GATE  >  GATE ECE (Electronics) 2023 Mock Test Series  >  GATE Mock Test Electronics Engineering (ECE)- 11 Download as PDF

GATE Mock Test Electronics Engineering (ECE)- 11


Test Description

65 Questions MCQ Test GATE ECE (Electronics) 2023 Mock Test Series | GATE Mock Test Electronics Engineering (ECE)- 11

GATE Mock Test Electronics Engineering (ECE)- 11 for GATE 2022 is part of GATE ECE (Electronics) 2023 Mock Test Series preparation. The GATE Mock Test Electronics Engineering (ECE)- 11 questions and answers have been prepared according to the GATE exam syllabus.The GATE Mock Test Electronics Engineering (ECE)- 11 MCQs are made for GATE 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for GATE Mock Test Electronics Engineering (ECE)- 11 below.
Solutions of GATE Mock Test Electronics Engineering (ECE)- 11 questions in English are available as part of our GATE ECE (Electronics) 2023 Mock Test Series for GATE & GATE Mock Test Electronics Engineering (ECE)- 11 solutions in Hindi for GATE ECE (Electronics) 2023 Mock Test Series course. Download more important topics, notes, lectures and mock test series for GATE Exam by signing up for free. Attempt GATE Mock Test Electronics Engineering (ECE)- 11 | 65 questions in 180 minutes | Mock test for GATE preparation | Free important questions MCQ to study GATE ECE (Electronics) 2023 Mock Test Series for GATE Exam | Download free PDF with solutions
1 Crore+ students have signed up on EduRev. Have you?
GATE Mock Test Electronics Engineering (ECE)- 11 - Question 1

Statement: All bags are cakes. All lamps are cakes.
Conclusions:
I. Some lamps are bags.
II. No lamp is bag.
Deduce which of the above conclusion logically follows statements:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 1

Since the middle term 'cakes' is not distributed even once in the premises, no definite conclusion follows. However, I and II involve only the extreme terms and form a complementary pair. So, either I or II follows.

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 2

Man does not live by __________ alone.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 2

Man does not live by bread alone.

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 3

Extreme focus on syllabus and studying for tests has become such a dominant concern of Indian students that they close their minds to anything ___________ to the requirements of the exam.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 3

Extraneous – irrelevant or unrelated to the subject being dealt with

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 4

Select the pair that best expresses a relationship similar to that expressed in the pair:
LIGHT : BLIND

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 4

As a blind person cannot see light and a Dumb person cannot speak. So option (a) is the correct answer.

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 5

If = log(a + b), then:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 5

log a/b + logb/a = log ( a + b)

log ( a/b * b/a ) = log ( a + b )

log 1 = log ( a + b)

So, a + b = 1 .

a = 1 - b

b = 1 - a.

Therefore, If log a/b + log b/a = log (a + b), then a+ b = 1

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 6

In an exam, the average was found to be 50 marks. After deducting computational errors the marks of the 100 candidates had to be changed from 90 to 60 each and average came down to 45 marks. Total number of candidates who took the exam was:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 6

Let number of candidates be n
50 × n – 100 × 30 = 45 × n
n = 600

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 7

A solid 4cm cube of wood is coated with red paint on all the six sides. Then the cube is cut into smaller 1cm cubes. How many of these 1cm cubes have no colour on any side?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 7

Number of cubes with 3 faces red = 8
Number of cubes with 2 faces red = 24
Number of cubes with 1 face red = 24
Number of cubes with no face red = 64-(24+24+8) = 8

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 8

One of the warmest winters on record has put consumers in the mood to spend money.Spending is to be the strongest in thirteen years. During the month of February, sales of existing single family homes hit an annual record rate of 4.75 million. This paragraph best supports the statement that:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 8

This clearly the best answer because the paragraph directly states that warm weather affects consumers inclination to spend. It furthers states that the sales of single-family homes was at an all-time high. There is no support for choice a or c. Choice b is wrong because even though there were high sales for a particular February, this does not mean that sales are not higher in other months. Choice d presents a misleading figure of 4 million. The paragraph states that the record of 4.75 million was at an annual, not a monthly, rate.

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 9

The age of father is 4 times more than the age of his son Amit. After 8 years, he would be 3 times older than Amit. After further 8 years, how many times will he be older than Amit?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 9

Let Amit’s age be n years
Age of his father = (4n + n) = 5n years
After 8 years:
Amit’s age = n + 8
Father’s age =
3(n + 8) = 5n + 8
3n + 24 = 5n + 8
16 = 2n
n = 8 years
After further 8 years:
Amit’s age = 24 years
Father’s age = 5n +16 = 56 years
So father is 2.33 times older than Amit.

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 10

Pipe A, B and C are kept open and together fill a tank in t minutes. Pipe A is kept open throughout, pipe B is kept open for the first 10 minutes and then closed. Two minutes after pipe B is closed, pipe C is opened and is kept open till the tank is full. Each pipe fills an equal share of the tank. Furthermore, it is known that if pipe A and B are kept open continuously, the tank would be filled completely in t minutes. Find t.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 10
  • A is kept open for all t minutes and fills one-third the tank. Or, A should be able to fill the entire tank in '3t' minutes.
  • A and B together can fill the tank completely in t minutes. A alone can fill it in 3t minutes.
  • A and B together can fill 1t of the tank in a minute. A alone can fill 13t of the tank in a minute. So, in a minute, B can fill 1t−13t=23t. Or, B takes 3t2 minutes to fill an entire tank.
  • To fill one-third the tank, B will take t2 minutes. B is kept open for t - 10 minutes.
    t2 = t - 10, t = 20 minutes.
  • A takes 60 minutes to fill the entire tank, B takes 30 minutes to fill the entire tank. A is kept open for all 20 minutes. B is kept open for 10 minutes.
  • C, which is kept open for 8 minutes also fills one-third the tank. Or, c alone can fill the tank in 24 minutes. Choice (D) is the correct answer.
GATE Mock Test Electronics Engineering (ECE)- 11 - Question 11

The Thevenin's equivalent resistance to the left of AB in figure given below is given by:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 11

For finding the Thevenin’s equivalent resistance, 8 A current source will be open circuited while V volt voltage source will be short circuited.

Hence 

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 12

What is the time period of the sinusoidal signal
x(n) = 10 cos(0.4n)  

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 12

Period N= K
=5K
= 5 {K = 1}

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 13

In a radar system, the range of R1 is achieved at a frequency f1. Then the range R2 at frequency R2 = 8f1 would be __________ [Neglect effect of 1 on the Beam width.]

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 13

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 14

Gate function is defined as ______________

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 14

A gate function is a rectangular function defined as

Where τ is pulse width.

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 15

A signal m(t) = 5 cos(2π100t) is frequency modulated a carrier. The resulting FM signal is:

The approximate bandwidth of the FM signal would be:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 15


  = 
  =  3.2 KHz

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 16

A system has a receiver noise resistance of It connected to an antenna with an input resistance of The noise figure of the system is:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 16

Noise figure = 1 +
Where, is the equivalent receiver noise resistance & is the resistance of antenna So, F = 1+ =2

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 17

A phase lead compensating network has its transfer function The maximum phase lead occurs at a frequency of:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 17

Comparing with the standard phase lead compensating network


= 50 rad /sec

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 18

The time taken for the output to settle within 2% of step input for the control system represented by is given by:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 18

Comparing the above transfer function with the standard second order transfer function



GATE Mock Test Electronics Engineering (ECE)- 11 - Question 19

If diameter of a circular cross-sectional antenna is 2m, and it is transmitted a signal at frequency 100MHz, determine the minimum distance for far field approximation.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 19

For far field approximation,

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 20

A load impedance is to be matched to a lossless transmission line by using a quarter wave line transformer. The characteristics impedance of the quarter wave transformer.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 20

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 21

What is the velocity of a plane wave in a lossless medium having a relative permittivity of 5 and relative permeability of unity?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 21

Velocity of propagation is given by 

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 22

A dipole antenna of/2 length has an equivalent total loss resistance is 0.5. The efficiency of antenna is:  

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 22

Radiation resistance of antenna,

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 23

If an electrical system is compared to a fluid system, the electrical current corresponds to the:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 23

Electric current is nothing but flow of charge. Similarly in fluid system flow of water is comparable with electric current.

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 24

For the circuit shown below, the collector to emitter voltage is:
 

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 24

Let the transistor be operating under active region.

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 25

A power supply has a full load voltage of 24V. What is its no-load voltage for 5% regulation?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 25

Voltage - Regulation,

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 26

Two 100 W, 220 V bulbs are required to be connected across a 400 volts supply. The value of resistance to be inserted in the line so that the voltage across the bulbs does not exceed 220 V is:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 26

Total power drawn from the circuit = 2 x 100 = 200 W 

Hence, supply current is:

Let R be the series resistance to be inserted in the circuit such that the voltage across the bulbs is 220 V.

Applying KVL, we get:

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 27

For a given op-amp, CMMR = 105 and differential gain Ad = 105. The common-mode gain of the op-amp is:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 27

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 28

The system of linear equations:
3x + 4y = 6
2x + 3y = 5
has:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 28

Auguemented matrix,


Since rank of matrix A as well as auguemented matrix is 2, therefore it has unique solution.

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 29

Which of the following is a solution to the differential equation ?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 29

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 30

The value of the integral dz  is: (where C is the circle z = 1)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 30



Since both the poles lie outside the unit circle therefore according to Cauchey – Integral,
the integration is zero.

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 31

A fair dice is tossed two times. The probability that the second results in a value that is higher than the first toss is:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 31

In the first toss, results can be 1, 2, 3, 4, 5
For 1, the second toss results can be 2, 3, 4, 5, 6
For 2, the second toss results can be 3, 4, 5, 6
For 3, the second toss results can be 4, 5, 6
For 4, the second toss results can be 5, 6
For 5, the second toss results can be 6
So the required probability

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 32

where P is a vector is equal to:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 32

∇ × ∇ × = ∇(∇.P) − ∇2P

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 33

A certain multiplexer can switch one of 32 data inputs to its output. The number of control inputs in this multiplexer:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 33

No. of input lines = 2m (where, m = No. of select/ control inputs)
or, 32 = 2m or m = 5.

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 34

The initial of MOD-16 down counter is 0110. After 37 clock pulses, the state of the counter will be:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 34


After 37 clock pulses, the state of MOD-16 DOWN counter will be five states below the present state.

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 35

12 MHz clock frequency is applied to a cascaded counter of modulus-3 counter, modulus-4 counter, and modulus-5 counter. Determine the output frequency.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 35

Modulus of cascaded counter = 3 * 4 * 5 = 60
So, frequency of the output signal,

= 200 KHz

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 36

The value of Z22 (Ω) for the circuit shown below is:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 36

Applying KVL, we get:

Putting the value of V1 from equation (ii) into equation (i), we get:

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 37

 An ac voltage of 220 V is applied to a pure inductance of 50 H. If the current is 5 A, the instantaneous value of voltage and current will be respectively given by:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 37

Max. value of current:

Assuming voltage as the reference phasor:

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 38

For the circuit shown below, a resistance R is to be connected across an active network, it is given that when

(i) R = infinity, V = 5 volts

(ii) R = 0, current output of the active source is 2.5 A.

What would be the current when the value of R is 5 Ω?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 38

When R = ∞, Voc= V = 5 volts
When R = 0 , ISC = 2 .5 A

Hence, the Thevenin's equivalent circuit will be drawn as shown below.

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 39

The output y(t) of a continuous – time system S for the input x(t) is given by

Which one of the following is correct?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 39


Therefore s(t) = u(t)
Which is linear and time-invariant

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 40

What is the output of the system with in response to the input  ?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 40


GATE Mock Test Electronics Engineering (ECE)- 11 - Question 41

The 3-dB bandwidth of the low-pass filter having impulse response is is:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 41

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 42

What is the inverse fourier transform of u(w)?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 42

By duality property

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 43

The signal flow graph shown below has M number of forward path and N number of individual loops.

Q. ​What are the values of M and N ?

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 44

The transfer functions of the sytem is given by  The impulse response of the system is: 

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 44

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 45

Determine the value of K for a unity feedback system with to have a peak overshoot of 50%.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 45

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 46

Assuming an operating temperature T = 300K and =26 mV. What is the change in semi-conductor silicon diode forward voltage to produce a 10:1 change in dioide current  while operating in the forward bias region (<25mA)?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 46

For silicon diode

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 47

An n-type semiconductor is illuminated by a steady flux of photons with energy greater than band gap energy. The change in conductivity obey which relation?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 47

When the semiconductor is illuminated by a steady flux of photons, the number of new hole-election pairs is proportional to the number of incident photons and also they are equal
i.e.
Therefore, change in conductivity

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 48

A single-phase full-bridge diode rectifier delivers a load current of 10 A, which is ripple free. The rms and average values of diode currents are respectively:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 48

Given, I0 = 10A
The rms value of diode current,

The average value of diode current,

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 49

What is the value of inductance to be used in the inductor filter connected to a full-wave rectifier operation at 60 Hz to provide a DC output with 4% ripple for a 100 Ω load?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 49

The ripple factor for an inductor filter is

∴  
or  

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 50

For the circuit shown below, the output voltage is given by

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 50

Using superposition theorem, the output voltage is given by

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 51

An amplifier using BJT has two identical stages each having a lower cut-off (3dB) frequency of 64Hz due to coupling capacitor. The emitter bypass capacitor also provides a lower cut-off (3dB) frequency due to emitter degeneration alone of 64Hz. The lower 3dB frequency of the overall amplifier is nearly?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 51

So overall lower cut-off frequency

= 100 Hz

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 52

What is the value of  where  
Here, s is the surface bounded by x = 0, x = 1, y = 0, y = 1, z = 0, z =1 and are unit vectors along x, yand z axes respectively.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 52

By divergence theorem,



Since, the surface s is bounded by x = 0, 1; y = 0, 1 and z = 0, 1 so, putting the limits, we have

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 53

For sea water with = 5 mho/m and = 80, what is the distance for which radio signal can be transmitted with 90% attenuation at 25 KHz?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 53


= 44938.6
Since
hence sea water is a acting as a good conductor Since attenuation is 90%, so transmission is 10%
So,


 x = 3.27 m

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 54

A lossless transmission line with = 50 and = 1.5m connects a voltage source  to a load  If  = 60V and = 50 then determine the current I supplied by the source at a frequency 100 MHz.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 54

Here,

So current supplied by the generator

= (0.48 – j 0.24) A

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 55

A pulse train with a frequency of 1MHz is counted using a mod-1024 ripple counter built with J–K flip-flops. For proper operation of the counter the maximum permissible propagation delay per flip-flop stage is:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 55

Here mod – 1024
So to count 1024 number of flip-flops requires
n = 10
of counter

So propagation delay per flip flop

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 56

In a binary adder with two inputs X and Y, the correct set of logical expression for the output S (sum) and C (carry) are respectively:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 56

The truth table for half adder is shown below

Here, sum output   and carry output is C= XY

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 57

Consider the following statements:

1. When a MUX is used to implement a logic function, the logic variables are applied to the MUX’s data inputs.

2. The circuit for a DEMUX is basically the same as that for a decoder.

Of these:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 57

When a MUX is used to implement a logic function, the logic variables are applied to the MUX's select or control input.
For example, if we want to implement a NOT gate using a 2 x 1 MUX, then the circuit will be represented as shown below:

Hence, statement-1 is not correct.

Statement-2 is correct because the internal circuit for a DEMUX (data distributor) and a decoder are basically same.

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 58

A liner system has the transfer function When it is subjected to an input white noise process with a constant spectral density A, the spectral density of the output will be:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 58

PSD of output:

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 59

A 3 x 8 decoder with two enable inputs is to be used to address 8 blocks of memory. What will be the size of each memory block when addressed from a sixteen bit bus with two MSBs used to enable the decoder?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 59
  • A 16-bit address space contains 64K addresses. 
  • Since you're using the two msb's for enabling inputs, you can select among 4 16K address regions. 
  • Since each of those regions is split into 8 blocks, each block contains 2K addresses.
GATE Mock Test Electronics Engineering (ECE)- 11 - Question 60

Four signals are to be multiplexed and transmitted.  and  have a bandwidth of 4 KHz, and remaining two signals have bandwidth of 8KHz. Each sample requires 8 bits for encoding. What is the minimum transmission bit rate of the system?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 60


=48 K samples / sec

= 384 Kbps

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 61

A telephone channel has bandwidth B of 3KHz and SNR of 30dB. It is connected to a telephone machine having 32 different symbols. The symbol rate required for errorless transmission is nearly:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 61

The symbol rate required for error tree transmission, i.e. nyquist sampling rate is

= 6000 symbols/sec

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 62

A modem is classified as low speed if data rate handled is:

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 63

The divergence of the vector field

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 63


at point (1, 2, 1)

      = 8

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 64

A solution for the differential equation with initial condition  =0 is:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 64

Taking Laplace transform both sides we get,

GATE Mock Test Electronics Engineering (ECE)- 11 - Question 65

For the transistor circuit shown below, the Q-point values are given as ICQ = 1 mA and VCEQ = 6V 

If VBE (on) = 0.7 V, the values of the RB and Rc will be respectively:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 11 - Question 65

The collector resistance is:

Base current is,

Use Code STAYHOME200 and get INR 200 additional OFF
Use Coupon Code
Information about GATE Mock Test Electronics Engineering (ECE)- 11 Page
In this test you can find the Exam questions for GATE Mock Test Electronics Engineering (ECE)- 11 solved & explained in the simplest way possible. Besides giving Questions and answers for GATE Mock Test Electronics Engineering (ECE)- 11, EduRev gives you an ample number of Online tests for practice