GATE EE 2020 Official Paper (Previous Year Paper)

The correct answer is option 3- is, are

The given question is on the subject-verb agreement rule in English grammar. 

In the first sentence, the subject is the singular noun 'book'. It will take a verb in the singular form. Thus, 'is' is the correct fit for the blank.

In the second sentence, the conjunction phrase 'as well as' has been used. It is followed by a verb agreeing with the first subject of the sentence- i.e. the subject present before the phrase.

Thus the correct fit for the blank is 'are', as it agrees with the plural subject 'students'. 

Note: There is no indication of the events to have occurred in the past. Thus we cannot consider any verb in the past tense for any of the blanks.

The correct answer is option 2- from parking

The verb prohibit means to forbid or prevent; it is followed by the preposition from.

The presence of the preposition from necessitates the use of the verb following it to be in its present participle form. 

Thus the correct sentence is: People were prohibited from parking their vehicles near the entrance of the main administrative building. 

Note: We cannot use 'to' after prohibit.

The correct answer is option 3- Distrust

In the given word problem, we have to find the link between the first and the second words in each set. 

When we look at 'Do' and 'Undo', it becomes easy for us to understand that we need to find the antonym of the word trust.

Trust: believe in the reliability, truth, or ability of

Distrust: doubt the honesty or reliability of; regard with suspicion

Thus we can see that the antonym of trust is distrust.

The meaning of the other words-

Entrust: assign the responsibility for doing something to (someone)

Intrust: it means the same as 'entrust' (archaic and obsolete usage)

Untrust: it means the same as 'distrust' (archaic and obsolete usage)

The correct answer is option 2- plunged

From the context, it is clear that we have to find out the reaction of the stock market to the news of the coup. A stock market can rise, fall or remain stable. 

Plunged: fell suddenly and uncontrollably; suffered a rapid decrease in value

Thus 'plunged' fits the blank both grammatically and contextually. 

The meaning of the other words-

Poised: been or caused to be balanced or suspended

Plugged: blocked or filled in (a hole or cavity)

Probed: physically explored or examined (something) with the hands or an instrument

Case 1: The number of teams with 2 members can be formed with Q as a member

Out of the 2 members, one member is Q and another member can be selected from the remaining three members.

The number of teams can be formed $= {3_{{C_1}}} = 3$

Case 2: The number of teams with 3 members can be formed with Q as a member

Out of the 3 members, one member is Q and another 2 members can be selected from the remaining three members.

The number of teams can be formed $= {3_{{C_2}}} = 3$

Case 3: The number of teams with 4 members can be formed with Q as a member

Out of the 4 members, one member is Q and another 3 members can be selected from the remaining three members.

The number of teams can be formed $= {3_{{C_3}}} = 1$

Now, the total number of teams can be formed = 3 + 3 + 1 = 7

Initial holding period = 360 days

Number of days for each quarter = 90 days

It is given that the holding period was reduced by one quarter each time i.e. reduced by 90 days each time and RBI has changed the definition of NPA thrice during 1993-2004.

After the first revision, the holding period of loans = 360 – 90 = 270

After the second revision, the holding period of loans = 270 – 90 = 180

After the third revision, the holding period of loans = 180 – 90 = 90

The given series: Z, WV, RQP

In the above series, each time the number of letters in the elements is increased by 1.

So, the next element will have four letters.

If we assign a number to each in alphabetical order i.e. A = 1, B = 2, C = 3, ….; the corresponding numbers for the above series will be

Z (26), WV (23, 22), RQP (18, 17, 16)

The difference between the first letters of the first two elements is 26 – 23 = 3

The difference between the first letters of the second and third elements is 23 – 18 = 5

So, the difference between the first letters of the third and fourth elements will be 7

So, the starting letter of the fourth element will be 11^th (18 – 7 = 11) letter in the alphabetical order i.e. the starting letter will be K.

The next element in the series will be KJIH

In four-digit integer, the digit group "37" (in the same sequence) appears as given below.

Case 1: _ _ 3 7

In this case, 37 appears once in each hundred i.e.

Once in 1101 to 1199 (1137) and

Once in 1201 to 1299 (1237) …

There are 90 such instances from 1001 to 9999.

Case 2: _ 3 7 _

In this case, 37 appears 10 times in each thousand i.e.

10 times in 1101 to 1999 (1370 to 1379) and

Once in 2101 to 2999 (2370 to 2379) …

There are 9 such instances from 1001 to 9999.

Total number of times that 37 have appeared = 10 × 9 = 90

Case 3:  3 7 _ _

In this case, 37 appears 100 times i.e. from 3700 to 3799

Therefore, total number of times the digit group "37" appears = 100 + 90 + 90 = 280

Common mistake:

The question is to find how many times the digit group "37" appears but not in how many numbers it appears.

Ex: 3737, in this number 37 appears two times. We need to count the 37 in this case twice but not once.

If the question is to find in how many numbers, the digit group "37" appears, then the answer will be 279.

Let the radius of the semicircle is r as shown in the diagram given below.

Now, OA = OC = OB = r

AB = OA + OB = 2r

$\overline {AC} = \sqrt {{{\left( {OA} \right)}^2} + {{\left( {OC} \right)}^2}} = \sqrt {{r^2} + {r^2}} = \sqrt 2 r$

$\overline {BC} = \sqrt {{{\left( {OB} \right)}^2} + {{\left( {OC} \right)}^2}} = \sqrt {{r^2} + {r^2}} = \sqrt 2 r$

The ratio $\frac{{\overline {AC} + \overline {CB} }}{{\overline {AB} }} = \frac{{\sqrt 2 r + \sqrt 2 r}}{{2r}} = \sqrt 2$

From the figure, the revenue of company Q in 2015 = 45 million rupees

Given that the revenue of company Q in 2015 was 20% more than that in 2014

The revenue of company Q in 2015 = 1.2 times the revenue of company Q in 2014

⇒ The revenue of company Q in 2014 = 45/1.2 = 37.5 million rupees

Given that the company Q had earned a profit of 10% on expenditure in 2014

% of profit = ((revenue – expenditure)/expenditure) × 100

Let the expenditure in 2014 is ‘x’ million rupees

$\Rightarrow 10 = \frac{{37.5 - x}}{x} \times 100$

⇒ 1.1 x = 37.5

⇒ x = 34.09 ≈ 34.1 million rupees

Common mistake:

It is mentioned that the company Q had earned a profit of 10% on expenditure in 2014.

If we calculate the profit on revenue, the answer will be 33.7 which is mentioned in one of the options given.

The given polynomial is: ax³ + bx² + cx + d = 0

Option 1:

At d = 0, the above equation becomes

ax³ + bx² + cx = 0

⇒ x (ax² + bx + c) = 0

Now, it is clear that x = 0 is a root for any given set a, b, c.

Therefore, the given statement is correct.

Option 2:

The given polynomial can be expressed as follows.

${x^3} + \frac{b}{a}{x^2} + \frac{c}{a}x + \frac{d}{a} = 0$

Let the above equation has three equal roots and x = r be a root.

Now, ${x^3} + \frac{b}{a}{x^2} + \frac{c}{a}x + \frac{d}{a} = {\left( {x - r} \right)^2} = 0$

$\Rightarrow {x^3} + \frac{b}{a}{x^2} + \frac{c}{a}x + \frac{d}{a} = {x^3} - 3r{x^2} + 3{r^2}x - {r^3} = 0$

By comparing on both sides,

$\Rightarrow \frac{b}{a} = - 3r,\frac{c}{a} = 3{r^2},\frac{d}{a} = - {r^3}$

By using the above relations, the conditions to get all the roots equal are

b² = 3ac and bc = 9ad

If we choose the values of a, b, c and d which satisfies the above relations, the roots will be equal, and the corresponding root will be $r = - \frac{b}{{3a}}$

Therefore, the given statement is incorrect.

Option 3:

The given polynomial is: ax³ + bx² + cx + d = 0

All the coefficients a, b, c, and d are real

As the coefficients are real, the complex roots must occur in conjugate.

The number of possible complex roots are either 0 or 2.

So, no choice of coefficients can make all roots complex.

Therefore, the given statement is incorrect.

Option 4:

c alone cannot ensure that all roots are real. It depends on all the coefficients.

Therefore, the given statement is incorrect.

Option 1:

\(\frac{1}{\pi }\mathop \smallint \limits_0^\pi \sin m\theta \sin n\theta d\theta \)

\( = \frac{1}{{2\pi }}\mathop \smallint \limits_0^\pi \cos \left( {m - n} \right)\theta - \cos \left( {m + n} \right)\theta d\theta \)

$= \frac{1}{{2\pi }}\left[ {\frac{{\sin \left( {m - n} \right)\theta }}{{m - n}} - \frac{{\sin \left( {m + n} \right)\theta }}{{m + n}}} \right]_0^\pi = 0$

Therefore, Option (1) is correct.

Option 2:

\(\frac{1}{{2\pi }}\mathop \smallint \limits_{ - \pi /2}^{\pi /2} \sin p\theta \sin q\theta d\theta \)

\( = \frac{1}{{4\pi }}\mathop \smallint \limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} \cos \left( {p - q} \right)\theta - \cos \left( {p + q} \right)\theta d\theta \)

$= \frac{1}{{4\pi }}\left[ {\frac{{\sin \left( {p - q} \right)\theta }}{{p - q}} - \frac{{\sin \left( {p + q} \right)\theta }}{{p + q}}} \right]_{ - \frac{\pi }{2}}^{\frac{\pi }{2}}$

$= \frac{1}{{2\pi }}\left[ {\frac{{\sin \left( {p - q} \right)\frac{\pi }{2};}}{{p - q}} - \frac{{\sin \left( {p + q} \right)\frac{\pi }{2}}}{{p + q}}} \right]$

The above expression not necessarily is zero. Therefore, Option (2) is incorrect.

Option 3:

\(\frac{1}{{2\pi }}\mathop \smallint \limits_{ - \pi }^\pi \sin p\theta \cos q\theta d\theta \)

\( = \frac{1}{{4\pi }}\mathop \smallint \limits_{ - \pi }^\pi \sin \left( {p + q} \right)\theta + \sin \left( {p - q} \right)\theta d\theta \)

$= \frac{1}{{4\pi }}\left[ { - \frac{{\cos \left( {p + q} \right)\theta }}{{p + q}} + \frac{{\cos \left( {p - q} \right)\theta }}{{p - q}}} \right]_{ - \pi }^\pi$

$= \frac{1}{{4\pi }}\left[ { - \frac{{\cos \left( {p + q} \right)\pi ;}}{{p + q}} + \frac{{\cos \left( {p + q} \right)\pi }}{{p + q}} + \frac{{\cos \left( {p + q} \right)\pi ;}}{{p + q}} - \frac{{\cos \left( {p + q} \right)\pi }}{{p + q}}} \right] = 0$

Therefore, Option (3) is correct.

Option 4:

\(\frac{1}{{2\alpha }}\mathop \smallint \limits_{ - \alpha }^\alpha \sin p\theta \sin q\theta d\theta \)

\( = \frac{1}{{4\alpha }}\mathop \smallint \limits_{ - \alpha }^\alpha \cos \left( {p - q} \right)\theta - \cos \left( {p + q} \right)\theta d\theta \)

$= \frac{1}{{4\alpha }}\left[ {\frac{{\sin \left( {p - q} \right)\theta }}{{p - q}} - \frac{{\sin \left( {p + q} \right)\theta }}{{p + q}}} \right]_{ - \alpha }^\alpha$

$= \frac{1}{{2\alpha }}\left[ {\frac{{\sin \left( {p - q} \right)\alpha }}{{p - q}} - \frac{{\sin \left( {p + q} \right)\alpha }}{{p + q}}} \right]$

\(\mathop {\lim }\limits_{\alpha \to \infty } \frac{1}{{2\alpha }}\mathop \smallint \limits_{ - \alpha }^\alpha \sin p\theta \sin q\theta d\theta \)

$= \mathop {\lim }\limits_{\alpha \to \infty } \frac{1}{{2\alpha }}\left[ {\frac{{\sin \left( {p - q} \right)\alpha }}{{p - q}} - \frac{{\sin \left( {p + q} \right)\alpha }}{{p + q}}} \right]$

$= \mathop {\lim }\limits_{\alpha \to \infty } \frac{1}{2}\left[ {\frac{{\sin \left( {p - q} \right)\alpha }}{{\left( {p - q} \right)\alpha }} - \frac{{\sin \left( {p + q} \right)\alpha }}{{\left( {p + q} \right)\alpha }}} \right] = 0$

Therefore, Option (4) is correct.

The Fourier transform of x(t) is, denoted by X(jω), is defined as:

\(X\left( {j\omega } \right) = \mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){e^{ - j\omega t}}dt\)

The Laplace transform of x(t), denoted by X(s), is defined as:

\(X\left( s \right) = \mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){e^{ - st}}dt\)

Where s is a continuous complex variable.

We can also express s as: s = σ + jω

Where σ and ω are the real and imaginary parts of s, respectively

The Laplace transform can be written as:

\(X\left( {\sigma + j\omega } \right) = \mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){e^{ - \left( {\sigma + j\omega } \right)t}}dt = \mathop \smallint \limits_{ - \infty }^\infty \left( {x\left( t \right){e^{ - \sigma t}}} \right){e^{ - j\omega t}}dt\)

By comparing the above Laplace and Fourier transform equations, it is clear that Laplace transform of x(t) is equal to the Fourier transform of $x\left( t \right){e^{ - \sigma t}}$.

When σ = 0 or s = jω, both are identical.

\({\left. {X\left( s \right)} \right|{s = j\omega }} = X\left( {j\omega } \right) = \mathop \smallint \limits{ - \infty }^\infty x\left( t \right){e^{ - j\omega t}}dt\)

That is, Laplace transform generalizes Fourier transform.

Option 1:

Region of Convergence (ROC):

ROC indicates when the Laplace transform of x(t) converges.

That is, if \(\left| {X\left( s \right)} \right| = \left| {\mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){e^{ - st}}dt} \right| \to \infty\) then the Laplace transform does not converge at point s.

Employing s = σ + jω and |e^jωt| = 1, Laplace transform exists if

\(\left| {X\left( {\sigma + j\omega } \right)} \right| \le \mathop \smallint \limits_{ - \infty }^\infty \left| {x\left( t \right){e^{ - \left( {\sigma + j\omega } \right)t}}} \right|dt = \mathop \smallint \limits_{ - \infty }^\infty \left| {x\left( t \right){e^{ - \sigma t}}} \right|dt < \infty\)

The set of values of σ which satisfies the above equation is called the ROC.

If \(\left| {X\left( {j\omega } \right)} \right| = \left| {\mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){e^{ - j\omega t}}dt} \right| \to \infty\) then the Fourier transform does not exist. While it exists if \(\left| {X\left( {j\omega } \right)} \right| \le \mathop \smallint \limits_{ - \infty }^\infty \left| {x\left( t \right){e^{ - j\omega t}}} \right|dt = \mathop \smallint \limits_{ - \infty }^\infty \left| {x\left( t \right)} \right|dt < \infty\)

Hence it is possible that the Fourier transform of x(t) does not exist. Also, the Laplace transform does not exist if there is no value of σ satisfies the required ROC equation as explained above.

Therefore, the given statement is incorrect.

Option 2:

Values of s for which X(s) = 0 are the Zeros of X(s)

Values of s for which X(s) = ∞ are the Poles of X(s)

For bounded inputs i.e. finite-amplitude finite width signals, the ROC is entire s plane and ROC never includes any pole. It implies for such signals there are no poles.

Therefore, the two-sided Laplace transform of any bounded input, will not have any poles

Therefore, the given statement is correct.

Option 3:

The number of finite pole and finite zero need not be equal.

Ex: ${e^{ - t}}u\left( t \right)\mathop \leftrightarrow \limits^{LT} \frac{1}{{s + 1}};ROC:;s > - 1$ and pole at s = -1

This has 1 finite pole and no zeroes.

Therefore, the given statement is incorrect.

Option 4:

If a signal can be expressed as a weighted sum of shifted one-sided exponential, then its Laplace Transform will have poles.

Ex: Let $x\left( t \right) = {e^{ - t}} + 2{e^{ - \left( {t - 1} \right)}} + 3{e^{ - \left( {t - 2} \right)}} + \cdots$

$x\left( s \right) = \frac{1}{{s + 1}} + \frac{{2{e^{ - s}}}}{{s + 1}} + \frac{{3{e^{ - 2s}}}}{{s + 1}} + \cdots$

$x\left( s \right) = \frac{{1 + 2{e^{ - s}} + 3{e^{ - 2s}}}}{{s + 1}}$

So, x(s) has one pole.

Therefore, the given statement is incorrect.

Concept:

Z transform:

The Z transform of x(t) is, denoted by X(z), is defined as:

\(X\left( z \right) = \mathop \sum \limits_{n = - \infty }^\infty x\left( n \right){z^{ - n}}\)

${a^n}u\left( n \right)\mathop \to \limits^{ZT} \frac{1}{{1 - a{z^{ - 1}}}}$; ROC:|z|>|a|

Shifting property:

If $x\left( n \right)\mathop \to \limits^{ZT} x\left( z \right)$ ; ROC: R

Then $x\left( {n - {n_0}} \right)\mathop \to \limits^{ZT} {z^{ - {n_0}}}x\left( z \right)$; ROC: R

The time-shifting will not affect ROC.

Calculation:

Given that, $x\left( n \right) = {\left( {\frac{1}{2}} \right)^n}1\left[ n \right]$

Z Transform (ZT) of x(n) will be,

$x\left( z \right) = \frac{z}{{z - \frac{1}{2}}} = \frac{1}{{1 - \frac{1}{2}{z^{ - 1}}}}$ ; ROC: $\left| z \right| > \frac{1}{2}$

And, $z\left( {x\left( {n - k} \right)} \right) = ;\frac{{{z^{ - k}}}}{{1 - \frac{1}{2}{z^{ - 1}}}}$ ; ROC: $\left| z \right| > \frac{1}{2}$

Concept:

Cauchy’s Theorem:

If f(z) is an analytic function and f’(z) is continuous at each point within and on a closed curve C, then

$\mathop \oint \limits_C f\left( z \right)dz = 0$

Cauchy’s Integral Formula:

If f(z) is an analytic function within a closed curve and if a is any point within C, then

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

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

Residue Theorem:

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

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

Formula to find residue:

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

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

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

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

Application:

\(\mathop \smallint \nolimits_C \frac{{{z^2} + 1}}{{{z^2} - 2z}}dz\)

\( = \mathop \smallint \nolimits_C \frac{{{z^2} + 1}}{{z\left( {z - 2} \right)}}dz\)

The simple poles are: z = 0, 2

The given region is a unit circle.

The residue at z = 2 is zero as it lies outside the given region.

The reside at z = 0, is given by

$= \mathop {\lim }\limits_{z \to 0} z\frac{{{z^2} + 1}}{{z\left( {z - 2} \right)}}dz = - \frac{1}{2}$

The value of the given integral $= 2\pi i \times \left( { - \frac{1}{2}} \right) = - \pi i$

Concept:

A function f(t) is said to be a periodic function, if f(t ± T) = f(t)

Where T is a time period

The average value of f(t) is given by,

\({f_{avg}} = \frac{1}{T}\mathop \smallint \nolimits_0^T f\left( t \right)dt\)

The RMS value of f(t) is given by,

\({f_{rms}} = \sqrt {\frac{1}{T}\mathop \smallint \nolimits_0^T {{\left( {f\left( t \right)} \right)}^2}} dt\)

Application:

Given that,

x_R and x_A are, respectively, the RMS and average values of x(t) = x(t – T)

y_R and y_A are, respectively, the RMS and average values of y(t) = kx(t)

k, T are independent of t.

The average value of x(t) is,

\({x_A} = \frac{1}{T}\mathop \smallint \nolimits_0^T x\left( t \right)dt\)

The average value of y(t) is,

\({y_A} = {\left( {kx} \right)_A} = \frac{1}{T}\mathop \smallint \nolimits_0^T kx\left( t \right)dt = k{x_A}\)

The RMS value of x(t) is,

\({x_R} = \sqrt {\frac{1}{T}\mathop \smallint \nolimits_0^T {x^2}\left( t \right)dt}\)

The RMS value of y(t) is,

\({y_R} = \sqrt {\frac{1}{T}\mathop \smallint \nolimits_0^T {{\left( {kx\left( t \right)} \right)}^2}dt}\)

\(= k\sqrt {\frac{1}{T}\mathop \smallint \nolimits_0^T {{\left( {x\left( t \right)} \right)}^2}dt} = k{x_R}\)

Given that the cylindrical synchronous generator with

Synchronous reactance = X_S and

Armature resistance = R_a = 0

Now, the simplified circuit of the synchronous generator looks like as shown below

Case 1: The load is lagging in nature.

The corresponding phasor diagram is

⇒ |V| < |E|

Case 2: The load is leading in nature.

The corresponding phasor diagram is shown below.

⇒ |V| > |E|

• A cylindrical rotor synchronous generator (with R_a = 0) has always positive voltage regulation for lagging p.f. loads i.e. E_A > V_A
• Whereas it has positive, zero, and negative regulation for leading loads i.e. all cases E_A > V_A, E_A = V_A, and E_A < V_A are possible.

 

Therefore, statement P is False, and Q is True.

Line reactance = 0.2 pu

Given that only 80% of the line is protected by distance relay.

So, the reactance of the line as seen by the relay will be

X = 0.8 × 0.2 = 0.16 pu

Terminal voltage, V_t = 1 Pu

The impedance is seen by impedance relay, $X = \frac{{{V_t}}}{{{I_f}}}$

The threshold pu current for operation of the distance relay ${I_f} = \frac{1}{{0.16}} = 6.25;pu$

• The most relevant information needed to specify P at PV buses is the solution of economic load dispatch.
• Economic load dispatch is a precursor to the load flow study (LFS), i.e. to perform the LFS the first step is to perform the economic load dispatch.
• The economic load dispatch means the real and reactive power of the generator varies within certain limits and fulfills the load demand with less fuel cost.
• The economic scheduling of the generators aims to guarantee at all times the optimum combination of the generator connected to the system to supply the load demand.
• The economic load dispatch problem involves two separate steps. These are the online load dispatch and unit commitment.
• The unit commitment selects that unit which will anticipate the load of the system over the required period at minimum cost.
• The online load dispatch distributes the load among the generating unit which is parallel to the system in such a manner as to reduce the total cost of supplying. It also fulfills the minute to the minute requirement of the system.

Note: Generator not always operates at rated power P because as per the demand by the load, P value changes. Base value can be different for different loads.

Concept:

A transfer function is defined as the ratio of Laplace transform of the output to the Laplace transform of the input by assuming initial conditions are zero.

TF = L[output]/L[input]

$TF = \frac{{C\left( s \right)}}{{R\left( s \right)}}$

For unit impulse input i.e. r(t) = δ(t)

⇒ R(s) = δ(s) = 1

Now transfer function = C(s)

Therefore, the transfer function is also known as the impulse response of the system.

Transfer function = L[IR]

IR = L-1 [TF]

Calculation:

Given the differential equation is,

$\frac{{{d^2}y\left( t \right)}}{{d{t^2}}} + 4y\left( t \right) = 6r\left( t \right)$

By applying the Laplace transform,

s² Y(s) + 4 Y(s) = 6 R(s)

$\Rightarrow \frac{{Y\left( s \right)}}{{R\left( s \right)}} = \frac{6}{{{s^2} + 4}}$

Poles are the roots of the denominator in the transfer function.

⇒ s² + 4 = 0

⇒ s = ±2j

Concept:

Fourier series of source current in a full bridge rectifier is given by

\({I_s} = \mathop \sum \limits_{n = 1,3,5 \ldots .}^\infty \frac{{4{I_0}}}{{n\pi }}\sin n\omega t\)

For n = 1, 3, 5, ......

The most dominant frequency components in the above expression are f, 3f

Application:

A single-phase, full-bridge diode rectifier fed from a 230 V, 50 Hz sinusoidal source

Dominant frequencies = f, 3f

= 50 and 150 Hz

Concept:

In a single-phase AC voltage controller, the range of variation in α is

Resistive load: 0 < α < π

Inductive load: ϕ < α < π

Where ϕ is the load angle and α is the firing angle

Calculation:

The given circuit is a single-phase full-wave AC voltage controller.

Load = 10∠60°

Load angle (ϕ) = 60°

The load is inductive and hence current is lagging the voltage by 60°.

It is possible to have control for a range of 60° to 180°.

Concept:

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

\(TF = \frac{{\mathop \sum \nolimits_{k = 1}^n {M_k}{{\rm{\Delta }}_k}}}{{\rm{\Delta }}}\)

Where, n = no of forward paths

Mk = kth forward path gain

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

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

Application:

In the given signal flow graph,

Forward paths: P₁ = ad

Loops: L₁ = cd, L₂ = ade

Δ = 1 – (cd + ade)

Δ₁ = 1

Transfer function $= \frac{{ad}}{{1 - \left( {cd + ade} \right)}}$

Now, let us check the options.

Option 1:

Forward paths: P₁ = a(d + c)

Loops: L₁ = ae(d + c)

Δ = 1 – ae(d + c)

Δ₁ = 1

Transfer function $= \frac{{a\left( {d + c} \right)}}{{1 - ae\left( {d + c} \right)}}$

Option 2:

Forward paths: P₁ = d(a + c)

Loops: L₁ = de(a + c)

Δ = 1 – de(a + c)

Δ₁ = 1

Transfer function $= \frac{{d\left( {a + c} \right)}}{{1 - de\left( {a + c} \right)}}$

Option 3:

Forward paths: ${P_1} = a\left( {\frac{d}{{1 - cd}}} \right)$

Loops: ${L_1} = ae\left( {\frac{d}{{1 - cd}}} \right)$

${\rm{\Delta }} = 1 - ae\left( {\frac{d}{{1 - cd}}} \right)$

Δ₁ = 1

Transfer function $= \frac{{a\left( {\frac{d}{{1 - cd}}} \right)}}{{1 - ae\left( {\frac{d}{{1 - cd}}} \right)}} = \frac{{ad}}{{1 - \left( {cd + ade} \right)}}$

Option 4:

Forward paths: ${P_1} = a\left( {\frac{c}{{1 - cd}}} \right)$

Loops: ${L_1} = ae\left( {\frac{c}{{1 - cd}}} \right)$

${\rm{\Delta }} = 1 - ae\left( {\frac{c}{{1 - cd}}} \right)$

Δ₁ = 1

Transfer function $= \frac{{a\left( {\frac{c}{{1 - cd}}} \right)}}{{1 - ae\left( {\frac{c}{{1 - cd}}} \right)}} = \frac{{ad}}{{1 - \left( {cd + ace} \right)}}$

Hence the signal graph in option (3) is the equivalent representation of the signal flow graph given in the question.

Concept:

From the AC model of a common source amplifier, the voltage gain is given by

$\frac{{{V_0}}}{{{V_i}}} = {A_v} = - {g_m}{R_D}$

g_m­ is the transconductance

R_D is the drain resistance

Calculation:

Given that, drain resistance, R_D = 4.7 kΩ

Transconductance, g_m = 520 × 10^-6 A/V

Power supply = 10 V

The voltage gain is given by

$\frac{{{V_0}}}{{{V_i}}} = - {g_m}{R_D} = - 0.52 \times 4.7 = - 2.44$

A sequence detector is a sequential circuit that outputs 1 when a particular pattern of bits sequentially arrives at its data input.

Given input data = 1,1,0,1,0,0,1,1,0,1,0,1,1,0

Overlapping sequences detectable.

The below table shows the output for each sequence.

 

The output = 0,1,0,0,0,0,0,1,0,1,0,0

Concept:

A point c in the domain of a function f at which either f'(c) = 0 or if f is nondifferentiable is called the critical point.

The slope of tangents at critical points are zero

Let f be a real valued function and c be an interior point in the domain of f such that f'(c)=0.

• If f'(x) > 0 at every point close to left of c and f'(x) < 0 at every point right of c, then is a local maximum.
• If f'(x) > 0 at every point close to right of c and f'(x) < 0 at every point left of c, then is a local minimum.
• If f'(x) does not change sign as x increases or decreases, then such a point is called point of inflection.

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Calculation:

y = 3x² + 3x + 1

f'(x) = 6x + 3 = 0

$\Rightarrow x = - \frac{1}{2}$

f'’(x) = 6 > 0

Therefore, the point $x = - \frac{1}{2}$ is the point of minima

For global maximum and minimum values we have to consider the extreme points also along with critical points.

Means we have to calculate function 'y' value at -2 and 0 also.

f(x = -2) = 3 (-2)² – 6 + 1 = 7   ------------------------(1)

f(x = 0) = 0 + 0 + 1 = 1       ______(2)

$f\left( {x = - \frac{1}{2}} \right) = 3{\left( { - \frac{1}{2}} \right)^2} + 3\left( { - \frac{1}{2}} \right) + 1 = \frac{1}{4}$  --------------(3)

From (1), (2) and (3) we can observe that,

The maximum value of y = 7 and the minimum value at y = 1/4.

Hence, the maximum, and minimum values of y are 7 and 1/4.

Concept:

For a vector F = F₁i + F₂j + F₃k

$Div = \nabla .F = \frac{{\partial {F_1}}}{{\partial x}} + \frac{{\partial {F_2}}}{{\partial y}} + \frac{{\partial {F_3}}}{{\partial z}}$

\(Curl = \nabla \times F = \left| {\begin{array}{*{20}{c}} i&j&k\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\ {{F_1}}&{{F_2}}&{{F_3}} \end{array}} \right|\)

For irrotational (or) conservative field $\nabla \times \vec F = 0$ (or) Null Vector.

Calculation:

Given that,

$\vec F = {\hat a_x};\left( {5y - {k_1}z} \right) + {\hat a_y}\left( {3z + {k_2}x} \right) + {\hat a_z}\left( {{k_3}y - 4x} \right)$ is a conservative field.

\(\left| {\begin{array}{*{20}{c}} {{{\hat a}_x}}&{{{\hat a}_y}}&{{{\hat a}_z}}\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\ {\left( {5y - {k_1}z} \right)}&{\left( {3z + {k_2}x} \right)}&{\left( {{k_3}y - 4x} \right){\rm{;}}} \end{array}} \right| = 0\)

$\Rightarrow {\hat a_x};\left( {{{\rm{k}}_3} - 3} \right) - {\hat a_y}\left( {; - 4 + {{\rm{k}}_1}} \right) + {\hat a_z}\left( {{{\rm{k}}_2} - 5} \right)$

k₃ – 3 = 0 ⇒ k₃ = 3

-4 + k₁ = 0 ⇒ k₁ = 4

k₂ – 5 = 0 ⇒ k₂ = 5

The required values are: k₁ = 4, k₂ = 5, k₃ = 3

Concept:

In a DC shunt motor, the back emf is given by

${E_b} = \frac{{NP\phi Z}}{{60A}}$

Where N is the speed

ϕ is the flux per pole

P is the number of poles

Z is the number of conductors

${E_b} \propto \frac{N}{\phi }$

$\Rightarrow \frac{{{N_2}}}{{{N_1}}} = \frac{{{E_{b2}}}}{{{E_{b1}}}} \times \frac{{{\phi _1}}}{{{\phi _2}}}$

Calculation:

At no-load condition:

Armature current, I_a1 = 5A

No-load speed, N₁ = 1200 rpm,

Brush drop = 1 V per brush

Total brush drop = 2V

E_b1 = V – I_a1 R_a – Brush drop

= 250 – 5 × 0.2 – 2 = 247 V

At load condition:

I_a2 = I_L – I_f = 47.5 A

ϕ₂ = 0.95 ϕ₁

E_b2 = V – I_a2 R_a – Brush drop

= 250 – 47.5 × 0.2 – 2 = 238.5 V

Now, $\frac{{{N_2}}}{{{N_1}}} = \frac{{{E_{b2}}}}{{{E_{b1}}}} \times \frac{{{\phi _1}}}{{{\phi _2}}}$

$\Rightarrow \frac{{{N_2}}}{{1200}} = \frac{{238.5}}{{247}} \times \frac{{{\phi _1}}}{{0.95{\phi _1}}} = 1.0164$

N₂ = 1200 × 1.0164 = 1219.688

Bus admittance matrix:

\(\left[ {\begin{array}{{20}{c}} {{{\rm{I}}{\rm{i}}}}\ {{{\rm{I}}{\rm{j}}}} \end{array}} \right] = \left[ {\begin{array}{{20}{c}} {{{\rm{Y}}{{\rm{ii}}}}}&{{{\rm{Y}}{{\rm{ij}}}}}\ {{{\rm{Y}}{{\rm{ji}}}}}&{{{\rm{Y}}{{\rm{jj}}}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{{\rm{V}}{\rm{i}}}}\ {{{\rm{V}}{\rm{j}}}} \end{array}} \right]\)

Two buses, i and j, are connected with a transmission line of admittance Y, at the two ends of which there are ideal transformers with turns ratios \(1:{{\rm{t}}{\rm{i}}}{\rm{;}}& {\rm{;}}1:{{\rm{t}}{\rm{j}}}\).

From the circuit shown above,

\(\frac{{{{\rm{I}}{\rm{i}}}}}{{{{\rm{t}}{\rm{i}}}}} = \left( {{{\rm{V}}{\rm{i}}}{{\rm{t}}{\rm{i}}} - {{\rm{V}}{\rm{j}}}{{\rm{t}}{\rm{j}}}} \right){\rm{Y}}\)

\({{\rm{I}}{\rm{i}}} = \left( {{{\rm{t}}{\rm{i}}}^2.{\rm{Y}}} \right){{\rm{V}}{\rm{i}}} - \left( { - {{\rm{t}}{\rm{i}}}{{\rm{t}}{{\rm{j}}.}}{\rm{Y}}} \right){{\rm{V}}{\rm{j}}}\)      ---(1)

Also, \(\frac{{{{\rm{I}}{\rm{j}}}}}{{{{\rm{t}}{\rm{j}}}}} = \left( {{{\rm{V}}{\rm{j}}}{{\rm{t}}{\rm{j}}} - {{\rm{V}}{\rm{i}}}{{\rm{t}}{\rm{i}}}} \right){\rm{Y}}\)

\({{\rm{I}}{\rm{j}}} = \left( { - {{\rm{t}}{\rm{i}}}{{\rm{t}}{{\rm{j}}.}}{\rm{Y}}} \right){{\rm{V}}{\rm{i}}} - {{\rm{V}}{\rm{j}}}({{\rm{t}}{\rm{j}}}^2.{\rm{Y}})\)      ---(2)

From equation (1) and (2), we get,

\(\left[ {{{\rm{Y}}{{\rm{Bus}}}}} \right] = \left[ {\begin{array}{*{20}{c}} {{{\rm{t}}{\rm{i}}}^2.{\rm{Y}}}&{ - {{\rm{t}}{\rm{i}}}{{\rm{t}}{{\rm{j}}.}}{\rm{Y}}}\ { - {{\rm{t}}{\rm{i}}}{{\rm{t}}{{\rm{j}}.}}{\rm{Y}}}&{{{\rm{t}}_{\rm{j}}}^2.{\rm{Y}}} \end{array}} \right]\)

Concept:

By using the voltage division rule,

The voltage applied across the resistor R1 is ${V_{R1}} = \frac{{V\left( {{R_1}} \right)}}{{{R_1} + {R_2}}}$

The voltage applied across the resistor R2 is ${V_{R2}} = \frac{{V\left( {{R_2}} \right)}}{{{R_1} + {R_2}}}$

Explanation:

V_D is the voltage across the diode, then by using voltage division

${V_D} = V \times \frac{{{R_2}}}{{{R_1} + {R_2}}}$

If R₂ is large, V_D is high

If R₁ is small, then V_D = V.

Therefore, to transfer maximum power, R₂ should be very large and R₁ should be very small.

Concept:

The current across a diode (I) is given by:

\(I={{I}{0}}\left( {{e}^{\frac{{{V}{D}}}{\eta {{V}_{T~}}}}}-1 \right)\)

VD = Applied Voltage.

VT = Thermal Voltage

η = Ideality factor.

Calculation:

\({{\rm{I}}{\rm{D}}} = {{\rm{I}}0}\left[ {{{\rm{e}}^{\frac{{{{\rm{V}}{\rm{D}}}}}{{{\rm{\eta }}{{\rm{V}}{\rm{T}}}}}}} - 1} \right]\)

I_D1 = I₁, I_D2 = 1.5I₁

\({{\rm{V}}{\rm{D}}} = - 0.03,;{\rm{\eta }} = \frac{{15}}{{13}},;{{\rm{V}}{\rm{T}}} = 26{\rm{mV}}\)

${\rm{\eta }} \times {{\rm{V}}_{\rm{T}}} = \frac{{15}}{{13}} \times 26 \times {10^{ - 3}} = 0.03$

\( \Rightarrow \frac{{{{\rm{I}}{{\rm{D}}2}}}}{{{{\rm{I}}{{\rm{D}}1}}}} = \frac{{{{\rm{e}}^{\frac{{{{\rm{V}}{{\rm{D}}2}}}}{{{\rm{\eta }}{{\rm{V}}{\rm{T}}}}}}} - 1}}{{{{\rm{e}}^{\frac{{{{\rm{V}}{{\rm{D}}1}}}}{{{\rm{\eta }}{{\rm{V}}{\rm{T}}}}}}} - 1}}\)

\( \Rightarrow \frac{{1.5 \times {{\rm{I}}_1}}}{{{{\rm{I}}1}}} = \frac{{{{\rm{e}}^{\frac{{{{\rm{V}}{{\rm{D}}2}}}}{{0.03}}}} - 1}}{{{{\rm{e}}^{\frac{{ - 0.03}}{{0.03}}}} - 1}}\)

$\Rightarrow 1.5 \times - 0.6321 = {{\rm{e}}^{\frac{{{{\rm{V}}_{{\rm{D}}2}}}}{{0.03}}}} - 1$

$\Rightarrow - 0.9481 + 1 = {{\rm{e}}^{\frac{{{{\rm{V}}_{{\rm{D}}2}}}}{{0.03}}}}$

$\Rightarrow 0.0518 = {{\rm{e}}^{\frac{{{{\rm{V}}_{{\rm{D}}2}}}}{{0.03}}}}$

$\Rightarrow \ln 0.0518 = \frac{{{{\rm{V}}_{{\rm{D}}2}}}}{{0.03}}$

$\Rightarrow {{\rm{V}}_{{\rm{D}}2}} = - 2.9599 \times 0.03 = - 0.09$

Dc power supply acts as an ideal 4 A current source as long as its terminal voltage is below 10 V.

Beyond the above point, it begins to behave as an ideal 10 V voltage source for all load currents going down to 0 A.

Now, the characteristics are as shown below.

So, 4² × R_L = 40 W

⇒ R_L = 2.5 Ω

So, the answer will be 2.5 Ω, 4 A, 10 V.

Alternate method:

If supply acts as a 4 A current source,

V = 4 × R, P = 4² × R

V_max = 10 V, R_min = 10/4 = 2.5 Ω

P_max = 16 × 2.5 = 40 W.

Now, if supply acts as 10 V source and R > $2.5;{\rm{\Omega }}$

${\rm{I}} = \frac{{10}}{{\rm{R}}},{\rm{;P}} = \frac{{{{10}^2}}}{{\rm{R}}}$

${{\rm{P}}_{{\rm{max}}}} = \frac{{{{10}^2}}}{{2.5}} = 40{\rm{;W}}$

R = 2.5 Ω, I = 4 A, V = 10 V.

Concept:

From Gauss’ Law, the volume charge density is given by

${\rho _\nu } = \nabla .\vec D = \nabla .{{\rm{\varepsilon }}_0}{{\rm{\varepsilon }}_r}\vec E$

In Cylindrical coordinate system, \(\nabla .\vec E = \left[ {\frac{\partial }{{\partial r}}\left( {{\rm{r}}{{\rm{E}}{\rm{r}}}} \right) + \frac{\partial }{{\partial \phi }}{{\rm{E}}\phi } + \frac{\partial }{{\partial \phi }}\left( {{\rm{r}}{{\rm{E}}_{\rm{z}}}} \right)} \right]\)

Calculation:

Given that,

Relative permittivity, ε_r = 2.25

$\vec E = ;2r\hat r + \left( {\frac{3}{r}} \right)\hat \phi + ;6\hat z$

${\rho _\nu } = {{\rm{\varepsilon }}_0}{{\rm{\varepsilon }}_r}\nabla .\vec E$

$= {{\rm{\varepsilon }}_0}{{\rm{\varepsilon }}_r}\frac{1}{r}\left[ {\frac{\partial }{{\partial r}}\left( {{\rm{r}}.2{\rm{r}}} \right) + \frac{\partial }{{\partial \emptyset }}\left( {\frac{3}{r}} \right) + \frac{\partial }{{\partial \emptyset }}\left( {{\rm{r}}.6} \right)} \right]$

$= {{\rm{\varepsilon }}_0} \times 2.25 \times \frac{1}{r}\left[ {4r + 0 + 0} \right]$

${\rho _\nu } = {{\rm{\varepsilon }}_0} \times 2.25 \times 4 = 9{\varepsilon _0}$

The standard transfer function is,

$\frac{{C\left( s \right)}}{{R\left( s \right)}} = T.F = \frac{K}{{1 + Ts}}$

Given that, DC input = 5 V

$\Rightarrow R\left( s \right) = \frac{5}{s}$

T = 0.5 s

Steady state speed = 10 rad/sec.

Now we have,

$C\left( s \right) = \frac{5}{s} \times \frac{K}{{1 + 0.5s}}$

Now according to final value theorem,

$\mathop {\lim }\limits_{s \to 0} \left[ {sC\left( s \right)} \right] = 10$

$\mathop {\lim }\limits_{s \to 0} \left[ {sC\left( s \right)} \right] = 5 \times K = 10 \Rightarrow K = 2$

Now, the required transfer function is,

$T.F = \frac{K}{{1 + Ts}} = \frac{2}{{1 + 0.5s}}$

From the block diagram,

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

$H\left( s \right) = \frac{{20}}{{\left( {s + 20} \right)}}$

As the given feedback is negative, the transfer function of the closed loop system is

$\frac{{Y\left( s \right)}}{{R\left( s \right)}} = \frac{{G\left( s \right)}}{{1 + G\left( s \right)H\left( s \right)}}$

$= \frac{{\frac{1}{{{s^2}\left( {s + 1} \right)}}}}{{1 + \frac{1}{{{s^2}\left( {s + 1} \right)}}\frac{{20}}{{\left( {s + 20} \right)}}}}$

$= \frac{{s + 20}}{{{s^2}\left( {s + 1} \right)\left( {s + 20} \right) + 20}}$

$= \frac{{s + 20}}{{{s^4} + 21{s^3} + 20{s^2} + 20}}$

The denominator of the above transfer function has the highest degree of 4. Therefore, the order of the system is 4.

The coefficient of ‘s’ term is zero in the characteristic equation (denominator of above transfer function). Therefore, the system is unstable.

Concept:

The characteristic equation for a given open-loop transfer function G(s) is

1 + G(s) H(s) = 0

According to the Routh tabulation method,

• The system is said to be stable if there are no sign changes in the first column of Routh array
• The number of poles lies on the right half of s plane = number of sign changes
• The system has the poles on the imaginary axis when the s¹ row becomes zero in the Routh array.

 

Calculation:

Give forward path transfer function is

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

As the feedback is unity, H(s) = 1

The characteristic equation: 1+ G(s) H(s) = 0

$1 + \frac{{{s^2} + s + 1}}{{{s^3} + 2{s^2} + 2s + K}} = 0$

⇒ s³ + 3 s² + 3s + (K + 1) = 0

By applying Routh tabulation method,

\(\begin{array}{{20}{c}} {{s^3}}\ {{s^2}}\ {{s^1}}\ {{s^0}} \end{array}\left| {\begin{array}{{20}{c}} 1&3\ 3&{1 + K}\ {9 - \left( {1 + K} \right)}&0\ {1 + K}&{} \end{array}} \right.\)

For the system to have poles on the imaginary axis,

9 – (1 + K) = 0

⇒ K = 8

Concept:

Let the input is x(t), the output is y(t) and the impulse response is h(t).

The convolution is given by

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

\(y\left( t \right) = \mathop \smallint \limits_{ - \infty }^\infty x\left( \tau \right)h\left( {t - \tau } \right)d\tau \)

Explanation:

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

Consider the signal x(t) = u(t) & h(t) = -u(t) for input x(t) with impulse response h(t)

y(t) = u(t) * (-u(t)) = -r(t)

For input |x(t)| with impulse response |h(t)|

z(t) = |u(t)| * |-u(t)| = r(t)

Hence,

Z(t) ≥ y(t) for all value of t.

Given that,

Poles = (2, -1) and (-2, 1)

Zeros = (2, 1), (-2, -1)

Now, drawing the Poles and Zeros to the Real and Imaginary axis-

The transfer function can be written as

$H\left( s \right) = \frac{{\left[ {s - \left( {2 + j} \right)} \right]\left[ {s - \left( { - 2 - j} \right)} \right]}}{{\left[ {s - \left( {2 + j} \right)} \right]\left[ {s - \left( { - 2 + j} \right)} \right]}}$

$H\left( s \right) = \frac{{{s^2} - 3 - 4j}}{{{s^2} - 3 + 4j}}$

Observations:

• Magnitude of H(jω) = 1. Therefore, the given system is an all-pass filter.
• Since one pole on the RHS, the system is unstable.
• Since the pole does not have complex conjugate poles and zeros present, the system is complex.

Given difference equation is,

y[n] – a y[n - 1] = b₀ x[n] – b₁ x[n - 1]

By applying Z-transform,

Y(z) – az^–1 Y(z) = b₀X(z) – b₁z^–1 X(z)

$H\left( z \right) = \frac{{Y\left( z \right)}}{{X\left( z \right)}} = \frac{{{b_0} - {b_1}{Z^{ - 1}}}}{{1 - a{Z^{ - 1}}}}$

By taking right sided Inverse Z- Transform-

h(n) = b₀aⁿ u(n) – b₁a^n-1 u(n – 1)

By taking left sided Inverse Z- Transform-

h(n) = -b₀aⁿ u(-n-1) – b₁a^n-1 u(-n)

Thus, the system is not necessarily causal.

The impulse response is non-zero at infinitely many instants.

Divergence theorem:

It states that the surface integral of a vector field over a closed surface, which is called the flux through the surface, is equal to the volume integral of the divergence over the region inside the surface.

$\psi =\mathop{{\int!!!!!\int}\mkern-21mu \bigcirc} \vec{D}.ds= \left( \iiint{\overrightarrow{\Delta }}.\vec{D} \right)dv$