴ IPF = 0.707

D_A = Q_C, D_B = Q_A, D_C = Q_B

\(D_A' = \bar Q_C',{\rm{;}}D_B' = \bar Q_A',{\rm{;}}D_C' = \bar Q_B'\)

\(Z = \left( {{Q_A} \oplus Q_A'} \right) + \left( {{Q_B} \oplus Q_B'} \right) + \left( {{Q_C} \oplus Q_C'} \right);\)

Z becomes zero when all the inputs of OR gates are zero

\(\Rightarrow {Q_A} \oplus Q_A' + {Q_B} \oplus Q_B' + {Q_C} \oplus Q_C' = 0\)

\(\Rightarrow {Q_A} = Q_A',{Q_B} = Q_B',{Q_C} = Q_C'\)

 

At 6^th clock pulse

\({Q_A}{Q_B}{Q_C} = Q_A'Q_B'Q_C' = 100\)

Concept:

Buck - Boost Converter: 

The buck–boost converter is a type of DC to DC converter that has an output voltage magnitude that is either greater than or less than the input voltage magnitude. It is equivalent to a Flyback converter using a single inductor instead of a Flyback transformer. The opposite voltage is the opposite polarity of that of the input.

The steady-state output voltage Vout of buck-boost converter is

\({V_{out}}; = -;{V_{in}}\left( {\frac{D}{{1; - ;D}}} \right)\)

Here D = duty cycle

• When Chopper is ON, Current flows from Sending Voltage, Vs to Chopper, CH, Inductor, L and back to Vs
• Energy is stored in the inductor during TON
• When the chopper is OFF, the inductor stored energy now discharges through the path, load, diode D, and L.

 

Explanation:

 

ripple current is given by,

\(\rm \Delta I_L = \frac{\Delta V_s}{fL}\)

\(\Rightarrow \rm \Delta I_L = \frac{50 \times 0.6}{10 \times 10^3 \times 0.6 \times 10^{-3}}= 5 A\)

using power balance equation,

V0I0 = VsIs

\(\rm \Rightarrow I_s = \frac{15 \times 75}{50} = 22.5 A\)

Then peak amplitude of source current is,

\(\rm I_{L(max)} = I_{s(max)} = I_L + \frac{\Delta I_L}{2}\)

\(\rm \Rightarrow I_{s(max)} = \frac{I_0}{1 - D}+ \frac{\Delta I_L}{2}\)

\(\rm \Rightarrow I_{s(max)} = \frac{15}{0.4}+ \frac{5}{2}\)

∴ Is(max) = 40 A

 

Additional Information

Identification of DC-DC Converter:

• Buck converter: Inductor is connected in series with the combination of the capacitor and resistor.
• Boost converter: Inductor is connected in series with the supply voltage source.
• Buck-Boost converter: Inductor is connected in parallel to the supply voltage source.

 

Under steady-state operating conditions:

• The average voltage across the inductor is zero. VL = 0
• The average current across the capacitor is zero. IC = 0

 

\(L = 2 \times {10^{ - 7}}In\left( {\frac{{GMD}}{{GMR}}} \right)H/m;\)

D_ab = D_bc = D_ca = 1m

GMD = 1 m

GMR = 0.01 m

frequency (f) = 50 Hz

\(L = 2 \times {10^{ - 7}}In\left( {\frac{1}{{0.01}}} \right)H/m\)

= 9.21 × 10^-7 H/m

X_L = 2πfL

= 2π × 50 × 9.21 × 10^-7

= 0.2893 × 10^-3 Ω/m

= 0.2893 Ω/km

Sending end voltage (V_s) = 17.87 kV

Receiving end voltage (V_r) = 17.32 kV

R = 0.25 Ω

X_L = 3.925 Ω

\(Z = \sqrt {{R^2} + X_L^2} = 3.933;{\rm{\Omega }}\)

\({P_r} = \frac{{\left| {{V_r}} \right|\left| {{V_s}} \right|}}{Z}\cos \left( {\theta - \delta } \right) - \frac{{V_r^2}}{Z}cos\theta \)

\(= \frac{{\left| {{V_r}} \right|\left| {{V_s}} \right|}}{Z}\cos \left( {\theta - \delta } \right) - \frac{{V_r^2}}{Z}\left( {\frac{R}{Z}} \right)\)

\(\Rightarrow 3 = \frac{{\left( {17.32} \right)\left( {17.87} \right)}}{{3.933}}\cos \left( {\theta - \delta } \right) - \frac{{{{\left( {17.32} \right)}^2} \times 0.25}}{{{{\left( {3.933} \right)}^2}}}\)

⇒ (θ - δ) = 87.276°

\({Q_r} = \frac{{\left| {{V_s}} \right|\left| {{V_r}} \right|}}{Z}\sin \left( {\theta - \delta } \right) - \frac{{V_r^2}}{Z}\sin \theta \)

\(= \frac{{\left| {{V_s}} \right|\left| {{V_r}} \right|}}{Z}\sin \left( {\theta - \delta } \right) - \frac{{V_r^2}}{Z}\left( {\frac{X}{Z}} \right)\)

= \(\frac{{\left( {17.87} \right)\left( {17.32} \right)}}{{3.933}}\sin \left( {84.276} \right) - \frac{{{{\left( {17.32} \right)}^2} \times 3.925}}{{{{\left( {3.933} \right)}^2}}} = 2.185;MVar\)

\({\rm{Power;factor}} = \cos \left( {{{\tan }^{ - 1}}\left( {\frac{{{Q_r}}}{{{P_r}}}} \right)} \right) = 0.8;lag\)

Generator 1: 300 MW, speed regulation of 6%

Generator 2: 400 MW, speed regulation of 4%

Let x is the load sharing by generator 1

Total load = 600 MW

load sharing by generator 2 = 600 – x

Form the property of triangles

\(\frac{{50 - {f_1}}}{3} = \frac{x}{{300}}\)

and \(\frac{{50 - {f_1}}}{2} = \frac{{600 - x}}{{100}}\)

⇒ x = 200 MW

Load shared by generator 2 = 600 – 200 = 400 MW

Rating of generator (S) = 250 MVA

kinetic Energy (K.E) = 1000 MJ

\(H = \frac{{KE}}{{Rating}} = \frac{{1000}}{{250}} = 4MJ/MVA\)

Power angle (δ) = 10°

P_m = P_s – P_e = 60 MW

\({\rm{\Delta }}\delta = \frac{{{{\left( {{\rm{\Delta }}t} \right)}^2}}}{2} - \frac{{{P_s} - {P_e}}}{M} = \frac{{{P_s} - {P_e}}}{{\frac{{SH}}{{180f}}}}.\frac{{{{\left( {{\rm{\Delta }}t} \right)}^2}}}{2} = \frac{{{{\left( {{\rm{\Delta }}t} \right)}^2}}}{2} \times \frac{{{P_m} \times 180f}}{{SH}}\)

t = 10 cycle \(= \frac{{10}}{{50}} = 0.2;sec\)

for t = 5 cycle = 0.1 sec

Δt = 0.1 sec

\({\rm{\Delta }}\delta = \frac{{{{\left( {0.1} \right)}^2}}}{2} \times \frac{{60 \times 180 \times 50}}{{4 \times 250}} = 2.7^\circ \)

⇒ δ = 10 + 2.7° = 12.7°

Volume of soap bubble = 4πR²a

Volume of soap drop \(= \frac{4}{3}\pi {r^3}\)

\(4\pi {R^2}a = \frac{4}{3}\pi {r^3}\)

⇒ Radius of soap drop \(= \sqrt[3]{{3{R^2}a}}\)

\(r = \sqrt[3]{{3{{\left( {1 \times {{10}^{ - 2}}} \right)}^2} \times 3.3 \times {{10}^{ - 6}}}}\)

= 0.0996 cm

Initial voltage (V_o) = 1 V

Initial charge, Q = (4πϵ_oR) × 1 = 4πϵ_oR 

After bursting, charge will remain same

⇒ Q = (4πϵ_or) (V₁)

⇒ 4 πϵ_or = (4 πϵ_or) V₁

\(\Rightarrow {V_1} = \frac{R}{r} = \frac{1}{{0.0996}} = 10.03;V\)