GATECounsellor RcPP2 Solutions · Solution: Here, Given X and Y are ... Please refer the concept of...

GATECounsellor RcPP2 Solutions

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1. (C) Concept and Solution: Roots | | are eigen values of

[ ] [

]

2. (B) Concept and Solution:

Let

∫

Then

∫ (

)

(

) (

)

by

∫

∫

∫

Hence by Adding (1) and (2)

∫

∫



3. (D) Concept and Solution:

[

]

(

) (

) (

)

4. (C) Concept and Solution: Since two point are chose such that one gives

negative function value and other positive function value .

5. (B) Concept and Solution: A BJT is generally doped as so the

base emitter junction is an abrupt or step graded junction, while the base collector

junction is a linearly graded junction.

6. (D) Concept and Solution: A MOS is voltage controlled current source. Option

(B) is wrong as even at Saturation different gate source voltages can give different

currents.

7. (D) Concept: If two random variables are independent, then the density of their

sum equal to the convolution of their densities.

Solution: Here, Given X and Y are independent uniform random variable in common

interval .

So, fz(z) will be the convolution of , fx(x) and fy(y).

Reference: Probability, Random Variables by Papoulis, Page Number 181.

fz(z)

fy(y)

fy(y)

fx(x)



8. (C) Concept: There are two frequencies namely that give at the

mixer output ( is the oscillator frequency). For commercial AM radio, =455kHz

and frequency range to 1605kHz.

Solution: The other (image frequency) that can be detected along with 600kHz is

600- = not possible and 600+ = 1510kHz.

Reference: Image interference, Communication Systems by Simon Haykin

9. (B) Concept: When a signal is multiplied by a single mode signal, it gets split

into two, each having half the original amplitude and centered at the positive and

negative frequency marks of the single mode signal.

Solution: Assume the signal is like this:

After multiplication with , it’ll become like:

Hence, after filtering with a filter of 15kHz, the signal will only be scrambled. No

information/energy will be lost (since it is also multiplied by 2), hence, same energy.

-15 15 F(kHz)

-15 F(kHz) 15



10. (D) Concept: Formula of error probability for QPSK given by

√

Solution: From this it is clear that depends on average signal energy and noise

variance. However, distance between points depends on the avg. energy of signal.

Also, in this analysis, noise is assumed to be zero-mean, if its effect is included, mean

will also come into play. Hence All.

Reference: Noise in QPSK, Principles of Communication systems, Taub and Schilling

11. (C) Concept and Solution:

where X is the no. of row address line and Y is the no of column

address lines. To make min or viceversa.

Thus a 32 line decoder are 64 line decoder is needed requiring a total of

NAND gates.


Previous State Next State FFC FFB FFA C B A C B A 0 0 0 0 0 1 0 X 0 X 1 X 0 0 1 0 1 0 0 X 1 X X 1 0 1 0 0 1 1 0 X X 0 1 X 0 1 1 1 0 0 1 X X 1 X 1 1 0 0 0 0 0 X 1 0 X 0 X

By using this table we can find the final expression using K-map.

13. (C) Concept: Different interrupts of 8085 Microprocessor and their priorities.

Solution:

TRAP has the highest priority. It is used in case of power failure

The priority of interrupts are

Reference:8085 Microprocessor by Gaonkar




In option both diodes are reverse biased so

{

In option one diode is forward biased & other is reverse biased

,

For negative the diodes exchange roles and the output will be inverted form of

the output for positive . So option (B) is correct.


When diode is ON,

Hence charge on capacitor

Now when diode is OFF. So output

So option (C) is correct.


Unit of potential vector : volt-seconds per metre (V·s·m−1)

Unit of current density vector : amperes per square metre (A.m-2)

Unit of ∫ : (V·s·m−1).(A.m-2).(m3) (V.s.A)=(Power.sec)= Energy



x1[ ] x[ ]


According to Gauss Law

18. (C)

Method (I)

Now we cannot shift [ ] by samples to get [ ]. Since 3 is not an

integer multiple of 2. So, for discrete time signals we cannot scale before shift. We

can do this only when is integer multiple of in [ ]. So Method (II) i.e.

shifts then scale is more useful.

Method (II)

x [ ] x[ ]

x1[ ] x[ ]




Causal signal: 1

Anti-causal signal:

Right sided signal:

Left sided signal:

All causal signals are right sided but the converse is not true. Similarly all the anti-

causal signals are left sided but not all the left sided signals are anti-causal.

, but it is RSS and not a causal, so statement (I) is false

, but it is a LSS and not an anti-causal. So statement (II) is also

false

20. (A) Concept and Solution: As the area under the curve decreases, energy also

decreases.

21. (C) Concept and Solution: Since the Polar plot is enclosing (-1,0) Thus system

is unstable. This encirclement is only once hence single pole is on the right hand side

of the plane.

22. (C) Concept: An actuator is a mechanical device for moving or controlling a

mechanism or system. It is operated by a source of energy, usually in the form of an

electric current, hydraulic fluid pressure or pneumatic pressure, and converts that

energy into some kind of motion. When a human body tries to approach an object,

his brain acts as Actuator as the brain controls the motion of the body.


Gain at should be less than 1 so & GM

GM = 1

|

24. (A) Concept: In case of capacitor current lead the voltage, but in case of

inductor current lag the voltage.

http://en.wikipedia.org/wiki/Hydraulic_fluid

http://en.wikipedia.org/wiki/Pneumatic



Solution: Here current lag voltage in phase so the two elements will be Resistor and

Inductor.


voltage across resistance

4H inductors acts like an open circuit at Total current at

Thus,

|

1

26. (C) Concept: Linear equation

Where P and Q are . This can be solved by

Integrating Factor ∫

∫ ∫ ∫

Solution:

1

1

Let 1

1

hence

∫

√ √ since

Hence correct solution is

√ ∫ √

√

, We can also write this as

∫ ∫

since



27. (B) Concept:

By Cauchy’s Integral Formula

1

∮

where is inside the circle

Solution:

is analytic within the circle | | and the two singular points

and lie inside circle.

Therefore, ∮

1 = ∮ (

1

1

1

)

∮ 1

1 ∮ (

1

)

Hence we have

1

∮ 1

1

1

∮ (

1

) , hence

By Cauchy’s Integral Formula

∮ 1

1 ∮ (

1

)


All possible number of cases= .

If , then tickets 1 and must come out of 14 tickets numbered 1 to 14.

This can be done in 1 ways.

Must come out of 15 tickets numbering 15 to 30 which can be done in 1 1 ways.

Therefore favorable number of cases 1 1 1

.

Hence probability of being 15 is


Step 1: find Eigen values

[ ] *

+



Hence

Step 2: find the Eigen vectors

For

*

+ *

+

*

+ * +

only one independent equation

1

Hence eigen vactor

For

*

+ *

+

*

+ * +

1

1 or

1

1

Hence eigen vectors are and

Hence for both eigen values, eigen vectors are ,

30. (C) Concept: For two regions A & B to be equilibrium, net current should be

zero. Now if then one of charge carries will be greater in ‘A’ than in ‘B’ &

it will cause diffusion. Similar argument holds for . So only way to have

equilibrium

Solution: Let be intrinsic Fermi level, then

Number of electron in ‘A’

⁄

Number of electron in ‘B’

⁄

For No Diffusion




We have 1

1

Now it is said that new minority concentration is 4 times the original one

So corresponding value of

1

Now for charge neutrality.

1 1

[For increase hole concentration we must add p type material only]

1

32. (A) Concept:

For a stationary process, cross-correlation between 1 and random

processes will depend only on the 1 .

1 1 1 [ 1 ]

Solution:

Given is a stationary random process, and

1

and, is uniformly distributed between [0,2π].

So, 1 [ 1 ] [

]

[ ]

Here, We have two random variable and , both are independent.

So, 1 [ ] [ ]

[ ]



1

[ ]

1

[ ] [ ]

[ ] ∫

∫

( Uniformly

Distributed)

1

[ ]


Combinational Circuits: Adder, Subtractor, Code Converter, Comparator, Encoder,

Decoder, Multiplexer (Data Selector), Demultiplexer (Data Distributors), etc.

Sequential Circuits: Shift Register, Flip Flop, Counters, etc.


is initial state, if ‘0’ comes, while the system is in , remains there, if a’1’ comes,

while the system is in , it goes to 1. When the system is in 1 it signifies odd no of

1’s are present. For another ‘1’ system goes back to and output is high because it

signifies presence of even no of 1’s. Also is the final state, hence option (B) is

correct.

35. (A) Concept and Solution:

Please refer the concept of Machine cycles and status indicating pins table in 8085

Microprocessor text book by Goankar.


(

)

[

]

*

+

Solving this we get




Gain =

( 1

)

Now slew rate maximum value of

[ ( 1

) ]

1 Vs-1

So option (C) is correct.

38. (B) Concept and Solution: Ratio of Transmitted to Incident power is 1 - | |2

where is Reflection Coefficient and is defined as

= 2 1

2 1

= 1/3. Hence Ratio is 1 – 1/9 = 8/9

39. (C) Concept and Solution: cosd where is angle of main beam and

α is progressive phase shift.

For end fire pattern =0.

Therefore .




11

⁄

⁄

1 11

1

Because network is reciprocal 11 and S matrix is also reciprocal.


[ ] , append zero to fix the length to

[ ] [ ] [

] [

]

[ ]

After discarding , [ ] will be [ ]

𝑍 Ω 𝑍 Ω Ω 𝑍 Ω Ω Ω

𝑍 Ω 𝑍 Ω Ω

𝑍𝑖𝑛1

𝑍 Ω 𝑍𝑖𝑛1 Ω Ω

𝜔

j2π

j

j



42. (C) Concept and Solution: Half wave symmetry. Any signal is said to have

half wave symmetry if where T is fundamental period.

For a signal with half wave symmetry, only odd harmonics exist.

Since the given signal satisfies half wave symmetry, the signal contains only odd

harmonics both sine and cosine. There will be no DC term since the average of given

signal is zero.

43. (B) Concept: Fourier Transform property of Convolution of signals.

Solution: Fourier Transform of sin(at)/πt is

Fourier Transform of is ;

Where is convolution and is multiplication.

When Input and Transfer function convolutes, there spectrum gets multiplied.

Reference: Signals and Systems by Oppenheim.

44. (B) Concept: Signal Flow Graphs and State Space Analysis

α

α

β

β

β

β

H : β 𝛼

γ x α β



1 1

1

1

[ 1

] [

] [

1

] [ ] [ ]

[ ] [ ] [

1

]

45. (C) Concept: State Space Analysis.

For system to be controllable:| |

| | Then the given system is uncontrollable

Controllability [ 1 ] | |

Observability: [ 1 ] | |

Solution:

[

]

| |

So Controllable

[

]

| |

So Observable



46. (A) Concept: Normally we calculate steady state errors for Unity Negative Feed

Back (UNFB) systems but the system shown is not a UNFB, when . So we

have to convert it into UNFB by adding one Unity Positive FB and one UNFB and

simplify the block diagram for

Solution:

When reference input is zero or , the transfer function between and

is given by

Hence

The steady state error due to a is given by

(

)


Use the concept of super node analysis

1

𝑠 𝑠

𝑅 𝑠

𝑇𝐿

𝐶 𝑠



1

Solving three equations we get

1


√ ( 1

)

√ ( 1

)

for max voltage across inductor

√ √


Ω

Ω

Ω



⁄

⁄


1 1

1



Ω



Minimum value of is needed when voltage across load is just equal to as

maximum current will be flowing through and negligible current through Zener

diode

Maximum value of is needed when the maximum current flows through Zener

diode i.e 3mA.

So by KCL at A

1

So option (B) is correct


Now if is changed from to the value of

Option (B) is correct

54. & 55. (B), (A)

Concept: General equation of PM [ ] and

modulation index, . Also,

Solution: In first part, and

⁄

In 2nd part, if is doubled, is same because modulation index of PM is

independent of message frequency.

Hence, and will be doubled = 24 kHz.

Reference: Angle-modulation, Communication Systems by Haykin



56. (A) dimish

distend means to expand or stretch or swell. Hence the opposite meaning is

diminish.

57. (D) economy

frugal means saving, not wasteful.

58. (C) stoic: perturbed

Buoyant: submerged (two different form of presence of body in water)

Obloquy: discredit (same meaning) Stoic (unaffected by pleasure or pain or impassive): perturbed (to disturb greatly, make uneasy or anxious (both are two types of negative emotion) Stealth: furtive (same meaning, stolen) Disaffected: rebel (same meaning, disloyal)

59. (A) somewhat nondescript, easily recognized

Common sense tells you that the markings described in the sentence — black bib, gray cap, and white lines trailing down from the mouth — would make a bird distinctive and readily identifiable. In the sentence, the connecting word while neatly sets up contrasting ideas as between what precedes it and what follows it. The word nondescript and the phrase easily recognized make for just the sort of contrast that lends coherence to the sentence as a whole.

60. (C) 10

If S is set of children who eat spinach, B is set of children who eat bean and C is the

set of children, who eat carrot then,

Again and Thus, Total number of children=

61. (A) Honeybees, unlike many other varieties of bees, are able to live through

the winter by clustering together in a dense ball for body warmth.

Main ideas are, unlike other bees, honey bees form cluster in winter to gain body

warmth for survival. The numbers, how they eat, how they move are secondary

ideas according to the passage.



62. (A) 46

Escalator is moving with its usual speed, addition to which the person steps down.

Now let us calculate the speed of escalator in terms of number of steps elapsed at

bottom. Let in 1 sec number of steps elapsed is x. In 30 sec, number of steps elapsed

is 30x and in 18 sec is 18x. Addition to this the man has stepped down in order to

reach earlier compared to normal case. Thus total number of steps elapsed in both

the case is same as speed of escalator is constant. Hence according to question,

steps

Thus required number of steps steps.

63. (D) 5:8

In 1 lt of mixture, amount of milk in vessel A=2/5 and in vessel B=9/16

To make ½ lt milk in vessel C, we need from vessel A, ½-2/5 =1/10 lt

And from vessel B, 9/16- 1/2 =1/16 lt.

Thus the ratio= 1/16 :1/10 =5:8

64. (C) 15 hours

Let 1st pipe fills the tank in x hrs, 2nd pipe in y hrs and 3rd in z hrs. Then in 1 hr 1st pipe fills ( )th of tank, 2nd pipe fills (1/y)th of tank and 3rd pipe fills (1/z)th of the tank. As time taken by 1st and 2nd pipe simultaneously to fill the tank = time taken by 3rd pipe to fill the tank. Hence in 1 hr the portion of tank filled by 1st and 2nd pipe simultaneously is equal that by 3rd pipe. Thus Again, and . From above 3 equations, we get . Solving this, or -2. Hence hrs, hrs

65. (C) 6

unit’s digit in the product of 37562156! and 67847675! = unit’s digit in the product of

2156! and 7675!

The unit digit for any power of 2 is one of the numbers, 2, 4, 8, 6. If n is a positive

number then unit digit of 24n = 6, 24n+1 = 2, 24n+2 = 4, 24n +3 = 8.

. Hence unit digit of 2156! =6.

Similarly . 74n gives unit digit 1. Try it.

Thus the the unit’s digit in the product of 37562156! and 67847675! is 6 1=6

GATECounsellor RcPP2 Solutions · Solution: Here, Given X and Y are ... Please refer the concept of...

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