Official Paper

GATE EC 2015 Official Paper: Shift 2 (Previous Year Paper)

65 questions · 180 minutes · with answers · free

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General Aptitude (10 questions)

1

Choose the appropriate word / phrase, out of the four options given below, to complete the following sentence:

Dhoni, as well as the other team members of Indian team, __________ present on the occasion

  1. ((a))

    were

  2. ((b))

    was

  3. ((c))

    has

  4. ((d))

    have

Show Answer
Answer: ((b))

was

Here, Dhoni is a singular noun and is connected to a plural noun ‘other team members’ by using a connector ‘as well as’. Whenever we use ‘as well as ‘ as a connector the verb follows first subject. Here, the verb will follow ‘Dhoni’ which is a singular noun and thus, singular verb must be used. Option A and D are omitted as they are plural. Option C will not be suitable as ‘present’ is used after it. Hence, option B is the correct one.

2

Choose the word most similar in meaning to the given word: Awkward

  1. ((a))

    Inept

  2. ((b))

    Graceful

  3. ((c))

    Suitable

  4. ((d))

    Dreadful

Show Answer
Answer: ((a))

Inept

‘Awkward’ means ‘hard to deal with or not graceful’. Hence, option B and C are omitted, as they are antonyms of awkward. Option D ‘dreadful’ is ‘extremely bad or causing great suffering’ which is too extreme to be a synonym to ‘awkward’. Hence, ‘Inept’ which means clumsy or not graceful is the perfect synonym. Thus, option A is the correct one.

3

What is the adverb for the given word below? Misogynous

  1. ((a))

    Misogynousness

  2. ((b))

    Misogynity

  3. ((c))

    Misogynously

  4. ((d))

    Misogynous

Show Answer
Answer: ((c))

Misogynously

The given form ‘misogynous’ is an adjective form and we need an adverb form. Thus, option D is omitted. Option A and B are not correct as per the English vocabulary. Hence, option C is the correct option.

4

An electric bus has onboard instruments that report the total electricity consumed since the start of the trip as well as the total distance covered. During a single day of operation, the bus travels on stretches M, N, O and P, in that order. the cumulative distances travelled and the corresponding electricity consumption are shown in the Table below:

StretchCumulative distance(km)Electricity used (kWh)
M2012
N4525
O7545
P10057
<br>

The stretch where the electricity consumption per km is minimum is

  1. ((a))

    M

  2. ((b))

    N

  3. ((c))

    O

  4. ((d))

    P

Show Answer
Answer: ((d))

P

Explanation:

On rearranging the given data in tabular form, we get,

StretchCumulative distance (km)Distance covered in given stretch A (km)Electricity used since beginning (kWh)Electricity used in given stretch B (kWh)Electricity consumption per km (B/A) kWh/km
M2020121212/20 = 0.6
N4545 – 20 = 252525 – 12 = 1313/25 = 0.52
O7575 – 45 = 304545 – 25 = 2020/30 = 0.66
P100100 – 75 = 255757 – 45 = 1212/25 = 0.48
<br>

Clearly, the electricity consumption per km is minimum in stretch P.

5

Ram and Ramesh appeared in an interview for two vacancies in the same department. The probability of Ram’s selection is 1/6 and that of Ramesh is 1/8. What is the probability that only one of them will be selected?

  1. ((a))

    47/48

  2. ((b))

    1/4

  3. ((c))

    13/48

  4. ((d))

    35/48

Show Answer
Answer: ((b))

1/4

Let E1 be the event that Ram gets selected and E2 be the event that Ramesh gets selected.

According to the given information: P(E1) = 1/6 and P(E2) = 1/8

Now, we need to find the probability that only one of them gets selected.

Let E3 be the event that only one of them gets selected.

E3 consists of two cases:

  1. Ram gets selected and Ramesh doesn’t get selected
  2. Ramesh gets selected and Ram doesn’t get selected

∴ Probability that only one gets selected = Probability of case(i) + Probability of case(ii)

⇒ Probability that only one gets selected = P(E1) × (1 – P(E2)) + P(E2) × (1 – P(E1))

⇒ Probability that only one gets selected =16×(1−18)+18×(1−16)= \frac{1}{6} \times \left( {1 - \frac{1}{8}} \right) + \frac{1}{8} \times \left( {1 - \frac{1}{6}} \right)

⇒ Probability that only one gets selected =16×(1−18)+18×(1−16) =16×78+18×56=14\begin{array}{l} = \frac{1}{6} \times \left( {1 - \frac{1}{8}} \right) + \frac{1}{8} \times \left( {1 - \frac{1}{6}} \right)\ = \frac{1}{6} \times \frac{7}{8} + \frac{1}{8} \times \frac{5}{6} = \frac{1}{4} \end{array}

6

In the following sentence, certain parts are marked in bold and marked P, Q and R. One of the parts may contain certain error or may not be acceptable is standard written communication. Select the part containing an error. Choose 4 as your answer if there is no error.

The student corrected all the errors (P) that the instructor marked (Q) on the answer book. (R)

  1. ((a))

    P

  2. ((b))

    Q

  3. ((c))

    R

  4. ((d))

    No error

Show Answer
Answer: ((b))

Q

Here, from the first part it is evident that the action happened in the past hence the word ‘corrected’ is used. But, since the students have already corrected the errors in the past it is clear that those errors were marked by the instructor much before they corrected it. Thus, the use of simple past is not enough here and so there is an error in this part. Past perfect tense must be used here and the sentence should read ‘The students corrected all the errors that the instructor had marked on the answer book.’ Thus, answer B is the correct one.

7

Given below are two statements followed by two conclusions. Assuming these statements to be true, decide which one logically follow.

Statements:

I. All film stars are playback singers

II. All film directors are film stars.

Conclusions:

I. All film directors are playback singers

II. Some film star are film directors

  1. ((a))

    Only conclusion I follows

  2. ((b))

    Only conclusion II follows

  3. ((c))

    Neither conclusions I nor II follows

  4. ((d))

    Both conclusion I and II follow

Show Answer
Answer: ((d))

Both conclusion I and II follow

The least possible Venn diagram for the given statements is as follows,

Conclusions:

I. All film directors are playback singers → Clearly true.

II. Some film star are film directors → Clearly true.

Thus both conclusion I and II follows.

8

A tiger is 50 leaps of its own behind a deer. The tiger takes 5 leaps per minute to the deer’s 4. If the tiger and the deer cover 8 metre and 5 metre per leap respectively, what distance in metres will the tiger have to run before it catches the deer?

9

If a2;+;b2;+;c2;=;1{{\rm{a}}^2}{\rm{;}} + {\rm{;}}{{\rm{b}}^2}{\rm{;}} + {\rm{;}}{{\rm{c}}^2}{\rm{;}} = {\rm{;}}1, then ab;+;bc;+;ac{\rm{ab;}} + {\rm{;bc;}} + {\rm{;ac}} lies in the interval

  1. ((a))

    [1,23]\left[ {1,\frac{2}{3}} \right]

  2. ((b))

    [−12,;1]\left[ { - \frac{1}{2},{\rm{;}}1} \right]

  3. ((c))

    [−1,12]\left[ { - 1,\frac{1}{2}} \right]

  4. ((d))

    [2,;−4]\left[ {2,{\rm{;}} - 4} \right]

Show Answer
Answer: ((b))

[−12,;1]\left[ { - \frac{1}{2},{\rm{;}}1} \right]

We know that, (a;+;b;+;c)2;=;a2;+;b2;+;c2;+;2(ab;+;bc;+;ca){\left( {{\rm{a;}} + {\rm{;b;}} + {\rm{;c}}} \right)^2}{\rm{;}} = {\rm{;}}{{\rm{a}}^2}{\rm{;}} + {\rm{;}}{{\rm{b}}^2}{\rm{;}} + {\rm{;}}{{\rm{c}}^2}{\rm{;}} + {\rm{;}}2\left( {{\rm{ab;}} + {\rm{;bc;}} + {\rm{;ca}}} \right)

Now, (a;+;b;+;c)2;≥;0{\left( {{\rm{a;}} + {\rm{;b;}} + {\rm{;c}}} \right)^2}{\rm{;}} \ge {\rm{;}}0

⇒;a2;+;b2;+;c2;+;2(ab;+;bc;+;ca);≥;0 ⇒;1;+;2(ab;+;bc;+;ca);≥;0\begin{array}{l} \Rightarrow {\rm{;}}{{\rm{a}}^2}{\rm{;}} + {\rm{;}}{{\rm{b}}^2}{\rm{;}} + {\rm{;}}{{\rm{c}}^2}{\rm{;}} + {\rm{;}}2\left( {{\rm{ab;}} + {\rm{;bc;}} + {\rm{;ca}}} \right){\rm{;}} \ge {\rm{;}}0\ \Rightarrow {\rm{;}}1{\rm{;}} + {\rm{;}}2\left( {{\rm{ab;}} + {\rm{;bc;}} + {\rm{;ca}}} \right){\rm{;}} \ge {\rm{;}}0 \end{array}

⇒;(ab;+;bc;+;ca);≥;−12\Rightarrow {\rm{;}}\left( {{\rm{ab;}} + {\rm{;bc;}} + {\rm{;ca}}} \right){\rm{;}} \ge {\rm{;}} - \frac{1}{2}     -----(A)

Now, from A.M.-G.M. inequality, we know that x+y2≥xy\frac{{{\rm{x}} + {\rm{y}}}}{2} \ge \sqrt {{\rm{xy}}}

∴a2+b22≥a2b2\therefore \frac{{{{\rm{a}}^2} + {{\rm{b}}^2}}}{2} \ge \sqrt {{{\rm{a}}^2}{{\rm{b}}^2}}

⇒;a2;+;b2;≥;2ab;\Rightarrow {\rm{;}}{{\rm{a}}^2}{\rm{;}} + {\rm{;}}{{\rm{b}}^2}{\rm{;}} \ge {\rm{;}}2{\rm{ab;}}    ---(i)

Similarly, we obtain:

⇒;b2;+;c2≥;2bc\Rightarrow {\rm{;}}{{\rm{b}}^2}{\rm{;}} + {\rm{;}}{{\rm{c}}^2} \ge {\rm{;}}2{\rm{bc}}    ----(ii)

⇒;c2+;a2;≥;2ca;\Rightarrow {\rm{;}}{{\rm{c}}^2} + {\rm{;}}{{\rm{a}}^2}{\rm{;}} \ge {\rm{;}}2{\rm{ca;}}   ----(iii)

Adding (i), (ii) and (iii), we get:

2;(a2;+;b2;+;c2);≥;2;(ab;+;bc;+;ca)2{\rm{;}}\left( {{{\rm{a}}^2}{\rm{;}} + {\rm{;}}{{\rm{b}}^2}{\rm{;}} + {\rm{;}}{{\rm{c}}^2}} \right){\rm{;}} \ge {\rm{;}}2{\rm{;}}\left( {{\rm{ab;}} + {\rm{;bc;}} + {\rm{;ca}}} \right)

⇒;1;≥;(ab;+;bc;+;ca)\Rightarrow {\rm{;}}1{\rm{;}} \ge {\rm{;}}\left( {{\rm{ab;}} + {\rm{;bc;}} + {\rm{;ca}}} \right)     -----(B)

From (A) and (B):

−1/2;≤;(ab;+;bc;+;ca);≤;1- 1/2{\rm{;}} \le {\rm{;}}\left( {{\rm{ab;}} + {\rm{;bc;}} + {\rm{;ca}}} \right){\rm{;}} \le {\rm{;}}1

10

Lamenting the gradual sidelining of the arts in school curricula, a goup of prominent artists wrote to the Chief Minister last year, asking him to allocate more funds to support arts education in schools. However, no such increase has been announced in this year`s Budget. The artists expressed their deep anguish at their request not being approved, but many of them remain optimistic about funding in the future.

Which of the statement(s) below is/are logically valid and can be inferred from the above statements?

(i) The artists expected funding for the arts to increase this year.

(ii) The Chief Minister was receptive to the idea of increasing funding for the arts.

(iii) The Chief Minister is a prominent artist.

(iv) Schools are giving less importance to arts education nowadays,

  1. ((a))

    (iii) and (iv)

  2. ((b))

    (i) and (iv)

  3. ((c))

    (i), (ii) and (iv)

  4. ((d))

    (i) and (iii)

Show Answer
Answer: ((b))

(i) and (iv)

Statement (iv) is clearly valid as the reason the group of prominent artists wrote to CM was a gradual sidelining of the arts in school curricula. Also, (i) is valid as the last line of the passage suggests that the students are still optimistic about the funding in the future. Statement (iii) is completely out of context and hence is invalid. Statement (ii) does not have any reference from the passage to be true as already the request of the artists has been rejected once which does not support the CM to be a ‘receptive’ one. Hence, option B is the correct one.

Electronics and Communication Engineering (55 questions)

11

The bilateral Laplace transform of a function \({\rm{f}}\left( {\rm{t}} \right) = \left{ {\begin{array}{*{20}{c}} {1{\rm{;if;a}} \le {\rm{t}} \le {\rm{b}}}\ {0{\rm{;otherwise}}} \end{array}} \right.\)

  1. ((a))

    a−bs\frac{{{\rm{a}} - {\rm{b}}}}{{\rm{s}}}

  2. ((b))

    e2(a−b)s{\rm{}}\frac{{{{\rm{e}}^2}\left( {{\rm{a}} - {\rm{b}}} \right)}}{{\rm{s}}}

  3. ((c))

    e−as−e−bss\frac{{{{\rm{e}}^{ - {\rm{as}}}} - {{\rm{e}}^{ - {\rm{bs}}}}}}{{\rm{s}}}

  4. ((d))

    es(a−b)s\frac{{{{\rm{e}}^{{\rm{s}}\left( {{\rm{a}} - {\rm{b}}} \right)}}}}{{\rm{s}}}

Show Answer
Answer: ((c))

e−as−e−bss\frac{{{{\rm{e}}^{ - {\rm{as}}}} - {{\rm{e}}^{ - {\rm{bs}}}}}}{{\rm{s}}}

Given \({\rm{f}}\left( {\rm{t}} \right) = \begin{array}{{20}{c}} 1\ 0 \end{array}{\rm{;}}\begin{array}{{20}{c}} {{\rm{;}}:{\rm{;a}} \le {\rm{t}} \le {\rm{b}}}\ {{\rm{;;}}:{\rm{otherwise}}} \end{array}\)

\(\begin{array}{l} {\rm{L}}\left{ {{\rm{f}}\left( {\rm{t}} \right)} \right} = \mathop \smallint \limits_{ - \infty }^\infty {{\rm{e}}^{ - {\rm{st}}}}{\rm{f}}\left( {\rm{t}} \right){\rm{dt}}\ = \mathop \smallint \limits_{ - \infty }^{\rm{a}} {{\rm{e}}^{ - {\rm{st}}}}{\rm{f}}\left( {\rm{t}} \right) + \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {{\rm{e}}^{ - {\rm{st}}}}{\rm{f}}\left( {\rm{t}} \right){\rm{dt}} + \mathop \smallint \limits_{\rm{b}}^\infty {{\rm{e}}^{ - {\rm{st}}}}{\rm{f}}\left( {\rm{t}} \right){\rm{dt}}\ = 0 + \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {{\rm{e}}^{ - {\rm{st}}}}{\rm{dt}} + 0\ = \left. {\frac{{{{\rm{e}}^{ - {\rm{st}}}}}}{{ - {\rm{s}}}}} \right|_{\rm{a}}^{\rm{b}} = \frac{{ - 1}}{{\rm{s}}}\left[ {{{\rm{e}}^{ - {\rm{bs}}}} - {{\rm{e}}^{ - {\rm{as}}}}} \right]\ = \frac{{{{\rm{e}}^{ - {\rm{as}}}} - {{\rm{e}}^{ - {\rm{bs}}}}}}{{\rm{s}}} \end{array}\)

12

The value of x  for which all the Eigen–values of the matrix given below are real is

\(\left[ {\begin{array}{*{20}{c}} {10}&{5 + {\rm{j}}}&4\ {\rm{x}}&{20}&2\ 4&2&{ - 10} \end{array}} \right]\)

  1. ((a))

    5;+;j5{\rm{;}} + {\rm{;j}}

  2. ((b))

    5;−;j5{\rm{;}}-{\rm{;j}}

  3. ((c))

    1;−;5j1{\rm{;}}-{\rm{;}}5{\rm{j}}

  4. ((d))

    1+;5;j1 + {\rm{;}}5{\rm{;j}}

Show Answer
Answer: ((b))

5;−;j5{\rm{;}}-{\rm{;j}}

Given that all Eigen values of A are real.

∴ A is a Hermition matrix

I.e. (Aˉ)T=A{\left( {{\rm{\bar A}}} \right)^{\rm{T}}} = {\rm{A}}

\(\begin{array}{l} \left[ {\begin{array}{{20}{c}} {10}&{{{\rm{x}}^{\rm{}}}}&4\ {5 - {\rm{j}}}&{20}&2\ 4&2&{ - 10} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {10}&{5 + {\rm{j}}}&4\ {\rm{x}}&{20}&2\ 4&2&{ - 10} \end{array}} \right]\ \therefore {\rm{;x;}} = {\rm{;}}\left( {5{\rm{;}} - {\rm{;j}}} \right) \end{array}\)

13

Let f(z)=az+bcz+d.{\rm{f}}\left( {\rm{z}} \right) = \frac{{{\rm{az}} + {\rm{b}}}}{{{\rm{cz}} + {\rm{d}}}}. If f(z1);=;f(z2){\rm{f}}\left( {{{\rm{z}}_1}} \right){\rm{;}} = {\rm{;f}}({{\rm{z}}_2}) for all z1;≠;z2,;a;=;2,;b;=;4;and;c;=;5{{\rm{z}}_1}{\rm{;}} \ne {\rm{;}}{{\rm{z}}_2},{\rm{;a;}} = {\rm{;}}2,{\rm{;b;}} = {\rm{;}}4{\rm{;and;c;}} = {\rm{;}}5, then d should be equal to  __________.

14

The general solution of the differential equation dydx=1+cos2y1−cos⁡2x\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{1 + {\rm{cos}}2{\rm{y}}}}{{1 - \cos 2{\rm{x}}}} is

  1. ((a))

    tan;y;−;cot;x;=;c(c;is;a;constant){\rm{tan;y;}}-{\rm{;cot;x;}} = {\rm{;c}}\left( {{\rm{c;is;a;constant}}} \right)

  2. ((b))

    tan;x;−;cot;y;=;c(c;is;a;constant){\rm{tan;x;}}-{\rm{;cot;y;}} = {\rm{;c}}\left( {{\rm{c;is;a;constant}}} \right)

  3. ((c))

    tan;y;+;cot;x;=;c(c;is;a;constant){\rm{tan;y;}} + {\rm{;cot;x;}} = {\rm{;c}}\left( {{\rm{c;is;a;constant}}} \right)

  4. ((d))

    tan;x;+;cot;Y;=;c(c;is;a;constant){\rm{tan;x;}} + {\rm{;cot;Y;}} = {\rm{;c}}\left( {{\rm{c;is;a;constant}}} \right)

Show Answer
Answer: ((c))

tan;y;+;cot;x;=;c(c;is;a;constant){\rm{tan;y;}} + {\rm{;cot;x;}} = {\rm{;c}}\left( {{\rm{c;is;a;constant}}} \right)

dydx=1+cos⁡2y1−cos⁡2x dydx=cos⁡2ysin⁡2x sec2ydy;=;cosec2x.dx tan⁡y=−cot⁡x+c tan⁡y+cot⁡x=c\begin{array}{l} \frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{1 + \cos 2{\rm{y}}}}{{1 - \cos 2{\rm{x}}}}\ \frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{{{\cos }^2}{\rm{y}}}}{{{{\sin }^2}{\rm{x}}}}\ {\rm{se}}{{\rm{c}}^2}{\rm{ydy;}} = {\rm{;cose}}{{\rm{c}}^2}{\rm{x}}.{\rm{dx}}\ \tan {\rm{y}} = - \cot {\rm{x}} + {\rm{c}}\ \tan {\rm{y}} + \cot {\rm{x}} = {\rm{c}} \end{array}

15

The magnitude and phase of the complex Fourier series coefficients aK{{\rm{a}}_{\rm{K}}} of a periodic signal x(t){\rm{x}}\left( {\rm{t}} \right) are shown in the figure. Choose the correct statement form the four choices given. Notation: C is the set of complex numbers, R is the set of purely real numbers, and P is the set purely imaginary numbers.

  1. ((a))

    x(t)∈;R{\rm{x}}\left( {\rm{t}} \right) \in {\rm{;R}}

  2. ((b))

    x(t)∈;P{\rm{x}}\left( {\rm{t}} \right) \in {\rm{;P}}

  3. ((c))

    x(t)∈;(C−R){\rm{x}}\left( {\rm{t}} \right) \in {\rm{;}}\left( {{\rm{C}} - {\rm{R}}} \right)

  4. ((d))

    The information given is not sufficient to draw any conclusion about x(t)

Show Answer
Answer: ((a))

x(t)∈;R{\rm{x}}\left( {\rm{t}} \right) \in {\rm{;R}}

\(\angle {{\rm{a}}{\rm{k}}}{\rm{;}} = {\rm{;}} - {\rm{\pi }}\) Only changes the sign of the magnitude. Since the magnitude spectrum \(\left| {{{\rm{a}}{\rm{k}}}} \right|\) is even then the corresponding time-domain signal is real.

16

The voltage (VC) across the capacitor (in Volts) in the network shown is ______

17

In the circuit shown the average value of the voltage Vab (in volts) in steady state condition is _____________.

18

The 2-port admittance matrix of the circuit shown is given by

  1. ((a))

    \(\left[ {\begin{array}{*{20}{c}} {0.3}&{ - 0.2}\ { - 0.2}&{0.3} \end{array}} \right]\)

  2. ((b))

    \(\left[ {\begin{array}{*{20}{c}} {15}&5\ 5&{15} \end{array}} \right]\)

  3. ((c))

    \(\left[ {\begin{array}{*{20}{c}} {3.33}&5\ 5&{3.33} \end{array}} \right]\)

  4. ((d))

    \(\left[ {\begin{array}{*{20}{c}} {0.3}&{0.4}\ {0.4}&{0.3} \end{array}} \right]\)

Show Answer
Answer: ((a))

\(\left[ {\begin{array}{*{20}{c}} {0.3}&{ - 0.2}\ { - 0.2}&{0.3} \end{array}} \right]\)

Given, that

for the π network

YA = 1/10 = 0.1

YB = 1/5 = 0.2

YC = 1/10 = 0.1 

The Y matrix for the π network is given by

\({\rm{Y}} = \left[ {\begin{array}{*{20}{c}} {{{\rm{Y}}{\rm{A}}} + {{\rm{Y}}{\rm{B}}}}&{ - {{\rm{Y}}{\rm{B}}}}\ { - {{\rm{Y}}{\rm{B}}}}&{{{\rm{Y}}{\rm{B}}} + {{\rm{Y}}{\rm{C}}}} \end{array}} \right]\)

\({\rm{Y}} = \left[ {\begin{array}{*{20}{c}} {0.3}&{ - 0.2}\ { - 0.2}&{0.3} \end{array}} \right]\)

Important Points

Z parameter for T network is given by

Z11 = ZA + ZB 

Z12 = Z21 = ZB 

Z22 = ZB + ZC

19

An n-type silicon sample is uniformly illuminated with light which generates 1020 electron-hole pairs per cm3 per second. The minority carrier lifetime in the sample is 1μs in the steady state, the hole concentration in the sample is approximately 10x, where x is an integer. The value of x is

20

A piece of silicon is doped uniformly with phosphorus with a doping concentration of 1016 /cm3 The expected value of mobility versus doping concentration for silicon full dopant ionization is shown below. The Charge of an electron is 1.6 × 10-19 C. The conductivity (in S cm-1) of the silicon sample at 300 K is_________

21

If the circuit shown has to function as a clamping circuit, which one of the following conditions should be satisfied for sinusoidal signal of period T?

  1. ((a))

    RC ≪ T

  2. ((b))

    RC = 0.35T

  3. ((c))

    RC ≈ T

  4. ((d))

    RC ≫ T

Show Answer
Answer: ((d))

RC ≫ T

Concept:

To clamp the signal, the voltage held at the capacitor must be held for a longer time, which requires high TCdis, so that the capacitor voltage does not discharge considerably throughout the non-conducting diode period (T2)\left( {\frac{T}{2}} \right).  

Discharging Time constant (TCdischarge) = RC

Where R is the resistance, C is capacitance and T is the time period of the input signal.

The voltage at the capacitor during discharging is given as:

Vc(t)=Voe−tRCdis{V_c}\left( t \right) = {V_o}{e^{ - \frac{t}{{R{C_{dis}}}}}}

Calculation:

Suppose at t = T/2:

Vc(t)=Vo=Voe−(T2)RC{V_c}\left( t \right) = {V_o} = {V_o}{e^{ - \frac{{\left( {\frac{T}{2}} \right)}}{{RC}}}}    ---(1)

And at t = T, the voltage will be:

Vc(t)=0.99Vo=Voe−TRC{V_c}\left( t \right) = 0.99{V_o} = {V_o}{e^{ - \frac{T}{{RC}}}}    ---(2)

∴ For equation (2), we get:

0.99=e−TRC0.99 = {e^{ - \frac{T}{{RC}}}}

−0.01=−TRC - 0.01 = - \frac{T}{{RC}}

RC = 100 T

This implies that

RC ≫ T

22

In the circuit shown, V0 = V0A for switch SW in position A and V0 = V0B for SW in position B. Assume that the Op-Amp is ideal. The value of \(\frac{{{{\rm{V}}{0{\rm{B}}}}}}{{{{\rm{V}}{0{\rm{A}}}}}}\) is ______.

23

In the bistable circuit shown, the ideal Op-Amp has a saturation level of ± 5V. The value of R1(in k Ω) that gives a hysteresis width of 500 mV is __________.

24

In the figure shown, the output Y is required to be Y = A.B + C̅.D̅. The gates G1 and G2 must be respectively.

  1. ((a))

    NOR, OR

  2. ((b))

    OR, NAND

  3. ((c))

    NAND, OR

  4. ((d))

    AND, NAND

Show Answer
Answer: ((a))

NOR, OR

Method: by option elimination

Option1:

Let G1 be NOR gate and G2 be OR gate

Output of G1 = (Aˉ+Bˉ)‾=A.B{\overline{(\bar A + \bar B)}} = A.B

1st Input of G2  = A. B

2nd Input of G2 = C+D‾=Cˉ.Dˉ\overline{C + D} = \bar C. \bar D

Output of G2 = Y = A.B + C̅.D̅

Hence option1 is correct.

25

In an 8085 microprocessor, which one of the following instructions changes the content of the accumulator?

  1. ((a))

    MOV B, M

  2. ((b))

    PCHL

  3. ((c))

    RNZ

  4. ((d))

    SBI BEH

Show Answer
Answer: ((d))

SBI BEH

In general, an arithmetic or logical operation update the data of accumulator and flag register. Here, only SBI BEH is an arithmetic operation. The instruction adds the content of accumulator with data BE H and stores the result in accumulator.

26

A mod–n counter using a synchronous binary up–counter with synchronous clear input is shown in the figure. The value of n is_________.

27

Let the signal f(t) = 0 outside the interval [T1, T2], where T1 and T2 are finite. Furthermore, ∣f(t)∣<∞\left| {{\rm{f}}\left( {\rm{t}} \right)} \right| < \infty. The region of convergence (ROC) of the signal’s bilateral Laplace transform F(s) is

  1. ((a))

    A parallel strip containing the jΩ axis

  2. ((b))

    A parallel strip not containing the jΩ axis

  3. ((c))

    The entire s - plane

  4. ((d))

    A half plane containing the jΩ axis

Show Answer
Answer: ((c))

The entire s - plane

ROC defines the region where the Laplace transform exists.

Laplace transform of f(t) is given as:

F(s) = ∫−∞∞f(t)e−stdt=∫−∞∞f(t)e−σte−jωtdt\int_{-\infty}^{\infty} f(t) e^{-st} dt=\int_{-\infty}^{\infty} f(t) e^{-\sigma t} e^{-j\omega t} dt

It is given that, f(t) < ∞ and the signal is zero outside in the Interval [T1, T2] where T1 & T2 are finite.

If f(t) is multiplied by a decaying Exponential (σ > 0) or by a growing exponential (σ < 0), this Exponential weighting is never unbounded.

Consequently, the Integrability of f(t) by this exponential weighting is not destroyed.

Hence, the ROC of signal f(t) is entire s-plane

Important Points

Properties of ROC: 

  1. Linearity:

If Laplace Transform of x1(t) is X1(s) with ROC R1 and Laplace transform of x2(t) is X2(s) with ROC R2. Then 

Laplace transform of a x1(t + b x2(t) ⇒ a X1(s) + b X2(s) with > ROC: R1 ∩ R2

  1. Time-shifting:

 x(t) ↔ X(s) with > ROC : R

Then, x(t - t0) ↔ e-st0 X(s) with > ROC: R

  1. Shift in S-domain:

 x(t) ↔ X(s) with ROC : R

Then, x(t) es0t ↔ X (s - s0) with > ROC: R + Re (s0).

  1. Time-reversal:

 x(t) ↔ X(s) with > ROC : R

Then, x(-t) ↔ X(-s) with ROC : - R

  1. Differentiation in time:

 x(t) ↔ X(s) with ROC : R

Then, dx(t) / dt ↔ s X(s) with > ROC : R

Note: For finite duration signal, the ROC is the entire s-plane.

28

Two causal discrete-time signal x[n] and y[n] are related as y[n]=∑m=0nx[m]{\rm{y}}\left[ {\rm{n}} \right] = \sum _{{\rm{m}} = 0}^{\rm{n}}{\rm{x}}\left[ {\rm{m}} \right] It the z-transform of y[n];is2z(z−1)2{\rm{y}}\left[ {\rm{n}} \right]{\rm{;is}}\frac{2}{{{\rm{z}}{{\left( {{\rm{z}} - 1} \right)}^2}}}, the value of x[2] is ________.

29

By performing cascading and / or summing / differencing operations using transfer function blocks G1 (s) and G2 (s), one CANNOT realize a transfer function of the form

  1. ((a))

    G1(s)G2(s){{\rm{G}}_1}\left( {\rm{s}} \right){{\rm{G}}_2}\left( {\rm{s}} \right)

  2. ((b))

    G1(s)G2(s)\frac{{{{\rm{G}}_1}\left( {\rm{s}} \right)}}{{{{\rm{G}}_2}\left( {\rm{s}} \right)}}{\rm{}}

  3. ((c))

    G1(s)(1G1(s)+G2(s)){{\rm{G}}_1}\left( {\rm{s}} \right)\left( {\frac{1}{{{{\rm{G}}_1}\left( {\rm{s}} \right)}} + {{\rm{G}}_2}\left( {\rm{s}} \right)} \right)

  4. ((d))

    G1(s)(1G1(s)−G2(s)){{\rm{G}}_1}\left( {\rm{s}} \right)\left( {\frac{1}{{{{\rm{G}}_1}\left( {\rm{s}} \right)}} - {{\rm{G}}_2}\left( {\rm{s}} \right)} \right)

Show Answer
Answer: ((b))

G1(s)G2(s)\frac{{{{\rm{G}}_1}\left( {\rm{s}} \right)}}{{{{\rm{G}}_2}\left( {\rm{s}} \right)}}{\rm{}}

G1(s)G2(s){{\rm{G}}_1}\left( {\rm{s}} \right){{\rm{G}}_2}\left( {\rm{s}} \right) can be realized directly by cascading the systems.

G1(s)(1G1(s)+G2(s))=1+G1(s)G2(s){{\rm{G}}_1}\left( {\rm{s}} \right)\left( {\frac{1}{{{{\rm{G}}_1}\left( {\rm{s}} \right)}} + {{\rm{G}}_2}\left( {\rm{s}} \right)} \right) = 1 + {{\rm{G}}_1}\left( {\rm{s}} \right){{\rm{G}}_2}\left( {\rm{s}} \right) can be realised by providing a unity feed forward path around the cascaded arrangement.

G1(s)(1G1(s)−G2(s))=1−G1(s)G2(s){{\rm{G}}_1}\left( {\rm{s}} \right)\left( {\frac{1}{{{{\rm{G}}_1}\left( {\rm{s}} \right)}} - {{\rm{G}}_2}\left( {\rm{s}} \right)} \right) = 1 - {{\rm{G}}_1}\left( {\rm{s}} \right){{\rm{G}}_2}\left( {\rm{s}} \right)  can be realised by providing a unity feed forward path around the cascaded arrangement and multiplying -1 to the cascaded arrangement of G1(s)G2(s){{\rm{G}}_1}\left( {\rm{s}} \right){{\rm{G}}_2}\left( {\rm{s}} \right).

Thus, only G1(s)G2(s)\frac{{{{\rm{G}}_1}\left( {\rm{s}} \right)}}{{{{\rm{G}}_2}\left( {\rm{s}} \right)}}{\rm{}} cannot be realised using the above systems.

Additional Information

Block diagram reduction rules:

Rule 1: Moving the branch point ahead of the block:

Rule 2: Moving the branch point before the block:

Rule 3: Moving the summing point ahead of the block

Rule 4: Moving the summing point before the block

Rule 5: Interchanging the summing points

Application:

Given Block diagramEquivalent block diagram
30

For the signal flow Graph shown in the figure, the value of C(s)R(s)\frac{{{\rm{C}}\left( {\rm{s}} \right)}}{{{\rm{R}}\left( {\rm{s}} \right)}}  is

  1. ((a))

    G1G2G3G41−G1G2H1−G3G4H2−G2G3H3+G1G2G3G4H1H2\frac{{{{\rm{G}}_1}{{\rm{G}}_2}{{\rm{G}}_3}{{\rm{G}}_4}}}{{1 - {{\rm{G}}_1}{{\rm{G}}_2}{{\rm{H}}_1} - {{\rm{G}}_3}{{\rm{G}}_4}{{\rm{H}}_2} - {{\rm{G}}_2}{{\rm{G}}_3}{{\rm{H}}_3} + {{\rm{G}}_1}{{\rm{G}}_2}{{\rm{G}}_3}{{\rm{G}}_4}{{\rm{H}}_1}{{\rm{H}}_2}}}

  2. ((b))

    G1G2G3G41+G1G2H1+G3G4H2+G2G3H3+G1G2G3G4H1H2\frac{{{{\rm{G}}_1}{{\rm{G}}_2}{{\rm{G}}_3}{{\rm{G}}_4}}}{{1 + {{\rm{G}}_1}{{\rm{G}}_2}{{\rm{H}}_1} + {{\rm{G}}_3}{{\rm{G}}_4}{{\rm{H}}_2} + {{\rm{G}}_2}{{\rm{G}}_3}{{\rm{H}}_3} + {{\rm{G}}_1}{{\rm{G}}_2}{{\rm{G}}_3}{{\rm{G}}_4}{{\rm{H}}_1}{{\rm{H}}_2}}}

  3. ((c))

    11+G1G2H1+G3G4H2+G2G3H3+G1G2G3G4H1H2\frac{1}{{1 + {{\rm{G}}_1}{{\rm{G}}_2}{{\rm{H}}_1} + {{\rm{G}}_3}{{\rm{G}}_4}{{\rm{H}}_2} + {{\rm{G}}_2}{{\rm{G}}_3}{{\rm{H}}_3} + {{\rm{G}}_1}{{\rm{G}}_2}{{\rm{G}}_3}{{\rm{G}}_4}{{\rm{H}}_1}{{\rm{H}}_2}}}

  4. ((d))

    11−G1G2H1−G3G4H2−G2G3H3+G1G2G3G4H1H2\frac{1}{{1 - {{\rm{G}}_1}{{\rm{G}}_2}{{\rm{H}}_1} - {{\rm{G}}_3}{{\rm{G}}_4}{{\rm{H}}_2} - {{\rm{G}}_2}{{\rm{G}}_3}{{\rm{H}}_3} + {{\rm{G}}_1}{{\rm{G}}_2}{{\rm{G}}_3}{{\rm{G}}_4}{{\rm{H}}_1}{{\rm{H}}_2}}}

Show Answer
Answer: ((b))

G1G2G3G41+G1G2H1+G3G4H2+G2G3H3+G1G2G3G4H1H2\frac{{{{\rm{G}}_1}{{\rm{G}}_2}{{\rm{G}}_3}{{\rm{G}}_4}}}{{1 + {{\rm{G}}_1}{{\rm{G}}_2}{{\rm{H}}_1} + {{\rm{G}}_3}{{\rm{G}}_4}{{\rm{H}}_2} + {{\rm{G}}_2}{{\rm{G}}_3}{{\rm{H}}_3} + {{\rm{G}}_1}{{\rm{G}}_2}{{\rm{G}}_3}{{\rm{G}}_4}{{\rm{H}}_1}{{\rm{H}}_2}}}

Concept:

Mason’s gain formula is

\(T = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{\mathop \sum \nolimits_{i = 1}^N {P_i}{{\rm{Δ }}_i}}}{{\rm{Δ }}}\)

Where,

C(s) is the output node

R(s) is the input node

T is the transfer function or gain between R(s) and C(s)

Pi is the ith forward path gain

Δ = 1−(sum of all individual loop gains) + (sum of gain products of all possible two non-touching loops) − (sum of gain products of all possible three non-touching loops) + ........

Δi is obtained from Δ by removing the loops which are touching the ith forward path.

Calculation:

Given signal flow graph,

There is only one forward path P1 = G1 G2 G3 G4

Number of loops = 3

L1 = - G1 G2 H1

L2 = - G3 G4 H2

L3 = - G2 G3 H3

Number of non-touching loops = L1 L2 = G1 G2 G3 G4 H1 H2

TF=G1G2G3G41+G1G2H1+G3G4H2+G2G3H3+G1G2G3G4H1H2 {\rm{TF}} = \frac{{{{\rm{G}}_1}{{\rm{G}}_2}{{\rm{G}}_3}{{\rm{G}}_4}}}{{1 + {{\rm{G}}_1}{{\rm{G}}_2}{{\rm{H}}_1} + {{\rm{G}}_3}{{\rm{G}}_4}{{\rm{H}}_2} + {{\rm{G}}_2}{{\rm{G}}_3}{{\rm{H}}_3} + {{\rm{G}}_1}{{\rm{G}}_2}{{\rm{G}}_3}{{\rm{G}}_4}{{\rm{H}}_1}{{\rm{H}}_2}}}

31

A unity negative feedback system has an open loop transfer function,

G(s)=Ks(s+10){\rm{G}}\left( {\rm{s}} \right) = \frac{{\rm{K}}}{{{\rm{s}}\left( {{\rm{s}} + 10} \right)}}

The gain K for which the system to have damping ratio of 0.250.25 is_________.

32

A sinusoidal signal of amplitude A is quantized by a uniform quantizer. Assume that the signal utilizes all the representation levels of the quantizer. If the signal to quantization noise ratio is 31.8 dB, the number of levels in the quantizer is ________.

33

The signal cos⁡(10πt+π4)\cos \left( {10{\rm{\pi t}} + \frac{{\rm{\pi }}}{4}} \right) is ideally sampled at a sampling frequency of 15 Hz\rm 15 \ Hz. the sampled signal is passed through a filter with impulse response

(sin⁡(πt)πt)cos⁡(40πt−π2).\left( {\frac{{\sin \left( {{\rm{\pi t}}} \right)}}{{{\rm{\pi t}}}}} \right)\cos \left( {40{\rm{\pi t}} - \frac{{\rm{\pi }}}{2}} \right). The filter output is

  1. ((a))

    152cos⁡(40πt−π4)\frac{{15}}{2}\cos \left( {40{\rm{\pi t}} - \frac{{\rm{\pi }}}{4}} \right)

  2. ((b))

    152(sin⁡(πt)πt)cos⁡(10πt+π4)\frac{{15}}{2}\left( {\frac{{\sin \left( {{\rm{\pi t}}} \right)}}{{{\rm{\pi t}}}}} \right)\cos \left( {10{\rm{\pi t}} + \frac{{\rm{\pi }}}{4}} \right)

  3. ((c))

    152cos⁡(10πt−π4)\frac{{15}}{2}\cos \left( {10{\rm{\pi t}} - \frac{{\rm{\pi }}}{4}} \right)

  4. ((d))

    152(sin⁡(πt)2)cos⁡(40πt−π2)\frac{{15}}{2}\left( {\frac{{\sin \left( {{\rm{\pi t}}} \right)}}{2}} \right)\cos \left( {40{\rm{\pi t}} - \frac{{\rm{\pi }}}{2}} \right)

Show Answer
Answer: ((a))

152cos⁡(40πt−π4)\frac{{15}}{2}\cos \left( {40{\rm{\pi t}} - \frac{{\rm{\pi }}}{4}} \right)

We have signal x(t)=cos;(10πt+π4);{\rm{x}}\left( {\rm{t}} \right) = {\rm{cos;}}\left( {10{\rm{\pi t}} + \frac{{\rm{\pi }}}{4}} \right){\rm{;}}

We can neglect the phase π/4 as it can be inserted again in the end

∴ If \({{\rm{x}}_1}\left( {\rm{t}} \right) = {\rm{cos}}\left( {10{\rm{\pi t}}} \right)\begin{array}{*{20}{c}} {{\rm{FT}}}\ \leftrightarrow \end{array}{{\rm{X}}_1}\left( {\rm{f}} \right) = \frac{1}{2}\left[ {{\rm{\delta }}\left( {{\rm{f}} - 5} \right) + {\rm{\delta }}\left( {{\rm{f}} + 5} \right)} \right]\)

Now for a sampling signal

\({\rm{P}}\left( {\rm{t}} \right) = \mathop \sum \limits_{{\rm{k}} = - \infty }^\infty {\rm{\delta }}\left( {{\rm{f}} - {\rm{k}}{{\rm{f}}_{\rm{s}}}} \right)\)

The result of sampling is x1(t)p(t){{\rm{x}}_1}\left( {\rm{t}} \right){\rm{p}}\left( {\rm{t}} \right). Now, multiplication in time domain is convolution in frequency domain. Thus

\({{\rm{x}}_{\rm{s}}}\left( {\rm{t}} \right) = {{\rm{x}}_1}\left( {\rm{t}} \right){\rm{p}}\left( {\rm{t}} \right)\begin{array}{*{20}{c}} {{\rm{FT}}}\ \leftrightarrow \end{array}{{\rm{X}}_1}\left( {\rm{f}} \right){\rm{*P}}\left( {\rm{f}} \right)\)

we have

\(\begin{array}{l} {{\rm{X}}{\rm{s}}}\left( {\rm{f}} \right) = {{\rm{X}}1}\left( {\rm{f}} \right){\rm{P}}\left( {\rm{f}} \right) = \frac{1}{2}\left[ {{\rm{\delta }}\left( {{\rm{f}} - 5} \right) + {\rm{\delta }}\left( {{\rm{f}} + 5} \right)} \right]{\rm{}}15\mathop \sum \limits{{\rm{k}} = - \infty }^\infty {\rm{\delta }}\left( {{\rm{f}} - 15{\rm{k}}} \right)\ = \frac{{15}}{2}\mathop \sum \limits{{\rm{k}} = - \infty }^\infty \left{ {{\rm{\delta }}\left( {{\rm{f}} - 15{\rm{k}} - 5} \right) + {\rm{\delta }}\left( {{\rm{f}} - 15{\rm{k}} + 5} \right)} \right} \end{array}\)

This can be simply be interpreted as that that the spectrum

X1(f)=12[δ(f−5)+δ(f+5)]{{\rm{X}}_1}\left( {\rm{f}} \right) = \frac{1}{2}\left[ {{\rm{\delta }}\left( {{\rm{f}} - 5} \right) + {\rm{\delta }}\left( {{\rm{f}} + 5} \right)} \right] reaps with frequency 15Hz and that each impulse magnitude is now 152\frac{{15}}{2}. Thus sampled signal spectrum is

Coming to the filter response

h(t)=(sin⁡(πt)πt)cos⁡(40πt−π2) =sinct.sin⁡(40;πt) ∴H(f)=rect;f∗12j[δ(f−20)−δ(f+20)] =12j[rect(f−20)−rect(f+20)]\begin{array}{l} {\rm{h}}\left( {\rm{t}} \right) = \left( {\frac{{\sin \left( {{\rm{\pi t}}} \right)}}{{{\rm{\pi t}}}}} \right)\cos \left( {40{\rm{\pi t}} - \frac{{\rm{\pi }}}{2}} \right)\ = {\rm{sinct}}.\sin \left( {40{\rm{;\pi t}}} \right)\ \therefore {\rm{H}}\left( {\rm{f}} \right) = {\rm{rect;f*}}\frac{1}{{2{\rm{j}}}}\left[ {{\rm{\delta }}\left( {{\rm{f}} - 20} \right) - {\rm{\delta }}\left( {{\rm{f}} + 20} \right)} \right]\ = \frac{1}{{2{\rm{j}}}}\left[ {{\rm{rect}}\left( {{\rm{f}} - 20} \right) - {\rm{rect}}\left( {{\rm{f}} + 20} \right)} \right] \end{array}

The filter spectrum is given by

We see that only the impulse at f=20{\rm{f}} = 20 of Xs(f){{\rm{X}}_{\rm{s}}}\left( {\rm{f}} \right) passes through the filter. Now, the output signal is given by

\(\begin{array}{l} {{\rm{X}}{\rm{s}}}\left( {\rm{f}} \right){\rm{H}}\left( {\rm{f}} \right)\ = \frac{{15}}{2}\mathop \sum \limits{{\rm{k}} = - \infty }^\infty \left{ {{\rm{\delta }}\left( {{\rm{f}} - 15{\rm{k}} - 5} \right) + {\rm{\delta }}\left( {{\rm{f}} + 15{\rm{k}} - 5} \right)} \right} \times \frac{1}{{2{\rm{j}}}}\left[ {{\rm{rect;}}\left( {{\rm{f}} - 20} \right) - {\rm{rect}}\left( {{\rm{f}} + 20} \right)} \right]\ = \frac{{15}}{{4{\rm{j}}}}\left[ {{\rm{\delta }}\left( {{\rm{f}} - 20} \right) - {\rm{\delta }}\left( {{\rm{f}} + 20} \right)} \right] \end{array}\)

Thus, recovered signal

\(\begin{array}{l} {{\rm{x}}{\rm{r}}}\left( {\rm{t}} \right) = \frac{{15}}{2}\sin \left( {40{\rm{\pi t}}} \right)\ \Rightarrow {{\rm{x}}{\rm{r}}}\left( {\rm{t}} \right) = \frac{{15}}{2}\cos \left( {40{\rm{\pi t}} - \frac{{\rm{\pi }}}{2}} \right) \end{array}\)

Inserting the phase π4\frac{{\rm{\pi }}}{4} again we have

\(\begin{array}{l} {{\rm{X}}{\rm{r}}}\left( {\rm{t}} \right) = \frac{{15}}{2}\cos \left( {40{\rm{\pi t}} - \frac{{\rm{\pi }}}{2} + \frac{{\rm{\pi }}}{4}} \right)\ \Rightarrow {{\rm{X}}{\rm{r}}}\left( {\rm{t}} \right) = \frac{{15}}{2}\cos \left( {40{\rm{\pi t}} - \frac{{\rm{\pi }}}{4}} \right) \end{array}\)

34

In a source free region in vacuum, if the electrostatic potential, ϕ=2x2+y2+cz2\phi = 2{{\rm{x}}^2} + {{\rm{y}}^2} + {\rm{c}}{{\rm{z}}^2}, the value of constant c must be____________.

35

The electric field of a uniform plane electromagnetic wave is

\({\rm{\vec E}} = \left( {{{{\rm{\vec a}}}{\rm{x}}} + {\rm{j}}4{{{\rm{\vec a}}}{\rm{y}}}} \right)\exp \left[ {{\rm{j}}\left( {2{\rm{\pi }} \times {{10}^7}{\rm{t}} - 0.2{\rm{z}}} \right)} \right]{\rm{;}}\)

The polarization of the wave is

  1. ((a))

    right handed circular

  2. ((b))

    right handed elliptical

  3. ((c))

    left handed circular

  4. ((d))

    left handed elliptical

Show Answer
Answer: ((d))

left handed elliptical

Concept: 

Ex = E1 sin (ωt – βz) âx

Ey = E2 sin (ωt – βz + δ) â­y

If δ = 0 linear polarization.

If δ = 90° and E1 = E2 circular

If δ ≠ 90°    δ ≠ 0    E1 ≠ E2 Elliptical        

Now left or right ??

Keep your left-hand thumb in the direction of propagation and then check the. The direction of the finger. If it follows the direction then it is left circular otherwise right circular.

Application: 

We have an electric field

\({\rm{\vec E}} = \left( {{{{\rm{\hat a}}}{\rm{x}}} + {\rm{j}}4{{{\rm{\hat a}}}{\rm{y}}}} \right)\exp \left[ {{\rm{j}}\left( {2{\rm{\pi }} \times {{10}^7}{\rm{t}} - 0.2{\rm{z}}} \right)} \right]\)

We see that \({{\rm{\hat a}}{\rm{x}}}\) and \({{\rm{\hat a}}{\rm{y}}}\) components are out of phase by 90° and have different magnitudes. Thus, the field is elliptical.

Now the field may be rewritten as

\({\rm{\vec E}} = \left( {{{{\rm{\hat a}}}{\rm{x}}} + {\rm{j}}4{{{\rm{\hat a}}}{\rm{y}}}} \right){{\rm{e}}^{ - {\rm{j\beta z}}}} \cdot {{\rm{e}}^{{\rm{j}}{{\rm{\omega }}_0}{\rm{t}}}}\)

At a constant z, we may leave e−jβz{{\rm{e}}^{ - {\rm{j\beta z}}}} out of the analysis. Thus,

\({\rm{\vec E}} = \left( {{{{\rm{\hat a}}}{\rm{x}}} + {\rm{j}}4{{{\rm{\hat a}}}{\rm{y}}}} \right){{\rm{e}}^{{\rm{j\omega t}}}}\)

At \({\rm{t}} = 0,{\rm{;\vec E}} = {{\rm{\hat a}}{\rm{x}}} + {\rm{j}}4{{\rm{\hat a}}{\rm{y}}}\)

At \({\rm{t}} = \frac{{\rm{T}}}{4},{\rm{;\vec E}} = {\rm{j}}{{\rm{\hat a}}{\rm{x}}} - 4{{\rm{\hat a}}{\rm{y}}}\)

At \({\rm{t}} = \frac{{\rm{T}}}{2},{\rm{\vec E}} = - {{\rm{\hat a}}{\rm{x}}} - {\rm{j}}4{{\rm{\hat a}}{\rm{y}}}\)

At \({\rm{t}} = \frac{{3{\rm{T}}}}{4},{\rm{\vec E}} = - {\rm{j}}{{\rm{\hat a}}{\rm{x}}} + 4{{\rm{\hat a}}{\rm{y}}}\)

Thus,

At t = 0

At \({\rm{t}} = \frac{{\rm{T}}}{8},{\rm{;}}{{\rm{\hat a}}{\rm{y}}}\) component turns from positive j4 to negative while \({{\rm{\hat a}}{\rm{x}}}\) component stays positive all the while. This is possible only if the vector rotates in clockwise sense with increasing time. For clockwise rotation, the rotation sense is left handed. Thus, the field is left handed circular.

36

Consider the differential equation dxdt=10;−;0.2x\frac{{{\rm{dx}}}}{{{\rm{dt}}}} = 10{\rm{;}}-{\rm{;}}0.2{\rm{x}} with initial condition ;x(0);=;1{\rm{;x}}\left( 0 \right){\rm{;}} = {\rm{;}}1. The response x(t) for t>0{\rm{t}} > 0 is

  1. ((a))

    2;−;e−0.2t2{\rm{;}}-{\rm{;}}{{\rm{e}}^{ - 0.2{\rm{t}}}}

  2. ((b))

    2;−;e0.2t2{\rm{;}}-{\rm{;}}{{\rm{e}}^{0.2{\rm{t}}}}

  3. ((c))

    50;−;49e−0.2t50{\rm{;}}-{\rm{;}}49{{\rm{e}}^{ - 0.2{\rm{t}}}}

  4. ((d))

    50;−;49e0.2t50{\rm{;}}-{\rm{;}}49{{\rm{e}}^{0.2{\rm{t}}}}

Show Answer
Answer: ((c))

50;−;49e−0.2t50{\rm{;}}-{\rm{;}}49{{\rm{e}}^{ - 0.2{\rm{t}}}}

\(\begin{array}{l} \frac{{{\rm{dx}}}}{{{\rm{dt}}}} = 10 - 0.2{\rm{x}}\ \frac{1}{{10 - \frac{{\rm{x}}}{5}}}.{\rm{dx}} = {\rm{dt}}\

  • 5\ln \left( {10 - \frac{{\rm{x}}}{5}} \right) = {\rm{t}} + {\rm{c}}\ \left( {10 - \frac{{\rm{x}}}{5}} \right) = {{\rm{e}}^{ - {{\frac{1}{5}}^{\left( {{\rm{c}} + {\rm{t}}} \right)}}}}\ \frac{{\rm{x}}}{5} = 10 - {{\rm{e}}^{ - \frac{{{\rm{c}} + {\rm{t}}}}{5}}}\ {\rm{x}} = 50 - 5{{\rm{e}}^{ - \frac{{{\rm{c}} + {\rm{t}}}}{5}}} \end{array}\)

Given x(0);=;1;;;⇒1;=;50;−;5e−c5;{\rm{x}}\left( 0 \right){\rm{;}} = {\rm{;}}1{\rm{;;;}} \Rightarrow 1{\rm{;}} = {\rm{;}}50{\rm{;}}-{\rm{;}}5{{\rm{e}}^{ - \frac{{\rm{c}}}{5}{\rm{;}}}}

5e−c5=49;⇒e−c5=9.8 ∴x(t)=50−5e−c5.;e−t5=50−49e−0.2t\begin{array}{l} 5{{\rm{e}}^{ - \frac{{\rm{c}}}{5}}} = 49{\rm{;}} \Rightarrow {{\rm{e}}^{ - \frac{{\rm{c}}}{5}}} = 9.8\ \therefore {\rm{x}}\left( {\rm{t}} \right) = 50 - 5{{\rm{e}}^{ - \frac{{\rm{c}}}{5}}}.{\rm{;}}{{\rm{e}}^{ - \frac{{\rm{t}}}{5}}} = 50 - 49{{\rm{e}}^{ - 0.2{\rm{t}}}} \end{array}

37

The value of the integral \(\mathop \smallint \limits_{ - \infty }^\infty 12\cos \left( {2{\rm{\pi t}}} \right)\frac{{\sin \left( {4{\rm{\pi t}}} \right)}}{{4{\rm{\pi t}}}}{\rm{dt}}\) is__________.

38

If C denotes the counter clockwise unit circle, the value of the contour integral \(\frac{1}{{2{\rm{\pi j}}}}\mathop \oint \limits_{\rm{c}}^{\rm{;}} {\rm{Re;}}\left{ {\rm{z}} \right}{\rm{dz}}\) is______.

39

Let the random variable X represent the number of times of fair coin needs to be tossed till two consecutive heads appear for the first time. The expectation of X is ________.

40

An LC tank circuit consists of an ideal capacitor C connected in parallel with a coil of inductance L having an internal resistance R. The resonant frequency of the tank circuit is

  1. ((a))

    12πLC\frac{1}{{2{\rm{\pi }}\sqrt {{\rm{LC}}} }}

  2. ((b))

    12πLC1−R2CL\frac{1}{{2{\rm{\pi }}\sqrt {{\rm{LC}}} }}\sqrt {1 - {{\rm{R}}^2}\frac{{\rm{C}}}{{\rm{L}}}}

  3. ((c))

    12πLC1−LR2C\frac{1}{{2{\rm{\pi }}\sqrt {{\rm{LC}}} }}\sqrt {1 - \frac{{\rm{L}}}{{{{\rm{R}}^2}{\rm{C}}}}}

  4. ((d))

    12πLC(1−R2CL)\frac{1}{{2{\rm{\pi }}\sqrt {{\rm{LC}}} }}\left( {1 - {{\rm{R}}^2}\frac{{\rm{C}}}{{\rm{L}}}} \right)

Show Answer
Answer: ((b))

12πLC1−R2CL\frac{1}{{2{\rm{\pi }}\sqrt {{\rm{LC}}} }}\sqrt {1 - {{\rm{R}}^2}\frac{{\rm{C}}}{{\rm{L}}}}

\(\begin{array}{l} {\rm{Y}} = {{\rm{Y}}{\rm{c}}} + {{\rm{Y}}{{\rm{LR}}}}\ {\rm{Y}} = {\rm{j\omega C}} + \frac{1}{{\left( {{\rm{j\omega L}} + {\rm{R}}} \right)}} = {\rm{j\omega C}} + \frac{{\left( {{\rm{R}} - {\rm{j\omega L}}} \right)}}{{\left( {{{\rm{R}}^2} + {{\rm{\omega }}^2}{{\rm{L}}^2}} \right)}} \end{array}\)

At resonance, we should have the Imaginary part to zero i.e.

lωC=ωLR2+ω2L2 ω2=L−CR2L2C ω=1LC1−R2CL f=12πLC1−R2CL{l} {\rm{ω C}} = \frac{{{\rm{ω L}}}}{{{{\rm{R}}^2} + {{\rm{ω }}^2}{{\rm{L}}^2}}}\ {{\rm{ω }}^2} = \frac{{{\rm{L}} - {{\rm{C}\rm{R}}^2}}}{{{{\rm{L}}^2}{\rm{C}}}}\ {\rm{ω }} = \frac{1}{{\sqrt {{\rm{LC}}} }}\sqrt {1 - {{\rm{R}}^2}\frac{{\rm{C}}}{{\rm{L}}}}\ {\rm{f }} = \frac{1}{{2\pi\sqrt {{\rm{LC}}} }}\sqrt {1 - {{\rm{R}}^2}\frac{{\rm{C}}}{{\rm{L}}}}

41

In the circuit shown, the Norton equivalent resistance (in Ω) across terminals a–b is ______________.

42

In the circuit shown, the initial voltages across the capacitors C1 and C2 are 1V and 3V, respectively. The switch is closed at time t = 0. The total energy dissipated (in Joules) in the resistor R until steady state is reached is ___________.

43

A dc voltage of 10;V10{\rm{;V}} is applied across an n – type silicon bar having a rectangular cross - section and a length of 1;cm1{\rm{;cm}} as shown in figure. The donor doping concentration \({{\rm{N}}{\rm{D}}}\) and the mobility of electrons \({{\rm{\mu }}{\rm{n}}}\) are 10−16cm−3{10^{ - 16}}{\rm{c}}{{\rm{m}}^{ - 3}} and 1000cm2V−1s−1,1000{\rm{c}}{{\rm{m}}^2}{{\rm{V}}^{ - 1}}{{\rm{s}}^{ - 1}}, respectively. The average time (in μsec) taken by the electrons to move from one end of the bar to other end is____________

44

In a MOS capacitor with an oxide layer thickness of 10 nm, the maximum depletion layer thickness is 100nm. The permittivities of the semiconductor and the oxide layer are ϵs and ϵox respectively. Assuming, ϵxϵox=3\frac{{{ϵ_{\rm{x}}}}}{{{ϵ_{{\rm{ox}}}}}} = 3, the ratio of the maximum capacitance to the minimum capacitance of this MOS capacitor is_____

45

The energy band diagram and the electron density profile n(x) in a semiconductor are shown in the figures. Assume that n(x)=10e15(qaxkT)cm−3{\rm{n}}\left( {\rm{x}} \right) = 10_{\rm{e}}^{15}\left( {\frac{{{\rm{qax}}}}{{{\rm{kT}}}}} \right){\rm{c}}{{\rm{m}}^{ - 3}}, with α=0.1V/cm{\rm{\alpha }} = 0.1{\rm{V}}/{\rm{cm}} and x expressed in cm. Given kTq=0.026V,;Dn=36cm2s−1,;andDμ=kTq\frac{{{\rm{kT}}}}{{\rm{q}}} = 0.026{\rm{V}},{\rm{;}}{{\rm{D}}_{\rm{n}}} = 36{\rm{c}}{{\rm{m}}^2}{{\rm{s}}^{ - 1}},{\rm{;and}}\frac{{\rm{D}}}{{\rm{\mu }}} = \frac{{{\rm{kT}}}}{{\rm{q}}}. The electron current density (in A/cm2) at x = 0 is

  1. ((a))

    −4.4×10−2- 4.4 \times {10^{ - 2}}

  2. ((b))

    −2.2×10−2- 2.2 \times {10^{ - 2}}

  3. ((c))

    00

  4. ((d))

    2.2×10−22.2 \times {10^{ - 2}}

Show Answer
Answer: ((c))

00

Concept:

electric field intensity is given by  \({{\rm{E}}{\rm{x}}} = \frac{1}{{\rm{e}}}\frac{{{\rm{d}}{{\rm{E}}{\rm{c}}}}}{{{\rm{dx}}}}\)

Application:

From the figure above, the electric field intensity, \({{\rm{E}}{\rm{x}}} = \frac{1}{{\rm{e}}}\frac{{{\rm{d}}{{\rm{E}}{\rm{c}}}}}{{{\rm{dx}}}}\)

\(\begin{array}{l} {{\rm{E}}{\rm{x}}} = \frac{1}{{\rm{e}}}\frac{{{\rm{d}}{{\rm{E}}{\rm{c}}}}}{{{\rm{dx}}}}\ \Rightarrow {{\rm{E}}{\rm{x}}} = = \frac{1}{{\rm{e}}}{\rm{x}}\left( {0.1} \right){\rm{e}}\frac{{\rm{V}}}{{{\rm{cm}}}}\ \Rightarrow {{\rm{E}}{\rm{x}}} = - 0.1\frac{{\rm{V}}}{{{\rm{cm}}}} \end{array}\)

Also, dn(x)dx=1015exp⁡(qαxkT)(qαkT);\frac{{{\rm{dn}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}} = {10^{15}}\exp \left( {\frac{{{\rm{q\alpha x}}}}{{{\rm{kT}}}}} \right)\left( {\frac{{{\rm{q\alpha }}}}{{{\rm{kT}}}}} \right){\rm{;}}

At x = 0

dn(x)dx∣x=0=1015⋅(qαkT){\left. {\frac{{{\rm{dn}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}}} \right|_{{\rm{x}} = 0}} = {10^{15}} \cdot \left( {\frac{{{\rm{q\alpha }}}}{{{\rm{kT}}}}} \right)

Now, electron current density at x = 0 is

\(\begin{array}{l} {{\rm{J}}{\rm{n}}} = {\rm{e}} \cdot {\rm{n}} \cdot {{\rm{\mu }}{\rm{n}}}{{\rm{E}}{\rm{x}}} + {\rm{e}}{{\rm{D}}{\rm{n}}}\frac{{{\rm{dn}}\left( {\rm{x}} \right)}}{{{{\rm{d}}{\rm{x}}}}}\ \Rightarrow {{\rm{J}}{\rm{n}}} = {\rm{e}} \cdot {{\rm{D}}{\rm{n}}}\left( {{\rm{n}} \cdot \frac{{{{\rm{\mu }}{\rm{n}}}}}{{{{\rm{D}}{\rm{n}}}}}{{\rm{E}}{\rm{x}}} + \frac{{{{\rm{d}}{\rm{n}}}\left( {\rm{x}} \right)}}{{{{\rm{d}}{\rm{x}}}}}} \right)\ = {\rm{e}} \cdot {{\rm{D}}{\rm{n}}}\left( {{\rm{n}} \cdot \frac{{\rm{q}}}{{{\rm{KT}}}}{{\rm{E}}{\rm{x}}} + {{10}^{15}} \cdot \frac{{{\rm{q\alpha }}}}{{{\rm{KT}}}}} \right) \end{array}\)

n(x);at;x=0,;n=1015,{\rm{n}}\left( {\rm{x}} \right){\rm{;at;x}} = 0,{\rm{;n}} = {10^{15}}, thus,

\(\begin{array}{l} {{\rm{J}}{\rm{n}}} = {\rm{e}} \cdot {{\rm{D}}{\rm{n}}} \cdot {10^{15}} \cdot \frac{{\rm{q}}}{{{\rm{KT}}}}\left( {{{\rm{E}}{\rm{x}}} + {\rm{\alpha }}} \right)\ = \frac{{1.6 \times {{10}^{ - 19}} \times 36 \times {{10}^{15}} \times \left( { - 0.1 + 0.1} \right)}}{{0.026}}\ \Rightarrow {{\rm{J}}{\rm{n}}} = 0\frac{{\rm{A}}}{{{\rm{c}}{{\rm{m}}^2}}} \end{array}\)

46

A function of Boolean variables x,y;and;z{\rm{x}},{\rm{y;and;z}} is expressed in terms of the min-term as  F(x,y,z);=;Σ;(1,2,5,6,7){\rm{F}}\left( {{\rm{x}},{\rm{y}},{\rm{z}}} \right){\rm{;}} = {\rm{;\Sigma ;}}\left( {1,2,5,6,7} \right). Which one of the product of sums given below is equal to F(x,y,z)

  1. ((a))

    (xˉ+;yˉ+;zˉ)(xˉ+;y;+;z)(x;+;yˉ+;zˉ)\left( {{\rm{\bar x}} + {\rm{;\bar y}} + {\rm{;\bar z}}} \right)\left( {{\rm{\bar x}} + {\rm{;y;}} + {\rm{;z}}} \right)\left( {{\rm{x;}} + {\rm{;\bar y}} + {\rm{;\bar z}}} \right)

  2. ((b))

    (x;+;y;+;z)(x;+;yˉ+;zˉ)(xˉ+;y;+;z)\left( {{\rm{x;}} + {\rm{;y;}} + {\rm{;z}}} \right)\left( {{\rm{x;}} + {\rm{;\bar y}} + {\rm{;\bar z}}} \right)\left( {{\rm{\bar x}} + {\rm{;y;}} + {\rm{;z}}} \right)

  3. ((c))

    (xˉ+;yˉ+;z)(xˉ+;y;+;zˉ)(x;+;yˉ+;z)(x;+;y;+;z)\left( {{\rm{\bar x}} + {\rm{;\bar y}} + {\rm{;z}}} \right)\left( {{\rm{\bar x}} + {\rm{;y;}} + {\rm{;\bar z}}} \right)\left( {{\rm{x;}} + {\rm{;\bar y}} + {\rm{;z}}} \right)\left( {{\rm{x;}} + {\rm{;y;}} + {\rm{;z}}} \right)

  4. ((d))

    (x;+;y;+;zˉ)(xˉ+;y;+;z)(xˉ+;y;+;zˉ)(xˉ+;yˉ+;z)(xˉ+;yˉ+;zˉ)\left( {{\rm{x;}} + {\rm{;y;}} + {\rm{;\bar z}}} \right)\left( {{\rm{\bar x}} + {\rm{;y;}} + {\rm{;z}}} \right)\left( {{\rm{\bar x}} + {\rm{;y;}} + {\rm{;\bar z}}} \right)\left( {{\rm{\bar x}} + {\rm{;\bar y}} + {\rm{;z}}} \right)\left( {{\rm{\bar x}} + {\rm{;\bar y}} + {\rm{;\bar z}}} \right)

Show Answer
Answer: ((b))

(x;+;y;+;z)(x;+;yˉ+;zˉ)(xˉ+;y;+;z)\left( {{\rm{x;}} + {\rm{;y;}} + {\rm{;z}}} \right)\left( {{\rm{x;}} + {\rm{;\bar y}} + {\rm{;\bar z}}} \right)\left( {{\rm{\bar x}} + {\rm{;y;}} + {\rm{;z}}} \right)

Concept:

A Boolean expression consisting purely of Minterms (product terms) is said to be in the canonical sum of products form.

A Boolean expression consisting purely of Maxterms (sum terms) is said to be in the canonical product of sums form.

a) (A + B)(C + D) – POS form

b) (A)B(C + D) – POS form

c) AB + CD – SOP form

Calculation:

Given logic, expression is minterm expression.

F;(x,;y,;z);=Σm(1,2,5,6,7);{\rm{F;}}\left( {{\rm{x}},{\rm{;y}},{\rm{;z}}} \right){\rm{;}} = {\rm{\Sigma m}}\left( {1,2,5,6,7} \right){\rm{;}}

The max term expression will be formed by the terms which are not present in the min-term expression.

By converting the above min-term expression into max term expression,

F;(x,;y,;z);=Σm(1,2,5,6,7);=;πM;(0,3,4){\rm{F;}}\left( {{\rm{x}},{\rm{;y}},{\rm{;z}}} \right){\rm{;}} = {\rm{\Sigma m}}\left( {1,2,5,6,7} \right){\rm{;}} = {\rm{;\pi M;}}\left( {0,3,4} \right)

We get

F=;(x;+;y;+;z)(x;+;yˉ+;zˉ)(xˉ+;y;+;z)F= {\rm{;}}\left( {{\rm{x;}} + {\rm{;y;}} + {\rm{;z}}} \right)\left( {{\rm{x;}} + {\rm{;\bar y}} + {\rm{;\bar z}}} \right)\left( {{\rm{\bar x}} + {\rm{;y;}} + {\rm{;z}}} \right)

47

The figure shows a binary counter with synchronous clear input with the decoding logic shown, the counter works as a

  1. ((a))

    mod – 2 counter

  2. ((b))

    mod – 5 counter

  3. ((c))

    mod – 4 counter

  4. ((d))

    mod – 6 counter

Show Answer
Answer: ((b))

mod – 5 counter

Concept:

CLR‾\overline {CLR} : It is an active low signal. It is activated when CLR‾=0\overline {CLR} = 0 and it resets the FF.

CLR: It is an active high signal. It is activated when CLR = 1 and it Resets the FF.

Synchronous: Synchronous clear is synchronized with the clock. It waits for a clock pulse to Reset FF output.

Asynchronous: Asynchronous Clear is not synchronized with the clock. It does not wait for a clock pulse to Reset FF output.

Application:

From given sequential circuit, we can write:

CLR‾=Q3⊙Q2\overline {CLR} = {Q_3} \odot {Q_2}

∴ If both Q3 & Q2 are the same, Output = 1 and

If both Q3 & Q2 are not the same, then the output = 0

Now,

CLKQ3Q2Q1Q0CLR‾\overline {CLR}
-00001
100011
200101
300111
401000
50000
<br>

It is given that the clear input is synchronous. So, the output of the flip-flop will reset at the arrival of Clock 5. So, it will count from 0 to 4.

∴ It is a mode - 5 counter.

If it was the case of Asynchronous clear, the counter would have reset at the 4th clock. So it would have counted from 0 to 3.

∴ For an asynchronous clear, the counter will be a mode-4 counter.

48

A 1 to 8 demultiplexer with data input Din, address inputs S0, S1, and S2, (with S0  as the LSB) and Y̅0 to Y̅ 7 as the eight de-multiplexed output, is to be designed using two 2 to 4 decoders (with enable input E and address input A0 and A1) as shown in the figure. Din, S0, S1, and S2 are to be connected to P, Q, R, and S but not necessarily in this order. The respective input connections to P, Q, R, and S terminals should be

  1. ((a))

    S2, Din, S0,  and S1

  2. ((b))

    S1, Din, S0,  and S2

  3. ((c))

    Din, S0, S1, and S2

  4. ((d))

    Din, S2, S0, and S1

Show Answer
Answer: ((d))

Din, S2, S0, and S1

The given question is the expansion of

2 : 4 decoder to  1 : 8 decoder 

We need to implement 1 : 8 DEMUX.

So, select lines of De-Mux should be mapped to address lines of the decoder.

The LSB of De-Mux should be connected to the LSB of address lines of the decoder

∴ R → S0 

and S→ S1

Input to both the decoder should be same so

∴ P → Din

∴ NOT gate along with OR gate in case to select one decoder at a time so Q → S2 

So,

P → Din

Q → S2

R → S0 

S→ S1

Hence option (4) is correct

49

The diode in the circuit given below has VON = 0.7V but is ideal otherwise. The current (in mA) in the 4k Ω resistor is __________.

50

Assuming that the Op-Amp in the circuit shown below is ideal, the output voltage V0 (in volts) is ___________________

51

For the voltage regulator circuit shown, the input voltage (Vin) is 20 V ± 20% and the regulated output voltage (Vout) is 10 V. Assume the opamp to be ideal. For a load RL drawing 200 mA, the maximum power dissipation in Q1 (in Watts) is _________.

52

In the ac equivalent circuit shown, the two BJTs are biased in active region and have identical parameters with β ≫ 1 . The open circuit small signal voltage gain is approximately ________________________

53

Input x(t){\rm{x}}\left( {\rm{t}} \right) and output y(t){\rm{y}}\left( {\rm{t}} \right) of an LTI system are related by the differential equation y"(t)−y′(t)−6y(t);=;x(t){\rm{y}"}\left( {\rm{t}} \right) - {\rm{y}'}\left( {\rm{t}} \right) - 6{\rm{y}}\left( {\rm{t}} \right){\rm{;}} = {\rm{;x}}\left( {\rm{t}} \right). If the system is neither causal nor stable, the impulse response h(t) of the system is

  1. ((a))

    15e3tu(−t)+15e−2tu(−t)\frac{1}{5}{{\rm{e}}^{3{\rm{t}}}}{\rm{u}}\left( { - {\rm{t}}} \right) + \frac{1}{5}{{\rm{e}}^{ - 2{\rm{t}}}}{\rm{u}}\left( { - {\rm{t}}} \right)

  2. ((b))

    −15e3tu(−t)+15e−2tu(−t)- \frac{1}{5}{{\rm{e}}^{3{\rm{t}}}}{\rm{u}}\left( { - {\rm{t}}} \right) + \frac{1}{5}{{\rm{e}}^{ - 2{\rm{t}}}}{\rm{u}}\left( { - {\rm{t}}} \right)

  3. ((c))

    ;15e3tu(−t)−15e−2tu(t){\rm{;}}\frac{1}{5}{{\rm{e}}^{3{\rm{t}}}}{\rm{u}}\left( { - {\rm{t}}} \right) - \frac{1}{5}{{\rm{e}}^{ - 2{\rm{t}}}}{\rm{u}}\left( {\rm{t}} \right)

  4. ((d))

    −15e3tu(−t)−15e−2tu(t)- \frac{1}{5}{{\rm{e}}^{3{\rm{t}}}}{\rm{u}}\left( { - {\rm{t}}} \right) - \frac{1}{5}{{\rm{e}}^{ - 2{\rm{t}}}}{\rm{u}}\left( {\rm{t}} \right)

Show Answer
Answer: ((b))

−15e3tu(−t)+15e−2tu(−t)- \frac{1}{5}{{\rm{e}}^{3{\rm{t}}}}{\rm{u}}\left( { - {\rm{t}}} \right) + \frac{1}{5}{{\rm{e}}^{ - 2{\rm{t}}}}{\rm{u}}\left( { - {\rm{t}}} \right)

Concept:

L{y''(t)] = s2Y(s);−;sy(0);−;y′(0);{{\rm{s}}^2}{\rm{Y}}\left( {\rm{s}} \right){\rm{;}}-{\rm{;sy}}\left( 0 \right){\rm{;}}-{\rm{;y'}}\left( 0 \right){\rm{;}}

L{y'(t)} = ;[s;Y(s);−;y(0)];{\rm{;}}\left[ {{\rm{s;Y}}\left( {\rm{s}} \right){\rm{;}}-{\rm{;y}}\left( 0 \right)} \right]{\rm{;}}

Calculation:

The given differential equation is  y"(t);−;y′(t);−;6y(t);=;x(t){\rm{y}"}\left( {\rm{t}} \right){\rm{;}}-{\rm{;y}'}\left( {\rm{t}} \right){\rm{;}}-{\rm{;}}6{\rm{y}}\left( {\rm{t}} \right){\rm{;}} = {\rm{;x}}\left( {\rm{t}} \right)

On applying Laplace transform both sides,

s2Y(s);−;sy(0);−;y′(0);−;[s;Y(s);−;y(0)];−6Y(s);=;X(s){{\rm{s}}^2}{\rm{Y}}\left( {\rm{s}} \right){\rm{;}}-{\rm{;sy}}\left( 0 \right){\rm{;}}-{\rm{;y'}}\left( 0 \right){\rm{;}}-{\rm{;}}\left[ {{\rm{s;Y}}\left( {\rm{s}} \right){\rm{;}}-{\rm{;y}}\left( 0 \right)} \right]{\rm{;}} - 6{\rm{Y}}\left( {\rm{s}} \right){\rm{;}} = {\rm{;X}}\left( {\rm{s}} \right)

To calculate the transfer function all initial conditions are taken as ‘0’.

∴;(s2−−s−6);Y(s);=;X(s) H(s)=1(s2−s−6)=1(s−3)(s+2)=15[1s−3−1s+2]\begin{array}{l} \therefore {\rm{;}}\left( {{{\rm{s}}^2}--{\rm{s}} - 6} \right){\rm{;Y}}\left( {\rm{s}} \right){\rm{;}} = {\rm{;X}}\left( {\rm{s}} \right)\ {\rm{H}}\left( {\rm{s}} \right) = \frac{1}{{\left( {{{\rm{s}}^2} - {\rm{s}} - 6} \right)}} = \frac{1}{{\left( {{\rm{s}} - 3} \right)\left( {{\rm{s}} + 2} \right)}} = \frac{1}{5}\left[ {\frac{1}{{{\rm{s}} - 3}} - \frac{1}{{{\rm{s}} + 2}}} \right] \end{array}

It is given that h(t){\rm{h}}\left( {\rm{t}} \right) is non-causal and un-stable.

To satisfy both the conditions ROC should be left of the left most poles. Using the following standard pair

1s+a↔−e−atu(−t);σ<−a 1s−a↔−eatu(−t);σ<a H(s)=15[1s−3−1s+2] =15[−e3tu(−t)+e−2tu(−t)] =−15e3tu(−t)+15e−2tu(−t)\begin{array}{l} \frac{1}{{{\rm{s}} + {\rm{a}}}} \leftrightarrow - {{\rm{e}}^{ - {\rm{at}}}}{\rm{u}}\left( { - {\rm{t}}} \right);{\rm{\sigma }} < - {\rm{a}}\ \frac{1}{{{\rm{s}} - {\rm{a}}}} \leftrightarrow - {{\rm{e}}^{{\rm{at}}}}{\rm{u}}\left( { - {\rm{t}}} \right);{\rm{\sigma }} < {\rm{a}}\ {\rm{H}}\left( {\rm{s}} \right) = \frac{1}{5}\left[ {\frac{1}{{{\rm{s}} - 3}} - \frac{1}{{{\rm{s}} + 2}}} \right]\ = \frac{1}{5}\left[ { - {{\rm{e}}^{3{\rm{t}}}}{\rm{u}}\left( { - {\rm{t}}} \right) + {{\rm{e}}^{ - 2{\rm{t}}}}{\rm{u}}\left( { - {\rm{t}}} \right)} \right]\ = \frac{{ - 1}}{5}{{\rm{e}}^{ 3{\rm{t}}}}{\rm{u}}\left( { - {\rm{t}}} \right) + \frac{1}{5}{{\rm{e}}^{ - 2{\rm{t}}}}{\rm{u}}\left( { - {\rm{t}}} \right) \end{array}

54

Consider two real sequences with time–origin marked by the bold value x1[n] = {1, 2, 3, 0}, x2[n] = {1, 3, 2, 1}. Let X1(k) and X2(k) be 4-point DFTs of x1[n] and x2[n] respectively. Another sequence x3[n] is derived by taking 4-point inverse DFT of X3(k) = X1(k) X2(k). The value of  x3[2] is _________.

55

Let x(t);=αs(t);+;s(−t){\rm{x}}\left( {\rm{t}} \right){\rm{;}} = {\rm{\alpha s}}\left( {\rm{t}} \right){\rm{;}} + {\rm{;s}}\left( { - {\rm{t}}} \right) with s(t);=βe−4tu(t){\rm{s}}\left( {\rm{t}} \right){\rm{;}} = {\rm{\beta }}{{\rm{e}}^{ - 4{\rm{t}}}}{\rm{u}}\left( {\rm{t}} \right), where u(t) is unit step function. If the bilateral Laplace transform of x(t) is \({\rm{X}}\left( {\rm{s}} \right) = \frac{{16}}{{{{\rm{s}}^2} - 16}}; - 4 < {\rm{Re}}\left{ {\rm{s}} \right} < 4\) .Then the value of β is __________.

56

The state variable representation of a system is given as

\(\begin{array}{l} {\rm{\dot x}} = \left[ {\begin{array}{{20}{c}} 0&1\ 0&{ - 1} \end{array}} \right]{\rm{x}};{\rm{x}}\left( 0 \right) = \left[ {\begin{array}{{20}{c}} 1\ 0 \end{array}} \right]\ {\rm{y}} = \left[ {\begin{array}{*{20}{c}} 0&1 \end{array}} \right]{\rm{x}} \end{array}\)

The response y(t) is

  1. ((a))

    sin t

  2. ((b))

    1 - et

  3. ((c))

    1 - cos t

  4. ((d))

    0

Show Answer
Answer: ((d))

0

X˙=AX{\rm{\dot X}} = {\rm{AX}}, where \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} 0&1\ 0&{ - 1} \end{array}} \right]\) is state transition matrix.

Thus, we have

\(\begin{array}{l} {\rm{X}}\left( {\rm{s}} \right) = {\left( {{\rm{sI}} - {\rm{A}}} \right)^{ - 1}}{\rm{X}}\left( 0 \right)\ \Rightarrow {\rm{X}}\left( {\rm{s}} \right) = {\left[ {\begin{array}{{20}{c}} {\begin{array}{{20}{c}} {{\rm{s;;;;;}}}&{ - 1} \end{array}}\ {\begin{array}{{20}{c}} 0&{{\rm{s}} + 1} \end{array}} \end{array}} \right]^{ - 1}}\left[ {\begin{array}{{20}{c}} 1\ 0 \end{array}} \right]\ \Rightarrow {\rm{X}}\left( {\rm{s}} \right) = \frac{1}{{{\rm{s}}\left( {{\rm{s}} + 1} \right)}}\left[ {\begin{array}{{20}{c}} {\begin{array}{{20}{c}} {{\rm{s}} + 1}&1 \end{array}}\ {\begin{array}{{20}{c}} 0&{{\rm{;;;;;;;;s;}}} \end{array}} \end{array}} \right]\left[ {\begin{array}{{20}{c}} 1\ 0 \end{array}} \right]\ \Rightarrow {\rm{X}}\left( {\rm{s}} \right) = \frac{1}{{{\rm{s}}\left( {{\rm{s}} + 1} \right)}}\left[ {\begin{array}{{20}{c}} {s + 1}\ 0 \end{array}} \right]\ \Rightarrow {\rm{X}}\left( {\rm{s}} \right) = \left[ {\begin{array}{{20}{c}} {\frac{1}{{\rm{s}}}}\ 0 \end{array}} \right]\ \Rightarrow {\rm{x}}\left( {\rm{t}} \right) = \left[ {\begin{array}{*{20}{c}} 1\ 0 \end{array}} \right]{\rm{;}} \end{array}\)

Now, 

\({\rm{y}}\left( {\rm{t}} \right) = \left[ {\begin{array}{{20}{c}} 0&1 \end{array}} \right]{\rm{x}} = \left[ {\begin{array}{{20}{c}} 0&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1\ 0 \end{array}} \right] = 0\)

57

The output of a standard second-order system for a unit step input is given as

y(t)=1−23e−tcos⁡(3t−π6){\rm{y}}\left( {\rm{t}} \right) = 1 - \frac{2}{{\sqrt 3 }}{{\rm{e}}^{ - {\rm{t}}}}\cos \left( {\sqrt {3{\rm{t}}} - \frac{{\rm{\pi }}}{6}} \right)

The transfer function is the system is

  1. ((a))

    2(s+2)(s+3)\frac{2}{{\left( {{\rm{s}} + 2} \right)\left( {{\rm{s}} + \sqrt 3 } \right)}}

  2. ((b))

    1(s2+2s+1)\frac{1}{{\left( {{{\rm{s}}^2} + 2{\rm{s}} + 1} \right)}}

  3. ((c))

    3s2+2s+3\frac{3}{{{{\rm{s}}^2} + 2{\rm{s}} + 3}}

  4. ((d))

    4s2+2s+4\frac{4}{{{{\rm{s}}^2} + 2{\rm{s}} + 4}}

Show Answer
Answer: ((d))

4s2+2s+4\frac{4}{{{{\rm{s}}^2} + 2{\rm{s}} + 4}}

Given the output of the second-order system is 

y(t)=1−23e−tcos⁡(3t−π6){\rm{y}}\left( {\rm{t}} \right) = 1 - \frac{2}{{\sqrt 3 }}{{\rm{e}}^{ - {\rm{t}}}}\cos \left( {\sqrt {3{\rm{t}}} - \frac{{\rm{\pi }}}{6}} \right)

Compare this with the standard equation with unit step input

 \({\rm{c}}\left( {\rm{t}} \right) = 1 - \frac{{{{\rm{e}}^{ - {\rm{\zeta }}{{\rm{\omega }}{{\rm{nt}}}}}}}}{{\sqrt {1 - {{\rm{\zeta }}^2}} }}{\rm{sin}}\left( {{{\rm{\omega }}{\rm{d}}}{\rm{t}} + \phi } \right)\)

∴;1−ζ2=32\therefore {\rm{;}}\sqrt {1 - {{\rm{\zeta }}^2}} = \frac{{\sqrt 3 }}{2}

By squaring on both sides

⇒1−ζ2=34\Rightarrow 1 - {{\rm{\zeta }}^2} = \frac{3}{4}

⇒ζ=12=0.5\Rightarrow {\rm{\zeta }} = \frac{1}{2} = 0.5

⇒ζωn=1\Rightarrow {\rm{\zeta }}{{\rm{\omega }}_{\rm{n}}} = 1

⇒ωn=1ζ=2\Rightarrow {{\rm{\omega }}_{\rm{n}}} = \frac{1}{{\rm{\zeta }}} = 2

The transfer function of the standard second-order system is:

TF=C(s)R(s)=ωn2s2+2ζωns+ωn2TF = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{ω _n^2}}{{{s^2} + 2ζ {ω _n}s + ω _n^2}}

Where

ζ is the damping ratio = 0.5

ωn is the natural frequency = 2

TF=4s2+2s+4{\rm{TF}} = \frac{4}{{{{\rm{s}}^2} + 2{\rm{s}} + 4}}

58

The transfer function of a mass – spring – damper system is given by

G(s)=1Ms2+Bs+K{\rm{G}}\left( {\rm{s}} \right) = \frac{1}{{{\rm{M}}{{\rm{s}}^2} + {\rm{Bs}} + {\rm{K}}}}

The frequency response data for the system are given in the following table.

ω{\rm{\omega }} in rad/s∣G(jω)∣\left| {{\rm{G}}\left( {{\rm{j\omega }}} \right)} \right| in dBarg⁡(G(jω))\arg \left( {{\rm{G}}\left( {{\rm{j\omega }}} \right)} \right) in deg
0.01-18.5-0.2
0.1-18.5-1.3
0.2-18.4-2.6
1-16-16.9
2-11.4-89.4
3-21.5-151
5-32.8-167
10-45.3-174.5

The unit step response of the system approaches a steady state value of ________.

59

A zero mean white Gaussian noise having power spectral density N02\frac{{{{\rm{N}}_0}}}{2} is passed through an LTI filter whose impulse response h(t){\rm{h}}\left( {\rm{t}} \right) is shown in the figure. The variance of the filtered noise at t=4{\rm{t}} = 4 is

  1. ((a))

    32A2N0\frac{3}{2}{{\rm{A}}^2}{{\rm{N}}_0}

  2. ((b))

    34A2N0\frac{3}{4}{{\rm{A}}^2}{{\rm{N}}_0}

  3. ((c))

    A2N0{{\rm{A}}^2}{{\rm{N}}_0}

  4. ((d))

    12A2N0\frac{1}{2}{{\rm{A}}^2}{{\rm{N}}_0}

Show Answer
Answer: ((a))

32A2N0\frac{3}{2}{{\rm{A}}^2}{{\rm{N}}_0}

Concept:

Convolution of a signal x(t) with unit impulse δ(t) is the signal itself. i.e. x(t) ⊕ δ(t) = x(t)

Fourier transform of auto-correlation function of a power signal x(t) is power spectral density Sx(f). i.e. RX(τ)↔FTSX(f){R_X}\left( \tau \right)\mathop \leftrightarrow \limits^{FT} {S_X}\left( f \right)

And E(x2 (t)) = RX (0)

The variance of the signal x(t) is defined as:

var(x(t))=E(x2(t))−(E(x(t))2var\left( {x\left( t \right)} \right) = E\left( {{x^2}\left( t \right)} \right) - (E{\left( {x\left( t \right)} \right)^2}

Fourier transform of unit impulse is 1.

δ(t)↔FT1\delta \left( t \right)\mathop \leftrightarrow \limits^{FT} 1

Calculation:

Let n(t) be the input white noise with zero mean and N02\frac{{{N_0}}}{2} power spectral density.

Mean of the white noise = E(n(t)) = 0

Power spectral density is:

Sn(f)=N02{S_n}\left( f \right) = \frac{{{N_0}}}{2} ;

And the auto-correlation function is:

Rn(τ)↔FTSn(f){R_n}\left( \tau \right)\mathop \leftrightarrow \limits^{FT} {S_n}\left( f \right)

N02→IFTN02δ(t)\frac{{{N_0}}}{2}\mathop \to \limits^{IFT} \frac{{{N_0}}}{2}\delta \left( t \right)  

Rn(τ)=N02δ(t){R_n}\left( \tau \right) = \frac{{{N_0}}}{2}\delta \left( t \right)

Let yn(t) is the output noise.

Mean of the output noise:

\( = E\left( {{y_n}\left( t \right)} \right) = E\left( {n\left( t \right) \times \mathop \smallint \nolimits_{ - \infty }^\infty h\left( t \right)dt} \right)\) 

\( = E\left( {n\left( t \right)} \right) \times \mathop \smallint \nolimits_{ - \infty }^\infty h\left( t \right)dt\) 

\( = 0 \times \mathop \smallint \nolimits_{ - \infty }^\infty h\left( t \right)dt = 0\) 

The variance of the output noise is:

Var(yn(t))=E(yn2(t))−(E(yn(t))2Var\left( {{y_n}\left( t \right)} \right) = E\left( {y_n^2\left( t \right)} \right) - (E{\left( {{y_n}\left( t \right)} \right)^2} 

=E(yn2(t)) = E\left( {y_n^2\left( t \right)} \right) 

E(yn2(t))=Ryn(0)E\left( {y_n^2\left( t \right)} \right) = {R_{{y_n}}}\left( 0 \right) 

\({R_{{y_n}}}\left( \tau \right) = h\left( \tau \right){h^}\left( { - \tau } \right)*{R_n}\left( \tau \right)\) 

\( = \left( {\mathop \smallint \nolimits_{ - \infty }^\infty h\left( t \right).h\left( {t + \tau } \right)dt} \right)*\frac{{{N_0}}}{2}{\rm{\delta }}\left( {\rm{\tau }} \right)\) 

\( = \left( {\mathop \smallint \nolimits_{ - \infty }^\infty h\left( t \right).h\left( {t + \tau } \right)dt} \right) \times \frac{{{N_0}}}{2};\) 

\({R_{{y_n}}}\left( 0 \right) = \left( {\mathop \smallint \nolimits_{ - \infty }^\infty h\left( t \right).h\left( t \right)dt} \right)\frac{{{N_0}}}{2} = \left( {\mathop \smallint \nolimits_{ - \infty }^\infty {h^2}\left( t \right)dt} \right)\frac{{{N_0}}}{2} = 3{A^2} \times \frac{{{N_0}}}{2}\) 

Var(yn(t))=32A2.N0Var\left( {{y_n}\left( t \right)} \right) = \frac{3}{2}{A^2}.{N_0}

60

\(\left{ {{{\rm{X}}{\rm{n}}}} \right}{{\rm{n}} = - \infty }^{{\rm{n}} = \infty }\) is an independent and identically distributed (i.i.d.)  random process with Xn equally likely to be +1+ 1 or −1- 1. \(\left{ {{{\rm{Y}}{\rm{n}}}} \right}{{\rm{n}} = - \infty }^{{\rm{n}} = \infty }\) is another random process obtained as \({{\rm{Y}}{\rm{n}}} = {{\rm{X}}{\rm{n}}} + 0.5{{\rm{X}}{{\rm{n}} - 1}}{\rm{;}}\). The autocorrelation function of \(\left{ {{{\rm{Y}}{\rm{n}}}} \right}{{\rm{n}} = - \infty }^{{\rm{n}} = \infty }\), denoted by \({{\rm{R}}{\rm{y}}}\left[ {\rm{k}} \right],\) is_______

  1. ((a))

  2. ((b))

  3. ((c))

  4. ((d))

Show Answer
Answer: ((b))

Concept: For two signals x(n) and y(n)  auto-correlation is given by \({{\rm{R}}{\rm{x}}}\left[ {\rm{k}} \right] = {\rm{E}}\left[ {{\rm{x}}\left[ {\rm{n}} \right]{\rm{x}}\left[ {{\rm{n}} + {\rm{k}}} \right]} \right]\)  and  \({{\rm{R}}{\rm{y}}}\left[ {\rm{k}} \right] = {\rm{E}}\left[ {{\rm{y}}\left[ {\rm{n}} \right]{\rm{y}}\left[ {{\rm{n}} + {\rm{k}}} \right]} \right]\)respectively.

Application: Let \({{\rm{Y}}{\rm{n}}} = {\rm{y}}\left[ {\rm{n}} \right]{\rm{;and;}}{{\rm{X}}{\rm{n}}} = {\rm{x}}\left[ {\rm{n}} \right]\)

Then Ry[k]=E[y[n]y[n+k]]{{\rm{R}}_{\rm{y}}}\left[ {\rm{k}} \right] = {\rm{E}}\left[ {{\rm{y}}\left[ {\rm{n}} \right]{\rm{y}}\left[ {{\rm{n}} + {\rm{k}}} \right]} \right]

Using, y[n]=x[n]+0.5x[n−1]{\rm{y}}\left[ {\rm{n}} \right] = {\rm{x}}\left[ {\rm{n}} \right] + 0.5{\rm{x}}\left[ {{\rm{n}} - 1} \right] we have

\(\begin{array}{l} {{\rm{R}}{\rm{y}}}\left[ {\rm{k}} \right] = {\rm{E}}\left{ {\left[ {{\rm{x}}\left[ {\rm{n}} \right] + 0.5{\rm{x}}\left[ {{\rm{n}} - 1} \right]} \right]\left[ {{\rm{x}}\left[ {{\rm{n}} + {\rm{k}}} \right] + 0.5{\rm{x}}\left[ {{\rm{n}} + {\rm{k}} - 1} \right]} \right]} \right}\ = {\rm{E}}\left[ {{\rm{x}}\left[ {\rm{n}} \right]{\rm{x}}\left[ {{\rm{n}} + {\rm{k}}} \right] + 0.5{\rm{x}}\left[ {\rm{n}} \right]{\rm{x}}\left[ {{\rm{n}} + {\rm{k}} - 1} \right] + 0.5{\rm{x}}\left[ {{\rm{n}} - 1} \right]{\rm{x}}\left[ {{\rm{n}} + {\rm{k}}} \right] + 0.25{\rm{x}}\left[ {{\rm{n}} - 1} \right]{\rm{x}}\left[ {{\rm{n}} + {\rm{k}} - 1} \right]} \right]\ = {\rm{E}}\left[ {{\rm{x}}\left[ {\rm{n}} \right]{\rm{x}}\left[ {{\rm{n}} + {\rm{k}}} \right]} \right] + 0.5{\rm{E}}\left[ {{\rm{x}}\left[ {\rm{n}} \right]{\rm{x}}\left[ {{\rm{n}} + {\rm{k}} - 1} \right]} \right] + 0.5{\rm{E}}\left[ {{\rm{x}}\left[ {{\rm{n}} - 1} \right]{\rm{x}}\left[ {{\rm{n}} + {\rm{k}}} \right]} \right] + 0.25{\rm{E}}\left[ {{\rm{x}}\left[ {{\rm{n}} - 1} \right]{\rm{x}}\left[ {{\rm{n}} + {\rm{k}} - 1} \right]} \right]\ = {{\rm{R}}{\rm{x}}}\left[ {\rm{k}} \right] + 0.5{{\rm{R}}{\rm{x}}}\left[ {{\rm{k}} - 1} \right] + 0.5{{\rm{R}}{\rm{x}}}\left[ {{\rm{k}} + 1} \right] + 0.25{{\rm{R}}{\rm{x}}}\left[ {\rm{k}} \right]\ \Rightarrow {{\rm{R}}{\rm{y}}}\left[ {\rm{k}} \right] = 1.25{{\rm{R}}{\rm{x}}}\left[ {\rm{k}} \right] + 0.5{{\rm{R}}{\rm{x}}}\left[ {{\rm{k}} - 1} \right] + 0.5{{\rm{R}}_{\rm{x}}}\left[ {{\rm{k}} + 1} \right] \end{array}\)

Now, Rx[k]=E[x[n]x[n+k]]{{\rm{R}}_{\rm{x}}}\left[ {\rm{k}} \right] = {\rm{E}}\left[ {{\rm{x}}\left[ {\rm{n}} \right]{\rm{x}}\left[ {{\rm{n}} + {\rm{k}}} \right]} \right]

For, k=;0,;Rx[0]=E[x2[n]];{\rm{k}} = {\rm{;}}0,{\rm{;}}{{\rm{R}}_{\rm{x}}}\left[ 0 \right] = {\rm{E}}\left[ {{{\rm{x}}^2}\left[ {\rm{n}} \right]} \right]{\rm{;}}

We have only two possible values of x[n], i.e. −1- 1 and +1+ 1

x[n]{\rm{x}}\left[ {\rm{n}} \right]P(x[n]){\rm{P}}\left( {{\rm{x}}\left[ {\rm{n}} \right]} \right)x2[n]{{\rm{x}}^2}\left[ {\rm{n}} \right]
-1½1
+1½1
<br>

Thus, E[x2[n]]=1.12+1.12=1{\rm{E}}\left[ {{{\rm{x}}^2}\left[ {\rm{n}} \right]} \right] = 1.\frac{1}{2} + 1.\frac{1}{2} = 1

For k;≠;0{\rm{k;}} \ne {\rm{;}}0

x[n]{\rm{x}}\left[ {\rm{n}} \right]x[n+k]{\rm{x}}\left[ {{\rm{n}} + {\rm{k}}} \right]P(x[n]x[n+k]){\rm{P}}\left( {{\rm{x}}\left[ {\rm{n}} \right]{\rm{x}}\left[ {{\rm{n}} + {\rm{k}}} \right]} \right)x[n].x[n+k]{\rm{x}}\left[ {\rm{n}} \right].{\rm{x}}\left[ {{\rm{n}} + {\rm{k}}} \right]
1-1¼-1
11¼1
-1-1¼1
-11¼-1
<br>

Thus, E[x[n]x[n+k]]=−1.14+1.14+1.14−1.14=0{\rm{E}}\left[ {{\rm{x}}\left[ {\rm{n}} \right]{\rm{x}}\left[ {{\rm{n}} + {\rm{k}}} \right]} \right] = - 1.\frac{1}{4} + 1.\frac{1}{4} + 1.\frac{1}{4} - 1.\frac{1}{4} = 0

Thus, \({{\rm{R}}_{\rm{x}}}\left[ {\rm{k}} \right] = \left{ {\begin{array}{*{20}{c}} 1&{{\rm{for}}}&{{\rm{k}} = 0}\ 0&{{\rm{for}}}&{{\rm{k}} \ne 0} \end{array}} \right.\)

Hence \({{\rm{R}}{\rm{y}}}\left[ {\rm{k}} \right] = 1.25{{\rm{R}}{\rm{x}}}\left[ {\rm{k}} \right] + 0.5{\rm{;}}{{\rm{R}}{\rm{x}}}\left[ {{\rm{k}} - 1} \right] + 0.5{{\rm{R}}{\rm{x}}}\left[ {{\rm{k}} + 1} \right]\)

For, k;=;0{\rm{k;}} = {\rm{;}}0

\(\begin{array}{l} {{\rm{R}}{\rm{y}}}\left[ 0 \right] = 1.25{{\rm{R}}{\rm{x}}}\left[ 0 \right] + 0.5{{\rm{R}}{\rm{x}}}\left[ { - 1} \right] + 0.5{{\rm{R}}{\rm{x}}}\left[ 1 \right] = 1.25\ {{\rm{R}}{\rm{y}}}\left[ { - 1} \right] = 1.25{{\rm{R}}{\rm{x}}}\left[ { - 1} \right] + 0.5{{\rm{R}}{\rm{x}}}\left[ { - 2} \right] + 0.5{{\rm{R}}{\rm{x}}}\left[ 0 \right] = 0.5\ {{\rm{R}}{\rm{y}}}\left[ 1 \right] = 1.25{{\rm{R}}{\rm{x}}}\left[ 1 \right] + 0.5{{\rm{R}}{\rm{x}}}\left[ 0 \right] + 0.5{{\rm{R}}{\rm{x}}}\left[ 2 \right] = 0.5 \end{array}\)

All other terms will be zero. Thus Ry[k]{{\rm{R}}_{\rm{y}}}\left[ {\rm{k}} \right] is given by

61

Consider a binary, digital communication system which used pulses g(t){\rm{g}}\left( {\rm{t}} \right) and −g(t)- {\rm{g}}\left( {\rm{t}} \right) for transmitting bits over an AWGN channel. If the receiver uses a matched filter, which one of the following pulses will give the minimum probability of bit error?

  1. ((a))

  2. ((b))

  3. ((c))

  4. ((d))

Show Answer
Answer: ((a))

Concept:

Probability of error for PSK is given as:

Pe=Q(2EN0;){P_e} = Q\left( {\sqrt {\frac{{2E}}{{{N_0}}}} ;} \right) 

Complementary error function Q(x) is defined as:

\(Q\left( x \right) = \mathop \smallint \nolimits_x^\infty \frac{1}{{\sqrt {2\pi } }}{e^{ - \frac{{{t^2}}}{2}}}dt\) 

Q(x) decreases as x increases.

Calculation:

Since:

Pe=Q(2EN0;){P_e} = Q\left( {\sqrt {\frac{{2E}}{{{N_0}}}} ;} \right) 

Q(2EN0;)Q\left( {\sqrt {\frac{{2E}}{{{N_0}}}} ;} \right) decreases as 2EN0\sqrt {\frac{{2E}}{{{N_0}}}}  increases.

Therefore, to minimize the probability of error, maximize the energy of the signal E.

Energy of the signal for option A = EA = 1 × 1 = 1

Energy of the signal for option B:

\({E_B} = 2 \times \mathop \smallint \nolimits_0^{1/2;} {\left( {2t} \right)^2}dt = 2 \times \frac{{4{t^3}}}{3}|\begin{array}{*{20}{c}} {1/2}\ 0 \end{array} = \frac{8}{3} \times \frac{1}{8} = \frac{1}{3}\) 

The energy of the signal for option C:

\({E_C} = \mathop \smallint \nolimits_0^{1;} {\left( {1 - t} \right)^2}dt = \mathop \smallint \nolimits_0^{1;} \left( {1 - 2t + {t^2}} \right)dt\) 

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

EC=1−1+13{E_C} = 1 - 1 + \frac{1}{3} 

The energy of the signal for option D:

\({E_D} = \mathop \smallint \nolimits_0^{1;} {\left( t \right)^2}dt = \mathop \smallint \nolimits_0^{1;} {t^2}dt\) 

\( = \left( {\frac{{{t^3}}}{3}} \right)\begin{array}{*{20}{c}} 1\ 0 \end{array} = \frac{1}{3}\) 

Since EA > EB = EC = ED

Therefore, the probability of error will be minimum for option A.

62

Let X ∈ {0,1} and Y ∈ {0,1} be two independent binary random variables. If P(X = 0) = p and P(Y = 0) = q, then P(X + Y) ≥ 1 is equal to 

  1. ((a))

    pq + (1 - p)(1 - q)

  2. ((b))

    pq

  3. ((c))

    p(1 - q)

  4. ((d))

    1 - pq

Show Answer
Answer: ((d))

1 - pq

Concept:

1) P(X = 0) + P(X = 1) = 1

2) If Z = X + Y, then pdf (Z) = convolution of pdf (x) and pdf (y)

Application:

Given:

P(X = 0) = p

∴ P(X = 1) = 1 - p

Also Given: P(Y = 0) = q

∴ P (Y = 1) = 1 - q

P(X + Y ≥ 1) = P (X = 0) . P(Y = 1) + P(X = 1) . P(Y = 0) + P(X = 1) . P(Y = 1)

= p . (1 - q) + (1 - p) . q + (1 - p) . (1 - q) = 1 - pq

Alternate Method:

P(X + Y ≥ 1) = 1 - P(X + Y < 1) = 1 - P(X = 0) . P(Y = 0) = 1 - pq

63

An air – filled rectangular waveguide of internal dimensions a;cm×b;cm;(a>b){\rm{a;cm}} \times {\rm{b;cm;}}({\rm{a}} > {\rm{b}}) has a cutoff frequency of 6GHz6{\rm{GHz}} for the dominant \({\rm{T}}{{\rm{E}}{10}}\) mode. For the same waveguide, if the frequency of the \({\rm{T}}{{\rm{M}}{11}}\) mode is 15GHz15{\rm{GHz}}, the cutoff frequency of the \({\rm{T}}{{\rm{E}}{01}}\) mode in GHz  is________.

64

Two half – wave dipole antennas placed as shown in the figure are excited with sinusoidal varying current of frequency 3MHz3{\rm{MHz}} and phase shift of π/2 between them (the element at the origin leads in phase.) If the maximum radiated E – field at the point P in the x-y plane occurs at an azimuthal angle of 60°, the distance d{\rm{d}} (in meters) between the antennas is__________

65

The electric field of a plane wave propagation in a lossless non – magnetic medium is given by the following expression

\({\rm{E}}\left( {{\rm{z}},{\rm{t}}} \right) = {{\rm{\hat a}}{\rm{x}}}5\cos \left( {2{\rm{\pi }} \times {{10}^9}{\rm{t}} + {\rm{\beta z}}} \right) + {{\rm{\hat a}}{\rm{y}}}3\cos \left( {2{\rm{\pi }} \times {{10}^9}{\rm{t}} + {\rm{\beta z}} - \frac{{\rm{\pi }}}{2}} \right)\)

The type of the polarization is

  1. ((a))

    Right Hand Circular

  2. ((b))

    Left Hand Elliptical

  3. ((c))

    Right Hand Elliptical

  4. ((d))

    Linear

Show Answer
Answer: ((b))

Left Hand Elliptical

Given the electric field,

\({\rm{E}}\left( {{\rm{z}},{\rm{t}}} \right) = 5\cos \left( {2{\rm{\pi }} \times {{10}^9}{\rm{t}} + {\rm{\beta z}}} \right){{\rm{\hat a}}{\rm{x}}} + 3\cos \left( {2{\rm{\pi }} \times {{10}^9}{\rm{t}} + {\rm{\beta z}} - \frac{{\rm{\pi }}}{2}} \right){{\rm{\hat a}}{\rm{y}}}\) we see that \({{\rm{\hat a}}{\rm{x}}}\) and \({{\rm{\hat a}}{\rm{y}}}\) components are out of phase by 90° and have unequal magnitude. Thus, the wave is elliptically polarized.

The plot of electric phasor with time is shown below

Now, since the wave is traveling along –z axis. For increasing time, we point the thumb along –z axis, then left hand fingers curl in the direction of vector rotation. Thus, the wave is left hand elliptical.

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