Practice Test: Electronics Engineering (ECE)- 14 - Electronics and Communication Engineering (ECE) MCQ

# Practice Test: Electronics Engineering (ECE)- 14 - Electronics and Communication Engineering (ECE) MCQ

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## 65 Questions MCQ Test - Practice Test: Electronics Engineering (ECE)- 14

Practice Test: Electronics Engineering (ECE)- 14 for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Practice Test: Electronics Engineering (ECE)- 14 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Practice Test: Electronics Engineering (ECE)- 14 MCQs are made for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Practice Test: Electronics Engineering (ECE)- 14 below.
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Practice Test: Electronics Engineering (ECE)- 14 - Question 1

### A invest 1/3 part of the capital for 1/6 of the time, B invest 1/4 part of the capital for 1/2 of the time and C invest rest of the capital for rest of the time. Out of a profit of Rs. 23000, B’s share is?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 1
Raio of their Investment

Practice Test: Electronics Engineering (ECE)- 14 - Question 2

### Pipe A can fill a tank three times as fast as pipe B. If together two pipes can fill the tank in 48 min, the slower pipe alone will be able to fill the tank in:

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 2
A = 3B

Ratio of efficiency, A:B = 3:1

Ratio of times, A:B = 1:3

Total capacity = Total efficiency × Total time = 4 × 48 = 192 unit

Time taken by slower pipe

Practice Test: Electronics Engineering (ECE)- 14 - Question 3

### In the following question, out of the four alternatives, select the word opposite in meaning to the given word.Gratuitous

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 3
The word “gratuitous” means given or done free of charge. Thus, the word “costly” would be the correct antonym of the given word.

Gratis means without charge; free. Thus, option B is the correct answer.

Practice Test: Electronics Engineering (ECE)- 14 - Question 4

Find the area bounded between parabola and the line y2 = x,y = 2.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 4

y = 2, y2 = x

⇒ x = 22 = 4

∴ Parabola and intersect at point (4,2)

Practice Test: Electronics Engineering (ECE)- 14 - Question 5

The bar graph shows the number of employees working under the six different Departments (A, B, C, D, E, F) of a certain company. Study the diagram and answer the following questions.

If departments F and D are merged to create a new department G, then which department will have the least number of employees?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 5
If departments F and D are merged to create a new department G, then

Employees in department A = 25

Employees in department B = 6

Employees in department C = 10

Employees in department E = 15

Employees in department G = 8

∴ Department B has the least number of employees

Practice Test: Electronics Engineering (ECE)- 14 - Question 6

In the following question, a sentence is given with a blank to be filled in with an appropriate word. Select the correct alternative out of the four and indicate it by selecting the appropriate option.

Confusion prevails in madrasas in Uttar Pradesh over the distribution of free NCERT textbooks as the academic session ____________ from August.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 6

The answer is ‘has begun’ because we use Present Perfect Tense, if the action is important and not the time of action or an action that has recently finished. Thus, option D is the correct answer.

Practice Test: Electronics Engineering (ECE)- 14 - Question 7

A sum of Rs.400 amounts to Rs.480 in 4 years. What will it amount to if the rate of interest is increased by 2 % for the same time?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 7

We know that,

Now rate is increased by 2 %

So, new rate is 7%

New Amount = S.I + P = 112 + 400 = Rs.512

Practice Test: Electronics Engineering (ECE)- 14 - Question 8

The question below consists of a set of labelled sentences. These sentences, when properly sequenced form a coherent paragraph. Select the most logical order of sentences from among the options.

P: The Information and Broadcasting Ministry plans to conduct an independent study to gauge the impact of government advertisements on people.

Q: The advertisements are carried on various platforms, including print and visual media.

R: The Directorate of Advertising and Visual Publicity (DAVP) is the nodal agency of the government for advertising on behalf of the various ministries.

S: The initiative comes ahead of the Lok Sabha election in 2019 for which the government is expected to reach out to the people and highlight the works done by it in the past 4 years.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 8

P is the first statement. The word ‘initiative’ given in the sentence S is talking about the plans.

Hence, S must follow P. Now, the introduction the advertising agency is given in the sentence R, which must be the next statement.

Thus, the sequence after rearrangement is PSRQ and option B is the correct answer.

Practice Test: Electronics Engineering (ECE)- 14 - Question 9

In the following question, some part of the sentence may have errors. Find out which part of the sentence has an error and select the appropriate option. If the sentence is free from error, select 'No error'.

The gold foil used liberal (1)/ in Thanjavur paintings serves (2)/ many objectives that makes the painting more attractive. (3)/ No error

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 9

The error is in part (1) of the sentence. Change ‘liberal’ to ‘liberally’ because in this sentence it is in adjective form while the proper usage of liberal is in its adverb form i.e. ‘liberally’ as it qualifies the gold foil here.

Practice Test: Electronics Engineering (ECE)- 14 - Question 10

A, B and C can do a job in 6 days, 12 days and 15 days respectively. C works till 1/8 of the work is completed and then leaves. Rest of the work is done by A and B together. Time taken to finish the remaining work by A & B together is how much?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 10

Remaining work

∴ Time taken in doing 7/8 part of work

Practice Test: Electronics Engineering (ECE)- 14 - Question 11

Given a network with values of components depicted in the figure.

Find the sum of current through 5Ω and 4Ω resistor.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 11
Assume the current in the first loop be I1 and in the second loop be I2.

Then, we can apply mesh analysis to solve it.

Mesh 1: 15I1 - 10I2 = 5

Mesh 2: -10I1 + 20I2 = 10

On solving them, we have

I1 = 1A

I2 = 1A

Thus, sum = 2A

Practice Test: Electronics Engineering (ECE)- 14 - Question 12

The number of possible distinct Boolean expression of 3 variables will be ___________.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 12
The number of distinct Boolean expression of n variables is 22n, where n is number of variables

Practice Test: Electronics Engineering (ECE)- 14 - Question 13

Given the following system

T{X[n]} = X[n] + 3u[n+1]

Where u[x] represents unit step functions-

Which of the following is a correct representation of the system?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 13
We have,

T{X2[n] + X1[n]} = X1[n] + X2[n] +3u[n+1]

And

T{X1[n]} = X1[n] + 3u[n+1]

T{X2[n]} = X2[n] + 3u[n+1]

Since,

T{X2[n] + X1[n]} ≠ T{X1[n]} + T{X2[n]}

Thus, system is non linear.

T{X[n-no]} = X[n-no] + u[n+1]

≠ y[n-no]

Thus, system is Time Variant.

Practice Test: Electronics Engineering (ECE)- 14 - Question 14

Consider the following sentences regarding a constant signal.

1) A constant signal is a periodic signal because it has no fundamental period.

2) A constant signal is an anti periodic signal because it never repeats itself.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 14
1 is true because a constant signal is an special type of periodic signal whose fundamental period is not defined because it repeats itself at each and every small increment of time and in continuous time domain small increment can’t be calculated.
Practice Test: Electronics Engineering (ECE)- 14 - Question 15

List-1 (pole location) with list-2(shown constant amplitude with impulse response).

Select the correct answer using the codes given below.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 15
For A

If we plot A then it is similar to the 4 which is followed by equation K1+K2e-at+K3eat

For B

If we plot B then it is similar to the 1 which is followed by equation (sinat + sinbt) u(t)

For C

If we plot c then it is similar to the 3 which is followed by equation eatsinbt u(t)

For D

If we plot D then it is similar to the 2 which is followed by equation sinat u(t)

Practice Test: Electronics Engineering (ECE)- 14 - Question 16

In the root locus for open loop transfer function is the ‘break away/in points are located at

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 16
1 +GH = 0

s2 + 8s + 15 + K(s+6) = 0

Practice Test: Electronics Engineering (ECE)- 14 - Question 17

In a PCM system, if the code word length is increased from 6 to 8 bits, the signal to quantization noise ratio improves by the factor

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 17
When word length is 6

When word length is 8

Thus it improves by a factor of 16.

Practice Test: Electronics Engineering (ECE)- 14 - Question 18

The op-amp configuration shown below has following transfer function The feedback resistance used has 1.5 MΩ , the value of capacitance will be ___________ µf.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 18

Practice Test: Electronics Engineering (ECE)- 14 - Question 19

A semiconductor sample at room temperature has intrinsic concentration of 2.5 X 1017 /m3. After doping what will be the minority carrier concentration if the majority carrier concentration is given as 5.5 X 1021 /m3.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 19
In a pure Semiconductor (Intrinsic Semiconductor), the electron and hole concentrations are n1p1 respectively. By doping impurity atoms the SC becomes extrinsic then the electrons and hole concentrations n2p2 respectively, then the following equations are acceptable

n1p1 = n2p2 = ni2For Intrinsic Semiconductor, n = p = ni2 and as per questions before doping n1p1 = ni2

Therefore,

Practice Test: Electronics Engineering (ECE)- 14 - Question 20

An LED is connected as shown in figure. It should glow when V1 is at logic ‘0’ state (0.2V). Calculate R( in Ω). (Assume in active state current as 15 mA)

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 20
V1 is at logic ‘0’ ⇒ V1 = 0.2

By KVL

5 - 1.7 - 0.2 - (15 + R)15mA = 0

3.1/ (15 × 10-3) = 15 + R

R = 191.67 Ω

Practice Test: Electronics Engineering (ECE)- 14 - Question 21

For an ideal p-channel MOSFET, μp = 300cm2/v-s, W = 15μm, L = 1.5μm, tox = 300A, Vt = -0.7V. If the transistor is non-saturation region at VSD = 0.5V, then what is the Transconductance gm?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 21
ID = (μpCox/2)(W/L)(2(VSG+VT)VSD-VSD2)2,

where

Cox = (3.9 x 85 x 1014)/(300 x 1010).

Cox = 1.15 x 10-7 F/m2. Also, gm = ∂(ID)/∂(VSG).

On substituting and solving, gm = 0.172mS.

Practice Test: Electronics Engineering (ECE)- 14 - Question 22

In the circuit shown below express the current I0 in terms of Vi

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 22
The current going in the op amp is shown above

As per virtual short,

Practice Test: Electronics Engineering (ECE)- 14 - Question 23

From the circuit given below, find out the operating region of the transistors T1 and T2

(VTH = -0.4)

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 23
For T1

VSD = VS – VD = 1.5 – 0 = 1.5 V

VSD(sat) = VSG + VTH = (1.5 − 0.5) − 0.4

= 1 – 0.4 = 0.6V

Here, VSG > (VTH) & VSD > VSD(sat)

So, T1 is in Saturation region

Similarly for T2,

VSD = VS − V0 = 0.9 − 0.9 = 0

VSD(sat) = VSG + VTH = (0.9 − 0) − 0.4 = 0.5V

Here VSD < VSD(sat) & VSG > (VTH)

∴ T2 in linear region

Practice Test: Electronics Engineering (ECE)- 14 - Question 24

In the circuit shown below, assume that the voltage drop across a forward biased diode is 0.7 V. The thermal voltage V1 = kT/q = 25 mV. The small signal input Vi = Vp cos(ωt) where Vp = 100 mV.

The bias current IDC through the diodes is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 24
IDC = [12.7-(0.7 + 0.7 + 0.7 + 0.7)]/9900 = 1mA

Hence option A is correct

Practice Test: Electronics Engineering (ECE)- 14 - Question 25

A communication channel having AWGN characteristics is operating in such a way that SNR >> 1. The bandwidth of signal being transmitted is B and capacity C1. Determine the capacity of channel if a signal with half the bandwidth is transmitted through the same channel with same quality.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 25
Here same quality implies that SNR for transmitted signals is kept equal. We have a formula for channel capacity as

Now, B2 = 0.5 × B1. SNR is same, thus C2 would be,

Practice Test: Electronics Engineering (ECE)- 14 - Question 26

A three stage amplifier with identical stages with lower cut-off frequency per stage 'f1' is given overall negative feedback. Depending on the overall gain, the system may oscillate at a low frequency fc given by

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 26

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 14 - Question 27

Consider the differential equation 3yn(x) + 27y(x) = 0 with initial conditions y(0) = 0 and y'(0) = 2000. The value of y at x = 1 is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 27
3y′′(x) + 27y(x) = 0, y(0) = 0, y′(0) = 2000

Auxiliary equation, 3m2 + 27 = 0 ⇒ m2 + 9 = 0

⇒ m = 0 + 3i

yc = c1 cos⁡ 3x + c2 sin⁡3x and yp = 0

∴ yc = c1 cos ⁡3x + c2 sin ⁡3x

y(0) = 0 ⇒ c1 + 0 = 0 ⇒ c1 = 0

∴ y = c2 sin⁡3x

y = 3c2 cos⁡3x

y′(0) = 2000 ⇒ 2000 = 3c2 ⇒ c2 = 2000/3

Here in sin⁡ 3, 3 is in radian as it is a value.

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 14 - Question 28

Let z be a complex variable. For a counter-clockwise integration around a unit circle C, centred at origin, the value of A is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 28
5z - 4 = 0

z = 4/5 lies circle

Practice Test: Electronics Engineering (ECE)- 14 - Question 29

In the given circuit, the equivalent impedance of the circuit between terminals A-B is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 29
To find out Thevenin’s equivalent, we put a test source between terminals A-B.

KCL at node A

Practice Test: Electronics Engineering (ECE)- 14 - Question 30

The ROCs of different impulse responses are shown below. Which of the following impulse responses are stable?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 30
For stability of a system, ROC must include imaginary axis.
Practice Test: Electronics Engineering (ECE)- 14 - Question 31

The current mirror circuit acts as a _________.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 31
The current mirror circuit gains its name because it copies or mirrors the current flowing in one active device in another, keeping the output current constant regardless of loading.

The current being mirrored can be a constant current, or it can be a varying signal dependent upon the requirement and hence the circuit.

Hence it acts as a current regulator circuit.

Practice Test: Electronics Engineering (ECE)- 14 - Question 32

The given signal is the sum of two-real exponentials:

What will be the region of convergence (ROC) in this case?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 32
The given signal which is the sum of two real exponentials,

The z-transform will be,

Now, for convergence of X(z), both sums must converge. This requires,

Thus, the region of convergence is |z| > 1/2

Practice Test: Electronics Engineering (ECE)- 14 - Question 33

In the circuit given below, if current through the resistance R is given by I = 1 + 5 cos 2t , then find the value of R.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 33
Redrawing the circuit diagram with the branch currents, we get

We have, VA = IR = R(1 + 5 cos⁡ 2t)

Separating the AC and DC parts, we get, Vddc = R,Vdac = 5 R cos 2t

Idc = 1,Iac = 5 cos ⁡2t

Taking into account the DC part and eliminating the AC part, we get,

VAde = supply dc = 1V

Comparting it with the previous equation we have R = 1Ω

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 14 - Question 34

A ‘dual slope integrating type’ A to D converter, converts analog voltage of 15 volt to (11001010)2. If the clock frequency is 2 MHz, the total conversion time is _______ μsec.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 34
Conversion time for ‘dual slope integrating type’,

= 2n+1 × TCLK

= (28+1) × ½ usec

= 28 usec

= 256 usec

Practice Test: Electronics Engineering (ECE)- 14 - Question 35

The open loop transfer function of a unity negative feedback control system is given by

Find the value of K for the phase margin of the system to be 450 is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 35

If ωg is the gain crossover frequency

Practice Test: Electronics Engineering (ECE)- 14 - Question 36

Consider a binary channel shown below, the value of P(Y1) and P(Y2), if P(X1) and P(X2) are 0.3 and 0.7 respectively are

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 36

Channel matrix

After solving, we get P(Y) = [0.135, 0.865]

Practice Test: Electronics Engineering (ECE)- 14 - Question 37

The random process Z(t) is defined as Z(t) = X +Y; where X and Y are independent random variables. X is uniformly distributed on (-1, 1) and Y is uniformly distributed on (0, 1)

The auto correlation function of Z(t) is Rz (0) is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 37

∵ X and Y are independent

Rz(0) = E[X2] + E[Y2]

Practice Test: Electronics Engineering (ECE)- 14 - Question 38

Consider the following program for an INTEL 8085 microprocessor

1) MVI A, 92 H

2) ORA A

3) JP PORT

4) XRI 66 H

5) PORT OUT F2 H

6) HLT

The content of output port F2 H after execution of the program is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 38

1) A = 92H

2) A = 92H Parity flag = 0

3) Checks for parity

4) A = 92 ⊕ 66 = F4H

5) Send the content of accumulator to port F2H

⇒ Content of output port F2 H = F4

Practice Test: Electronics Engineering (ECE)- 14 - Question 39

Consider a GaAs sample at T = 300 K. Let the Hall effect device is fabricated with the following geometry: d = 0.01cm, W = 0.05 cm and L = 0.5 cm. The electrical parameters are : Ix = 2.5 mA, Vx = 2.2 V and Bz = 2.5 × 10-2 tesla. The hall voltage is VH = -4.5 mV. What will be the resistivity of the sample?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 39

Majority carrier concentration, n is given by

In Hall Effect, mobility μn is given by

= 8182cm2/V−sec

Hence, we obtain the conductance as

σ = eμnn

= 1.6×10−19× 8.68 × 1014 × 8182

= 0.88Ω - cm

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 14 - Question 40

Consider an abrupt PN junction (at T = 300 K) shown in the figure. The depletion region width Xn on the N-side of the junction is 0.2 μm and the permittivity of silicon (εsi) is 1.044 x 10-12 F/cm At the junction, the approximate value of the peak electric field (in kV/cm) is _________.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 40

Practice Test: Electronics Engineering (ECE)- 14 - Question 41

Given an arrangement of memories

1) 8 kB ROM

2) 8 kB RAM – 1

3) 4 kB RAM -2

4) Free space of memory of 4 KB after ROM.

Then the value of chip select (active low) for RAM-2 is ________?.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 41

ROM: 8 kB ⇒ P = 13, N = 8

Free space: 4 kB ⇒ P = 12, N = 8

RAM -1: 8 kB ⇒ P = 13, N = 8

RAM -2: 4 kB ⇒ P = 12, N = 8

Rom:

Free space:

RAM 1:

RAM -2:

Same as free space

0000 to oFFF.

The arrangement:

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 14 - Question 42

For an abrupt p-n junction diode, the doping concentrations on p-side is 9 × 1016 cm3 and an n-side is 5 × 1016/cm3. The diode is in reverse bias operating mode with total depletion width of 4 μm. Then the depletion width on n-side is __________ μm.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 42
Given

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 14 - Question 43

The Nyquist interval of signal

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 43

Hence here maximum frequency terms have frequency

Practice Test: Electronics Engineering (ECE)- 14 - Question 44

The input impedance of lossless λ/8 transmission line with other end short circuited is given by ______. (Assume characteristic impedance Zo = 50Ω)

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 44

Practice Test: Electronics Engineering (ECE)- 14 - Question 45

For a certain number system (51/4)b = (13)b is true

Then b is:

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 45

(51/4)b = (13)b

i.e., (51)b = (4)b x (13)b

∴ 5b + 1 = (4) · (b + 3) = 4b + 12

b = 12 - 1 = 11

Hence b = 11

Practice Test: Electronics Engineering (ECE)- 14 - Question 46

Following system of equation has

4x + 5y + 9z = 4

2x + y + 3z = 2

X + 2y + 3z = 1

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 46

Writing the Augmented matrix [A : B.

Rank (A) = Rank⁡(A : B) = 2 < 3="" (number="" of="" />

Hence the given system of equations have infinite solution.

Practice Test: Electronics Engineering (ECE)- 14 - Question 47

A 400W, 220V bulb is supplied with 110V; then the power consumed by the bulb will be____?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 47

Hence option C is correct.

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 14 - Question 48

For the PNP transistor circuit, given RB = 500kΩ, VCC = 12 volt, ICBO = 18 μA, β = 60. The value of ICQ = ____ mA?

(Assume |VBE| = 0.7 V)

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 48
Given β = 60, VCC = 12 volt, RB = 500 k, ICBO = 18 μA

By KVL to Base-emitter loop 0 + 0.7 + (500k) IBQ - 12 = 0

Now we have

ICQ = βIBQ + (1 + β) ICBO

∴ ICQ = (60)(22.6) + (1 + 60) (18)

= 2.454 mA

Practice Test: Electronics Engineering (ECE)- 14 - Question 49

A waveform of second derivative of a certain signal is shown below.

The Fourier transform of given signal f(t) is.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 49

We have

taking laplace transform,

Hence option (C)

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 14 - Question 50

For the signal

The ratio of power content in 1st harmonics and 8th harmonics is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 50
We have

the period of signal is = T1 = 8T2

i.e., T = 1

∴ ωm = 2πrad/sec - ... Fundamental frequency,

Now

Comparing with

Power in 8th harmonic

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 14 - Question 51

Let Q√γ be the BER of a BPSK system over an AWGN channel with two-sided noise power spectral density N0/2. The parameter γ is a function of bit energy and noise power spectral density.

A system with two independent and identical AWGN channels with noise power spectral density N0/2 is shown in the figure. The BPSK demodulator receives the sum of outputs of both the channels.

If the BER of this system is Q b√γ, then the value of b is _____________.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 51
Bit error rate for BPSK

Function of bit energy and noise

Counterllation diagram of BPSK

Channel is AWGN which implies noise sample as independent

Let 2x + n1 + n2 = x' + n'

where x' = 2x

n′ = n1 + n2

Now Bit error rate

E1 is energy in x2

is PSD of h1

E1 = 4E[as amplitudes are getting doubled]

= N0[ independent and identical channel

Practice Test: Electronics Engineering (ECE)- 14 - Question 52

An LTI system with transfer function H(f) is to be obtained in such a manner that it can generate a random signal x(t) having autocorrelation function Rx(τ) = 3ɳe-3|τ|, by passing white noise n(t) having PSD Sn(f) = ɳ/2 watts/Hz. What would be the suitable transfer function H(f)?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 52

We have autocorrelation function

Then spectral density at output is given by

This will give

We have a relation while filtering through an LTI system given by Sx(f) = |H(f)|2Sn(f)

This relation would give impulse response the LTI system as

Practice Test: Electronics Engineering (ECE)- 14 - Question 53

A biliary symmetric channel (BSC) has a transition probability of 1/8. If the binary transmit symbol X is such that P(X = 0) = 9/10, then the probability of error for an optimum receiver will be

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 53

Py = 1)/(x = 0 P(x = 0)/Py = 1)/(x = 1P(x = 1)

P(x = 0) = 9/10

P(x = 0)+P(x = 1) = 1 So P⁡(x = 1) = 1/10

Transition Probability is nothing but Probability of changing the value from 0 ta 1 and 1 to 0

i.e.

P(0/1) = P(1/0) = 1/8

Pe = P(x = 0)Py = 1)/(x = 0 + P(x = 1)Py = 0)/(x = 1

= (9/10) × (1/8) + (1/10) × (1/8)

= 1/8

But this nat the Pe for optimum Receiver. So,Pe = 1 − Pc

where Pc is Probability of correct detection Use MAP detection Rule:

The received symbal m = m1 if P(y/x1)P(x1)/P(y/xj)P(xj) ≥ 1

This is obtained from following:

Pc(m = mi) = P(mi/y)⋅P(y)=P(xi/y)⋅P(y)

M is the estimated output signal Using Baye's theorem

P(xi/y)⋅P(y)=P(y/xi)⋅P(xi)

when o/py = 0 then Py = 0)/(x = 0P(x = 0)/Py = 0)/(x = 1P(x = 1)

= (7/8⋅9/10)/(1/8⋅1/10) = 63 ≥ 1

So when y = 0, m = 0

Similarly when y = 1 then =(1/8⋅9/10)/(7/8⋅1/10) = 9/7 ≥ 1

So when y = 0 then alsa

So if we choose the rule such that the message signal is 0 for y =0 (or) 1 then the Probability of error is nothing but P(x = 1) = (1/10)

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 14 - Question 54

If an electric field is given as:

Determine the work (in nJ) that is to be done in moving a 3μC charge along this path if the path is located at P(-0.3, -4, 0.6).

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 54
Here, length and work is differential in nature, and hence, no integration is required.

(Differential because charge is moved through ΔL).

Therefore,

= -Q[-8y2z+40xyz -24xy2] x 10-6 (in μm)

Now it is given that Q = 3 x 10-6

So, we get: dW= -3 x 10-6 [-8y2z +40xyz -24xy2] x 10-6

Now, we have point P = (-0.3, -4, 0.6)

x = -0.3, y = -4 and z = 0.6

Therefore, dW= -3 x 10-6[ (-8 x (-4)2 x 0.6) +40(-0.3) (-4)(0.6) - 24(-0.3)(-4)2] x 10-6

= 4.89 x 10-10 J.

Practice Test: Electronics Engineering (ECE)- 14 - Question 55

A 15 A source is operating at 100 MHz and feeds a Hertzian dipole of length 6mm situated at origin of space, determine electric field at P (4, 300, 900)

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 55

since it is a current source driving a hertzian dipole, We need to get Magnetic field intensity and then Electric field intensity as,

Here, I0 = 15A,r = 4m,η = 377Ω, dl = 6mm,θ = 30 and β can be determined,

Practice Test: Electronics Engineering (ECE)- 14 - Question 56

In SSB modulation system, if signal (cos 30t) is modulated with a carrier signal of frequency 850 rad /sec, then the expression for the lower side band is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 56

m(t) = cos⁡ 30t

m(t) = cos⁡(30t − 90) = sin⁡30t

For lower side band

s(t) = cos⁡30t cos ⁡850t + sin⁡ 30t sin ⁡850t

=cos⁡(850−30)t

= cos⁡ 820t

Practice Test: Electronics Engineering (ECE)- 14 - Question 57

The divide by N counter is shown below. If initially Q0 = 0, Q1 = 1, Q2 = 0 the value of N is _______.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 57

State is repeating after 5 clock pulses.

N = 5

Practice Test: Electronics Engineering (ECE)- 14 - Question 58

The logic function implemented by the circuit given below is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 58

Practice Test: Electronics Engineering (ECE)- 14 - Question 59

Consider following connection to the memory

The accessible range of address from memory is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 59

From the given data A00 − A11 are 12 address lines while remaining 4 address lines are decided by OR gate If A12A13A14A15 all are zero then only chip select will active.

Hence A12 − A15 will remain fixed at zero A15A14A13A12A11A10A9A8A7A6A5A4A3A2A1A0

Accessible range is 0000− OFFF H

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 14 - Question 60

A box contains 2 identical bags A and B . Bag A contains 2 Red and 5 Green balls. Bag B contains 2 Red and 6 Green Balls. A person draws a ball at random. If the drawn ball was Red what is the probability that it was from bag A?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 60

Practice Test: Electronics Engineering (ECE)- 14 - Question 61

An intrinsic semiconductor bar of Si is doped with donor type impurity to the extent of 1 atom per 108 silicon atoms, then the resistivity of silicon crystal will be (atomic density of Si crystals =5 × 1022 atoms /cm3 and μn = 1300cm2N − sec)

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 61

∵ It is an n -type semiconductor

Practice Test: Electronics Engineering (ECE)- 14 - Question 62

The system of equations x - 4y + 7z = 12 , 3x + 8y - 2z = 10, 26 z - 8y = 6

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 62

Here, Rank [A : B] = Rank⁡[A] = 3 = n

And n = 3

Therefore, the system has a unique solution.

Practice Test: Electronics Engineering (ECE)- 14 - Question 63

(D2 − 5D + 6)y = ex cos x is a Partial differential equation whose solution is given by

y = C1e2x + C2e3x − Kex(3sin2x + cos2x)

The value of K is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 63

Practice Test: Electronics Engineering (ECE)- 14 - Question 64

Evaluate using Simpson’s ⅓ rd rule. Using h = 0.5

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 64

Simpson's ⅓ rd rule

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 14 - Question 65

In a certain animal husbandry weight of animals had a distribution as shown in table below:

Mean weight of animals in the husbandry is______.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 65

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