RcPP_SolutionsSet_1

GATECounsellor RcPP1 Solution

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1. (D) Concept: If determinant then rank is else maximum of order of any non-

zero determinant cofactor of the matrix.

Solution:

by substituting values from each option we get

2. (B) Concept: Definition of linear equation: Where P and Q are .

Solution: Replacing B and C from options.

Only option (B) results in standard form of Linear equation

3. (A) Solution: These are standard formulae for solving Linear Differential Equations.

4. (B) Concept and Solution:

As Zener breakdown takes place because of the existence of high electric which initiate direct

rapture of bonds without the collision of particles. This occurs at fields approx equal to

, for highly doped diodes. Hence, both p & n side are very heavily doped to obtain

high electric field.

5. (C) Concept and Solution:

Due to doping barrier potential exists and hence in order to align the conduction bands on both

sides we need to apply a voltage equal to built-in voltage .

6. (B) Concept: CDF function should be monotonically non-decreasing and right-continuous.

Solution: CDF is not a monotonically increasing (strictly increasing) since there is a possibility of

it being constant in some interval.

Reference: Properties of cdf, Principles of Communication Systems, Taub and Schilling.

7. (D) Concept: The given condition is required for distortion-less detection using synchronous

detection method.

8. (B) Concept: Modulation index,

= frequency deviation which is proportional to amplitude of message signal as

, K is constant.

= maximum frequency in the message signal



9. (D) Concept: Quantization noise power = where = step-size.

Solution: Since step-size is reduced by a factor of 8, quantization noise power will be reduced by

by proportionality relation.

Reference: Pulse Code Modulation, Principles of Communication Systems, Taub and Schilling

10. (A)

11. (D) Concept: Simplify the expression using Boolean laws

12. (B) Solution:

Starting Address of Memory: 0000H

4kB = 212 Bytes => ROM will take 212 bytes space

So, Address Range: 0000H (0 decimal) to 0FFFH (212-1 decimal)

16kB = 214 Bytes => RAM1 will take 214 Bytes space

So, Address Range: (ROM max address +1) to (ROM max address + 0x3FFF)

= 1000H to 4FFFH

4kB = 212 Bytes => RAM2 will take 212 bytes space

So, Address Range: (RAM1 max address +1) to (RAM1 max address + 0x0FFF)

= 5000H to 5FFFH

13. (A) Solution:

When diode is forward biased, so

When diode is reverse biased, so

Reference: Electronics Devices and Circuit Theory by Boylestead or any basic Electronics book



14. (C) Concept and Solution: Half-wave rectifier

Here current flows only for half wave cycle,

Maximum current is = maximum input voltage/resistance =

So rms value =

15. (D) Concept: Cut off frequency of rectangular waveguide is fmn =

2 2

2

c m n

a b

where a and b are the dimensions of waveguide (a>b), m and n are the mode number, c is the

speed of light.

Solution:

For TE10, f10 = c/2a and f01 = c/2b. so, f01/ f10 = a/b =2

16. (B) Concept: Directivity in the case of end fire array is D = 4L/λ where L = (N-1)d.

N stands for number of isotropic antennas.

Solution: 4L/λ = 30. L/λ= 7.5.

So BWFN = .

17. (a) Concept: Input Impedance of a Transmission line is Zin= Z00

0

tan

tan

L

L

Z jZ l

Z jZ l

where Z0 is

characteristic Impedance, ZL is load impedance, β = 2π/λ is phase shift constant, l is length of

waveguide.

Solution: Given l = λ/8, βl = π/4. ZL = 0 since short circuited. So Zin= jZ0.

Even if the frequency is changed, βl remains same. So no change in input impedance.



18. (A) Concept: Transformation of signal parameters.

Solution:

Concept-I : Scale, reversal then shift.

Scaling: x1(t)=x(nt)

All time axis(divide by n) will either compress or expand. Here Compress.

Reversal: x1(t)=x(-t)

Mirror image wrt y-axis

Shifting: x1(t)=x(t+n)

Shift the signal by n units on left side if x1(t)=x(t+n) and right side if x1(t)=x(t-n).



Concept-II: Shift then Scale and reversal.

19. (C) Concept : is an unit impulse function which defined as,

And,

Scaling Property:

Solution:

I. =

II. : if we substitute 0 directly into , ; which is

meaningless.

Since is an impulse defined between to . So here we use L'Hospital's rule.

III.

IV.



20. (D) Concept: Shifting property of Fourier Transforms.

Solution:

21. (B) Concept: Harmonic(Sinusoidal) inputs always produces harmonic outputs with scaling

factor of system response.

Let x(t)= ,

y(t)= x(t) h(t) = =

Solution:

Hence, Verified.

22. (B) Concept: Break points in Root locus diagrams,

Solution:



23. (a) Concept: Signal Flow Graphs and State Space Analysis

Solution:

So, state equation is,

Standard equation:

Comparing above equations, we get A as [-1]

The above system is of the order 1. So the order of the state matrix is also 1.

24. (D) Concept: Series RC circuit solving using Laplace or time-domain.

Solution:

Voltage across capacitor:

at t=0,



25. (c) Solution:

Given, I= 2U(-t)+5U(t) For T<0 , I = 2 amp and for T>0, I=5 amp

At T<0, when switch is at position 1 we draw the circuit diagram into two parts.

Left hand side of the circuit:

The voltage across the resistance (or two series capacitors) is 5 volt after a long time.

Let voltage across capacitor 1F be V1 , V1+Vx = 5 When capacitor are connected in series charge remain same so we can write

Q1 = Q2 i.e C1V1 = C2Vx 4 V1 = Vx Solving above equation we get, Vx =4 volts So,

Right Hand side of the circuit:

By solving Right side of the circuit we can calculate the value of (0-) as,

After a long time, inductor will act as short circuit and capacitor act as open circuit, so 2A

current equally divide into 3ohm resistance and

And,

2Ω

=

1A



Now switch is at position 2, for T>0, I=5amp

KCL at node , at ,

And,

KCL at node B,

,

, (Using equation )

Differentiate the equations wrt t,

Circuit at T( ): In this case, inductor will act as short circuit and capacitors as open circuit. So, current 5A will equally divide into 3ohm resistance. So, Voltage across 3ohm resistance is 3*5/2 = 7.5 V. Now, This voltage appears across two capacitors of 1F connected in series. So, Voltage across 1F capacitor = 7.5/2 = 3.75V

T

>

0

th

e

cu

rr

en

t

I=

5a

m

p



i.e.

2 Mark Questions

26. (D) Concept: A vector is said to be Solenoidal Function if

Solution:

27. (A) Concept:

If has a pole of order at then

Solution:

has simple pole at ,

hence

Also since is finite and nonzero.

Hence residues are

28. (D) Solution:



Taking a,b,c, common from and from column

Passing over in second matrix

Hence since

29. (B) Concept:

General Formula for nth stage solution of by Regula falsi method is

Solution:

Let

Negative, Negative and = Positive

Root lies between 2 and 3

, , -0.59794 and = 0.23136

2.72102

Now,

The root lies between 2.72102 and 3

Taking and

We get,

= 2.74021

Reference: Numerical methods, Higher Engg. Mathematics, B. S. Grewal



30. (B) Concept: Zener breakdown takes place at a lower voltage than the avalanche breakdown

voltage . So Since in Zener breakdown rapture of bonds takes place directly due to

high electric field so the electric field needed for Zener breakdown is higher than avalanche

breakdown . So .

31. (A) Concept: As

and width of depletion region in material

So in emitter base junction, width of depletion region in base is more than of emitters as

Similarly for base collector junction, width of depletion region in collector side is more than that

present in base as

32. (C)Concept: Bandwidth is defined as frequency band in which a signal can be transmitted

without distortion (Practically not possible). We can also say mathematically it is the reciprocal

of the total duration of the signal.

Practical transmission bandwidth is equal to Roll off factor * 1/Total duration of the signal.

Solution:

Here, In T1 system(it implies sampling frequency is 8KHz) 24 independent voice inputs are time

multiplexed. Given that each sample is represented by 8bits.

Bandwidth required= 24 * 8000Sample/second * 8 bits/sample = 1.536*106 bits/second.

33. (A) Concept: For matched filter, impulse response .

Where for

Output of matched filter =

Solution: g(-t) will be like option (c), h(t) = g(T-t) will be like option (b). After convolution,

output will be triangular positive as in (a)

Reference: Matched filter, Communication systems, Simon Haykin

34. (A) Concept: Given circuit is Ring Oscillator with 3 inverters.

Number of inverters is always odd in Ring Oscillator. The fundamental Frequency of Ring

Oscillator = where N = No of inverters and = propagation delay of an inverter.

35. (B)Concept and Solution: 5 Bit ring counter is a MOD-5, so it divides the 100MHz by 5,

therefore after Ring counter frequency becomes 20MHz, 2 Bit parallel counter is a MOD-4

counter, it will divide it by 4 so we have 5MHz frequency after parallel counter, again MOD-5

ripple counter = MHz =1MHz and 5 Bit Johnson is a MOD-10, This gives frequency as MHz =

100kHz.



36. (a) Concept: Instruction set of 8085 microprocessor

Solution: SIM Instruction interprets the Accumulator as follows

D7 D6 D5 D4 D3 D2 D1 D0

SOD SDE XXX R7.5 MSE M7.5 M6.5 M5.5 0 0 1 0 1 0 1 0

Since A is loaded with 2A H.

If SDE (D6) =1, it enables the serial output. So serial output is disabled.

If MSE (D3) =1, it enables the functions of bits D2 ,D1 ,D0. So functions are enabled.

If Mx= 0 then the corresponding interrupt is enabled else masked or disabled. Here M6.5(D1)=1. So

RST 6.5 is disabled.

Reference: 8085 Microprocessor by Gaonkar

37. (A) Concept: Bypass capacitor acts as short in ac analysis & act as open in dc analysis .So

they are generally used to short out degeneration resistance (at emitter/source) to have a large

gain.

38. (C) Concept: Voltage divider bias makes the current almost independent of variation in β

and hence least sensitive.

39. (B) Concept: VSWR = ZL/Z0 = 2 (from the given data). We will measure the minimum

impedance of the line at which VSWR is ½. (Required VSWR = 1/ VSWR calculated from data).

Solution:

So, ZL,min = Required VSWR * Z0 = .5*50 =25.




Intrinsic Impedance of B

2 = / =

0 0/r r = /r r *120π = 80π

Reflection Coefficient

= 2 1

2 1

where

2 = 2 2/ and

1 = 1 1/ .

Similar to above calculations 1 = 60 π. Hence = 1/7.

Transmission coefficient

T = 1 + = 8/7.

Phase Shift constant of A = βA = 2π/λ. Here λ is not directly given. But equation of E

is given.

General equation is of form E

= E0 cos(wt-βx) z

. Hence β = 6π.

41. (A) Concept: The mapping from s to z-plane is done by using the equation z = sTe .

Solution: Here T=1/3 since the sampling frequency is 3Hz.

Pole x: s=0 => z=10.

s= -3 ± j(π/2) => z= = ( / 6) .

Similarly zeroes also can be mapped.

42. (A) Solution:

43. (A) Solution:

Circular convolution of two functions is defined as,

Here, we need to do circular shifting operation over a period N.

Given, and



1,-7,-5,3 Circularly mirror image wrt y axis

3,1,-7,-5 Circularly shift signal to 1 unit right hand side

-5,3,1,-7

,-7,-5,3,1

So, -36, -6, 44, 14

44. (A) Concept: Stability criterion in Nyquist plots.

Solution: As the Open Loop system is stable, P=0

From Nyquist plot N=0

Number of poles on RHS plane =Z=0

Therefore system is stable


State Space equation:

For system to be controllable,

Then the given system is uncontrollable



46. (D) Concept:

Pole Moment figures: (See the directions of curve)

(A) Here, Real part is constant and increases so decreases.

Also, increases increases decreases (see above relations)i.e. transient

response improved.

As, decreases increases Less Stable

(B) Real Part is increasing, increases decreases settling time decreases

constant => constant => constant but and have slight variations.

As, increases decreases More Stable

(C) cos = constant constant No effect on Stability

But increases increases decreases

(D) increases from zero to one decreases More stable

But decreases increases and has slight variation.

47. (A) Concept:

1. Transmission Parameter is represented as

2. For Cascade Network Overall T parameter is Multiplication of Individual N/W parameter.

Solution:

T parameter of network in dotted line is (calculated) and

Overall T Parameter is (given)

Hence

And,



48. (A) Concept and Solution:

Let the current flowing into the circuit at the node be . Since the infinite network is symmetrical about current I in going from x to infinity, will be equally distributed in all branches connected with x node xQ, xT, xP, xY. The current ' ' then return from infinity and is taken from the network at the node .

Again by symmetry, the current flowing along , , , are each .

Hence, the total current flowing along xy is

So, the voltage between and

So, the effective Resistance =

49. (b) Concept: At resonance

Solution: The frequency at which and are taking equal values is the resonance point and i.e. at which

50. (A) Solution:

T

R S

Q T

P



Here

Here, from equivalent diagram,

(Using potential divider rule)

51. (B) Solution: From above figures,

Input Resistance:

For output resistance set

So resistance seen at output is

Rout = 4kΩ

Reference: Microelectronics Circuits by Sedra and Smith or Integrated Electronics by Millman-

Halkias

52. (A) Concept and Solution: By Poisson’s equation

Here, , So by putting in above equation and integrate,



At electric field must be zero at the end of depletion region, So

Maximum magnitude of E = EMAX =

53. (C) Solution: Now

Integrating from x=-d to x=+d we get

dx

54. (C) Concept: Implement Huffman coding,

Don’t forget to arrange all messages in decreasing order of probabilities at each step.

Solution:

Messages Probabilities S0 0.4 1 0.4 1 0.4 1 0.6 0 S1 0.2 01 0.2 01 0.4 00 0.4 1 S2 0.2 000 0.2 000 0.2 01 S3 0.1 0010 0.2 001 S4 0.1 0011

Average code word length L= = 0.4*1+0.2*2+0.2*3+0.1*4+0.1*4 = 2.2 bits

55. (C) Concept:

Coding efficiency, where = minimum possible value of L.

By source coding theorem, where is source entropy. Hence,

Average minimum possible length = Entropy= H(X)

H(X)=

=

= 2.122

So, = H(X)/L = 2.122/2.2 = 0.9645



56. (D) Solution: obsessed: attracted

Intimate: close (nearly same use)

Evanescent(vanish): permanency (opposite meaning)

Articulate (clear and distinct /capable of speech): speech

Enclose: parenthesis (parenthesis encloses)

Obsessed: attracted(nearly same use)

57. (C) Solution: conciliatory

Polemical means controversial argument or against some opinion. Lavish- expended or limitless Imitative-derivative, not original Conciliatory-compromising Attractive- adorable Hence most opposite meaning word is conciliatory 58. (C) Solution: harmful

Deleterious means harmful.

59. (B) Solution: not, apart

Disperse means scatter or not together. Hence the answer is apart.

60. (B) Solution: Unit’s digit in the product = unit’s digit of

gives unit digit as Thus 7116 gives unit digit 1 and gives unit digit Thus the required unit digit is 61. (D) Solution: I, II and III

Federal law requires hospitals to treat anyone who walks in.

The finance committee balked at the hefty price and killed the bill, another casualty of a

failed legislative session.

Unfortunately, the problem of access to medical care for those of limited means is not

going to go away anytime soon and, despite the well-intended regulations, too-full

hospitals compromise everyone's welfare

The above three statements from the passage clearly indicate that, all the three options

are one of the factors which has contributed to the overburdening of hospitals.

62. (D) Solution: 6.00pm

Let speed of Ram is “x” and that of Hari is “y”. After 3hrs of Ram started from A (6.00

am) and after 2hrs Hari started from B (7.00 am), both meet at 9.00 am. So during their

meet distance travelled by Ram is 3x and by Hari is 2y. Again they meet at 11.00 am

2hrs from 9.00 am. During this 2 hr distance travelled by Ram is 2x.



And distance travelled by Hari is 2y. As Ram has reached B and again turns back

towards A and then they both meet at 11.00am, the distance covered in terms of y is

2y(from 1st meeting to B) +2y(from B to 1st meeting point)+2y(from 1st meeting point to

2nd meeting point)= 6y.

Thus 2x=6y => x=3y

Thus total distance between A and B is 3x+2y=11y

Hari has already covered 4y distance. For covering remaining 7y distance, he needs 7

hrs. So the time is 11.00 am +7hrs= 6pm.

63. (B) Solution: 250 Starting from zero time (time=0min.), the person thinks for first 5min. (time=5min.) then types 50 lines of code in next 5 min.(time=10min.), then take brake for 5 min.(time=15min.), then again types the remaining 50 lines (of 10 min. typing) for next 5min.(time=20min.). The procedure goes. In first 20 min. he writes 100 lines. Then he thinks for 5 min. (time=25 min.). Again takes break for next 5 min. (as its already 10 min. to take break)(time=30min.), then types for 10 min. (time=40min.). Now he has completed 200 lines of codes in 40 min.. We have 20 min. remaining. Then he takes break for 5 min. (time=45min.) thinks for 5 min. (time=50min.) and got time to type for next 5 min.(time=55min.) as last 5min. is for break time(time=60min.). Thus he can write maximum 250 lines. 64. (A) Solution: 24

There are 8 corners cubes are of three sided coloured. Central 4 cubes in all the six faces

are 1 sidedcoloured cubes. Remaining visible border cubes are two side coloured cubes.

Hence, 8 cubes from front side +8 cubes from back side + 4 from left side + 4 cubes from



right side = 24 cubes (all two side colored cubes are covered in these 24 cubes including

top and bottom faces)

65. (B) Solution: 3

Let's mark the balls using numbers from 1 to 12 and these special symbols:

x? means I know nothing about ball number x;

xL means that this ball is maybe lighter than the others;

xH means that this ball is maybe heavier than the others;

x. means this ball is "normal".

At first, lay on the left pan balls 1? 2? 3? 4? and on the right pan balls 5? 6? 7? 8?.If there

is equilibrium, then the wrong ball is among balls 9-12. Put 1. 2. 3. on the left and 9? 10?

11? on the right pan.If there is equilibrium, then the wrong ball is number 12 and

comparing it with another ball it can find out if it is heavier or lighter.If the left pan is

heavier, 12 is normal and 9L 10L 11L. weigh 9L and 10L. If they are the same weight,

then ball 11 is lighter than all other balls.If they are not the same weight, then the

lighter ball is the one up.If the right pan is heavier, then 9H 10H and 11H and the

procedure is similar to the former text.

If the left pan is heavier, then 1H 2H 3H 4H, 5L 6L 7L 8L and 9. 10. 11. 12. Now lay on

the left pan 1H 2H 3H 5L and on the right pan 4H 9. 10. 11.If there is equilibrium, then

the suspicious balls are 6L 7L and 8L. Identifying the wrong one is similar to the former

case of 9L 10L 11L. If the left pan is lighter, then the wrong ball can be 5L or 4H.

Compare for instance 1. and 4H. If they weigh the same, then ball 5 is lighter than all the

others. Otherwise ball 4 is heavier. If the left pan is heavier, then all balls are normal

except for 1H 2H and 3H. Identifying the wrong ball among 3 balls was described

earlier.

In all possible way at least 3 times we have to weigh to find out which one is odd one.

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