gatecounsellor rcpp4 solution · here r is not given. but we know, when i/p is analog o/p is 1...

GATECounsellor RcPP4 Solution

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1. (B) Concept and Solution:

Poles are the values of z at which system becomes unstable or unbounded.

Mathematically we also define poles are defined values of z by equating denominator to zero,

Hence,

,

Hence, Poles are 0, -2,

2. (C) Concept and Solution:

if then is Orthogonal Matrix

Properties of the Orthogonal matrix are,

1. is also an orthogonal matrix

2. is also an orthogonal matrix

3.

4.


Definition of Order: Highest order of differentiation in Ordinary Differential Equation. Here it is 1

Definition of degree: Degree of highest differential order. Here it is 2

4. (D) Concept and Solution: In Newton Raphson Method, general formula for roots of the

equation is

If replace in Options (C) it is same as (A). Hence D is answer.

5. (D) Concept:

Common mode current gain

Emitter Injection Efficiency

Base transport factor

For BJT

Solution:

Here

6. (C) Concept:

Unity gain frequency of a MOSFET,

As length of transistor decreases, frequency increases.



7. (B) Concept:

Nyquist rate for sampling: the minimum sampling rate is twice the maximum frequency present

in the signal.

Solution:

Taking Fourier Transform of , we have

( Frequency of is shifted by

because of ) Therefore we have

Maximum frequency in is 30kHz.

Now , where . denotes multiplication and * denotes convolution.

Therefore, the limits of frequency in are , or

So, minimum sampling frequency = Nyquist rate = (By Nyquist Theorem)

8. (C)

Reference: Gaussian Random variables, Principles of Communication systems, Taub and

Schilling

9. (B) Concept: Remember the general equation of an angle-modulated wave.

]

Where A= amplitude of angle-modulated wave

= carrier frequency

=message signal

=frequency deviation/sensitivity

Solution: Compare with general equation. Since the second term is sin wave, must be

cosine so that the integral becomes sin.

Hence, is of the type:

Then,

Since, ,

Hence, on comparison with x(t)

Reference: Angle-modulation, Communication Systems by Simon Haykins

5

1

f(kHz)



10. (A) Concept:

An 8 bit DAC converts the 8 bit binary value in to analog value

The number of steps or difference level in N-bit DAC would be , where N signifies the size

of DAC( here N=8).

Hence steps size or Resolution= (let say)

Analog output value for input binary value=

Solution:-

Here R is not given. But we know, when i/p is analog o/p is 1 volt.

Hence =

………………………………….(.1)

Similarly

…………………………………….(2)

Dividing (2) by (1),

So, for input code , output voltage V

11. (A) Concept and Solution: All the statements are correct

a. HTL is designed from DTL gate with use of one zener diode. Zener diode is used to couple the

two transistor which increases noise immunity of gate.

b. DCTL exhibits current hogging phenomena. This is change in current at fixed output voltage

level because of absence of resistance in DCTL.

c. Problem of variable load is overcome by the totem pole arrangement.

12. (C) Concept: Instructions set of Microprocessor 8085 has following addressing modes

a. Immediate addressing mode ( ex. MVI B,FF))

b. Register direct addressing mode. (MOVE A,B)

c. Register indirect (XTHL, LDAX B)

d. Implicit (RRC, LLC)

Memory direct is type of Register addressing mode.

XTHL: Means Exchange HL with top of the stack pointer. Because of the pointer involvement this

instruction is called as Register indirect instruction.

Reference: Instruction set, 8085 Microprocessor by Gaonkar

13. (C) Concept: Static resistance is defined as .

14. (B)Concept: Class B has crossover distortion due to nonlinear current-voltage

characteristics of transistor.




The phase shift constant of wave propagating in waveguide is

= β 21 ( / )cf f , Where fc is cut off frequency, β is phase shift constant of wave propagating

in free space.

We know that waveguide acts as high pass filter. So f > fc . Therefore < β.

It is also known that

, where is guide wavelength and is free space wavelength and is cutoff

wavelength of waveguide.

>λ.


For a homogeneous medium,2 /V

For Inhomogeneous medium, .( )V

Where is charge density and is permittivity.

17. (C) Solution: Image theory is applicable to problem involving Both Electrostatic field and

Magneto static field.

18. (B) Concept:

When a periodic signal in time domain is converted in to freq. domain the resultant transform

will be discrete (i.e. frequency is discrete.)

Periodicity TransformDiscreteness; Aperiodicity Transformcontinuousness

19. (B) Concept:- Duality Theorem of Fourier Transform

If then by duality (Replacing t by – and by t)

Solution: We know that,

by duality

As is an odd signal.



20. (A) Concept:

I. Wave Explanation

i. Speech Speech is an 1-D varying signal .

ii. Image Image is a combination of 2-D pixels .

iii. T.V. Picture TV signal is a 3-D signal as 2-D image should be moved sequentially so

other dimension is required .

iv. Temperature Requires both space coordinates and time coordinate so .

21. (A) Concept: Herwitz Routh Criterion- For stable system first column of Herwitz table

should not change its sign. Number of times sign change is equal to number of poles in RHS.

Solution: Write the characteristic equation,

For no sign change we must have 3

Hence

22. (C) Concept:

The gain margin is a factor by which the gain of the system is allowed to increase before the

system reaches instability.

The phase margin of a stable system is the amount of additional phase lag required to bring the

system to bring it to the point of instability.

For stability both GM and PM should be positive. A system which has more positive GM and PM is

relatively more stable to system having lesser GM and PM.

; Phase margin is positive: Stable

or dB, PM : Marginally Stable

or dB (-ve), PM : Unstable ( -ve)

23. (B) Concept:

Low pass system:- System which blocks higher frequency value inputs

High pass system:- System which blocks lower frequency value inputs

All pass system:- System which allows all range of frequency value inputs to pass through or it

has symmetrical zero-pole about imaginary axis



Minimum phase system:- Transfer function of these systems have neither poles nor zero on the

right hand side of the s-plane.

Solution:- By signal flow graph method transfer function of the given block diagram is

So All pass system. Since magnitude is 1 or since it has symmetrical pole-zero about

imaginary axis

24. (A) Concept:

i) Current and voltage are in phase in purely resistive circuit or in circuit where capacitive and

inductive reactance nullifies each other i.e. they are of equal magnitude.

ii) Current leads voltage in RC circuit or in circuit where capacitive reactance is higher than

inductive reactance.

iii) Current lags voltage in RL circuit or in circuit where inductive reactance is higher than

capacitive reactance.

iv) Z=impedance R=resistance

Solution: Current is leading then combination ,

Hence

Thus, this is correct combination. 25. (C) Concept: KVL and KCL

Solution: V be voltage across 5Ω resistor

The sum of current should be zero at or

So

26. (C) Concept: A is coefficient matrix and K is augmented matrix ( it has LHS as last column in

coefficient matrix)

Let the rank of A be and that of K be and n is order of A

i) If the equations are inconsistent or it has no solution

ii) If the equations are consistent or it has unique solution

iii) If the equation are consistent or there are infinite number of solution



Solution:-

We have

Hence, If so Infinite solution.

27. (C) Concept: If probability of any event is very small it is assumed that it follows Poisson

probability distribution

Solution:

It follows a Poisson distribution as the probability of occurrence is very small.

Mean

Probability that more than 2 will get a infection

= 1- (Probability that no one get gets infection + Probability that one gets a infection +

Probability that two gets a infection)

=

=

=



28. (D) Concept and Solution:

Writing the given LDE in symbolic form

Complete solution of the LDE is given by = Complimentary Function +Particular Integral

Complimentary Function: Writing the auxiliary equation

Thus, Complementary Function is

Particular Integral

.

= —

Hence

29. (A) Concept: For a BJT, Common Base Configuration

Common Emitter Configuration

Hence BJT is current controlled current source

In BJT current is due to both hole and electrons

For a FET

. Hence it is voltage controlled current source.

In FET current is due to electron for n channel FET and hole for p channel FET.

So statement I is correct

But statement II is wrong as BJT is bipolar device but FET is unipolar as it carries current due to

only one type of charge carrier.

30. (C)Concept: Average energy of a signal constellation depends on the distance of signal points

from origin, while max. Probability of error depends on the min. distance between constellation

points. Please write the expressions

Solution: Prob of error is same in both cases, however, average energy of constellation (ii) is

greater.

Reference: Signal space, Principles of Communication systems, Taub and Schilling



31. (B) Concept: The extended source with source alphabet has Sn distinct blocks, where S is

the number of distinct symbols in source alphabet of the original source. In case of discrete

memoryless system, symbols are statistically independent. Hence, the probability of source

symbol in is equal to the product of the probabilities of the n source symbols in constituting

the particular source symbol in

And,

Solution:

Here, S0, S1, S2 having probabilities 0.25, 0.25 and 0.5 respectively.

So,

Symbols of source alphabet are,

S0S0 S0S1 S0S2 S1 S0 S1S1 S1S2 S2S0 S2S1 S2S2

0.25*0.25 0.25*0.25 0.25*0.5 0.25*0.25 0.25*0.25 0.25*0.5 0.5*0.25 0.5*0.25 0.5*0.5

bits

32. (C) Concept: Asynchronous counter

If N JK flip-flops are used counter repeats after . But if combinational circuit is also used

counter repeating after value less than can be designed. In this case counter is designed

so.

Solution:-Since NAND gate (combinational) is used counter repeats before 31. We have to look

for value which clears clear the JK flip flop.

At 10100 output of NAND gate will be ‘0’ which will clear the JK flip flop. Then it counts from

00000 again hence the Mod number is

33. (B) Concept: 2’s Compliment representation:- Two methods are used

1. When number is positive, magnitude is represented in true binary form and 0 is placed in MSB

to represent a positive number.

110 is +6

2. When the number is negative. The magnitude is written by 2’s compliment of the number and

1 is written in the place of MSB to represent it a negative number.

1010 is -6. How?

110 is +6 its 2’s compliment is 010, keeping 1 at MSB 1010 which is -6

By moving in reverse direction decimal equivalent of 2’s compliment number can be found.



Solution:

In 2’s complement, the first bit represents the sign.

Take negative positional decimal equivalent of MSB bit.

Add with positional decimal value of other bits present. We get the value of numbers. (This is

shortcut)

1010 = -8+2 = -6

111010 = -32+16+8+2 = -6

11010 = -16+8+2 = -6

34. (D) Concept:

1. Objective is to select the chip

2. Check whether chip is active low or High

3. If active low then output of combinational circuit (here NAND gate) should be zero

4. If active high then output of combinational circuit (here NAND gate) should be 1.

Solution:

Our objective is to select the chip. To select the chip, we need to get the output as logic ‘0’ for

NAND gate as chip is active low.

8085 has 16 bit address bus.

Since the gate is NAND gate, the output will be logic ‘0’ only when all the inputs are logic ‘1’. But

here A15 is inverted before applying to the gate. So it should be logic ‘0’. So

A15=0,A14=1,A13=1;

A12=0 or 1 since it is given as don’t care. We need to check with the options given.

The remaining inputs A11……A0 are as follows

A11A10A9A8A7A6A5A4A3A2A1 A0 are varied as

0 0 0 0 0 0 0 0 0 0 0 0 which is equivalent to 000 H

1 1 1 1 1 1 1 1 1 1 1 1 which is equivalent to FFF H

Therefore, 6000 H-6FFF H or 7000 H -7FFF H satisfy the above conditions

Hence address range is 6000 H-7FFF H.

Reference: 8085 Microprocessor by Gaonkar



35. (C) Concept: As is very large & finite, current in base can be assumed to be very small

since . Becomes significant because is very large.

Thevenin equivalent of such circuit has

Solution:- Thevenin equivalent of given circuit is

Now Since is very large

Now

Also in loop Collector to emitter




Range of current

Transistors Q1 and Q2 are matched transistors with

is in active region if

Hence we have

Range of is

37. (B) Concept: AC analysis

Solution:

Short the capacitors & short the voltage sources



Equivalent resistance seeing from emitter side

(Equivalent resistance seen from base side)

Angular frequency due to (since we have )

38. (C)Solution:

CMRR depends upon on both mismatch between transistors and resistance of the bias current

source.


The distance between adjacent voltage maxima is λ/2.

From the given data, λ/2 =15 cm. This implies λ= 30 cm.

Now, f = c/ λ = 1GHz.


Reflection Coefficient where and , ,

(assumption since nothing is given about this).

Therefore (air),

41. (B) Concept: Parseval’s Theorem

Solution: Since the Fourier transform is of form (2*a) / (a2 + 2). Its Inverse Fourier Transform

is exp(-a|t|) .

By

Hence by Parseval’s Theorem,



42. (D) Concept :

When no information about system is given, we assume it is LTI (Linear,Time Invariant) system.

The output in LTI system is convolution of input and system function .

Therefore, where stands for convolution.

……………………………………(1)

Solution:

From the first block diagram

From the second block diagram

By eqn (1)

comparing with

Reference: Signals and Systems by Oppenheim.

43. (D) Concept and Solution: Application of Final value theorem and its constraints.

Final value theorem is applicable only when magnitude of poles is less than 1 except a simple

pole at .

Since one pole lies outside the unit circle, the system is unstable. Hence Final value theorem is

not applicable.

44. (C) Concept and solution: By Laplace transform of RLC circuit and KVL

So By comparative study we get



45. (A) Concept and Solution: By KCL at each node

(When only are acting )

When and is acting

When and

When only

Solving we have,

46. (C) Concept:

Use Reciprocity theorem,

It is applicable only for single source network. It states that in a single source, linear bilateral

network the ratio of excitation to response remains same when position of the excitation and

response are interchanged.

47. (D)


If BJT is npn then in base region, minority carriers are electrons, whose concentration is given by

Now 10% of emitter doping concentration

By law of junction, we have,




In order to have punch through the depletion region in base at base, collector junction must start

touching the depletion region in base at base emitter junction.

Now as width in a region

Initially we have

Now if we increase to then, in base depletion region of 45 is needed to cause punch

through,

So we must have

From (1) and (2)


Magnitude Criteria:

From fig (b), for all points on root locus

Given, P(-6,0) is on the root locus, it implies satisfy the above equation



51. (B) Solution:

52. (B)Concept and Solution:

Information bit duration =

PN Chip duration =

Processing gain = = 4000

53. (C)Concept and Solution:

Jamming Margin (in dB) = Processing gain(in dB) -

Given,

Jamming Margin (in dB) =

Reference: Communication Systems by Simon Haykin, Chapeter 7, Page no: 479

54. (A) Concept and Solution: Equivalent Circuit

In general, this is a band pass filter.



transfer function of a high pass filler

transfer function of a low pass filler

55. (A) Solution:

Clearly it will be a band reject filter.

56. (C)juggernaut: crush

Quisling (betrayer, who betray): betray

Taunt (to provoke): provoke

Juggernaut (over powerful, destructive object, who destroy): crush (to destroy)

Inception (commencement, beginning): termination (opposite meaning)

Obstinate (firmly or stubbornly adhering to one's purpose, opinion): preserve (to keep)

Hence close relation to given word is (C) juggernaut: crush

57. (B) yielding

Recalcitrant- hard to deal, not obedient

Yielding-flexible, hence the opposite word.

58. (D) New

Novice means a person new to some work or circumstances.

59. (A) an eloquent The second clause describes the discourses of a person as informative and inspirational, which gives us a clue that the missing blank should relate to informative and inspirational discourses. An eloquent speaker is the one whose discourses are informative and inspirational. Hence (A) is the answer. 60. (B) 0.38

Condition for contradiction: If true then lie or if lie true.

Thus answer is

61. (A) has never been applied, Congress is required to call a convention to consider

possible amendments to the document when formally asked to do so

Corrections are was never applied=>has never applied has been required=> is required for considering=> to consider

62. (B) 5min. late

Total travel time including rest time according to watch at home=from 2:35 pm to

4:00pm. =1hr 25 min

Total travel time excluding rest time = 1hr 25 min - 25 min=1hr=60min



The speed during return to home was twice as that while going from home and hence the

time taken by the person to go office from home was twice that of going home from

office.

Thus ratio of time for going tooffice and coming from office and total time taken

only for travelling =60 min. min and min.

He took 40 min to go to office. Thus the time while he reached at office must be 2:35+40

min=3:15. But the time at office watch was 3:10. Hence the office watch was 5 min late.

63. (B) 28 ltr.

Let original milk in container is

After 4th operation quantity of milk present in the container lt. According

to the question, the ratio of quantity of milk to water after 4th operation is .

Thus, ratio of milk after 4th operation to the original milk amount

64. (C) 440

Considering relative speed, as A meets C every 88 seconds, then if C is at a constant point

then A covers the circular track in each 88 seconds. Same for B also which covers the

distance in 110 seconds.

With respect to C, speed of A, and that of B, . Relative speed of

A and B is

Where t is the time after which A meets B.

Thus

65. (C) 79

One line divides the whole space in to two and two lines into 4.

2 regions 4 regions 7 regions 11 regions

If no two lines are parallel and no three are concurrent, then 3rd line can cut the existing lines at most two places or three extra spaces are added. Thus 3 lines divide the whole space in to . One more line can intersect at most 3 points and will provide additional 4 spaces. The nth line can increase the region by k if and only if it divides k of the old regions and it divides k regions if and only if it intersects the existing lines at atmost points. 1 line region

2 lines regions

3 lines regions

4 lines regions



n lines=2+2+3+4+…..+n regions=1+1+2+……+n-1+n regions=Sn+1

Where Sn is sum of n natural numbers number of non-overlapping regions inside the circle (as all intersects are inside

the circle, the total regions are inside the circle)=1+S12=1+78=79

gatecounsellor rcpp4 solution · here r is not given. but we know, when i/p is analog o/p is 1...

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