Official Paper

GATE EC 2016 Official Paper: Shift 2 (Previous Year Paper)

65 questions · 180 minutes · with answers · free

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General Aptitude (10 questions)

Based on the given statements, select the appropriate option with respect to grammar and usage.

Statements

(i) The height of Mr. X is 6 feet.

(ii) The height of Mr. Y is 5 feet.

((a))
Mr. X is longer than Mr. Y.
((b))
Mr. X is more elongated than Mr. Y.
((c))
Mr. X is taller than Mr. Y.
((d))
Mr. X is lengthier than Mr. Y.

Show Answer

Answer: ((c))

Mr. X is taller than Mr. Y.

Since the heights of two people are being compared, the correct adjective in the comparative degree to be used is “taller”.

The students ___________ the teacher on teachers’ day for twenty years of dedicated teaching.

((a))
facilitated
((b))
felicitated
((c))
fantasized
((d))
fecilitated

Show Answer

Answer: ((b))

felicitated

Let us look at the meanings of the words:

Facilitated: Helps improve something

Felicitated: Congratulate

Fantasize: Dream about something

The other option is not a word. Thus, the correct word to be used in the blank is ‘felicitated’ that will give the meaning of the teacher being congratulated.

After India’s cricket world cup victory in 1985, Shrotria who was playing both tennis and cricket till then, decided to concentrate only on cricket. And the rest is history.

What does the underlined phrase mean in this context?

((a))
history will rest in peace
((b))
rest is recorded in history books
((c))
rest is well known
((d))
rest is archaic

Show Answer

Answer: ((c))

rest is well known

The phrase ‘rest is history’ means something that has happened in the past and has become very famous. Thus ‘the rest is well known’ is the correct answer.

Given ${\left( {9{\rm{;inches}}} \right)^{\frac{1}{2}}} = {\left( {0.25{\rm{;yards}}} \right)^{\frac{1}{2}}}$ . Which one of the following statements is TRUE?

((a))
3 inches = 0.5 yards
((b))
9 inches = 1.5 yards
((c))
9 inches = 0.25 yards
((d))
81 inches = 0.0625 yards

Show Answer

Answer: ((c))

9 inches = 0.25 yards

${\left( {9{\rm{;inches}}} \right)^{\frac{1}{2}}} = {\left( {0.25{\rm{;yards}}} \right)^{\frac{1}{2}}}$

Taking roots we have

$3\sqrt {Inches} = 0.5\sqrt {Yards}$

$\frac{{\sqrt {inches} }}{{\sqrt {yards} }} = \frac{{0.5}}{3}$

squaring both sides and cross multiplying we have

9 inches = 0.25 yards

Method 2:

simply squaring both sides of the equation given in the question

we have

9 inches = .25 yards

S, M, E and F are working in shifts in a team to finish a project. M works with twice the efficiency of others but for half as many days as E worked. S and M have 6 hour shifts in a day, whereas E and F have 12 hours shifts. What is the ratio of contribution of M to contribution of E in the project?

((a))
1:1
((b))
1:2
((c))
1:4
((d))
2:1

Show Answer

Answer: ((b))

1:2

Given:

M works with twice the efficiency of others but for half as many days as E worked

S and M have 6 hour shifts in a day, whereas E and F have 12 hours shifts

Calculation:

Let the efficiency of E be x then the efficiency of M is 2x.

Let M work for y day. Since E works for twice as long, E must work for 2y days.

Thus, amount of work done by M = 2x × y × 6

and the amount of work done by E = x × 2y × 12

The ratio of work done by M to E is

⇒ $\frac{{\rm{M}}}{{\rm{E}}} = \frac{{ {\rm{2x}} \ × \ {\rm{y} \ \times 6}}}{{{\rm{x}} \ × \ 2{\rm{y} \ \times 12}}} = 1:2$ .

∴ The contribution of M and E in the project is 1 : 2.

The Venn diagram shows the preference of the student population for leisure activities.

From the data given, the number of students who like to read books or play sports is ____.

((a))
44
((b))
51
((c))
79
((d))
108

Show Answer

Answer: ((d))

108

According to the given Venn diagram number of students who like to read books are.

13 + 12 + 44 + 7 = 76

Number of students who like to play sports are,

44 + 7 + 15 + 17 = 83

But the number of students who like both to read books and play sports are,

44 + 7 = 51

Hence the number of students who like to read books or play sports are,

76 + 83 – 51 = 108.

Here formula used is:

(No. of students Read books ⋃ no. of students play sports) ⋂ No of students read books 2 play sports.

Social science disciplines were in existence in an amorphous form until the colonial period when they were institutionalized. In varying degrees, they were intended to further the colonial interest. In the time of globalization and the economic rise of postcolonial countries like India, conventional ways of knowledge production have become obsolete.

Which of the following can be logically inferred from the above statements?

Social science disciplines have become obsolete.
Social science disciplines had a pre- colonial origin.
Social science disciplines always promote colonialism
Social science must maintain disciplinary boundaries.

((a))
(ii) only
((b))
(i) and (iii) only
((c))
(ii) and (iv) only
((d))
(iii) and (iv) only

Show Answer

Answer: ((a))

(ii) only

The passage does not state the boundaries of discipline for social sciences. Neither can this be inferred. Again, the passage only states that the social sciences were institutionalized because of colonial motives. This is not the same as promoting colonialism. Thus this cannot be inferred either. The knowledge production methods have become obsolete and not the discipline. Thus, point (i) is eliminated.

The passage states that social sciences discipline were in an amorphous form till the colonial period. This means that it existed before the colonial age. Thus point (ii) can be inferred.

Two and a quarter hours back, when seen in a mirror, the reflection of a wall clock without number markings seemed to show 1:30. What is the actual current time shown by the clock?

((a))
8:15
((b))
11:15
((c))
12:15
((d))
12:45

Show Answer

Answer: ((d))

12:45

When reflection of a wall clock without number markings seemed to show 1:30 in a mirror, the actual time was 10:30 as shown in figure.

This was a time two and a quarter hours back.

Hence the actual current time is 10:30 + 2:15 = 12:45

M and N start from the same location. M travels 10 km East and then 10 km North-East. N travels 5 km South and then 4 km South-East. What is the shortest distance (in km) between M and N at the end of their travel?

((a))
18.60
((b))
22.50
((c))
20.61
((d))
25.00

Show Answer

Answer: ((c))

20.61

Let both ${\rm{M;and;N}}$ start from origin. The position vector of M will be

${{\rm{\vec P}}_{\rm{M}}} = 10{\rm{\hat i}} + 10\cos {45^ \circ }{\rm{\hat i}} + 10\cos {45^ \circ }{\rm{\hat j}} = \left( {10 + \frac{{10}}{{\sqrt 2 }}} \right){\rm{\hat i}} + \frac{{10}}{{\sqrt 2 }}{\rm{\hat j}}$

Similarly, position vector of N will be,

${{\rm{\vec P}}_{\rm{N}}} = 4\cos {45^ \circ }{\rm{\hat i}} - 5{\rm{\hat j}} - 4\sin {45^ \circ }{\rm{\hat j}} = \frac{4}{{\sqrt 2 }}{\rm{\hat i}} - \left( {5 + \frac{4}{{\sqrt 2 }}} \right){\rm{\hat j}}$

Displacement vector between M and N, ${{\rm{\vec P}}{{\rm{NM}}}} = {{\rm{\vec P}}{\rm{M}}} - {{\rm{\vec P}}_{\rm{N}}}$

$\Rightarrow {{\rm{\vec P}}_{{\rm{NM}}}} = \left( {10 + \frac{{10}}{{\sqrt 2 }} - \frac{4}{{\sqrt 2 }}} \right){\rm{\hat i}} + \left( {\frac{{10}}{{\sqrt 2 }} + 5 + \frac{4}{{\sqrt 2 }}} \right){\rm{\hat j}}$

Distance between them, $\left| {{{{\rm{\vec P}}}_{{\rm{NM}}}}} \right| = \sqrt {{{\left( {10 + \frac{{10}}{{\sqrt 2 }} - \frac{4}{{\sqrt 2 }}} \right)}^2} + {{\left( {\frac{{10}}{{\sqrt 2 }} + 5 + \frac{4}{{\sqrt 2 }}} \right)}^2}}$

$\Rightarrow \left| {{{{\rm{\vec P}}}_{{\rm{NM}}}}} \right| = 20.612{\rm{;m}}$

A wire of length 340 mm is to be cut into two parts. One of the parts is to be made into a square and the other into a rectangle where sides are in the ratio of 1:2. What is the length of the side of the square (in mm) such that the combined area of the square and the rectangle is a MINIMUM?

((a))
30
((b))
40
((c))
120
((d))
180

Show Answer

Answer: ((b))

Let the side of square formed be y. And the length and breadths of the rectangle formed by 2x and x as they are in the ratio 2:1. Then, Perimeter of square ${\rm{P}}\left( {\rm{S}} \right) = 4{\rm{y}}$ and perimeter of rectangle ${\rm{P}}\left( {\rm{R}} \right) = 6{\rm{x}}$ . Since this all length is cut form the 340 mm wire, we have ${\rm{P}}\left( {\rm{S}} \right) + {\rm{P}}\left( {\rm{R}} \right) = 340$

$\begin{array}{l} \Rightarrow 4{\rm{y}} + 6{\rm{x}} = 340\ \Rightarrow {\rm{y}} = 85 - \frac{3}{2}{\rm{x}} \end{array}$

Now the combined are of square and rectangle, ${\rm{A}} = {{\rm{y}}^2} + 2{{\rm{x}}^2}$

Substituting the value of y we have,

$\begin{array}{l} {\rm{A}} = {\left( {85 - \frac{3}{2}{\rm{x;}}} \right)^2} + 2{{\rm{x}}^2}\ \Rightarrow {\rm{A}} = 7225 + \frac{9}{4}{{\rm{x}}^2} - 255{\rm{x}} + 2{{\rm{x}}^2}{\rm{;}}\ \Rightarrow {\rm{A}} = \frac{{17{{\rm{x}}^2} - 1020{\rm{x}} + 28900}}{2}{\rm{;}} \end{array}$

Form minimum area, $\frac{{{\rm{dA}}}}{{{\rm{dx}}}} = 0$

$\begin{array}{l} \Rightarrow \frac{{{\rm{dA}}}}{{{\rm{dx}}}} = 34{\rm{x}} - 1020 = 0\ \Rightarrow {\rm{x}} = \frac{{1020}}{{34}} = 30 \end{array}$

Now, length of the side of the square

${\rm{y}} = 85 - \frac{3}{2} \times 30 = 85 - 45 = 40{\rm{mm}}$

Electronics and Communication Engineering (55 questions)

The value of x for which the matrix ${\rm{A}} = \left[ {\begin{array}{*{20}{c}} 3&2&14\ 9&7&{13}\ { - 6}&{ - 4}&{ - 9 + {\rm{x}}} \end{array}} \right]$ has zero as an eigenvalue is ________

Consider the complex valued function $\rm f(z) = 2z^3 + b |z|^3$ where $\rm z$ is a complex variable.

The value of $\rm b$ for which the function ${\rm{f}}\left( {\rm{z}} \right)$ is analytic is ________

As $\rm x$ varies from $\rm −1\ to \ +3$ , which one of the following describes the behaviour of the function $\rm f(x) = x^3 – 3x^2 + 1$ ?

((a))
$\rm f(x)$ increases monotonically.
((b))
$\rm f(x)$ increases, then decreases and increases again.
((c))
$\rm f(x)$ decreases, then increases and decreases again.
((d))
$\rm f(x)$ increases and then decreases.

Show Answer

Answer: ((b))

$\rm f(x)$ increases, then decreases and increases again.

Concept:

Whenever we have to analyse the graph for a given function generally substitute consecutive numbers and define. Here the range is already specified simply we can substitute those values in a given function and then we can draw a graph based on results

Calculation:

Given f(x) = x³ – 3x² + 1, x varies from -1 to 3

x = -1, f(X) = (-1)³ – 3(-1)² + 1 = -1 -3 + 1 = -3

x = 0, f(x) = (0)³ – 3(0)² + 1 = 0 – 0 + 1 = 1

x = 1, f(x) = (1)³– 3(1)² + 1 = 1 – 3 + 1 = -1

x = 2, f(x) = (2)³ – 3(2)² + 1 = 8 – 12 + 1 = -3

x = 3, f(x) = (3)³ – 3(3)²+ 1 = 27 – 27 +1 = 1

X	-1	0	1	2	3
f(x)	-3	1	-1	-3	1

Graph Result:

From the calculation graph is first increases, then decreases and increases again

Alternate Method:

Given f(x) = x³ – 3x² + 1

Let us first find the stationary points by making f’(x) = 0 & then we will decide for maxima & minima by the sign of f”(x) at that point

f'(x) = 3x² – 6x

3x²– 6x = 0

3x (x - 2) = 0

x = 0, 2

x = 0 & x = 2 are stationary points.

f"(x) = 6 × -6

f”(0) = -6 < 0 Hence, maxima

f”(2) = 6 > 0 Hence, minima

f(-1) = 3

f(0) = 1

f(2) = -3

f(3) 1

f(x) increases, then decreases & increases again.

So, option B is true.

How many distinct values of $\rm x$ satisfy the equation $\rm sin(x) = x/2$ , where $\rm x$ is in radians?

((a))
1
((b))
2
((c))
3
((d))
4 or more

Show Answer

Answer: ((c))

Given ${\rm{sinx}} = \frac{{\rm{x}}}{2}$

let $\rm f(x) = sin x$ and $\rm g(x) = x / 2$

∴ The given function has 3 distinct roots

Consider the time-varying vector ${\bf{I}} = {\rm{\hat x;}}15\cos \left( {{\rm{wt}}} \right) + {\rm{\hat y;}}5$ sin(𝜔𝑡) in Cartesian coordinates, where 𝜔 > 0 is a constant. When the vector magnitude |𝐈| is at its minimum value, the angle 𝜃 that 𝐈 makes with the 𝑥 axis (in degrees, such that 0 ≤ 𝜃 ≤ 180) is ________

In the circuit shown below, V_S is a constant voltage source and I_L is a constant current load.

The value of $\rm I_L$ that maximizes the power absorbed by the constant current load is

((a))
$\frac{{{{\rm{V}}_{\rm{s}}}}}{{4{\rm{R}}}}$
((b))
$\frac{{{{\rm{V}}_{\rm{s}}}}}{{2{\rm{R}}}}$
((c))
$\frac{{{{\rm{V}}_{\rm{s}}}}}{{\rm{R}}}$
((d))
$\rm\infty$

Show Answer

Answer: ((b))

$\frac{{{{\rm{V}}_{\rm{s}}}}}{{2{\rm{R}}}}$

For the given circuit,

The equivalent Thevenin circuit is given as

The maximum power transfer for the resistive circuit is possible when

R_L = R_Th = R

Front the above figure the load current can be derived as

$ {{\rm{I}}{\rm{L}}} = \frac{{{{\rm{V}}{\rm{s}}}}}{{{\rm{R}} + {\rm{R}}}}$

I_L = V_s/2R

Hence option (2) is correct

The power across Load R_L = R is given by

$P_{max} = {(\frac{V_s}{2R})}^2.R$

$P_{max} = \frac{{V_s}^2}{4R}$

Load variable	The load impedance for maximum power transfer
RL and XL are variable	RL = RS XL = -XS ZL = ZS*
RL only varied and XL = Constant	RL =√(RS2 + (XL + XS)2)
RL only varied and XL = 0	RL =√(RS2 + XS2)

The switch has been in position 1 for a long time and abruptly changes to position 2 at t = 0.

If time $\rm t$ is in seconds, the capacitor voltage $\rm V_C$ (in volts) for $\rm t > 0$ is given by

((a))
$\rm 4(1 − exp(−t/0.5))$
((b))
$\rm 10 − 6\ exp(−t/0.5)$
((c))
$\rm 4(1 − exp(−t/0.6))$
((d))
$\rm 10 − 6 exp(−t/0.6)$

Show Answer

Answer: ((d))

$\rm 10 − 6 exp(−t/0.6)$

Concept:

For t > 0, capacitor voltage V_c(t) is given as,

${V_C}\left( t \right) = \left[ {{V_C}\left( {{0^ + }} \right) - {V_C}\left( \infty \right)} \right]{e^{ - t/\tau }} + {V_C}\left( \infty \right)$ ---(1)

Where,

V_C(0⁺): Voltage across capacitor at t = 0⁺

V_C(0): Voltage across capacitor at steady state

τ :Time constant = R_eq C_eq

Calculation:

At t = 0^-, switch is at position – 1

Capacitor behaves open circuit at t = 0^- (steady state)

${V_C}\left( {{0^ - }} \right) = \frac{{2 \times 10}}{{3 + 2}}$

V_C(0^-) = 4 V

We know that, capacitor opposes sudden change in voltage

So,

V_C(0^-) = V_C(0⁺) = 4V

Now, at t = ∞, the switch is at position – 2

V_C(∞) = 5 × 2 = 10 V

Calculation of time constant:

Draw the Circuit at t > 0 and open circuit the current source,

R_eq = 4 + 2 = 6Ω

So,

τ = R_eq⋅ C_eq = 6 Ω × .1F = 0.6 sec

Now from equation (1), we get:

${V_C}\left( t \right) = {V_C}\left( \infty \right) + \left[ {{V_C}\left( {{0^ + }} \right) - {V_L}\left( \infty \right)} \right]{e^{ - t/\tau }}$

${V_C}\left( t \right) = \left[ {10 - 6{e^{ - \frac{t}{{0.6}}}}} \right]V$

The figure shows an RLC circuit with a sinusoidal current source:

At resonance, the ratio |𝐈_𝐋|/|𝐈_𝐑|, i.e., the ratio of the magnitudes of the inductor current phasor and the resistor current phasor, is ________

The z-parameter matrix for the two-port network shown is

$\left[ {\begin{array}{*{20}{c}} {2{\rm{jω }}}&{{\rm{jω }}}\ {{\rm{jω }}}&{3 + 2{\rm{jω }}} \end{array}} \right]$

where the entries are in Ω. Suppose Z_b(jω) = R_b + jω.

Then the value of R_b (in Ω) equals ________.

The energy of the signal $x(t) = \frac{{{\rm{sin}}\left( {4{\rm{\pi t}}} \right)}}{{4{\rm{\pi t}}}}$ is______

The Ebers-Moll model of a BJT is valid

((a))
only in active mode
((b))
only in active and saturation modes
((c))
only in active and cut-off modes
((d))
in active, saturation and cut-off modes

Show Answer

Answer: ((d))

in active, saturation and cut-off modes

Ebers-Moll model is valid for all of the active, saturation and cutoff regions.

A long-channel NMOS transistor is biased in the linear region with $\rm V_{DS}=50\ mV$ and is used as a resistance. Which one of the following statements is NOT correct?

((a))
If the device width W is increased, the resistance decreases.
((b))
If the threshold voltage is reduced, the resistance decreases.
((c))
If the device length L is increased, the resistance increases.
((d))
If V_GS is increased, the resistance increases.

Show Answer

Answer: ((d))

If V_GS is increased, the resistance increases.

Concept:

For VDS < VGS – VT, n-channel enhancement type MOSFET will operate in triode/linear region:

${I_D} = \frac{{{\rm{W\mu_n }}{C_{ox}}}}{{2L}}\left[ {2\left( {{V_{GS}} - {V_T}} \right){V_{DS}} - V_{DS}^2} \right]$

For VDS ≥ VGS – VT, the MOSFET will be in saturation with current given by:

${I_D} = \frac{{{\rm{W\mu_n }}{C_{ox}}}}{{2L}}\left[ {{{\left( {{V_{GS}} - {V_T}} \right)}^2}} \right]$

Application:

The current in an NMOS in the linear region, therefore:

${I_D} = \frac{{{\rm{W\mu_n }}{C_{ox}}}}{{2L}}\left[ {2\left( {{V_{GS}} - {V_T}} \right){V_{DS}} - V_{DS}^2} \right]$

The transistor operates in the linear region for small values of V_DS

${I_D} \approx \frac{{{\rm{W\mu_n }}{C_{ox}}}}{{L}}\left[ {\left( {{V_{GS}} - {V_T}} \right){V_{DS}}} \right]$

$R_d=\frac{V_{DS}}{I_{DS}}$

$R_d=\frac{1}{\frac{{{\rm{W\mu_n }}{C_{ox}}}}{{L}}\left[ {\left( {{V_{GS}} - {V_T}} \right){V_{DS}}} \right]}$

Observations:

As W increases, the resistance decreases.
As V_T decreases, the resistance decreases.
As L increases, the resistance increases.
However, as V_GS increases, the resistance decreases. So, Option D is the required incorrect statement

Assume that the diode in the figure has V_on = 0.7 V, but is otherwise ideal.

The magnitude of the current i₂ (in mA) is equal to ________

Resistor R₁ in the circuit below has been adjusted such that I₁ = 1 mA. The bipolar transistors Q₁ and Q₂ are perfectly matched and have very high current gain, so their base currents are negligible. The supply voltage V_CCis 6 V. The thermal voltage kT/q is 26 mV.

The value of R₂ (in Ω) for which I₂ = 100 μA is ________

Which one of the following statements is correct about an ac-coupled common-emitter amplifier operating in the mid-band region?

((a))
The device parasitic capacitances behave like open circuits, whereas coupling and bypass capacitances behave like short circuits.
((b))
The device parasitic capacitances, coupling capacitances and bypass capacitances behave like open circuits.
((c))
The device parasitic capacitances, coupling capacitances and bypass capacitances behave like short circuits.
((d))
The device parasitic capacitances behave like short circuits, whereas coupling and bypass capacitances behave like open circuits.

Show Answer

Answer: ((a))

The device parasitic capacitances behave like open circuits, whereas coupling and bypass capacitances behave like short circuits.

We know that:

$\left| {{X_C}} \right| = \frac{1}{{\omega C}} = \frac{1}{{2\pi fC}}$

The parasitic capacitances are in PF (10^-12 f) So, |X_C| becomes too large (≈ ∞) for the mid freq. band so parasitic capacitance act line open circuit.
The bypass and coupling capacitances are in μf (10^-6f) so, X_c becomes very small (≈ 0) for the mid-frequency band so, they behave as a short circuit.

Note:

Capacitors	Mid Freq.	Low Freq.	High Freq.
1. Large-cap. (μf)	Short circuit	Effect is considered	Short circuit
2. Small-cap (PF)	Open circuit	Open circuit	Effect is considered

Transistor geometries in a CMOS inverter have been adjusted to meet the requirement for worst-case charge and discharge times for driving a load capacitor C. This design is to be converted to that of a NOR circuit in the same technology, so that its worst-case charge and discharge times while driving the same capacitor are similar. The channel lengths of all transistors are to be kept unchanged. Which one of the following statements is correct?

((a))
Widths of PMOS transistors should be doubled, while widths of NMOS transistors should be halved.
((b))
Widths of PMOS transistors should be doubled, while widths of NMOS transistors should not be changed.
((c))
Widths of PMOS transistors should be halved, while widths of NMOS transistors should not be changed.
((d))
Widths of PMOS transistors should be unchanged, while widths of NMOS transistors should be halved.

Show Answer

Answer: ((b))

Widths of PMOS transistors should be doubled, while widths of NMOS transistors should not be changed.

Generally the area of PMOS is double to NMOS to maintain equal rise time and fall time.

Because mobility of electron is double that of mobility of hole. The $\left( {\frac{{\rm{W}}}{{\rm{L}}}} \right)$ of PMOS can be calculated by dividing its resistance with 2R whereas the $\left( {\frac{{\rm{W}}}{{\rm{L}}}} \right)$ of NMOS can be calculated by dividing its resistance with R.

The $\left( {\frac{{\rm{W}}}{{\rm{L}}}} \right)$ of first circuit:

${\left( {\frac{{\rm{W}}}{{\rm{L}}}} \right){\rm{p}}} = \frac{{2{\rm{R}}}}{{\rm{R}}} = 2$ and ${\left( {\frac{{\rm{W}}}{{\rm{L}}}} \right){\rm{p}}} = \frac{{\rm{R}}}{{\rm{R}}} = 1$

The $\left( {\frac{{\rm{W}}}{{\rm{L}}}} \right)$ of second circuit:

${\left( {\frac{{\rm{W}}}{{\rm{L}}}} \right){\rm{p}}} = \frac{{2{\rm{R}}}}{{{\rm{R}}/2}} = 4$ and ${\left( {\frac{{\rm{W}}}{{\rm{L}}}} \right){\rm{n}}} = \frac{{\rm{R}}}{{\rm{R}}} = 1$

So the width of PMOS is doubled and width of NMOS remains same.

Assume that all the digital gates in the circuit shown in the figure are ideal, the resistor R = 10 kΩ, and the supply voltage is 5 V. The D flip-flops D1, D2, D3, D4, and D5 are initialized with logic values 0, 1, 0, 1, and 0 respectively. The clock has a 30% duty cycle.

The average power dissipated (in mW) in the resistor R is ________

A 4:1 multiplexer is to be used for generating the output carry of a full adder. A and B are the bits to be added while C_in is the input carry and C_out is the output carry. A and B are to be used as the select bits with A being the more significant select bit.

((a))
I₀ = 0, I₁ = C_in, I₂ = C_in and I₃ = 1
((b))
I0 = 1, I1 = Cin, I2 = Cin and I3 = 1
((c))
I0 = C_in, I1 = 0, I2 = 1 and I3 = C_in
((d))
I0 = 1, I1 = Cin, I2 = 1 and I3 = C_in

Show Answer

Answer: ((a))

I₀ = 0, I₁ = C_in, I₂ = C_in and I₃ = 1

Concept:

The functional table is mentioned below:

S0	S1	Y
0	0	1
0	1	1
1	0	1
1	1	1

The general output equation is:

$Y = \overline {{S_1}} \overline {{S_0}} ;{I_0} + \overline {{S_1}} {S_0}{I_1} + {S_1}\overline {{S_0}} ;{I_2} + {S_1}{S_0}{I_3}$

The truth table for a full adder is as shown:

A	B	Ci	Cout	S
0	0	0	0	0
0	0	1	0	1
0	1	0	0	1
0	1	1	1	0
1	0	0	0	1
1	0	1	1	0
1	1	0	1	0
1	1	1	1	1

The expression for the sum bit and the output carry will be:

Cout = ∑ m (3, 5, 6, 7) and

S = ∑ m (1, 2, 4, 7)

Another equation for Cout is:

$C_{out}=\left( {\bar A;B + A;\bar B} \right){C_i} + AB$

Calculation:

Given S1 = A and S0 = B. The output expression becomes:

$Y = ;\bar A;\bar B;{C_i} + \bar A;B;{C_i} + A\bar B;{C_i} + ABC$ ---(iii)

The equation for Cout is:

$C_{out}=\left( {\bar A;B + A;\bar B} \right){C_i} + AB$

$C_{out}=\bar A;B;{C_i} + A\bar B;{C_i} + AB$ ---(iv)

Comparing equations (iii) and (iv), we get:

I0 = 0, I1 = Cin, I2 = Cin, I3 = 1

The response of the system ${\rm{G}}\left( {\rm{s}} \right) = \frac{{{\rm{s}} - 2}}{{\left( {{\rm{s}} + 1} \right)\left( {{\rm{s}} + 3} \right)}}$ to the unit step input $\rm u(t)$ is $\rm y(t)$ .

The value of $\frac{{{\rm{dy}}}}{{{\rm{dt}}}}$ at $\rm t = 0^+$ is _________

The number and direction of encirclements around the point $\rm −1 + j0$ in the complex plane by the Nyquist plot of ${\rm{G}}\left( {\rm{s}} \right) = \frac{{1 - {\rm{s}}}}{{4 + 2{\rm{s}}}}$ is

((a))
zero.
((b))
one, anti-clockwise.
((c))
one, clockwise.
((d))
two, clockwise.

Show Answer

Answer: ((a))

zero.

Concept:

The Nyquist plot is equal to the polar plot & mirror Image of the polar plot with respect to the real axis, with opposite direction + semicircle of the infinite radius in a clockwise direction as many as the type of system.

Calculation:

Given:

$G\left( s \right) = \frac{{1 - s}}{{4 + 2s}}$

$G\left( {j\omega } \right) = \frac{{1 - j\omega }}{{4 + 2j\omega }}$

So,

$\left| {G\left( {j\omega } \right)} \right| = \frac{{\sqrt {1 + {\omega ^2}} }}{{\sqrt {16 + 4{\omega ^2}} }}$

$\angle G\left( {j\omega } \right) = ; - {\tan ^{ - 1}}\omega - {\tan ^{ - 1}}\frac{\omega }{2}$

ω	\|G(jω)\|	∠G(jω)
0	.25	0°

∞	-0.5	-180°

Polar plot of Given G(s) is

Now,

Nyquist plot of Given G(s) is:

Hence, the number of Encirclements of (-1 + j0) = 0

A discrete memoryless source has an alphabet {a₁, a2, a3, a4} with corresponding probabilities $\left{ {\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{8}} \right}$. The minimum required average codeword length in bits to represent this source for error-free reconstruction is ________

A speech signal is sampled at 8 kHz and encoded into PCM format using 8 bits/sample. The PCM data is transmitted through a baseband channel via 4-level PAM. The minimum bandwidth (in kHz) required for transmission is ________

A uniform and constant magnetic field B ẑ. B̅ exists in the ${\rm{̂ z}}$ direction in vacuum. A particle of mass m with a small charge q is introduced into this region with an initial velocity v = x̂v_x + ẑv_z. Given that B, m, q, v_x, and v_z are all non-zero, which one of the following describes the eventual trajectory of the particle?

((a))
Helical motion in the ẑ direction.
((b))
Circular motion in the xy-plane.
((c))
Linear motion in the ẑ direction.
((d))
Linear motion in the x̂ direction.

Show Answer

Answer: ((a))

Helical motion in the ẑ direction.

Concept:

When any non-zero charge enters in a magnetic field with some velocity at an angle other than 90 degrees, then charge follow helical trajectory.

Application:

$\vec B = B{\hat z_0}$

$\vec V = {v_X}\hat x + {v_Z}\hat z$

x-component of $\vec V$ is perpendicular to the magnetic field $\vec B$ .

A charge moving perpendicular to the magnetic field will experience a radial force causing circular motion shown in figure (a).

z-component of $\vec V$ is parallel to the magnetic field $\vec B$ . A change moving parallel to the field generates no force shown in figure (b).

Motion with components perpendicular and parallel to the field causes the change to move in a helical path along +z direction as shown in figure (c).

Let the electric field vector of a plane electromagnetic wave propagating in a homogenous medium be expressed as $\rm \vec E={\rm{\hat x}}E_xe^{−(\omega t-\beta z)}$ , where the propagation constant $\rm \beta$ is a function of the angular frequency $\rm \omega$ . Assume that $\rm \beta(\omega)$ and $\rm E_x$ are known and are real. From the information available, which one of the following CANNOT be determined?

((a))
The type of polarization of the wave.
((b))
The group velocity of the wave.
((c))
The phase velocity of the wave.
((d))
The power flux through the $\rm z = 0$ plane.

Show Answer

Answer: ((d))

The power flux through the $\rm z = 0$ plane.

Let us analyze each option.

Option A: Because polarization is nothing but the orientation of the electric field vector. we can determine the polarization of the wave.

So option A is incorrect.

Option B:

The group velocity (v_g) of the wave is given by:

${v_g} = \frac{{d\omega }}{{d\beta }}$

Since β(ω) is known, we can find $\frac{{d\beta }}{{d\omega }}$ and hence the group velocity will be:

${v_g} = \frac{1}{{\frac{{d\beta }}{{d\omega }}}}$

So option B is incorrect.

Option C:

The phase velocity (v_p) is defined as:

${v_P} = \frac{\omega }{\beta }$

This can also be determined

So, option C is incorrect.

Option D:

The power flux through the z = 0 plane is:

$P = \frac{1}{2}\frac{{\varepsilon _0^2}}{\eta }$

Since only E is given, we cannot determine the intrinsic impedance of the medium from the given data. Hence power flux through any plane cannot be determined.

Hence option D is Correct.

Light from free space is incident at an angle θ₁to the normal of the facet of a step-index large core optical fibre. The core and cladding refractive indices are n₁ = 1.5 and n₂ = 1.4, respectively.

The maximum value of θ₁(in degrees) for which the incident light will be guided in the core of the fibre is ________.

The ordinary differential equation $\frac{{{\rm{dy}}}}{{{\rm{dt}}}} = - 3{\rm{x}} + {\rm{}}2,{\rm{with\ x}}\left( 0 \right){\rm{;}} = {\rm{;}}1$

is to be solved using the forward Euler method. The largest time step that can be used to solve the equation without making the numerical solution unstable is ________.

Suppose C is the closed curve defined as the circle $\rm x^2 + y^2 = 1$ with $\rm C$ oriented anti-clockwise. The value of $\rm ∮(xy^2 dx + x^2y dy)$ over the curve $\rm C$ equals ________

Two random variables $\rm X$ and $\rm Y$ are distributed according to

${\rm{f_{X,{{\rm{Y}}}}}{\left( {{\rm{x}},{\rm{y}}} \right)}} = \left{ {\begin{array}{*{20}{c}} {\left( {{\rm{x}} + {\rm{y}}} \right)}&{{\rm{0}} \le {\rm{x}} \le 1}&{0 \le {\rm{y}} \le 1}\ {0,}&{{\rm{otherwise}}.}&{\rm{;}} \end{array}} \right.$

The probability $\rm (X+Y \le 1)$ is ________

The matrix ${\rm{A}} = \left[ {\begin{array}{*{20}{c}} {\rm{a}}&0&3&7\ 2&5&1&3\ 0&0&2&4\ 0&0&0&{\rm{b}} \end{array}} \right]$ has $\rm det(A) = 100$ and $\rm trace(A) = 14$ .

The value of $\rm |a-b|$ is ________.

In the given circuit, each resistor has a value equal to 1 Ω.

What is the equivalent resistance across the terminals 𝑎 and 𝑏?

((a))
1/6 Ω
((b))
1/3 Ω
((c))
9/20 Ω
((d))
8/15 Ω

Show Answer

Answer: ((d))

8/15 Ω

Concept:

To convert Star from Delta, we follow the following procedure:

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

For branch resistance, PS, Ra, and Rb are connected, so:

${R_A} = \frac{{{R_a} \times {R_b}}}{{{R_a} + {R_b} + {R_c}}}$

For branch resistance QS, Ra and Rc are connected, so:

${R_B} = \frac{{{R_a} \times {R_c}}}{{{R_a} + {R_b} + {R_c}}}$

For branch resistance, SR, Rb, and Rc are connected, so:

${R_C} = \frac{{{{R_c} \times {R_b}}}}{{{R_a} + {R_b} + {R_c}}}$

Calculation:

Covert Delta Network to star Network we get

Covert star Network to delta Network we get

${{\rm{R}}_{{\rm{eq}}}} = \frac{{\frac{4}{5} \times \frac{8}{5}}}{{\frac{4}{5} + \frac{8}{5}}} = \frac{{\frac{{32}}{{25}}}}{{\frac{{12}}{{5}}}} ~\Omega$

R_eq = 32/60 = 8/15 Ω

In the circuit shown in the figure, the magnitude of the current (in amperes) through R2 is ________.

A continuous-time filter with transfer function ${\rm{H}}\left( {\rm{s}} \right) = \frac{{2{\rm{s}} + 6}}{{{{\rm{s}}^2} + 6{\rm{s}} + 8}}{\rm{;}}$ is converted to a discretetime filter with transfer function ${\rm{G}}\left( {\rm{z}} \right) = \frac{{2{{\rm{z}}^2} - 0.5032{\rm{z}}}}{{{{\rm{z}}^2} - 0.5032{\rm{z}} + {\rm{k}}}}$ so that the impulse response of the continuous-time filter, sampled at $\rm 2\ Hz$ , is identical at the sampling instants to the impulse response of the discrete time filter. The value of $\rm k$ is ________

The Discrete Fourier Transform (DFT) of the 4-point sequence

$\rm x[n] = {x[0], x[1], x[2], x[3]} = {3, 2, 3, 4}$ is

$\rm X[k] = {X[0], X[1], X[2], X[3]} = {12, 2j, 0, −2j}$ .

If $\rm X_1[k]$ is the DFT of the 12-point sequence $\rm x_1[n] = {3, 0, 0, 2, 0, 0, 3, 0, 0, 4, 0, 0}$ , the value of $\left| {\frac{{{{\rm{X}}_1}\left[ 8 \right]}}{{{{\rm{X}}1{\left[ {11} \right]}}{\rm{;}}}}} \right|$ is ________

The switch S in the circuit shown has been closed for a long time.

It is opened at time t = 0 and remains open after that. Assume that the diode has zero reverse current and zero forward voltage drop.

The steady-state magnitude of the capacitor voltage V_c (in Volts) is ________.

A voltage $\rm V_G$ is applied across a MOS capacitor with metal gate and p-type silicon substrate at $\rm T=300 \ K$ . The inversion carrier density (in number of carriers per unit area) for $\rm V_G = 0.8 \ V$ is $\rm 2 × 10^{11} cm^{−2}$ . For $\rm V_G = 1.3 \ V$ , the inversion carrier density is $\rm 4 × 10^{11} cm^{−2}$ . What is the value of the inversion carrier density for $\rm V_G = 1.8 \ V$ ?

((a))
$\rm 4.5 × 10^{11} cm^{−2}$
((b))
$\rm 6 × 10^{11} cm^{−2}$
((c))
$\rm 7.2 × 10^{11} cm^{−2}$
((d))
$\rm 8.4 × 10^{11} cm^{−2}$

Show Answer

Answer: ((b))

$\rm 6 × 10^{11} cm^{−2}$

Concept:

In MOS capacitor charges Accumulated near the Interface is known as the Inversion carrier density (Q).

It is given as,

Q ∝ (V_G - V_T)

Where,

Q – Inversion carrier density (cm^-2)

V_G – Applied Gate voltage

V_T – Thermal voltage.

Calculation:

Given:

${V_{{G_1}}} = .8;V$ , Q₁ = 2 × 10¹¹,

${V_{{G_2}}} = 1.3;V$ , Q₂ = 4 × 10¹¹

${V_{{G_3}}} = 1.8;V$ , Q₃ = ?

We know that,

Q ∝ (V_G - V_T)

So,

$\frac{{{V_{{G_1}}} - {V_T}}}{{{V_{{G_2}}} - {V_T}}} = \frac{{{Q_1}}}{{{Q_2}}} \Rightarrow \frac{{.8 - {V_T}}}{{1.3 - {V_T}}} = \frac{{2 \times {{10}^{11}}}}{{4 \times {{10}^{11}}}}$

On solving, we’ll get

V_T = 0.3 V

Now,

$\frac{{{V_{{G_2}}} - {V_T}}}{{{V_{{G_3}}} - {V_T}}} = \frac{{{Q_2}}}{{{Q_3}}}$

$\frac{{1.3 - 0.3}}{{1.8 - 0.3}} = \frac{{4 \times {{10}^{11}}}}{{{Q_3}}}$

Hence,

Inversion carrier density with V_G = 1.8 V is

Q₃ = 6 × 10¹¹ /cm²

Consider avalanche breakdown in a silicon $\rm p^+n$ junction. The n-region is uniformly doped with a donor density $\rm N_D$ . Assume that breakdown occurs when the magnitude of the electric field at any point in the device becomes equal to the critical field $\rm E_{crit}$ . Assume $\rm E_{crit}$ to be independent of $\rm N_D$ . If the built-in voltage of the $\rm p^+n$ junction is much smaller than the breakdown voltage, $\rm V_{BR}$ , the relationship between $\rm V_{BR}$ and $\rm N_D$ is given by

((a))
${{\rm{V}}{{\rm{BR}}}} \times \sqrt {{{\rm{N}}{\rm{D}}}}$ constant
((b))
${{\rm{N}}{\rm{D}}} \times \sqrt {{{\rm{V}}{{\rm{BR}}}}}$ constant
((c))
${{\rm{N}}{\rm{D}}} \times {{\rm{V}}{{\rm{BR}}}}$ constant
((d))
${{\rm{N}}{\rm{D}}}/{{\rm{V}}{{\rm{BR}}}}$ constant

Show Answer

Answer: ((c))

${{\rm{N}}{\rm{D}}} \times {{\rm{V}}{{\rm{BR}}}}$ constant

Analysis:

In any type of PN junction;

E_critical $= \frac{{{V_{bi}} + {V_{RB}}}}{{W/2}}$

E_critical $= \frac{{2\left[ {{V_{bi}} + {V_{RB}}} \right]}}{W}$ ---(1)

Where,

V_bi: Built-in potential.

V_RB: Reverse bias voltage.

$W = \sqrt {\frac{{2\varepsilon }}{q}\left( {\frac{1}{{{N_A}}} + \frac{1}{{{N_D}}}} \right)\left( {{V_{bi}} + {V_{RB}}} \right)}$ ---(2)

P⁺N junction is given so

${N_A} \gg {N_D};;So,;;\frac{1}{{{N_A}}} \ll \frac{1}{{{N_D}}}$ ---(3)

From equation (1) and (2)

$\left[ {{V_{bi}} + {V_{RB}}} \right] = \frac{{\varepsilon ;E_{crit}^2}}{{2q}}\left[ {\frac{1}{{{N_D}}} + \frac{1}{{{N_A}}}} \right]$

From equation (3),

${V_{RB}} = \frac{{\varepsilon ;E_{crit}^2}}{{2q}}\left[ {\frac{1}{{{N_D}}}} \right]$

∵ V_bi = 0 (Given)

${V_{RB}} \propto \frac{1}{{{N_D}}}$

V_RB . N_D = constant

Consider a region of silicon devoid of electrons and holes, with an ionized donor density of ${\rm{N}}_{\rm{d}}^ + = {10^{17}}{\rm{;c}}{{\rm{m}}^{ - 3}}.$ The electric field at x = 0 is 0V/cm and the electric field at x = L is 50 kV/cm in the positive x-direction. Assume that the electric field is zero in the y and z directions at all points.

Given $\rm q = 1.6 × 10^{−19}$ coulomb, $\rm \epsilon_o = 8.85 × 10^{−14} F/cm, \epsilon_r = 11.7$ for silicon, the value of $\rm L( in \ nm)$ is ________.

Consider a long-channel NMOS transistor with source and body connected together. Assume that the electron mobility is independent of 𝑉_𝐺𝑆 and 𝑉_𝐷𝑆. Given, g_m = 0.5 𝜇A/V for 𝑉_𝐷𝑆 = 50 mV and 𝑉_𝐺𝑆 = 2 V,

g_d = 8 𝜇A/V for 𝑉_𝐺𝑆 = 2 V and 𝑉_𝐷𝑆 = 0 V,

Where ${{\rm{g}}{\rm{m}}} = \frac{{\partial {{\rm{I}}{\rm{D}}}}}{{\partial {{\rm{V}}{{\rm{GS}}}}}}$ and ${{\rm{g}}{\rm{d}}} = \frac{{\partial {{\rm{I}}{\rm{D}}}}}{{\partial {{\rm{V}}{{\rm{DS}}}}}}$

The threshold voltage (in volts) of the transistor is ________.

The figure shows a half-wave rectifier with a 475 𝜇F filter capacitor. The load draws a constant current 𝐼_𝑂 = 1 A from the rectifier. The figure also shows the input voltage 𝑉_𝑖, the output voltage 𝑉_𝐶 and the peak-to-peak voltage ripple 𝑢 on 𝑉_𝐶. The input voltage 𝑉_𝑖 is a triangle-wave with an amplitude of 10 V and a period of 1 ms.

The value of the ripple 𝑢 (in volts) is ________

In the opamp circuit shown, the Zener diodes $\rm Z_1$ and $\rm Z_2$ clamp the output voltage $\rm V_o$ to $\rm +5 \ V$ or $\rm −5 \ V$ . The switch S is initially closed and is opened at time t = 0.

The time $\rm t = t_1$ (in seconds) at which $\rm V_o$ changes state is ________.

An opamp has a finite open-loop voltage gain of 100. Its input offset voltage V_ios (= +5mV) is modeled as shown in the circuit below. The amplifier is ideal in all other respects. V_input is 25 mV.

The output voltage (in millivolts) is _________.

An 8 Kbyte ROM with an active low Chip Select input $\left( {\overline {{\rm{CS}}} } \right)$ is to be used in an 8085 microprocessor-based system. The ROM should occupy the address range 1000H to 2FFFH. The address lines are designated as 𝐴₁₅ to 𝐴₀, where 𝐴₁₅ is the most significant address bit. Which one of the following logic expressions will generate the correct $\overline {{\rm{CS}}}$ signal for this ROM?

((a))
A₁₅ + A₁₄ + (A ₁₃. A₁₂ + A̅₁₃. A̅ ₁₂)
((b))
A₁₅. A₁₄ . (A₁₃ + A₁₂)
((c))
A̅15 . A̅ 14 . (A 13 . A̅12 + A̅13 . A12)
((d))
A̅15 + A̅14 + (A13 . A12)

Show Answer

Answer: ((a))

A₁₅ + A₁₄ + (A ₁₃. A₁₂ + A̅₁₃. A̅ ₁₂)

Given address range: 1000 H to 2FFF

∴ Total No. of address lines = 13 i.e. $\begin{array}{*{20}{c}} {2{\rm{FFF}}}\ {\underline {1000} }\ {1{\rm{FFF}}} \end{array}$

(1FFF)_H = (0001 1111 1111 1111)₂ = 13 address lines

i.e. $\begin{array}{*{20}{c}} {{{\rm{A}}{15}}}&{{{\rm{A}}{14}}}&{{{\rm{A}}{13}}}&{{{\rm{A}}{12}}}&{{{\rm{A}}{11}}}&{{{\rm{A}}{10}}}& \cdots &{{{\rm{A}}_2}}&{{{\rm{A}}1}}&{{{\rm{A}}0}}&{}\ 0&0&0&1&0&0&{}&0&0&0&{ = {{\left( {1000} \right)}{\rm{H}}}}\ {}&{}& \vdots &{}&{}& \vdots &{}&{}&{}&{}&{}\ 0&0&1&0&1&1&{}&1&1&1&{ = {{\left( {2{\rm{FFF}}} \right)}{\rm{H}}}} \end{array}$

i.e. $\begin{array}{*{20}{c}} {{{\rm{A}}{15}}}&{{{\rm{A}}{14}}}&{{{\rm{A}}{13}}}&{{{\rm{A}}{12}}}\ 0&0&0&1\ 0&0&1&0 \end{array}$

To provide $\overline {{\rm{CS}}}$ as low, the condition is

A₁₅= A₁₄ = 0 and A₁₃ = A₁₂ = (01) or (10)

i.e. A15 = A14 = 0 and A13 , A12 shouldn't be (11) , (00)

Thus it is A15 + A14 + (A 13 . A12 + A̅13 . A̅ 12)

Hence option (1) is correct

In an N bit flash ADC, the analog voltage is fed simultaneously to 2^𝑁 − 1 comparators. The output of the comparators is then encoded to a binary format using digital circuits. Assume that the analog voltage source V_in (whose output is being converted to digital format) has a source resistance of 75 Ω as shown in the circuit diagram below and the input capacitance of each comparator is 8 pF.

The input must settle to an accuracy of 1/2 LSB even for a full-scale input change for proper conversion. Assume that the time taken by the thermometer to the binary encoder is negligible.

If the flash ADC has 8-bit resolution, which one of the following alternatives is closest to the maximum sampling rate?

((a))
1 megasamples per second
((b))
6 megasamples per second
((c))
64 megasamples per second
((d))
256 megasamples per second

Show Answer

Answer: ((a))

1 megasamples per second

For an N bit flash type converter, the number of comparators required is ${2^{\rm{N}}} - 1$

So, number of capacitors required for an 8 bit flash convertor is ${2^8} - 1 = 255.$

Since all capacitors are in parallel, total capacitance ${\rm{C}} = 255 \times 8{\rm{;pF}} = 2.04{\rm{;nF}}.$

So the input time constant ${\rm{\tau }} = {\rm{RC}} = 75{\rm{\Omega }} \times 2.04{\rm{;nF}} = 153{\rm{;nsec}}$

The input must settle within $\frac{1}{2}$ LSB. Now, $\frac{1}{2}{\rm{LSB}} = \frac{1}{2} \times \frac{1}{{{2^{\rm{N}}} - 1}}{{\rm{V}}{{\rm{in}}}} = {\left. {\frac{{{{\rm{V}}{{\rm{in}}}}}}{{510}}} \right|_{{\rm{N}} = 8}}$

So, voltage required by the capacitor to settle within $\frac{1}{2}{\rm{LSB}}$ is ${{\rm{V}}{\rm{c}}} = {{\rm{V}}{{\rm{in}}}} - \frac{{{{\rm{V}}_{{\rm{in}}}}}}{{510}}$

Now voltage across capacitor is given by ${{\rm{V}}{\rm{c}}}\left( {\rm{t}} \right) = {{\rm{V}}\infty } + \left( {{{\rm{V}}{\rm{o}}} - {{\rm{V}}\infty }} \right){{\rm{e}}^{ - \frac{{\rm{t}}}{{\rm{\tau }}}}}$.

We have ${{\rm{V}}\infty } = {{\rm{V}}{{\rm{in}}}}$, ${{\rm{V}}{\rm{o}}} = 0$. Substituting we have, ${{\rm{V}}{\rm{c}}}\left( {\rm{t}} \right) = {{\rm{V}}{{\rm{in}}}} + \left( {0 - {{\rm{V}}{{\rm{in}}}}} \right){{\rm{e}}^{ - \frac{{\rm{t}}}{{\rm{\tau }}}}}$

$\begin{array}{l} \Rightarrow {{\rm{V}}{{\rm{in}}}} - \frac{{{{\rm{V}}{{\rm{in}}}}}}{{510}} = {{\rm{V}}_{{\rm{in}}}}\left( {1 - {{\rm{e}}^{ - \frac{{\rm{t}}}{{\rm{\tau }}}}}} \right)\ \left( {1 - {{\rm{e}}^{ - \frac{{\rm{t}}}{{\rm{\tau }}}}}} \right) = \frac{{509}}{{510}}\ \Rightarrow {\rm{t}} = - {\rm{\tau }}\ln \left( {1 - \frac{{509}}{{510}}} \right)\ \Rightarrow {\rm{t}} = 953.87{\rm{;nsec}} \end{array}$

This is the time required to convert on set of data applied at the input.

So, sampling rate ${{\rm{f}}_{\rm{s}}} = \frac{1}{{\rm{t}}} = 1.04{\rm{;M}}\frac{{{\rm{samples}}}}{{{\rm{sec}}}}$

The state transition diagram for a finite state machine with states A, B and C, and binary inputs X,

Y and Z, is shown in the figure.

Which one of the following statements is correct?

((a))
Transitions from State A are ambiguously defined.
((b))
Transitions from State B are ambiguously defined.
((c))
Transitions from State C are ambiguously defined.
((d))
All of the state transitions are defined unambiguously.

Show Answer

Answer: ((c))

Transitions from State C are ambiguously defined.

From the state diagram we can write state table.

X	Y	Z	PS	NS	PS	NS	PS	NS
0	0	0	A	B	B	A	C	C
0	0	1	A	A	B	C	C	not specified
0	1	0	A	A	B	B	C	C
0	1	1	A	A	B	B	C	B
1	0	0	A	C	B	A	C	C
1	0	1	A	C	B	C	C	A
1	1	0	A	A	B	B	C	C
1	1	1	A	A	B	B	C	A/B (ambiguous)

<br>

So state C is ambiguously defined.

In the feedback system shown below $\rm G(s)=\frac{1}{{\left( {{{\rm{s}}^2} + 2{\rm{s}}} \right)}}$

The step response of the closed-loop system should have a minimum settling time and have no overshoot.

The required value of gain 𝑘 to achieve this is ________

In the feedback system shown below $\rm G(s)=\frac{1}{{\left( {{\rm{s}} + 1} \right)\left( {{\rm{s}} + 2} \right)\left( {{\rm{s}} + 3} \right)}}$ :

The positive value of 𝑘 for which the gain margin of the loop is exactly 0 dB and the phase margin of the loop is exactly zero degrees is ________.

The asymptotic Bode phase plot of $\rm G(s)=\frac{{\rm{K}}}{{\left( {{\rm{s}} + 0.1} \right)\left( {{\rm{s}} + 10} \right)\left( {{\rm{s}} + {{\rm{p}}_1}} \right)}}$ , with K and 𝑝₁ both positive, is shown below.

The value of 𝑝₁ is ________.

An information source generates a binary sequence {𝛼_𝑛}. 𝛼_𝑛 can take one of the two possible values −1 and +1 with equal probability and are statistically independent and identically distributed. This sequence is precoded to obtain another sequence {𝛽_𝑛}, as $\rm 𝛽_𝑛 = 𝛼_𝑛 + 𝑘 𝛼_{𝑛−3}$ . The sequence {𝛽_𝑛} is used to modulate a pulse $\rm X(t)$ to generate the baseband signal

${\rm{X}}\left( {\rm{t}} \right) = \mathop \sum \limits_{\rm{n= - \infty}}^\infty {\rm{;\beta_ng}}\left( {{\rm{t}} - {\rm{nT}}} \right),$ where ${\rm{g}}\left( {\rm{t}} \right) = \left{ {\begin{array}{*{20}{c}} {1,}&{0 \le {\rm{t}} \le {\rm{T}}}\ {0,}&{{\rm{othervise}}.} \end{array}} \right.$

If there is a null at 𝑓 = $\frac{1}{{3{\rm{T}}}}$ in the power spectral density of $\rm X(t)$ , then 𝑘 is ________

An ideal band-pass channel 500 Hz - 2000 Hz is deployed for communication. A modem is designed to transmit bits at the rate of 4800 bits/s using 16-QAM. The roll-off factor of a pulse with a raised cosine spectrum that utilizes the entire frequency band is ________

Consider a random process $\rm X(t) = 3V(t) − 8$ , where (𝑡) is a zero mean stationary random process with autocorrelation $\rm R_{V}(\tau)=4e^{-5\left|\tau\right|}$ . The power in $\rm X(t)$ is ________

A binary communication system makes use of the symbols “zero” and “one”. There are channel errors. Consider the following events:

𝑥₀ : a "zero" is transmitted

𝑥₁: a "one" is transmitted

𝑦₀ : a "zero" is received

𝑦₁ : a "one" is received

The following probabilities are given:

$\rm P(x_0) =\frac{1}{2},{\rm{;P}}\left( {{{\rm{y}}_0}{\rm{|}}{{\rm{x}}_0}} \right) = \frac{3}{4}\ and;p\left( {{y_0}{\rm{|}}{x_1}} \right) = \frac{1}{2}.$ .

The information in bits that you obtain when you learn which symbol has been received (while you know that a “zero” has been transmitted) is ________

The parallel-plate capacitor shown in the figure has movable plates. The capacitor is charged so that the energy stored in it is 𝐸 when the plate separation is 𝑑. The capacitor is then isolated electrically and the plates are moved such that the plate separation becomes 2𝑑.

At this new plate separation, what is the energy stored in the capacitor, neglecting fringing effects?

((a))
2𝐸
((b))
√2𝐸
((c))
𝐸
((d))
𝐸/2

Show Answer

Answer: ((a))

2𝐸

Concept:

Capacitance with the separation is r is given by ${{\rm{C}}} = \frac{{{\rm{\epsilon A}}}}{{{\rm{r}}}}$

Application:

Capacitance when the separation is 2d.

${{\rm{C}}_1} = \frac{{{\rm{\epsilon A}}}}{{2{\rm{d}}}}$

Let the charge on plates be Q coulomb. Then, the energy of the capacitor with separation 2d is

$\begin{array}{l} {{\rm{E}}_1} = \frac{{{{\rm{Q}}^2}}}{{2{{\rm{C}}_1}}}\ {{\rm{E}}_1} = \frac{{{{\rm{Q}}^2}}}{{2{{\rm{C}}_1}}} = \frac{{{{\rm{Q}}^2}2{\rm{d}}}}{{2{\rm{\epsilon A}}}} = \frac{{{{\rm{Q}}^2}{\rm{d}}}}{{{\rm{\epsilon A}}}} \end{array}$

The energy of the capacitor with separation d is

Now, ${\rm{C}} = \frac{{{\rm{A}}}}{{\rm{d}}} \Rightarrow {\rm{E}} = \frac{{{{\rm{Q}}^2}{\rm{d}}}}{{2{\rm{\epsilon A}}}}$

$\Rightarrow {\rm{E}} = \frac{{{{\rm{Q}}^2}{\rm{d}}}}{{2{\rm{\epsilon A}}}}$

Thus, we see ${\rm{E}} = \frac{1}{2}{{\rm{E}}_1}$

$\Rightarrow {{\rm{E}}_1} = 2{\rm{E}}$

A lossless microstrip transmission line consists of a trace of width 𝑤. It is drawn over a practically infinite ground plane and is separated by a dielectric slab of thickness 𝑡 and relative permittivity 𝜀_𝑟 > 1. The inductance per unit length and the characteristic impedance of this line are 𝐿 and 𝑍₀, respectively.

Which one of the following inequalities is always satisfied?

((a))
${Z_0} > \sqrt {\frac{{Lt}}{{{\varepsilon _0}{\varepsilon _r}w}}}$
((b))
${Z_0} < \sqrt {\frac{{Lt}}{{{\varepsilon _0}{\varepsilon _r}w}}}$
((c))
${Z_0} > \sqrt {\frac{{Lw}}{{{\varepsilon _0}{\varepsilon _r}t}}}$
((d))
${Z_0} < \sqrt {\frac{{Lw}}{{{\varepsilon _0}{\varepsilon _r}t}}}$

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Answer: ((b))

${Z_0} < \sqrt {\frac{{Lt}}{{{\varepsilon _0}{\varepsilon _r}w}}}$

Concept:

A lossless microstrip transmission line consists of a trace of width ‘w’ as shown:

The characteristic impedance of the transmission line is given as:

${Z_0} = \sqrt {\frac{L}{C}}$ , where $C = \frac{{\varepsilon A}}{d}$

Application: As we know ${Z_0} = \sqrt {\frac{L}{C}}$ where $C = \frac{{\varepsilon A}}{d}$

here d = t, A ≅ w, and ε = εeff

$\therefore C = \frac{{{\varepsilon _{eff}}w}}{t}$

The above is the actual capacitance, with no fringing taken into account.

$\therefore {Z_0} = \sqrt {\frac{L}{{\frac{{{\varepsilon _{eff}} \cdot w}}{t}}}} = \sqrt {\frac{{L.t}}{{w;{\varepsilon _{eff}}}}}$

Let the characteristic impedance of a practical transmission line be Z0’.

Since in practice, the capacitance C will be smaller than the actual capacitance due to the fringing effects, i.e.since εeff < ε0 εr, we can write:

Z0’ > Z0, i.e.

${Z_0} < \sqrt {\frac{{Lt}}{{{\varepsilon _0}{\varepsilon _r}w}}}$

This is because the capacitance is inversely related to the characteristic impedance, i.e. smaller the C, larger is the impedance.

A microwave circuit consisting of lossless transmission lines T₁ and T₂is shown in the figure. The plot shows the magnitude of the input reflection coefficient Γ as a function of frequency 𝑓. The phase velocity of the signal in the transmission lines is 2 × 10⁸ m/s.

The length 𝐿 (in meters) of T₂ is ________

A positive charge $\rm q$ is placed at $\rm x = 0$ between two infinite metal plates placed at $\rm x = −d$ and at $\rm x = +d$ respectively. The metal plates lie in the yz plane.

The charge is at rest at $\rm t=0$ , when a voltage $\rm +V$ is applied to the plate at $\rm – d$ and voltage $\rm −V$ is applied to the plate at $\rm x = +d$ . Assume that the quantity of the charge $\rm q$ is small enough that it does not perturb the field set up by the metal plates. The time that the charge $\rm q$ takes to reach the right plate is proportional to

((a))
$\rm d/V$
((b))
$\rm \sqrt d/V$
((c))
$\rm d/\sqrt V$
((d))
$\sqrt {{\rm{d}}/{\rm{V}}}$

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Answer: ((c))

$\rm d/\sqrt V$

Concept:

from Newton’s equations of motion ${\rm{s}} = {\rm{\vec ut}} + \frac{1}{2}{\rm{\vec a}}{{\rm{t}}^2}$

Application:

We have potential difference in between plates is given by ${\rm{V}} - \left( { - {\rm{V}}} \right) = 2{\rm{V}}$

Electric field in between plates, ${\rm{E}} = \frac{{2{\rm{V}}}}{{2{\rm{d}}}} = \frac{{\rm{V}}}{{\rm{d}}}$

Force on change Q, ${\rm{F}} = {\rm{QE}} = \frac{{{\rm{QV}}}}{{\rm{d}}}$

Let the charge $\rm Q$ be of mass $\rm m$ . then acceleration experienced by it,

${\rm{a}} = \frac{{\rm{F}}}{{\rm{M}}} = \frac{{{\rm{QV}}}}{{{\rm{md}}}}$

Now, the charge has to cover distance $\rm d$ before it reaches the right plate with initial velocity zero. So, from Newton’s equations of motion we have,

${\rm{s}} = {\rm{\vec ut}} + \frac{1}{2}{\rm{\vec a}}{{\rm{t}}^2}$

${\rm{u}} = 0,{\rm{;s}} = {\rm{d}}$ and ${\rm{a}} = \frac{{{\rm{qV}}}}{{{\rm{md}}}}$

$\begin{array}{l} \Rightarrow {\rm{d}} = \frac{1}{2}\left( {\frac{{{\rm{qV}}}}{{{\rm{md}}}}} \right){{\rm{t}}^2}\ \Rightarrow {{\rm{t}}^2} = \frac{{2{\rm{qmd}}}}{{{\rm{qV}}}}\ {{\rm{t}}^2} = \frac{{2{\rm{d}}.{\rm{md}}}}{{9{\rm{V}}}}\ \Rightarrow {\rm{t}} = \sqrt {\frac{{2{\rm{m}}}}{{\rm{q}}}} \cdot \frac{{\rm{d}}}{{\sqrt {\rm{V}} }} \end{array}$

Thus, ${\rm{t}} \propto \frac{{\rm{d}}}{{\sqrt {\rm{V}} }}$

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