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28A/11, Ground Floor- Jia Sarai New Delhi16 Ph.: 09811382221, 011-32662945 www.panaceainstitute.org.in 1

+

+

S

R

Q

Q 3

4

RB

A

R

12V

8 7

6

5

2

12V

R

8

7

6

5

1

2

3

4

GND

Trigger

O/P

Reset

VCC

Discharge

Threshold

Control

Conventional Paper-II- 2013 Solutions :( ECE)

Sol.1 (a)

CCB

CCA

2V 2R 24 2V reference 16V

3R 3

2V R 24 1V reference 8V

3R 3

Sol.1 (b)

In a PNP Transistor EBV 0.7volt

in NPN Transistor BEV 0.7volt

Here base currents are neglected because value of is large.

12 0.7 22.6I 0.7 12 0

22.6 22.6I I 1mA

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00 01 11 10

00

01

11

10

CDAB

0 0 0

0 0 0

0 0

0

So value of 02 01I 1mA I

Sol.1 (c) Y 0, 1, 3, 5, 6, 7, 10, 14, 15

Y A B C D A B C A C D

Sol.1 (d) Character equation is 2s as k 0

n n2 a k

a

2 k

Herer

2

1M

2 1

2

r n 1 2

by solving both n 21.98 Rad/sec

by solving k 483.12

a 26.45

Sol.1 (e)8

14

6

3 10f 3 10 Hz

1 10

Stimulated emission rate 1

hfSpontaneous emission rateexp 1

KT

Where 34h 6.6 10 j sec

23

T 200 273 473K

K 1.38 10 joule/ K

f frequency

34 14

23

1

6.6 10 3 10exp

1.38 10 473

14

3

19.36 10

exp 0.03 10

Sol.1 (f)

2

c1

n 1.46

sin n 1.48

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LowerValley

MiddleValley

UpperValley

E = 1.33eVG

0.6eV0.8eV

CB

FB

VB

c

2 2

1 2

80.57

N.A. n n 0.2425

Sol.1 (g)

Comparison of peak to valley current ratio:

In GaAs diode electron transfer process from lower valley to upper valley is very slow. It is

found that Ga As at a voltage above Threshold, current flow is due to large concentration of

es in lower valley as compared to upper valley. Due to this Low peak to valley ratio is

achieved becoz of large contribution from lower valley.

In InP e transfer takes place faster because of increase in field. This is due to coupling b/w

upper & lower valley is weaker in InP than that in GaAs. To prevent break down due to high

energy lower valley is weakly attached to middle but strongly coupled with upper valley.

Thus In P diode has large peak to valley current ratio than GaAs.

This is the adv. of In power GaAs which leads to high efficiency.

Sol.1 (h) (A 5 F 1)

001010010111110001

8

122761

Sol.2 (a) For M1:

GS in out

DS DD out

V V V

V V V

For M1to be saturation:

1

DS GS T

2

D 1 in out T

V V V

I K V V V

2

2

D 2 B SS TI K V V V

Here1 2D D 1 2

I I but K 2K

So

2 2

in out T B SS T

in out T B SS T

2 V V V V V V

1V V V V V V

2

So out in T B SS T1

V V V V V V2

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x 0in xL

1 2

v vv vI (1)

R R

0 xx4 3

v vv2

R R

by equation (1)

0 x in xL

2 1

v v v vI 3

R R

by equation (2) 3 x0 x4

R .vv v 4

R

0 x 3x

2 4 2

v v Rv

R R R

.

For ILto be independence of Lv

if 3 x x

4 2 1

R v vthen

R R R

by equation (3)

x in xL

1 1

v v vI

R R

inL

1

vI

R

So Condition is 3 2

4 1

R R

R R

Sol.3 (a) (PS) (NS)

2 2 1 1 0 0

2 1 0 2 1 0

A B C A B C J K J K J K

Q Q Q Q Q Q

0 0 0 0 0 1 0 0 1

0 0 1 0 1 0 0 1 1

0 1 0 1 0 0 1 1 0

1 0 0 1 0 1 0 0 1

1 0 1 1 1 0 0 1 1

1 1 0 0 0 0 1 1 0

0 1 1Not used

1 1 1

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00 01 11 10

0

1

Q2

Q1 Q0

1

J = Q2 1

00 01 11 10

0

1

Q2

Q1 Q0

1

K = Q2 1

00 01 11 10

0

1

Q2

Q1 Q0

1

J = Q1 0

1

00 01 11 10

0

1

Q2

Q1 Q0

1

1

K = 11

00 01 11 10

0

1

Q2

Q1 Q0

1

J = Q0 1

00 01 11 10

0

1

Q2

Q1 Q0

1

K = 10

1

J0

K0

Q0

Q0

J0

K0

Q0

Q0

J0

K0

Q0

Q011

Sol.3 (b) NS

PS x 0 x 1

a a, 0 b, 0

b c, 0 d, 0

c a, 0 d, 0

d e, 0 f ,1

e a, 0 f ,1

f g, 0 f ,1

g a, 0 f ,1

Now it can be reduced further last is redundant

For Reducing State table

Let a = 000

b = 001c = 010

d = 011

e = 100

f = 101

This is solved problem of Morris meno. So can check their also.

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G(0 )+

G(0 ) Real

Im

01T

Reducing the state table:

PS NS Z

2 1 0Q Q Q

0 0 0

0 0 1

0 1 0

0 1 1

1 0 0

1 0 1

x 0

0 0 0

0 1 0

0 0 0

1 0 0

0 0 0

1 0 0

x 1

0 0 1

0 1 1

0 1 1

1 0 1

1 0 1

1 0 1

x 0

0

0

0

0

0

0

x 1

0

0

0

1

1

1

Sol.4 (a)

KG s H s

s 1 Ts

K > 0

T > 0

O.L.T.F

Open loop poles s = 0,1

sT

P 0 Open loop poles at RHS if Im axis = 0

O.LTF

s

KG s H s

s 1 T

At 0

KG j0 H j0 90

j0 1 Tj0

At

KG j H j 0 180

j 1 Tj

Rough sketches Possibilities

KG j H j

j 1 Tj

Multiply & divide by 1 Tj

2 2

K 1 TjG j H j

j 1 T

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G(0 )

Im

Real

Separate Real & Imaginary Parts

2 22 2

RealImag

jK KTG j H j

1 T1 T

For all +ve values of Real part is () ve.

At phase crossover frequency P

Imaginary part = 0

P2 2

jK0

1 T

Since p system is either highly stable or highly unstable

Put pin G j H j

p p KTj H j 01

Graph or Nyquist plot not touches negative real axis

Actual Nyquist Plot

Since 1 pole at origin ;

It will take 1. ie clockwise encirclement from G 0 to a 0

G &G are short circuit

Absolute stability

N P Z

N = No. of encirclement around 1,0 point

P+= No of open loop poles on RHS of Imaginer axis

Z+= For equation zeroes or closed loop poles on RHS of Imaginer axis

Since no encirclement around (1, 0) point

N 0

P 0

Z 0

system is highly stable

Relative Stability

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w= 2

0.25Real

Im

p pk

G j H j 0

p p1 1

GM0G j H j

Gain margin = 10 log 0 = dB

Highly stable hence proved

Sol.4 (b) At pphase cross over frequency Im = 0

21 0.5 2 rad/ sec

Atp p p

1.125G j H j

1 2 1 0.25 2

1.1250.25

3 1.5

Gain margin =1

40.25

GMin dB 12.04dB

Sol.4 (c)

k1 G s H s 1 0

s s 1 s 2

KG s H s

s s 1 s 2

Char equation is :

2

3 2

1 G s H s 0

s s 1 s 2 K 0

s s s 2 K 0

s 3s 2s K 0

(1) Angle of Asymptotes

k

2K 1 180whereK 0,1,2

P Z

0

1

2

60

180

300

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60

K = 0

0K = 0

K = 02

CentroidP Z

P Z

0 1 2

13

Centroid (1, 0)

Break away point

dK0

ds

3 2

2

K s 3s 2s

dK0 3s 6s 2 0

ds

1

2

s 0.422

s 1.577

These are 2 breakaway pts.

Check validity:

s1= valid

s2= not valid

Sol.6 (a):

(i) Cut off frequency is mCO

gs

gf

2 C

CO 12

0.05f

2 0.6 10

13.26GHz

(ii) Maximum operating frequency is

12

CO dmax

s g i

f Rf

2 R R R

19 213.26 10 450

2 2.5 3 2.5

49.73 GHz

Sol.6 (b) Power gain of horn antenna e2 24 4A

4.5 4.5

So here pG 72

2

2

parabolic

DDG 6 1600 610

D 163.2 cm

70 70 10H.P.B.W

D 163.2

4.28

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IonisedLayer

Virtualheight

Actualheitht

ActualPath

Projected Path

(2)

(1)

Skip distance

F Layer2

F Layer1

E Layer

D Layer

Drawback of parabolic reflectors:

(1) Wave reflected from parabola back to feed antenna produces interaction & mismatching.

(2) The feed antenna acts as an obstruction which results in increasing side lobes and decreasing

gain.

Sol.6(c)

(1) Critical frequency: It is the maximum value of frequency below which if a ray is incident at

normal then it will return back to ground. Only frequency above (fc) at normal incident will

go into atmosphere.(2) Maximum usable frequency: It is same as fcwith only difference that incident is other than

vertical incidence. If angle of incidence is then

cMUF f sec Secant Law

(3) Skip distance:

It is the distance from transmitter where HF wave after reflecting from Ionosphere reaches at

the earths surface back. It depends upon angle of incidence, degree of ionization and

frequency of wave. No communication is possible by sky wave for points nearer than skip

distance. Important layers are D, E, F1& F2layers

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r

i

P 10 1K

P 300 30

K 0.182

1 K 1 0.182VSWR 1.451 K 1 0.182

Sol.7 (a) # incldue

# include

Void main ( )

{

int arr [8] = {7, 10, 13, 8, 6, 1, 19, 5}

int i, j, temp;

clrscr ( );

printf (Bubble sort\n);

printf (\n bfore sorting\n);

for (i = o; i < = 7; i + +)

printf (%d\t, arr [i]);

for (i = o i < = 6; i + +)

{

for (j = o; j < = 5i ; j + +)

{

if (arr [j] > arr [j + 1]){

temp = arr [j];

arr [j] = arr [j + 1];

arr [j + 1] = temp;

}

}

}

printf (\n\n array after sorting : \n);

for (i = o; i < = 7; i + +)

printf (%d\t, arr [i]);

getch ( ) ;

}

Output :

Bubble sort

Array before sorting

7 10 13 8 6 1 19 5

Array after sorting

1 5 6 7 8 10 13 19

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SecondaryStorage

PrimaryStorage

Controlunit

ALU

Outputunit

Inputunit

Information

CPU

Sol.7 (b) Basic block diagram of a computer:

Role of Control Unit:

1. How does input device know that it is time for it to feed data into memory.

2. How does ALU know what should be done with data once they are received.

3. How is that only final results are sent to output device and not intermediate results.

All above are done by control unit of the computer.

It does not perform any actual processing of data it acts as a central nervous system for other

components of computer system.

It manages and co-ordinates the entire computer system. It obtains instructions from programstored in main memory, interprets the instructions and issue signals which cause other units

of system to execute them.

Role of Main memory / Primary Storage:

It is used to hold pieces of program instruction and data, immediate results of processing and

recently produced results of processing of jobs, which computer system is currently working

on.

These pieces of information are represented electronically in main memory and while it

remains in main memory the CPU unit can access it directly at a fast speed.

Primary storage cr main memory holds its information while computer is ON Main memory

has limited storage capacity and is expensive.

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D Q

Q

RST

7.5

R 7.5

RST 7.5Interrupt

recognised

RST6.5

RST5.5

M

7.5

M6.5

M5.5

QS

R

INTR

DI EI

RESETAny interruptrecognised

GETRSTcodefrom

ext. H/W

002C

I6.50034

003CI7.5

TRAP

VectorLocation

0024

Microprocessor

Memory

Micr

oprocessor

IOP

Sol.7 (c)

INTEL 8085 has 5 interrupts Namely RST 7.5, TRAP, RST 5.5, RST 6.5 and INTR.

TRAP has highest priority followed by RST 7.5, RST 6.5 RST 5.5 and INTR.

EIEnables all interrupts& DI Disable all interrupts

TRAP is a NMI it need not be enabled. It cant be disabled. It is not accessible to user it is

used for emergency operation.

Sol.7 (d)

Buses are namely Address bus, data bus and control bus.

Date post:	02-Jun-2018
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