GATECounsellor RcPP2 Solutions
E-mail id: [email protected] Page 1 of 22
1. (C) Concept and Solution: Roots | | are eigen values of
[ ] [
]
2. (B) Concept and Solution:
Let
∫
Then
∫ (
)
(
) (
)
by
∫
∫
∫
Hence by Adding (1) and (2)
∫
∫
GATECounsellor RcPP2 Solutions
E-mail id: [email protected] Page 2 of 22
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)
GATECounsellor RcPP2 Solutions
E-mail id: [email protected] Page 3 of 22
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
GATECounsellor RcPP2 Solutions
E-mail id: [email protected] Page 4 of 22
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.
12. (B) Concept and Solution:
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
GATECounsellor RcPP2 Solutions
E-mail id: [email protected] Page 5 of 22
14. (B) Concept and Solution:
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.
15. (C) Concept and Solution:
When diode is ON,
Hence charge on capacitor
Now when diode is OFF. So output
So option (C) is correct.
16. (C) Concept and Solution:
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
GATECounsellor RcPP2 Solutions
E-mail id: [email protected] Page 6 of 22
x1[ ] x[ ]
17. (C) Concept and Solution:
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[ ]
GATECounsellor RcPP2 Solutions
E-mail id: [email protected] Page 7 of 22
19. (D) Concept and Solution:
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.
23. (C) Concept and Solution:
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.
GATECounsellor RcPP2 Solutions
E-mail id: [email protected] Page 8 of 22
Solution: Here current lag voltage in phase so the two elements will be Resistor and
Inductor.
25. (B) Concept and Solution:
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
GATECounsellor RcPP2 Solutions
E-mail id: [email protected] Page 9 of 22
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
)
28. (D) Concept and Solution:
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
29. (D) Concept and Solution:
Step 1: find Eigen values
[ ] *
+
GATECounsellor RcPP2 Solutions
E-mail id: [email protected] Page 10 of 22
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
GATECounsellor RcPP2 Solutions
E-mail id: [email protected] Page 11 of 22
31. (C) Concept and Solution:
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 [ ] [ ]
[ ]
GATECounsellor RcPP2 Solutions
E-mail id: [email protected] Page 12 of 22
1
[ ]
1
[ ] [ ]
[ ] ∫
∫
( Uniformly
Distributed)
1
[ ]
33. (C) Concept and Solution:
Combinational Circuits: Adder, Subtractor, Code Converter, Comparator, Encoder,
Decoder, Multiplexer (Data Selector), Demultiplexer (Data Distributors), etc.
Sequential Circuits: Shift Register, Flip Flop, Counters, etc.
34. (B) Concept and Solution:
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.
36. (A) Concept and Solution:
(
)
[
]
*
+
Solving this we get
GATECounsellor RcPP2 Solutions
E-mail id: [email protected] Page 13 of 22
37. (C) Concept and Solution:
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 .
GATECounsellor RcPP2 Solutions
E-mail id: [email protected] Page 14 of 22
40. (C) Concept and Solution:
11
⁄
⁄
1 11
1
Because network is reciprocal 11 and S matrix is also reciprocal.
41. (D) Concept and Solution:
[ ] , 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
GATECounsellor RcPP2 Solutions
E-mail id: [email protected] Page 15 of 22
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
GATECounsellor RcPP2 Solutions
E-mail id: [email protected] Page 16 of 22
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
GATECounsellor RcPP2 Solutions
E-mail id: [email protected] Page 17 of 22
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
(
)
47. (A) Concept and Solution:
Use the concept of super node analysis
1
𝑠 𝑠
𝑅 𝑠
𝑇𝐿
𝐶 𝑠
GATECounsellor RcPP2 Solutions
E-mail id: [email protected] Page 18 of 22
1
Solving three equations we get
1
48. (A) Concept and Solution:
√ ( 1
)
√ ( 1
)
for max voltage across inductor
√ √
49. (C) Concept and Solution:
Ω
Ω
Ω
GATECounsellor RcPP2 Solutions
E-mail id: [email protected] Page 19 of 22
⁄
⁄
50. (D) Concept and Solution:
1 1
1
51. (C) Concept and Solution:
52. (B) Concept and Solution:
Ω
GATECounsellor RcPP2 Solutions
E-mail id: [email protected] Page 20 of 22
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
53. (B) Concept and Solution:
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
GATECounsellor RcPP2 Solutions
E-mail id: [email protected] Page 21 of 22
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.
GATECounsellor RcPP2 Solutions
E-mail id: [email protected] Page 22 of 22
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