Official Paper

GATE EC 2021 Official Paper (Previous Year Paper)

65 questions · 180 minutes · with answers · free

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General Aptitude (10 questions)

1

The current population of a city is 1,10,250. If it has been increasing at the rate of 5% per annum, what was its population 2 years ago?

  1. ((a))

    9,95,006

  2. ((b))

    9,92,500

  3. ((c))

    1,25,150

  4. ((d))

    1,00,000

Show Answer
Answer: ((d))

1,00,000

Given:

The present population of the city = 1,10,250

Increasing rate, (R) = 5% P.a.

Time, (n) = 2 years

Formula used:

Amount, (A) = P(1 + R/100)n

Here A is the present population of the city and P is the population of the city 2 years ago

Concept used:

This problem can be solved by the use of the compound interest formula

Calculation:

(A) = P(1 + R/100)n

⇒ 110250 = P(1 + 5/100)2

⇒ 110250 = P(105/100)2

⇒ 110250 = P(21/20)2

⇒ 110250 = P × (441/400)

⇒ P = (110250 × 400)/441

⇒ P = 100000

∴ The population of the city 2 years ago was 1,00,000

2

p and q are positive integers and pq+qp=3\frac{p}{q}+\frac{q}{p}=3

then p2q2+q2p2=\frac{p^2}{q^2} + \frac{q^2}{p^2} =

  1. ((a))

    7

  2. ((b))

    3

  3. ((c))

    9

  4. ((d))

    11

Show Answer
Answer: ((a))

7

Concept:

(a + b)2 = a2 + b2 + 2ab ---(1)

Calculation:

Given:

pq + qp=3\frac{p}{q} \ + \ \frac{q}{p}=3

Squaring both sides:

(pq + qp)2=9(\frac{p}{q} \ + \ \frac{q}{p})^2=9

Using the formula given in equation (1):

p2q2 + q2p2 + 2=9\frac{p^2}{q^2} \ + \ \frac{q^2}{p^2} \ + \ 2=9

p2q2 + q2p2 =7\frac{p^2}{q^2} \ + \ \frac{q^2}{p^2} \ =7

Hence option (1) is the correct answer.

3

The least number of squares that must be added so that the line P - Q becomes the line of symmetry is ____

  1. ((a))

    6

  2. ((b))

    4

  3. ((c))

    7

  4. ((d))

    3

Show Answer
Answer: ((a))

6

 Shows the number of squares added so that P-Q becomes a line of symmetry.

Hence, 6 squares added to the figure.

4

Nostalgia is to anticipation as _____ is to ____ 

Which one the following options maintains a similar logical relation in the above sentence?

  1. ((a))

    Past, future

  2. ((b))

    Future, present

  3. ((c))

    Future, past

  4. ((d))

    Present, past

Show Answer
Answer: ((a))

Past, future

The correct answer is Past, future.

Key Points

Nostalgia means a feeling of pleasure and also slight sadness when you think about things that happened in the past.

Anticipation means a feeling of excitement about something that is going to happen in the near future.

In short, nostalgia means thinking about the past and anticipation means thinking about the future.

Thus, Nostalgia is to anticipation as past is to future.

5

Consider the following sentences:

(i) I woke up from sleep

(ii) I woked up from sleep

(iii) I was woken up from sleep

(iv) I was wakened up from sleep

Which of the above sentences are grammatically CORRECT?

  1. ((a))

    (ii) and (iii)

  2. ((b))

    (i) and (ii)

  3. ((c))

    (i) and (iii)

  4. ((d))

    (i) and (iv)

Show Answer
Answer: ((c))

(i) and (iii)

The correct answer is (i) and (iii).

Key Points

Wake means to stop sleeping; to make somebody stop sleeping.

The three forms of the verb wake:

  • Present - Wake.
  • Past - Woke.
  • Past participle -  Woken/ waked.

 

​Thus, sentence (i) I woke up from sleep and sentence (iii) I was woken up from sleep is correct.

Hence, option (i) and (iii) follows.

6

Given below are two statements and ttwo conclusions.

Statement 1: All purple are green.

Statement 2: All black are green

Conclusion I: Some black are purple.

Conclusion II: No black is purple

Based on the above statements and conclusions, which one of  the following options is logically CORRECT?

  1. ((a))

    Both conclusion I and II are correct

  2. ((b))

    Either conclusion I or II is correct

  3. ((c))

    Only conclusion I is correct

  4. ((d))

    Only conclusion II is correct

Show Answer
Answer: ((b))

Either conclusion I or II is correct

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

Conclusion I: Some black are purple → False (As there is no direct relation between purple and back)

Conclusion II: No black is purple → False (As there is no direct relation between purple and back)

Both conclusions are complementary to each other.

Hence, Either conclusion I or II follows.

Important Points 

Either and or conditions:

  1. One conclusion is positive and one is negative.
  2. The elements are the same in both the conclusions.
  3. Individually both conclusions must be wrong.

Thus, Either I or II follows.

7

Computers are ubiquitous. They are used to improve efficiency in almost all fields from agriculture to space exploration. Artificial intelligence (AI) is currently a hot topic. AI enables computers to learn given enough training data. For humans, sitting in front of a computer for long hours can lead to health issues.

Which of the following can be deduced from the above passage?

(i) Nowadays, computers are present in almost all places.

(ii) Computers cannot be used for solving problems in engineering.

(iii) For humans, there are both positive and negative effects of using computers.

(iv) Artificial intelligence can be done without data.

  1. ((a))

    (i) and (iii)

  2. ((b))

    (ii) and (iii)

  3. ((c))

    (ii) and (iv)

  4. ((d))

    (i), (iii) and (iv)

Show Answer
Answer: ((a))

(i) and (iii)

The correct answer is (i) and (iii).

Key Points

Look at the lines:

Computers are ubiquitous. 

  • ​Ubiquitous means present, appearing or found everywhere. ​
  • From this line, it can be deduced that Nowadays, computers are present in almost all places.
  • ​They are used to improve efficiency in almost all fields from agriculture to space exploration. For humans, sitting in front of a computer for long hours can lead to health issues.

 

​Upon the perusal of the above lines, it can be concluded that computers help in increasing efficiency but can lead to health issues as well. Thus, For humans, there are both positive and negative effects of using computers.

Option (i) and (iii) can be found in the passage.

None of the other options follows.

8

Consider a square sheet of side 1 unit In the first step, it is cut along the main diagonal to get two triangles. In the next step, one of the cut triangles is revolved about its short edge to form a solid cone. The volume of resulting cone, in cubic units, is _____

  1. ((a))

    2π3\frac{{2\pi }}{3}

  2. ((b))

    π3\frac{{\pi }}{3}

  3. ((c))

    3π 

  4. ((d))

    3π2\frac{{3\pi }}{2}

Show Answer
Answer: ((b))

π3\frac{{\pi }}{3}

Concept:

If the side of a square is a unit then its diagonal length will be:

D=a2 + a2D=\sqrt {a^2 \ + \ a^2}

D = √2a units  ---(1)

The volume of a cone is given by:

V=πr2h3V=\frac{\pi r^2h}{3}  ---(2)

Where r =radius of the cone

h = height of the cone

Calculation:

When the square is cut along the main diagonal then the square will be divided into two equal triangles.

The length of the main diagonal can be calculated from equation (1)

D = √2 × 1 = √2  

Then one of the cut triangles is revolved about its short edge to form a solid cone:

According to figure

r = 1

h = 1

The volume of come can be calculated using equation (2):

V=π × 1 × 13V=\frac{\pi \ \times \ 1 \ \times \ 1}{3}

V=π3V=\frac{\pi}{3}

Hence option (2) is the correct answer.

9

The number of minutes spent by two students, X and Y, exercising every day in a given week is shown in the bar chart above.

The number of the day in the given week in which one of the students spent a minimum of 10% more than the other students, on a given day, is

  1. ((a))

    7

  2. ((b))

    6

  3. ((c))

    4

  4. ((d))

    5

Show Answer
Answer: ((b))

6

Given:

Minimum 10% more

Formula used:

Percentage = (Obtained value/Maximum value) × 100

Concept used:

Here the fraction part is always the smaller number because comparing number is the smaller number

Calculation: 

DayXYDifference% increase
Monday45702555.55%
Tuesday55651018.18%
Wednesday60501020%
Thursday605559.09%
Friday20351575%
Saturday60501020%
Sunday55651018.18%

 

∴ The number of the day in the given week in which one of the students spent a minimum of 10% more than the other students is 6 days

10

Corners are cut from an equilateral triangle to produce a regular convex hexagon as shown in the figure above.

The ratio of the area of the regular convex hexagon to the area of the original equilateral triangle is

  1. ((a))

    5 : 6

  2. ((b))

    2 : 3

  3. ((c))

    3 : 4

  4. ((d))

    4 : 5

Show Answer
Answer: ((b))

2 : 3

Formula used:

The area of the equilateral triangle = (√3/4) × a2

The area of the regular hexagon = (3√3/2) × b2

Here a → side of the equilateral triangle and b → Side of the regular hexagon

Concept used:

When the corner is cut from the equilateral triangle to produce the regular hexagon then the side of the triangle is divided into 3 equal parts and one part is the side of the hexagon.

Then, b = a/3

Calculation:

The area of the equilateral triangle = (√3/4) × a2

The area of the regular hexagon = (3√3/2) × b2

⇒ (3√3/2) × (a/3)2

⇒ (√3/6) × (a)2

The ratio of the regular hexagon to the equilateral triangle = {(√3/6) × (a)2}/{(√3/4) × a2}

⇒ 4/6 = 2 ∶ 3

∴ The required ratio is 2 ∶ 3

Electronics and Communication Engineering (55 questions)

11

The vector function F(r) = -x î + yĵ is defined over a circular are C shown in the figure.

The line integral of ∫C F(r) ⋅ dr is

  1. ((a))

    16\frac{1}{6}

  2. ((b))

    14\frac{1}{4}

  3. ((c))

    12\frac{1}{2}

  4. ((d))

    13\frac{1}{3}

Show Answer
Answer: ((c))

12\frac{1}{2}

Concept:

∫C F(r) dr can be solved by putting:

x = r cos θ

y = r sin θ

dx = -r sin θ dθ

dy = r cos θ dθ

Application:

Given r = 1

θ = 0 to 45°

The required integral can be written as:

\(\mathop \smallint \nolimits_C \vec F \cdot dr = \mathop \smallint \nolimits_0^{45^\circ } \left[ {\left( { - r\cos \theta } \right)\left( { - r\sin \theta } \right)d\theta + \left( {r\sin \theta } \right)\left( {r\cos \theta } \right)d\theta } \right]\)

\(\mathop \smallint \nolimits_0^{45^\circ } \left( {{r^2}\cos \theta \sin \theta + {r^2}\sin \theta \cos \theta } \right)d\theta \)

With r = 1, the above integral becomes:

\( = \frac{1}{2}\mathop \smallint \nolimits_0^{45^\circ } \left( {\sin 2\theta + \sin 2\theta } \right)d\theta \)

r = 1

\( = \mathop \smallint \nolimits_0^{45^\circ } \sin 2\theta ;d\theta \)

=(−cos⁡2θ)045∘2 = \frac{{\left( { - \cos 2\theta } \right)_0^{45^\circ }}}{2}

=−0−(−1)2=12 = \frac{{ - 0 - \left( { - 1} \right)}}{2} = \frac{1}{2}

12

Consider the differential equation given below:

dydx+x1−x2y=xy\dfrac{dy}{dx} + \dfrac{x}{1-x^2} y= x\sqrt{y}

The integrating factor of the differential equation is:

  1. ((a))

    (1 - x2)-1/4

  2. ((b))

    (1 - x2)-1/2

  3. ((c))

    (1 - x2​)-3/4

  4. ((d))

    (1 - x2​)-3/2

Show Answer
Answer: ((a))

(1 - x2)-1/4

Concept:

dydx+P(x)y=Q(x)\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right)

The solution of a linear differential equation of a general form shown above is:

y(I.F) = ∫Q(x) (IF) dx + C

Where:

IF = Integrating factor calculated as:

I.F=e∫PdxI.F = {e^{\smallint Pdx}}

Calculation:

Given:

dydx+x1−x2y=xy\frac{{dy}}{{dx}} + \frac{x}{{1 - {x^2}}}y = x\sqrt y

The above differential equation is not in a general form. Converting it first in the general form of a linear differential equation, we divide the equation by √y to get:

1ydydx+x1−x2;y=x\frac{1}{{\sqrt y }}\frac{{dy}}{{dx}} + \frac{x}{{1 - {x^2}}};\sqrt y = x

Let √y = u

[12y;;dydx]=dudx\left[ {\frac{1}{{2\sqrt y }};;\frac{{dy}}{{dx}}} \right] = \frac{{du}}{{dx}}

1y;dydx=2dudx\frac{1}{{\sqrt y }};\frac{{dy}}{{dx}} = 2\frac{{du}}{{dx}}

<br>

2dudx+x1−x2u=x2\frac{{du}}{{dx}} + \frac{x}{{1 - {x^2}}}u = x

dudx+12(x1−x2)u=x2\frac{{du}}{{dx}} + \frac{1}{2}\left( {\frac{x}{{1 - {x^2}}}} \right)u = \frac{x}{2}

dudx+P(x)u=Q(x)\frac{{du}}{{dx}} + P\left( x \right)u = Q\left( x \right)

IF=e∫PdxIF = {e^{\smallint Pdx}}

IF=e∫x2(1−x2)dxIF = {e^{\smallint \frac{x}{{2\left( {1 - {x^2}} \right)}}dx}}

Let 1 – x2 = t

-2xdx = dt

xdt=−dt2xdt = \frac{{ - dt}}{2}

IF=e∫−1t;dt4IF = {e^{\smallint - \frac{1}{t};\frac{{dt}}{4}}}

=e−14ln⁡t = {e^{\frac{{ - 1}}{4}\ln t}}

=eln⁡t−14 = {e^{\ln {t^{\frac{{ - 1}}{4}}}}}

=t−14 = {t^{\frac{{ - 1}}{4}}}

IF=(1−x2)−14IF = {\left( {1 - {x^2}} \right)^{\frac{{ - 1}}{4}}}

13

Two continuous random variables X and Y are related as

Y = 2X + 3

Let σX2\sigma_X^2 and σY2\sigma_Y^2 denote the variances of X and Y, respectively. The variances are related as

  1. ((a))

    σY2=4σX2\sigma_Y^2 = 4 \sigma_X^2

  2. ((b))

    σY2=2σX2\sigma_Y^2 = 2 \sigma_X^2

  3. ((c))

    σY2=25σX2\sigma_Y^2 = 25 \sigma_X^2

  4. ((d))

    σY2=5σX2\sigma_Y^2 = 5 \sigma_X^2

Show Answer
Answer: ((a))

σY2=4σX2\sigma_Y^2 = 4 \sigma_X^2

Concept:

Variance of a random variable ‘y’ is given by:

Var[y] = E[y2] – E2[y]

Properties of mean:

  1. E[K] = K, Where K is some constant

  2. E[c X] = c. E[X], Where c is some constant

  3. E[a X + b] = a E[X] + b, Where a and b are constants

  4. E[X + Y] = E[X] + E[Y]

Application:

Variance of y = E[(2x + 3)2] – (E[2x + 3])2

= E[4x2 + 12x + 9] – (E[2x + 3])2

= 4E[x2] + 12E[x] + 9 – (E[2x] + E[3])2

= 4E[x2] + 12E[x] + 9 – (2E[x] + 3)2

= 4E[x2] + 12E[x] + 9 – (4E2[x] + 9 + 12E[X])

= 4E[x2] + 12E[x] + 9 – 4E2[x] – 9 – 12E[X]

= 4E[x2] – 4E2[x]

= 4[E[x2] – E2[x]]

This can be written as:

= 4 (variance of x), i.e.

The variance of y = 4 times the variance of x

σy2=4;σx2\sigma _y^2 = 4;\sigma _x^2

Properties of Variance:

  1. V[K] = 0, Where K is some constant.

  2. V[cX] = c2 V[X]

  3. V[aX + b] = a2 V[X]

  4. V[aX + bY] = a2 V[X] + b2 V[Y] + 2ab Cov(X,Y)

Cov.(X,Y) = E[XY] - E[X].E[Y]

14

Consider a real-value based-band signal x(t), band limited to 10 kHz. The Nyquist rate for the signal y(t) = x(t) x(1+t2)x(1+ \frac{t}{2}) is

  1. ((a))

    60 kHz

  2. ((b))

    30 kHz

  3. ((c))

    15 kHz

  4. ((d))

    20 kHz

Show Answer
Answer: ((b))

30 kHz

Concept:

Consider x(t) → f(kHz) signal

Then:

x(t2);→;;is;;;f2(kHz)x\left( {\frac{t}{2}} \right);\mathop \to \limits^{;;is;;} ;\frac{f}{2}\left( {kHz} \right) signal

x(t2+1)→f2x\left( {\frac{t}{2} + 1} \right) \to \frac{f}{2} (kHz) signal; as shifting in the time domain does not affect the frequency spectrum.

x(t)x(t2+1)→(f+f2)kHzx\left( t \right)x\left( {\frac{t}{2} + 1} \right) \to \left( {f + \frac{f}{2}} \right)kHz

=3f2kHz = \frac{{3f}}{2}kHz signal

Since multiplication in one domain is the convolution in another domain, and in convolution, due to the time-invariant system, the duration of the signal gets added.

Hence the frequency of both signals gets added in the frequency domain.

More clearly, we can understand from the following specturem diagram:

Application:

y(t)=x(t)⋅x(t2+1);;↔;;F.T;;;y(f)=x(f)∗x1(f)y\left( t \right) = x\left( t \right) \cdot x\left( {\frac{t}{2} + 1} \right);;\mathop \leftrightarrow \limits^{;;F.T;;} ;y\left( f \right) = x\left( f \right)*{x_1}\left( f \right)

Given x(t) = 10 kHz signal

Then, X(t2+1)⇒5;kHzX\left( {\frac{t}{2} + 1} \right) \Rightarrow 5;kHz signal

y(t)=x(t)⋅x(t2+1)y\left( t \right) = x\left( t \right) \cdot x\left( {\frac{t}{2} + 1} \right)

∴ y(t) will be (10 + 5) = 15 kHz signal

Nyquist rate of y(t) = 2 [frequency of y(t)]

= 2 × 15 kHz

= 30 kHz

15

Consider two 16-point sequences x[n] and h[n]. Let the linear convolution of x[n] and h[n] be denoted by y[n], while z[n] denotes the 16-point inverse discrete Fourier transform (IDFT) of the product of the 16-point DFTs of x[n] and h[n]. The value(s) of k for which z[k] = y[k] is/are

  1. ((a))

    k = 0, 1, 2, ..., 15

  2. ((b))

    k = 0 and k = 15

  3. ((c))

    k = 15

  4. ((d))

    k = 0

Show Answer
Answer: ((c))

k = 15

Concept:

Linear convolution = x[n] * h[n]

y[n] = x[n] * h[n]

Y[k] = X[k] H[k]

Also, the function which is circularly convoled can be written as:

z[n] = IDFT [x(k) y(k)]

i.e. z[n] = x[n] ⊗ h(n)

⊗ → circular convolution

We have 16 point sequence x(n) and h(n) but for the case of understanding and solving, we will

consider 4-point sequence as:

x(n) = {1, 1, 2, 2}

h(n) = {1, 2, 3, 4}

Linear convolution:

y(n) = {1, 3, 7, 13, 16, 14, 8}

Circular convolution:

z[n] = x[n] h(n)

\(\left[ {\begin{array}{{20}{c}} 1&2&2&1\ 1&1&2&2\ 2&1&1&2\ 2&2&1&1 \end{array}} \right]\left[ {\begin{array}{{20}{c}} 1\ 2\ 3\ 4 \end{array}} \right]\)

z[n] = [15, 17, 15, 13]

z[3] = y[3]

Clearly, if N point sequence is given for x(n) and h(n), then:

\({\left[ {x\left( n \right)*h\left( n \right)} \right]{at;n = N - 1}} = {\left[ {x\left( n \right)h\left[ n \right]} \right]{at;n = N - 1}}\)

Application:

The given sequence is a 16 point sequence.

Hence for N = 16 – 1 = 15:

Z[k] = y [k]

16

A bar of silicon is doped with a boron concentration of 1016 cm-3 and assumed to be fully ionized. It is exposed to light such that electron-hole pairs are generated throughout the volume of the bar at the rate of 1020 cm-3s-1. If the recombination lifetime is 100 μs, the intrinsic carrier concentration of silicon is 1010 cm-3 and assuming 100% ionization of boron, then the approximate product of steady-state electron and hole concentrations due to this light exposure is

  1. ((a))

    1032 cm-6

  2. ((b))

    1020 cm-6

  3. ((c))

    2 × 1032 cm-6

  4. ((d))

    2 × 1020 cm-6

Show Answer
Answer: ((c))

2 × 1032 cm-6

Concept:

The number of minority carriers generated because of light is given by the generation rate as:

G=excess;minority;carrier;generatedminority;carrier;life;timeG = \frac{{excess;minority;carrier;generated}}{{minority;carrier;life;time}}

G=Δpτp=ΔnτnG= \frac{{{\rm{\Delta }}p}}{{\tau_p}} = \frac{{{\rm{\Delta }}n}}{{\tau_n}}

Analysis:

NA = 1016 / cm3

G = 1020 / cm3 – sec

τ = 100 μsec

ni = 1010 / cm3

before shining light:

Hole concertration (p) ≃ NA = 1016 / cm3

Electron concentration (n) =ni2p=10201016=104/cm3= \frac{{n{i^2}}}{p} = \frac{{{{10}^{20}}}}{{{{10}^{16}}}} = {10^4}/c{m^3} 

After illumination of light minority carrier will be generated.

So after light illumination hole conc. (p’) = p + Δp

p’ = 1016 + 1020 × 10-6 × 100

= 2 × 1016 / cm3

After illumination e- conc. (n’) = n + Δn

= 104 + 1020 × 10-6 × 100

≃ 1016 / cm3

Product of e- and hole concertration = n’ × p’

= 2 × 1032 / cm-6

17

The energy band diagram of a p-type semiconductor bar of length L under equilibrium condition (i.e., the Fermi energy level EF is constant) is shown in the figure. The valance band EV is sloped since doping is non-uniform along the bar. The difference between the energy levels of the valence band at the two edges of the bar is Δ.

 

If the charge of an electron is q, then the magnitude of the electric field developed inside this semiconductor bar is

  1. ((a))

    Δ2qL\dfrac{\Delta}{2qL}

  2. ((b))

    ΔqL\dfrac{\Delta}{qL}

  3. ((c))

    3Δ2qL\dfrac{3\Delta}{2qL}

  4. ((d))

    2ΔqL\dfrac{2\Delta}{qL}

Show Answer
Answer: ((b))

ΔqL\dfrac{\Delta}{qL}

Concept:

Electric field

It is defined as the ratio of the applied voltage to the total length of the semiconductor specimen.

E=VLE = \frac{V}{L}

Calculation:

Electric field developed will be calculated as:

E=−∂V∂xE = \frac{{ - \partial V}}{{\partial x}}

V=work;donecharge;transferred=dWdqV = \frac{{work;done}}{{charge;transferred}}=\frac{dW}{dq}

Work done is calculated in terms of energy used. From the energy diagram given the total energy is Δ and the charge is 'q'

V=ΔqV=\frac{\Delta}{q}

The electric field generated is:

∣E∣=VL=ΔqL\left| E \right| =\frac{V}{L} = \frac{{\bf{\Delta }}}{{qL}}

18

In a circuit shown in the figure, the transistors M1 and M2 are operating in saturation. The channel length modulation coefficients of both the transistors are non-zero. The transconductance of the MOSFETs M1 and M2 are gm1 and gm2, respectively, and the internal resistance of the MOSFETs M1 and M2 are r01 and r02, respectively.

Ignoring the body effect, the ac small-signal voltage gain (∂Vout / ∂Vin) of the circuit is:

  1. ((a))

    −gm2(1gm1∣∣r02)- {g_{m2}}\left( {\frac{1}{{{g_{m1}}}}||{r_{02}}} \right)

  2. ((b))

    −gm2(1gm1∣∣r01∣∣r02)- {g_{m2}}\left( {\frac{1}{{{g_{m1}}}}||{r_{01}}||{r_{02}}} \right)

  3. ((c))

    −gm1(1gm2∣∣r01∣∣r02)- {g_{m1}}\left( {\frac{1}{{{g_{m2}}}}||{r_{01}}||{r_{02}}} \right)

  4. ((d))

    −gm2(r01∣∣r02)- {g_{m2}}\left( {r_{01}||{r_{02}}} \right)

Show Answer
Answer: ((b))

−gm2(1gm1∣∣r01∣∣r02)- {g_{m2}}\left( {\frac{1}{{{g_{m1}}}}||{r_{01}}||{r_{02}}} \right)

Concept:

Whenever gate and drain are shorted of MOSFET, we can replace directly MOSFET by resistance which is equal to

Analysis:

Since the gate and drain of MOSFET 1 are short-circuited.

It will work like a resistor.

→ drawing AC - equivalent model of MOSFET - 2

Vgs2=Vin{V_{gs_2}} = {V_{in}}

\({V_0} = - {g_{{m_2}}};{V_{g{s_2}}}\left{ {{r_{{0_2}}}||{r_{{0_1}}}||\frac{1}{{{g_{{m_1}}}}}} \right}\)

\(\frac{{{V_0}}}{{{V_{in}}}} = - {g_{{m_2}}}\left{ {{r_{{0_2}}}||{r_{{0_1}}}||\frac{1}{{{g_{{m_1}}}}}} \right}\)

Important Points

When gate and source of MOSFET are short-circuited:

19

For the circuit with an ideal OPAMP shown in the figure. VREF is fixed.

If VOUT = 1 for VIN = 0.1 volt and VOUT = 6 volt for VIN = 1 volt, where VOUT is measured across RL connected at the output of this OPAMP, the value of RF / RIN is

  1. ((a))

    3.285

  2. ((b))

    5.555

  3. ((c))

    2.860

  4. ((d))

    3.825

Show Answer
Answer: ((b))

5.555

Concept:

When VIN is active and VREF = 0, the circuit is drawn as:

VOUT1=−RFRINVIN{V_{OUT1}} = - \frac{{{R_F}}}{{{R_{IN}}}}{V_{IN}}  

When VIN = 0 and VREF is active, the circuit is drawn as:

VOUT2=(VREF×R2R1+R2)(1+RFRIN){V_{OU{T_2}}} = \left( {\frac{{{V_{REF}} \times {R_2}}}{{{R_1} + {R_2}}}} \right)\left( {1 + \frac{{{R_F}}}{{{R_{IN}}}}} \right)

Total Vout = Vout1 + Vout2

Vout=(VREF×R2R1+R2)(1+RFRIN)−RERIN;VIN{V_{out}} = \left( {\frac{{{V_{REF}} \times {R_2}}}{{{R_1} + {R_2}}}} \right)\left( {1 + \frac{{{R_F}}}{{{R_{IN}}}}} \right) - \frac{{{R_E}}}{{{R_{IN}}}};{V_{IN}}

Given VREF is fixed then V+ is fixed, i.e.

V+=VREF×R2R1+R2{V^ + } = \frac{{{V_{REF}} \times {R_2}}}{{{R_1} + {R_2}}}

VOUT=(1+RFRIN)(VREF;R2R1+R2)−RFRINVIN{V_{OUT}} = \left( {1 + \frac{{{R_F}}}{{{R_{IN}}}}} \right)\left( {\frac{{{V_{REF}};{R_2}}}{{{R_1} + {R_2}}}} \right) - \frac{{{R_F}}}{{{R_{IN}}}}{V_{IN}}

For VIN = 0.1 V, we have VOUT = 1 V

1=(1+RFRIN)(VREF;R2R1+R2)−RFRIN(0.1)1 = \left( {1 + \frac{{{R_F}}}{{{R_{IN}}}}} \right)\left( {\frac{{{V_{REF}};{R_2}}}{{{R_1} + {R_2}}}} \right) - \frac{{{R_F}}}{{{R_{IN}}}}\left( {0.1} \right)        ---(i)

For VIN = 10 V, we have VOUT = 6 V

6=(1+RFRIN)(VREF;R2R1+R2)−RFRIN(1)6 = \left( {1 + \frac{{{R_F}}}{{{R_{IN}}}}} \right)\left( {\frac{{{V_{REF}};{R_2}}}{{{R_1} + {R_2}}}} \right) - \frac{{{R_F}}}{{{R_{IN}}}}\left( 1 \right)      ---(ii)

From (i)

(1+RFRIN)(VREF⋅R2R1+R2)=1+RFRIN(0.1)\left( {1 + \frac{{{R_F}}}{{{R_{IN}}}}} \right)\left( {\frac{{{V_{REF}} \cdot {R_2}}}{{{R_1} + {R_2}}}} \right) = 1 + \frac{{{R_F}}}{{{R_{IN}}}}\left( {0.1} \right)

Putting above value in eq (ii)

6=1+RFRIN(0.1)−RFRIN(1)6 = 1 + \frac{{{R_F}}}{{{R_{IN}}}}\left( {0.1} \right) - \frac{{{R_F}}}{{{R_{IN}}}}\left( 1 \right)

5=RFRIN(−0.9)5 = \frac{{{R_F}}}{{{R_{IN}}}}\left( { - 0.9} \right)

RFRIN=−50.9=−5.55\frac{{{R_F}}}{{{R_{IN}}}} = - \frac{5}{{0.9}} = - 5.55

Since only positive values are given in the options:

RFRIN=5.55\frac{{{R_F}}}{{{R_{IN}}}} = 5.55

20

Consider the circuit with and ideal OPAMP shown in the figure.

Assuming |VIN| << |VCC| and |VREF| << |VCC|, the condition at which VOUT equals to zero is

  1. ((a))

    VIN = 0.5 VREF

  2. ((b))

    VIN = VREF

  3. ((c))

    VIN = 2VREF

  4. ((d))

    VIN = 2 + VREF

Show Answer
Answer: ((b))

VIN = VREF

Concept:

In the below circuit  V1 is connected to the ground, so V1 = 0.

Thus V2 also will be at ground potential.

Note:

  1. This concept is valid only when negative feedback is applied to op-amp like in inverting amplifiers

2.When op-amp gain has a low value then also this concept is not valid.

Analysis:

Using virtual ground concept:

VA = VB = 0 → ground

Applying KVL at node A:

Vi−0R+−VRef−0R=0−V0RF\frac{{{V_i} - 0}}{R} + \frac{{ - {V_{Ref}} - 0}}{R} = \frac{{0 - {V_0}}}{{{R_F}}}

V0 = 0

VIN = VREF

21

If (1235)x = (3033)y, where x and y indicate the bases of the corresponding numbers, then

  1. ((a))

    x = 9 and y = 7

  2. ((b))

    x = 8 and y = 6

  3. ((c))

    x = 7 and y = 5

  4. ((d))

    x = 6 and y = 4

Show Answer
Answer: ((b))

x = 8 and y = 6

(1235)x = (3033)y

Converting both the RHS and LHS into its decimal equivalent, we get:

(x3 + 2x2 + 3x + 5)10 = (3y3 + 0y2 + 3y + 3)10

Substituting the values given in the options in above equation, we get:

x = 8 and y = 6

Proof:

Putting x = 8 in x3 + 2x2 + 3x + 5

= (8)3 + 2(8)2 + 3(8) + 5

= (669)10

Putting y = 6 in 3y3 + 0y2 + 3y + 3

= 3(6)3 + 3(6) + 3

= 3(216) + 18 + 3

= (669)10

22

Addressing of a 32K × 16 memory is realized using a single decoder. The minimum number of AND gates required for the decoder is

  1. ((a))

    215

  2. ((b))

    219

  3. ((c))

    232

  4. ((d))

    28

Show Answer
Answer: ((a))

215

Concept of memory:

Any memory size is given by = 2K × m

K = address line

m = data line

Eg: 1 KB memory = 210 × 8

Concept of decoder:

For n × 2n decoder no. of AND gates required are 2n.

Eg: 2:4 decoder,  4 AND gates are required.

i / po / p
ENABY3Y2Y1Y0
0000000
1000001
1010010
1100100
1111000
<br>

Analysis:

Given

For this memory 15 × 215 decoders required

Since n = 15

So no. of AND gate are required = 215

23

The block diagram of a feedback control system is shown in the figure.

The transfer function Y(s)X(s)\frac{Y(s)}{X(s)} of the system is

  1. ((a))

    G1+G21+G1H\frac{{{G_1} + {G_2}}}{{1 + {G_1}H}}

  2. ((b))

    G1+G2+G1G2H1+G1H\frac{{{G_1} + {G_2} + {G_1}{G_2}H}}{{1 + {G_1}H}}

  3. ((c))

    G1+G21+G1H+G2H\frac{{{G_1} + {G_2} }}{{1 + {G_1}H}+{G_2}H}

  4. ((d))

    G1+G2+G1G2H1+G1H+G2H\frac{{{G_1} + {G_2} + {G_1}{G_2}H}}{{1 + {G_1}H + {G_2}H}}

Show Answer
Answer: ((a))

G1+G21+G1H\frac{{{G_1} + {G_2}}}{{1 + {G_1}H}}

Concept:

Mason's Gain Formula is used to evaluate an overall transmittance (gain), which can be expressed as,

T=∑PkΔkΔT = \frac{{\sum {P_k}{{\rm{\Delta }}_k}}}{{\rm{\Delta }}}

Where

Pk = forward path transmittance of kth path

Δ = graph determinant comprising closed-loop transmittances & mutual interactions between non-touching loops.

ΔK = path factor consisting of all isolated closed loops from the forward path in the graph.

Analysis:

No of forward path = 2

Forward paths:- G2, G1

No of Loops = 1

Loop: -G1H

Using mason gain formula:

Y(s)X(s)=G1+G21+G1H\frac{{Y\left( s \right)}}{{X\left( s \right)}} = \frac{G_1+ G_2}{{1 + G_1 H }}

Note:-

Loop:- G1H is touching the forward path G2, hence option 2 is incorrect.

24

The complete Nyquist plot of the open-loop transfer function G(s) H(s) of a feedback control system is shown in the figure.

If G(s) H(s) has one zero in the right-half of the s-plane. the number of poles that the closed-loop system will have in the right-half of the s-plane is

  1. ((a))

    3

  2. ((b))

    1

  3. ((c))

    0

  4. ((d))

    2

Show Answer
Answer: ((a))

3

We will observe the encirclement around the origin: (for open loop transfer function → G(s)H(s)= 0 )

N = P - Z

where,

P = the no. of open-loop poles lie in the right half of plane

Z = the no. of closed-loop poles or open-loop zeros lie in the right half of plane

N = no. of encirclement about (-1 + j0) in G(S) H(S) plane

OR,

no. of encirclement about the origin in 1 + G(S) H(S) plane.

Given;

Z = 1 (for open loop system)

N = 2 (at the origin in ACW direction)

then

2 = P - 1

∴ P = 3 

For the analysis of closed loop system:

observe the encirclement around the (-1 , j0) [∵ 1 + GH = 0 → GH = -1]

N = 0

P = 3

Then,

N = P - Z

Z= P - N

∴ Z= 3 - 0 = 3

There are 3 poles in the right half-plane.

25

Consider a rectangular coordinate system (x, y, z) with unit vectors ax, ay, and az. A plane wave traveling in the region z ≥ 0 with electric field vector E = 10 cos (2 × 108 t + βz) ay is incident normally on the plane at z = 0, where β is the phase constant. The region z ≥ 0 is in free space and the region z < 0 is filled with a lossless medium (permittivity ε = ε0, permeability μ = 4μ0, where ε0 = 8.85 × 10-12 F/m  and μ0 = 4π × 10-7 H/m). The value of the reflection coefficient is

  1. ((a))

    23\dfrac{2}{3}

  2. ((b))

    35\dfrac{3}{5}

  3. ((c))

    13\dfrac{1}{3}

  4. ((d))

    25\dfrac{2}{5}

Show Answer
Answer: ((c))

13\dfrac{1}{3}

Concept:

Normal Incidence is as shown:

The Reflection coefficient is defined as:

Γ=ErEi{\rm{\Gamma }} = \frac{{{E_r}}}{{{E_i}}}

Γ=η2−η1η2+η1{\rm{\Gamma }} = \frac{{{\eta _2} - {\eta _1}}}{{{\eta _2} + {\eta _1}}}

Where:

η1 = intrinsic impedance of medium 1

η2 = intrinsic impedance of medium 2

Application:

Given medium 1 = free space

ϵ1 = ϵ0 and μ1 = μ0

Medium 2 = lossless medium

ϵ2 = ϵ0 and μ2 = 4μ0

Reflection coefficient will be:

η=μϵ\eta = \sqrt {\frac{\mu }{\epsilon}}

Γ=η2−η1η2+η1{\rm{\Gamma }} = \frac{{{\eta _2} - {\eta _1}}}{{{\eta _2} + {\eta _1}}}

  Γ=μ2ϵ2−μ1ϵ1μ2ϵ2;+μ1ϵ1{\rm{\Gamma }} = \frac{{\sqrt {\frac{{{\mu _2}}}{{{\epsilon _2}}}} - \sqrt {\frac{{{\mu _1}}}{{{\epsilon _1}}}} }}{{\sqrt {\frac{{{\mu _2}}}{{{\epsilon _2}}}} ; + \sqrt {\frac{{{\mu _1}}}{{{\epsilon _1}}}} }}

Γ=4μ0ϵ0−μ0ϵ04μ0ϵ0;+μ0ϵ0{\rm{\Gamma }} = \frac{{\sqrt {\frac{{4{\mu _0}}}{{{\epsilon_0}}}} - \sqrt {\frac{{{\mu _0}}}{{{\epsilon_0}}}} }}{{\sqrt {\frac{{4{\mu _0}}}{{{\epsilon_0}}}} ; + \sqrt {\frac{{{\mu _0}}}{{{\epsilon_0}}}} }}

Γ=2−12+1{\rm{\Gamma }} = \frac{{2 - 1}}{{2 + 1}}

Γ=13{\rm{\Gamma }} = \frac{1}{3}

Extra Information: The Transmission coefficient is given by:

τ = 1 + Γ

26

If the vectors (1.0, -1.0, 2.0), (7.0, 3.0, x) and (2.0, 3.0, 1.0) in R3 are linearly dependent, the value of x is ______

27

Consider the vector field F = ax (4y - c1z) + ay (4x + 2z) +az (2y +z) in rectangular coordinated system (x, y, z) with unit vectors ax, ay, and az, If the filed F is irrotational (conservative), then the constant c1 (in integer) is _____

28

Consider the circuit shown in the figure.

The current I flowing through the 7 Ω resistor between P and Q (rounded off to one decimal place) is _______ A.

29

Consider the circuit shown in the figure.

The value of v0 (rounded off to one decimal place) is ______ V.

30

An 8-bit unipolar (all analog output values are positive) digital-to-analog converter (DAC) has a full-scale voltage range from 0 V to 7.68 V. If the digital input code is 10010110 (the leftmost bit is MSB), then the analog output voltage of DAC (rounded off to one decimal place) is _______ V.

31

The auto correlation function RX(τ) of a wide-sense stationary random process X(t) is shown in the figure.

The average power of X(t) is _______

32

Consider a carrier signal which is amplitude modulated by a single-tone sinusoidal message signal with a modulation index of 50%. If the carrier and one of the sidebands are suppressed in the modulated signal, the percentage of power saved (rounded off to one decimal place) is _______.

33

A speech signal, band-limited to 4 kHz, is sampled at 1.25 times the Nyquist rate. The speech samples assumed to be statistically independent and uniformly distributed in the range -5 V to +5 V, are subsequently quantized in an 8-bit uniform quantizer and then transmitted over a voice-grade AWGN telephone channel. If the ratio of the transmitted signal power to channel noise power is 26 dB, the minimum channel bandwidth required to ensure reliable transmission of the signal with an arbitrarily small probability of transmission error (rounded off to two decimal places) is _______ kHz.

34

A 4 kHz sinusoidal message signal having amplitude 4 V is fed to a delta modulator (DM) operating at a sampling rate of 32 kHz. The minimum step size required to avoid slope overload noise in the DM (rounded off to two decimal places) is _______ V.

35

The refractive indices of the core and cladding of an optical fiber are 1.50 and 1.48, respectively. The critical propagation angle, which is defined as the maximum angle that the light beam makes with the axis of the optical fiber to achieve the total internal reflection, (rounded off to two decimal places) is ______ degree.

36

Consider the integral ∮Csin⁡(x)x2(x2+4)dx\displaystyle\oint_C \dfrac{\sin (x)}{x^2 (x^2 + 4)}dx

Where C is a counter-clockwise oriented circle defined as |x - i| = 2. The value of the integral is

  1. ((a))

    π4sin⁡(2i)\dfrac{\pi}{4} \sin (2i)

  2. ((b))

    πi2−π8sin⁡(2i)\frac{\pi i}{2}-\dfrac{\pi}{8} \sin (2i)

  3. ((c))

    π8sin⁡(2i)\dfrac{\pi}{8} \sin (2i)

  4. ((d))

    −π4sin⁡(2i)-\dfrac{\pi}{4} \sin (2i)

Show Answer
Answer: ((b))

πi2−π8sin⁡(2i)\frac{\pi i}{2}-\dfrac{\pi}{8} \sin (2i)

Concept:

Residue Theorem:

If f(z) is analytic inside and on a closed curve ‘C’ except at a finite number of singularities inside ‘C’

Then:

∮f(z) dz = 2πi (sum of residues)

Singularity: A point where the function f(z) fails to be analytic

Analysis:

Given f(x)=sin⁡(x)x2(x2+4)f\left( x \right) = \frac{{\sin \left( x \right)}}{{{x^2}\left( {{x^2} + 4} \right)}} 

f(x)=sin⁡(x)x2(x+2i)(x−2i)f\left( x \right) = \frac{{\sin \left( x \right)}}{{{x^2}\left( {x + 2i} \right)\left( {x - 2i} \right)}} 

x = 0 pole of order 2

x = 2i, -2i

Given curve or region is |x - i| = 2

x = -2i lies outside |x - i| = 2

So the singular points are x = 0 & x = 2i

Residue at x = 0

lim⁡x→0ddx[sin⁡x(x2+4)]\mathop {\lim }\limits_{x \to 0} \frac{d}{{dx}}\left[ {\frac{{\sin x}}{{\left( {{x^2} + 4} \right)}}} \right]

=lim⁡x→0[cos⁡x(x2+y)−sin⁡x(2x)(x2+4)2] = \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\cos x\left( {{x^2} + y} \right) - \sin x\left( {2x} \right)}}{{{{\left( {{x^2} + 4} \right)}^2}}}} \right]

=lim⁡x→0cos⁡xx2+y=14=0.25 = \mathop {\lim }\limits_{x \to 0} \frac{{\cos x}}{{{x^2} + y}} = \frac{1}{4} = 0.25

Residue at x = 2i

⇒lim⁡x→2i(x−2i)sin⁡xx2(x+2i)(x−2i) \Rightarrow \mathop {\lim }\limits_{x \to 2i} \left( {x - 2i} \right)\frac{{\sin x}}{{{x^2}\left( {x + 2i} \right)\left( {x - 2i} \right)}}

⇒sin⁡(2i)(2i)2(4i) \Rightarrow \frac{{\sin \left( {2i} \right)}}{{{{\left( {2i} \right)}^2}\left( {4i} \right)}}

⇒sin⁡(2i)−16;i \Rightarrow \frac{{\sin \left( {2i} \right)}}{{ - 16;i}}

 ∮Cf(x)dx=∮Csin⁡(x)x2(x2+4);dx=2πi\mathop \oint \nolimits_C f\left( x \right)dx = \mathop \oint \nolimits_C \frac{{\sin \left( x \right)}}{{{x^2}\left( {{x^2} + 4} \right)}};dx = 2\pi i (sum of residues)

=2πi[14−sin⁡(2i)16i] = 2\pi i\left[ {\frac{1}{4} - \frac{{\sin \left( {2i} \right)}}{{16i}}} \right]

=πi2−πsin⁡(2i)8 = \frac{{\pi i}}{2} - \frac{{\pi \sin \left( {2i} \right)}}{8}

37

A box contains the following three coins.

I. A fair coin with head on one face and tail on the other face.

II. A coin with heads on both the faces.

III. A coin with tails on both the faces.

A coin is picked randomly from the box and tossed. Out of the two remaining coins in the box, one coin is then picked randomly and tossed. If the first toss results in a head, the probability of getting a head in the second toss is

  1. ((a))

    25\frac{2}{5}

  2. ((b))

    13\frac{1}{3}

  3. ((c))

    23\frac{2}{3}

  4. ((d))

    12\frac{1}{2}

Show Answer
Answer: ((b))

13\frac{1}{3}

Concept:

Conditional probability is defined as:

P(AB)=P(A∩B)P(B)P\left( {\frac{A}{B}} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}

P(BA)=P(A∩B)P(A)P\left( {\frac{B}{A}} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}

Application:

Let event A is defined as:

A = Getting head in the first toss

Event B is defined as:

B = Getting head in the second toss

According to the question, we need to find the probability of getting a head in the second toss when

Already a head has occurred in the first toss, i.e.So, the probability of getting head in the first toss will be:

P(BA)=P(A∩B)P(A)P\left( {\frac{B}{A}} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}

P(A) = P(fair coin is selected) × P(Getting head) + P(double-headed coin is selected) × P(getting head)

=13×12+13×1=16+13=12 = \frac{1}{3} \times \frac{1}{2} + \frac{1}{3} \times 1 = \frac{1}{6} + \frac{1}{3} = \frac{1}{2}

P(A)=12P\left( A \right) = \frac{1}{2}

Now, finding the probability of getting head in both tosses will be: 

\(P\left( {A \cap B} \right) = \mathop {\mathop {\underbrace {\begin{array}{{20}{c}} {First;toss}\ {\left( {\frac{1}{3} \cdot \frac{1}{2}} \right)} \end{array};;\begin{array}{{20}{c}} {second;toss}\ {\left( {\frac{1}{2} \cdot 1} \right)} \end{array}}{\begin{array}{*{20}{c}} {when;first;fair}\ {coin;is;tossed} \end{array}};;}\limits{} }\limits_; + \underbrace {\begin{array}{{20}{c}} {First;toss}\ {\left( {\frac{1}{3} \cdot 1} \right)} \end{array};;\begin{array}{{20}{c}} {second;toss}\ {\left( {\frac{1}{2} \cdot \frac{1}{2}} \right)} \end{array}}_{\begin{array}{*{20}{c}} {When;first;double}\ {headed;coin;is;tossed} \end{array}}\)

P(A∩B)=112+112=16P\left( {A \cap B} \right) = \frac{1}{{12}} + \frac{1}{{12}} = \frac{1}{6}

P(BA)=P(A∩B)P(A)P\left( {\frac{B}{A}} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}

P(BA)=1612=13P\left( {\frac{B}{A}} \right) = \frac{{\frac{1}{6}}}{{\frac{1}{2}}} = \frac{1}{3}

38

The switch in the circuit in the figure is in position P for long time and then moved to position Q at time t = 0.

The value of dv(t)dt\frac{{dv\left( t \right)}}{{dt}} at t = 0+ is

  1. ((a))

    -5 V / s

  2. ((b))

    3 V / s

  3. ((c))

    -3 V / s

  4. ((d))

    0 V / s

Show Answer
Answer: ((c))

-3 V / s

Concept:

Under steady-state:

  • When a capacitor, is present with a D.C supply, it behaves as an open circuit.
  • When an inductor is present with D.C supply, it behaves as a short circuit
<br>

At t = 0+ :

Inductor replace with current source IL (0+) = IL (0-)

The inductor does NOT allow the sudden change in current”

The capacitor is replaced with the voltage source

Vc (0+) = Vc (0-)

The capacitor does NOT allow the sudden change in voltage.

Calculation:

At t = 0- → switch is at P position network is in steady-state

Capacitor → open circuit

Inductor → S.C.

iL(0-) = 20/20 = 1 mA

iL(0+) = iL(0-) = 1 mA

Vc (0-) = 10 × 1 = 10 V

Vc (0+) = Vc (0-) = 10 V

At t = 0+ → switch is at Q position.

Applying KCL at node A

10/5  + ic(0+) + 1 = 0

ic (0+) = -3 mA

CdVc(t)dt∣t=0+=−3mAC{\left. {\frac{{d{V_c}\left( t \right)}}{{dt}}} \right|_{t = {0^ + }}} = - 3mA

∵ ic=CdVcdt{i_c} = C\frac{{d{V_c}}}{{dt}}

dVc(t)dt∣t=0+=−3V/sec{\left. {\frac{{d{V_c}\left( t \right)}}{{dt}}} \right|_{t = {0^ + }}} = - 3V/sec

39

Consider the two-port network shown in the figure.

The admittance parameters, in siemens, are

  1. ((a))

    y11 = 2, y12 = -4, y21 = -1, y22 = 2

  2. ((b))

    y11 = 1, y12 = -2, y21 = -1, y22 = 3

  3. ((c))

    y11 = 2, y12 = -4, y21 = -4, y22 = 2

  4. ((d))

    y11 = 2, y12 = -4, y21 = -4, y22 = 3

Show Answer
Answer: ((a))

y11 = 2, y12 = -4, y21 = -1, y22 = 2

Concept:

Y parameters

These are also called the admittance parameters.

The Y parameters for the two-port network are shown as:

\(\left[ {\begin{array}{{20}{c}} {{I_1}}\ {{I_2}} \end{array}} \right] = \left[ {\begin{array}{{20}{c}} {{Y_{11}}}&{{Y_{12}}}\ {{Y_{21}}}&{{Y_{22}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_1}}\ {{V_2}} \end{array}} \right]\)

I1 = Y11V1 + Y12V2

I2 = Y21V1 + Y22V2

Calculation:

Consider the circuit with nodes specified as V1 and V2

Writing the node equation at the node1

−I1+V11−3V2+V1−V21=0 - {I_1} + \frac{{{V_1}}}{1} - 3{V_2} + \frac{{{V_1} - {V_2}}}{1} = 0

I1 = 2V1 – 4V2  ---- (i)

Writing the node equation at the node2

−I2+V2−V11+V21=0 - {I_2} + \frac{{{V_2} - {V_1}}}{1} + \frac{{{V_2}}}{1} = 0

I2 = - V1 + 2V2 ---- (ii)

From the equations (i) and (ii) the Y parameters for the given network are:

\(Y = \left[ {\begin{array}{*{20}{c}} 2&{ - 4}\ { - 1}&2 \end{array}} \right]\)

40

For an n-channel silicon MOSFET with 10 nm gate oxide thickness, the substrate sensitivity (∂VT/∂|VBS|) is found to be 50 mV/V at a substance voltage |VBS| = 2 V, where VT is the threshold voltage of the MOSFET. Assume that, |VBS| >> 2ϕB, where qϕB is the separation between the Fermi energy level EF and the intrinsic level Ei in the bulk. Parameters given are

Electron charge (q) = 1.6 × 10-19 C

Vacuum permittivity (ε0) = 8.85 × 10-12 F/m

Relative permittivity of silicon (εSi) = 12

Relative permittivity of oxide (εox) = 4

The doping concentration of the substrate is

  1. ((a))

    7.37 × 1015 cm-3

  2. ((b))

    9.37 × 1015 cm-3

  3. ((c))

    2.37 × 1015 cm-3

  4. ((d))

    4.37 × 1015 cm-3

Show Answer
Answer: ((a))

7.37 × 1015 cm-3

Concept:

The threshold value for the transistor is given by:

Vt=Vt0+γ[√VSB+∣2ϕF∣−√∣2ϕF∣]{V_t} = {V_{t0}} + γ \left[ {√ {{V_{SB}} + \left| {2{\phi _F}} \right|} - √ {\left| {2{\phi _F}} \right|} } \right]

VBS for the n channel transistor

Calculation:

Given that substrate, sensitivity is 50 mV/V

∂Vt∂∣VBS∣=γ2√VBS+2∣ϕF∣\frac{{\partial {V_t}}}{{\partial \left| {{V_{BS}}} \right|}} = \frac{γ }{{2√ {{V_{BS}} + 2\left| {{\phi _F}} \right|} }}

50mVV=γ2√2(∵∣VBS∣≫2ϕB)\frac{{50mV}}{V} = \frac{γ }{{2√ 2 }}\left( {\because\left| {{V_{BS}}} \right| \gg 2{\phi _B}} \right)

Consider ϕB = ϕF 

γ = 0.1 √2

γ is defined as:

\(γ = \frac{{√ {2q{N_A}{ \in {Si}}} }}{{{C{ox}}}}\)

\({C_{ox}} = \frac{{{ \in {ox}}}}{{{t{ox}}}} = \frac{{4{ \in _0}}}{{10nm}}\)

\(0.1√ 2 = \frac{{√ {2q{N_A}{ \in {Si}}} }}{{{C{ox}}}}\)

Squaring on both sides, we get:

0.01×16∈02100×10−18=1.6×10−19NA×12∈00.01 × \frac{{16 \in _0^2}}{{100 × {{10}^{ - 18}}}} = 1.6 × {10^{ - 19}}{N_A} × 12{ \in _0}

NA=16×8.85×1021.6×10−19×12{N_A} = \frac{{16 × 8.85 × {{10}^2}}}{{1.6 × {{10}^{ - 19}} × 12}}

NA=8.85×102212;m3{N_A} = \frac{{8.85 × {{10}^{22}}}}{{12;{m^3}}}

NA = 0.7375 × 1022 m-3

NA = 7.375 × 1015 cm-3

Hence option (1) is correct

41

The propagation delays of the XOR gate, AND gate, and multiplexer (MUX) in the circuit shown in the figure are 4 ns, 2 ns, and 1 ns, respectively.

If all the inputs P, Q, R, S, and T are applied simultaneously and held constant the maximum propagation delay of the circuit is

  1. ((a))

    6 ns

  2. ((b))

    7  ns

  3. ((c))

    3 ns

  4. ((d))

    5 ns

Show Answer
Answer: ((a))

6 ns

In the given question both MUX select line is ‘T’ input.

So MUX output depends only on T.

For input ‘T’ there is only two value possible T = 0 or T = 1

At T = 0 for MUX 1 → XOR gate o/p will propagate to next stage.

For MUX 2 → AND gate 1 (A1), o/p will propagate which is available to MUX 2 at t = 2 nsec

So for Y o/p only, AND 1 (A1) + MUX (2) is necessary.

Propagation delay

= 2 + 1

= 3 nsec

Case 2: T = 1

When T = 1 then for o/p Y: 

AND gate (A3) + MUX (1 )+ AND gate (A2) + MUX (2)

Propagation delay (tpd) = 2 + 1 + 2 + 1

= 6 nsec

So maximum propagation delay of the circuit is 6 nsec.

42

The content of the registers are R1 = 25H, R2 = 30H and R3 = 40H. The following machine instructions are executed.

PUSH{R1}

PUSH{R2}

PUSH{R3}

POP{R1}

POP{R2}

POP{R3}

After execution, the content of registers R1, R2, R3 are

  1. ((a))

    R1 = 40H, R2 = 30H and R3 = 25H

  2. ((b))

    R1 = 40H, R2 = 25H and R3 = 30H

  3. ((c))

    R1 = 30H, R2 = 40H and R3 = 25H

  4. ((d))

    R1 = 25H, R2 = 30H and R3 = 40H

Show Answer
Answer: ((a))

R1 = 40H, R2 = 30H and R3 = 25H

Concept:

PUSH: Store or push the content of the register pair into stock memory. Content is always pushed on top of the stock.

POP: Read or occurs the data from the top of the stock memory into register pair.

Content is always popped from the top of the stack.

Application:

R1 = 25 H, R2 = 30 H, R3 = 40 H

Push R1:

 

Push R2:

Push R3:

POP R1:

POP R2:

POP R3:

 

Therefore

R1 = 40 H

R2 = 30 H

R3 = 25 H

43

The electrical system shown in the figure converts input source current is(t) to output voltage v0(t).

Current iL(t) in the inductor and voltage vc(t) acros the capacitor are taken as the state variables, both assumed to be initially equal to zero, i.e., iL(0) = 0 and vc(0) = 0. Then system is

  1. ((a))

    neither state controllable nor observable

  2. ((b))

    completely state controllable as well as completely observable

  3. ((c))

    completely observable but not state controllable

  4. ((d))

    completely state controllable but not observable

Show Answer
Answer: ((a))

neither state controllable nor observable

Concept:

For controllability and observability following condition must be satisfied:

Controllability:

Qc = [B  AB]

If |Qc| = 0 then system is UNCONTROLLABLE

If |Qc| ≠ 0 then the system is CONTROLLABLE

Observability:

\({Q_0} = \left[ {\begin{array}{*{20}{c}} C\ {CA} \end{array}} \right]\)

If |Q0| = 0 then system is UNOBSERVABLE

If |Q0| ≠ 0 then the system is OBSERVABLE

Application:

Let us first develop the state model of the given circuit:

Let VL(t) be the voltage across the inductor and VR(t) be the voltage across resistor 1 Ω.

VR(t)=VL(t)=LdiL(t)dt{V_R}\left( t \right) = {V_L}\left( t \right) = L\frac{{d{i_L}\left( t \right)}}{{dt}}

With L =  1 H:

VL(t)=diL(t)dt{V_L}\left( t \right) = \frac{{d{i_L}\left( t \right)}}{{dt}}

is(t) = iL(l) + iR(t)

iR(t)=VR(t)1;Ω=VL(t){i_R}\left( t \right) = \frac{{{V_R}\left( t \right)}}{{1;{\rm{\Omega }}}} = {V_L}\left( t \right)

is(t)=iL(t)+diL(t)dt{i_s}\left( t \right) = {i_L}\left( t \right) + \frac{{d{i_L}\left( t \right)}}{{dt}}

diL(t)dt=−i1(t)+is(t)\frac{{d{i_L}\left( t \right)}}{{dt}} = - {i_1}\left( t \right) + {i_s}\left( t \right)    ---(1)

Let ic(t) be the current through the capacitor, then:

ic(t)=CdVc(t)dt{i_c}\left( t \right) = C\frac{{d{V_c}\left( t \right)}}{{dt}}

C = 1 F

ic(t)=dVc(t)dt{i_c}\left( t \right) = \frac{{d{V_c}\left( t \right)}}{{dt}}

And ic(t)=is(t)−V0(t)1;Ω;{i_c}\left( t \right) = {i_s}\left( t \right) - \frac{{{V_0}\left( t \right)}}{{1;{\rm{\Omega }}}};

And V0(t) = Vc(t)      ---(2)

∴;dVc(t)dt=is(t)−Vc(t)\therefore ;\frac{{d{V_c}\left( t \right)}}{{dt}} = {i_s}\left( t \right) - {V_c}\left( t \right)

dVc(t)dt=−Vc(t)+is(t)\frac{{d{V_c}\left( t \right)}}{{dt}} = - {V_c}\left( t \right) + {i_s}\left( t \right)     ---(3)

From (1) and (3) equation:

diL(t)dt=−iL(t)+0.Vc(t)+is(t)\frac{{d{i_L}\left( t \right)}}{{dt}} = - {i_L}\left( t \right) + 0.{V_c}\left( t \right) + {i_s}\left( t \right)

dVc(t)dt=0;iL(t)−VC(t)+is(t)\frac{{d{V_c}\left( t \right)}}{{dt}} = 0;{i_L}\left( t \right) - {V_C}\left( t \right) + {i_s}\left( t \right)

\(\left[ {\begin{array}{{20}{c}} {{i_L}}\ {{{\dot v}_c}} \end{array}} \right] = \left[ {\begin{array}{{20}{c}} { - 1}&0\ 0&{ - 1} \end{array}} \right]\left[ {\begin{array}{{20}{c}} {{i_L}}\ {{V_c}} \end{array}} \right] + \left[ {\begin{array}{{20}{c}} 1\ 1 \end{array}} \right]{i_s}\left( t \right)\)

\({V_0} = \left[ {0;;1} \right]\left[ {\begin{array}{*{20}{c}} {{i_L}}\ {{V_C}} \end{array}} \right]\)

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

C = [0  1]

Now, the Controllability matrix will be:

\({Q_c} = \left[ {B;;AB} \right] = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}\ 1&{ - 1} \end{array}} \right]\)

|Qc| = 0

Hence the system is not controllable.

Observability matrix:

\({Q_0} = \left[ {\begin{array}{{20}{c}} C\ {CA} \end{array}} \right] = \left[ {\begin{array}{{20}{c}} 0&1\ 0&{ - 1} \end{array}} \right]\)

|Q0| = 0

Hence the system is Not observable.

∴ The system is neither controllable nor observable.

44

A digital transmission system uses a (7, 4) systematic linear Hamming code for transmitting data over a noisy channel. If three of the message-codeword pairs in this code (mi ; ci), where ci is the codeword corresponding to the ith message mi, are known to be (1 1 0 0 ; 0 1 0 1 1 0 0), (1 1 1 0 ; 0 0 1 1 1 1 0) and (0 1 1 0 ; 1 0 0 0 1 1 0), then which of the following is a valid codeword in this code?

  1. ((a))

    1 0 1 1 0 1 0

  2. ((b))

    0 1 1 0 1 0 0

  3. ((c))

    0 0 0 1 0 1 1 

  4. ((d))

    1 1 0 1 0 0 1

Show Answer
Answer: ((c))

0 0 0 1 0 1 1 

Concept:

Let d = (d1, d2 …. dk)

Codeword C can be written as:

\(C = \left( {\underbrace {({C_1},;{C_2},; \ldots .{C_k}}{message;bits},;\underbrace {;{C{k + 1}} \ldots .{C_{k + m}})}_{parity;bits}} \right)\) 

Cd = d1

C2 = d2

C3 = d3

⋮

Ck = dk

\(\begin{array}{l} {C_{k + 1}} = {P_{11}};{d_1}; \oplus ;;{P_{12}};;{d_2}; \oplus ;; \ldots ..;;; \oplus {P_{1k}};{d_k}\ {C_{k + 1}} = {P_{11}};{d_1}; \oplus ;;{P_{12}};;{d_2}; \oplus ;; \ldots ..;;; \oplus {P_{1k}};{d_k}\ \vdots ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;} ;;;;;;;;;;;;;;;;;;;\ {C_{k + m}} = {P_{{m_1}}};{d_1}; \oplus ;;{P_{{m_2}}};;{d_2}; \oplus ; \ldots .;.;; \oplus {P_{mk}};{d_k}\ \end{array}\) 

Note: The parity bits are generated by a linear combination of message bits.

Application:

Given (7, 4) linear Hamming code.

(mi, ci) given are:

(1100 ; 0101100)

(1110 ; 0011110)

(0110 ; 1000110)

Clearly, we can observe that the first 3 bits are parity bits and the last 4 bits are message bits.

P0P1P2d0d1d2d3
C1 →0101100
C2 →0011110
C3 →1000110

 

Observe when P0, P1, and P2 are one, and write their combination, i.e.

P0 = d0 ⊕ d1 ⊕ d3

P1 = d1 ⊕ d2 ⊕ d3

P2 = d0 ⊕ d1 ⊕ d2

Now checking each of the options:

Option (1): 1011010

Message bits are:

\(\begin{array}{*{20}{c}} 1&0&1&0\ {{d_0}}&{{d_1}}&{{d_2}}&{{d_3}} \end{array}\)­

 P0 = d0 ⊕ d1 ⊕ d3 = 1

P1 = d1 ⊕ d2 ⊕ d3 = 1

P­2 = d0 ⊕ d1 ⊕ d2 = 0

So, the codeword is 1101010.

Hence option (1) is invalid codeword.

Option (2): 0110100

Message bits are:

 \(\begin{array}{*{20}{c}} 0&1&0&0\ {{d_0}}&{{d_1}}&{{d_2}}&{{d_3}} \end{array}\)

P0 = d0 ⊕ d1 ⊕ d3 = 1

P1 = d1 ⊕ d2 ⊕ d3 = 1

P­2 = d0 ⊕ d1 ⊕ d2 = 1

So, the valid codeword is 1110100

Hence option (2) is an invalid codeword.

Option (3): 0001011

Message bits are:

\(\begin{array}{*{20}{c}} 1&0&1&1\ {{d_0}}&{{d_1}}&{{d_2}}&{{d_3}} \end{array}\) 

P0 = d0 ⊕ d1 ⊕ d3 = 0

P1 = d1 ⊕ d2 ⊕ d3 = 0

P2 = d0 ⊕ d1 ⊕ d2 = 0

So the valid codeword is 0001011

Hence option (3) is a valid codeword.

45

The impedance matching network shown in the figure is to match a lossless line having characteristic impedance Z0 = 50 Ω with a load impedance ZL. A quarter-wave line having a characteristic impedance Z1 = 75 Ω is connected to ZL. Two stubs having characteristic impedance of 75 Ω each are connected to this quarter-wave line. One is a short-circuited (S.C.) stub of length 0.25 λ connected across PQ and the other one is an open-circuited (O.C.) stub of length 0.5 λ connected across RS.

  

The impedance matching is achieved when the real part of ZL is

  1. ((a))

    112.5 Ω 

  2. ((b))

    75.0 Ω

  3. ((c))

    33.3 Ω

  4. ((d))

    50.0 Ω

Show Answer
Answer: ((a))

112.5 Ω 

Concept:

Consider the figure shown below which shows a lossless transmission line being driven from the left and which is terminated by the ZL at the end.

Zin: Input impedance

ZL: Load impedance

Zo: Characteristic impedance

The interface between the source and transmission line is located at z = - l.

zin(l)=z0[1+Γe−j2βl1−e−j2βl]{z_{in}}\left( l \right) = {z_0}\left[ {\frac{{1 + {\rm{\Gamma }}{e^{ - j2\beta l}}}}{{1 - {e^{ - j2\beta l}}}}} \right]

The result also depends on the length and phase propagation of the line.

Note that Zin(l) is periodic in l. Since the argument of the complex exponential factors is 2βl, the frequency at which Zin(l) varies is β/π; and since β=2π/λ, the associated period is λ/2.

zin(l)=z0[zL+jz0tanβlz0+jzLtanβl]{z_{in}}\left( l \right) = {z_0}\left[ {\frac{{{z_L} + j{z_0}tan\beta l}}{{{z_0} + j{z_L}tan\beta l}}} \right]

Calculation:

For the length λ/2, the input impedance will be the same as that of the load.

Zin1=z0[zL+jz0tan2πλλ2z0+jzLtan2πλλ2]{Z_{in1}} = {z_0}\left[ {\frac{{{z_L} + j{z_0}tan\frac{{2\pi }}{\lambda }\frac{\lambda }{2}}}{{{z_0} + j{z_L}tan\frac{{2\pi }}{\lambda }\frac{\lambda }{2}}}} \right]

zin1=zL{z_{in1}} = {z_L}

For short circuit stub length is λ/4

zin2=z02zL{z_{in2}} = \frac{{z_0^2}}{{{z_L}}}

zin2=7520{z_{in2}} = \frac{{75^2}}{{{0}}}

Zin2 = ∞

The final structure of the transmission line is:

This is similar to that of the quarter-wave transformer. So, the impedance will be

z0′=z0zLz_0' = \sqrt {{z_0}{z_L}}

zL=z0′2z0=75250{z_L} = \frac{{z_0^{'2}}}{{{z_0}}} = \frac{{{{75}^2}}}{{50}}

ZL = 112.5 Ω

46

A real 2 × 2 non-singular matrix A with repeated eigenvalue is given as

\(A = \left[ {\begin{array}{*{20}{c}} x&{ - 3.0}\ {3.0}&{4.0} \end{array}} \right]\)

where x is a real positive number. The value of x (rounded off to one decimal place) is _______

47

For a vector field D = ρ cos2 ϕ aρ + z2 sin2 ϕ aϕ in a cylindrical coordinate system (ρ, ϕ, z) with unit vectors aρ, aϕ and az, the net flux of D leaving the closed surface of the cylinder (ρ = 3, 0 ≤ z ≤ 2) (rounded off to two decimal places) is ______.

48

In the circuit shown in the figure, the switch is closed at time t = 0, while the capacitor is initially charged to - 5 V (i.e. vc(0) = -5V)

The time after which the voltage across the capacitor becomes zero (rounded off to three decimal places) is ______ ms.

49

The exponential Fourier series representation of a continuous-time periodic signal x(t) is defined as:

x(t)=∑k=−∞∞akejkω0tx(t)=\displaystyle\sum_{k=-\infty}^\infty a_k e^{jkω_0 t}

Where ω0 is the fundamental angular frequency of x(t) and the coefficients of the series are ak. The following information is given about x(t) and ak.

I. x(t) is real and even having a fundamental period of 6

II. The average value of x(t) is 2

III. ak={k,  1≤k≤30,   k>3a_k = \left\lbrace \begin{matrix} k, \ \ 1 \le k \le 3 \\ 0, \ \ \ k > 3 \end{matrix} \right.

The average power of the signal x(t) (rounded off to one decimal place) is _______.

50

For a unit step input u[n], a discrete-time LTI system produces an output signal (2δ[n + 1] + δ[n] + δ[n - 1]). Let y[n] be the output of the system for an input ((12)nu[n])\left(\left(\dfrac{1}{2}\right)^n u[n]\right). The value of y[0] is _____

51

Consider the signals x[n] = 2n-1 u[-n + 2] and y[n] = 2-n+2 u[n + 1], where u[n] is the unit step sequence. Let X(ejω) and Y(ejω) be the discrete-time Fourier transform of x[n] and y[n], respectively. The value of the integral

12π∫02πX(ejω)Y(e−jω)dω\dfrac{1}{2\pi}\displaystyle\int_0^{2\pi} X(e^{j\omega}) Y(e^{-j\omega})d\omega

(rounded off to one decimal place) is _____

52

A silicon P-N junction is shown in the figure. The doping in the P region is 5 × 1016 cm-3 and doping in the N region is 10 × 1016 cm-3. The parameters given are

Built-in voltage (Φbi) = 0.8 V

Electron charge (q) = 1.6 × 10-19 C

Vacuum permittivity (ε0) = 8.85 × 10-12 F / m

Relative permittivity of silicon (εSi) = 12

The magnitude of reverse bias voltage that would completely deplete one of the two regions (P or N) prior to the other (rounded off to one decimal place) is _______ V.

53

An asymmetrical periodic pulse train vin of 10 V amplitude with on-time TON = 1 ms and off-time TOFF = 1 μs is applied to the circuit shown in the figure. The diode D1 is ideal. 

The difference between the maximum voltage and minimum voltage of the output waveform v0 (in integer) is _____V

54

For the transistor M` in the circuit shown in the figure, μn Cox = 100μA/V^2 and (W/L) = 10, where μn is the mobility of electron, Cox is the oxide capacitance per unit area, W is the width and L is the length.

 

The channel length modulation coefficient is ignored. If the gate-to-source voltage VGS is 1 V to keep the transistor at the edge of saturation, then the threshold voltage of the transistor (rounded off to one decimal place) is _____ V.

55

A circuit with an ideal OPAMP is shown in the figure. A pulse VIN of 20 ms duration is applied to the input. The capacitors are initially uncharged,

The output voltage VOUT of this circuit t = 0+ (in integer) is ______V.

56

The propagation delay of the exclusive-OR (XOR) gate in the circuit in the figure is 3 ns. The propagation delay of all the flip-flops is assumed to be zero. The clock (Clk) frequency provided to the circuit is 500 MHz.

Starting from the initial value of the flip-flop outputs Q2 Q1 Q0 = 1 1 1 with D2 = 1, then the minimum number of triggering clock edges after which the flip-flop outputs Q2 Q1 Q0 becomes 1 0 0 (in integer) is ______

57

The circuit in the figure contains a current source driving a load having an inductor and a resistor in series, with a shunt capacitor across the load. The ammeter is assumed to have zero resistance. The switch is closed at time t = 0.

Initially, when the switch is open, the capacitor is discharged and the ammeter reads zero ampere. After the switch is closed, the ammeter reading keeps fluctuating for some time till it settles to a final steady value. The maximum ammeter reading that one will observe after the switch is closed (rounded off to two decimal places) is _______ A.

58

A unity feedback system that uses proportional-integral (PI) control is shown in the figure.

The stability of the overall system is controlled by tuning the PI control parameters KP and KI. The maximum value of KI that can be chosen so as to keep  the overall system stable or, in the worst case, marginally stable (rounded off to three decimal places) is ________.

59

A sinusoidal message signal having root mean square value of 4 V and frequency of 1 kHz is fed to a phase modulator with phase deviation constant 2 rad / volt. If the carrier signal is c(t) = 2cos(2π106t). the maximum instantaneous frequency of the phase modulated signal (rounded off to one decimal place) is _______ Hz.

60

Consider a superheterodyne receiver tuned to 600 kHz. If the local oscillator feeds a 1000 kHz signal to the mixer, the image frequency (in integer) is _______ kHz.

61

In a high school having equal number of boy students and girl students, 75% of the students study Science and the remaining 25% students study Commerce. Commerce students are two times more likely to be a boy than are Science students. The amount of information gained in knowing that a randomly selected girl student studies Commerce (rounded off to three decimal places) is _______ bits.

62

A message signal having peak-to-peak value of 2 V, root mean square value of 0.1 V and bandwidth of 5 kHz is sampled and fed to a pulse code modulation (PCM) system that uses a uniform quantizer. The PCM output is transmitted over a channel that can support a maximum transmission rate of 50 kbps. Assuming that the quantization error is uniformly distributed, the maximum signal to quantization nose ratio that can be obtained by the PCM system (rounded off to two decimal places) is _____

63

Consider a polar non-return to zero (NRZ) waveform, using +2 V and -2 V for representing binary '1' and '0' respectively, is transmitted in the presence of additive zero-mean white Gaussian noise with variance 0.4 V2. If the a priori probability of transmission of a binary '1' is 0.4, the optimum threshold voltage for a maximum a posteriori (MAP) receiver (rounded off to two decimal places) is _____ V.

64

A standard air-filled rectangular waveguide with dimensions a = 8 cm, b = 4cm, operates at 3.4 GHz, For the dominant mode of wave propagation, the phase velocity of the signal is vp. The value (rounded off two decimal places) of vp / C, where C denotes the velocity of light, is _____

65

An antenna with a directive gain of 6 dB is radiating a total power of 16 kW. The amplitude of the electric field in free space at a distance of 8 km from the antenna in the direction of 6 dB gain (rounded off to three decimal places) is _______ V / m.

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