Exp. EMF & Antenna Practice Paper 1.pdf

7/28/2019 Exp. EMF & Antenna Practice Paper 1.pdf

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EX Page No. 1 of 8

LOGIC GATE Studies, Head Office : Raipur GATE For Details Contact : 098261-61828www.logicgate.co.in email : [email protected]

LOGIC GATE ElectronicsAndCommunicationEngineering

Explanation EMFand Antenna

Propagation in Lossless Medium

Q.1, 2, 3, 4, 5 Explanation :

(1) Positive y-direction

(2) 810 rad/s, 0.5 rad/mk = =

8810 2 10 m/s

0.5pu

k

= = =

(3) 2 / 2 / 0.5 12.6mk = = =

(4)

2 28

8

3 102.25

2 10r

p

c

u

= = =

(5) ,E k H=

( )

120 120

251.33 ,1.5r

= = = =

( )0.5 3 , and 30 10 A/mj yk y H z e = =

Hence,

( )0.5 3 0.5 251.33 30 10 7.54 V/mj y j yE y z e x e = =

and

( ) ( ) ( ) ( )8, Re 7.54cos 10 0.5 V/mj tE y t Ee x t y= =

Q.6 Explanation :

For 1GHz, 1, and 9r rf = = =

92 2 10 rad/sf = = ,

9

80

2 2 2 2 109 20 rad/m,

3 10r r

fk

c

= = = = =

( ) ( ) ( )9 0, 6cos 2 10 20 V/mE y t x t y = +

At t= 0 and y= 2 cm, E = 4 V/m :

( ) ( )2 0 04 6cos 20 2 10 6cos 0.4 = + = +

Hence,

10

40.4 cos 0.84 rad,

6

= =

Which gives

o0 2.1 rad 120.19 = =

Propagation in Lossless Medium

1 - A 2 - B 3 - C 4 - B 5 - D 6 - D 7 - D 8 - B 9 - A 10 A11 - B 12 - D 13 - C 14 - C 15 - C 16 - A 17 - B 18 - B 19 A 20 - D21 - D 22 - B

WavePolarization

1 B 2 A 3 A 4 D 5 C 6 B 7 B 8 D 9 D 10 D

11 D 12 B 13 A 14 A 15 B 16 B 17 A 18 D

Propagation in a Lossy Medium

1 - B 2 - C 3 - C 4 - A 5 - C 6 - B 7 - C 8 - D 9 - C 10 - B

11 - A 12 - A 13 - C 14 - C

EM Power Density

1 - D 2 - B 3 - C 4 - B 5 - A 6 - D 7 - C 8 - A 9 - B 10 - A

11 - D 12 - B 13 - C 14 - D 15 - D 16 - A 17 - B


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ExPracticePaperLOGIC GATE

EX Page No. 2 of 12


and

( ) ( ) ( )9 o, 6cos 2 10 20 120.19 V/mE y t x t y = +

Q.7, 8, 9. 10 Explanation :

(7) From ( )0.210 V/mj zE y e= , we deduce that

k= 0.2 rad/m. Hence

2 210 31.42m

0.2k

= = = =

(8)

861.5 10 4.77 10 Hz 4.77 MHz

31.42

puf

= = = =

(9) From

2 21 1 3

, 1.672.4 1.5

p rr pr r

c cu

u

= = = =

(10) ( )2.4

120 120 451.941.67

r

r

= = = ,

( ) ( ) 2.01 1

10j zH z E z y e

= =

= ( )0.222.13 mA/mj zx e ,

( ) ( ) ( ), 22.13cos 0.2 mA/mH z t x t z= + ,

With 62 9.54 10 rad/sf = =

Q.11, 12, 13, 14 Explanation :

(11) Since 0.2k = ,

2 210m

0.2k

= = =

(12)

7710 5 10 m/s

0.2pu

k

= = =

But pr

cu

=

Hence,

2 28

7

3 1036

5 10r

p

c

u

= = =

(13)

[ ( )71 1 3sin 10 0.2H k E x y t x

= =

( )74cos 10 0.2z t x +

= ( ) ( )7 73 4 sin 10 0.2 cos 10 0.2z t x y t x

(A/m),

with

( )0120

20 62.836r

= = =

(14) From 0 / ,r =

2 20 60 9

20r

= = =

Q. 15, 16, 17, 18, 19, 20 Explanation :

(15) 92 6 10 rad/s,f = =

93 10 Hz 3GHzf = =

(16)8

83 10 1.875 10 m/s2.56

pr

cu

= = =

(17)8

9

1.875 103.12cm,

6 10

pu

f

= = =

(18)2

2 2201.4rad/m,

3.12 10k

= = =

(19)0 377 377 235.62

1.62.56r

= = = =

(20) ( )920

cos 6 10H x t kz

=

= ( )920 cos 6 10 201.4235.62

x t z

= ( ) ( )2 98.49 10 cos 6 10 201.4 A/mx t z


3/12

EX Page No. 3 of 8


LOGIC GATE ElectronicsAndCommunicationEngineeringQ.21 Explanation :

Frequency of wave remain unchanged in both the

mediums

Wave length in free space1

1

v

f =

1v c=

phase velocity of wave is equal to velocity of light in

free space

wave length in medium2

2

v

f =

1

2 2

c

v

=

1

2

r r

c

c

=

1

2

& 1r r r

= =

nonmagnetic medium

2

1

2r

=

2

30 415

r = =

Q.22 Explanation :

Wave number is phase constant

9 rad2 8 10sec

f = =

94 10 4GHzf Hz= =

Phase velocity

1 1p

o r o r

v

= =

1, 1, nonmagnetic mediump r

r

v = =

883 10 1.875 10 m/s

2.56Pv

= =

8

9

1.875 104.69cm

4 10

Pv

f

= = =

2

2 2134.04rad/m

4.69 10

= = =

Wave Polarization

Q.1 Explanation :

For an RHC wave traveling in ,z let us try thew

following :

( ) ( ) cos sinE xa t kz ya t kz = + + +

Modulus ( )2 2 2 2 V/m .Hence,E a a a= + = =

22

2a = =

Next, we need to check the sign of the y -component

relative to that of the x-component. We do this by

examining the locus of E versus tat z= 0 : Since the

wave is traveling along z when the thumb of the right

hand is along z (into the page ), the other

four fingers point in the direction shown (clockwise asseen from above). Hence, we should reverse the sign of

the y -component :

( ) ( ) ( ) 2 cos 2 sin V/mE x t kz y t kz = + +

with

( )2

2 2104.72 rad/m ,

6 10k

= = =

and

( )8 102 3 10 10 rad/skc

= = =


4/12


EX Page No. 4 of 12


Q.2, 3, 4, 5 Explanation :

Polarization angles ( ),

( )10 tan / ,y xa a =

( )0

tan 2 tan 2 cos =

( )0sin 2 sin 2 sin =

Case ax ay 0 Polarization State(2)

(3)

(4)

(5)

3

3

3

3

4

4

4

3

0

180o

45o

-135o

53.13o

53.13o

45o

53.13o

53.13o

-53.13o

53.13o

-56.2o

22.5o

-21.37o

0

0

Linear

Linear

Left elliptical

Right elliptical

(2) ( ) ( ) ( ) , 3cos 4cosE z t x t kz y t kz = +

y

x

4

321

1 2 3 4-1-2-3-4 -1

-2-3-4

Plot of the locus ofE(0, t)

(3) ( ) ( ) ( ) , 3cos 4cosE z t x t kz y t kz =

y

x

4

321

1 2 3 4-1-2-3-4 -1

-2-3-4


(4) ( ) ( ) ( )o , 3cos 3cos 45E z t x t kz y t kz = + +

y

x

4

32

1

2 4-2-4 -1

-2-3-4

Plots of the locus ofE(0, t)

(5) ( ) ( ) ( )o , 3cos 4cos 135E z t x t kz y t kz = +

y

x

4321

2 4-2-4 -1

-2-3-4


Q.6, 7, 8 Explanation :

( ), Re j tE z t Ee =

( ) / 6 Re 20 j z j tx jy e e = +

( )/ 2 / 6 Re 20j j z j tx ye e e = +

( ) ( ) 20cos / 6 20cos / 6 / 2x t z y t z = + +

( ) ( ) ( ) 20cos / 6 20sin / 6 V/mx t z y t z = ,

( )

1 22 2

20 V/m ,x yE E E

= + = direction of electric field intensity is

( )1tan / 6y

x

Et z

E

= =


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EX Page No. 5 of 8


LOGIC GATE ElectronicsAndCommunicationEngineeringFrom,

87/ 6 3 10 2.5 10 Hz

2 2

c kcf

= = = = ,

72 5 10 rad/sf = =

At z= 0

7 o

o

0 at 0

5 10 0.25 45 at 5ns

0.5 90 at 10

t

t t t

t ns

=

= = = = =

= =

Therefore, the wave is LHC polarized

Q.9, 10 Explanation :

,jkzxE xa e=

RHC wave :

( )

/ 2 j jkz

R R

E a x ye e = +

( ) jkzRa x jy e=

LHC wave : ( )/ 2 j jkzL LE a x ye e = +

( ) jkzLa x jy e= +

,R LE E E= +

( ) ( ) x R Lxa a x jy a x jy= + +

By equating real and imaginary parts, ,x R La a a= +

0 , or / 2, / 2R L L x R xa a a a a a= + = =

Q.11, 12 13 Explanation :

( ) ( ) ( )o , 10sin 60 30cosE z t x t kz y t kz = +

( ) ( ) ( )o 10cos 30 30cos V/mx t kz y t kz = + + Phasor form :

o30

10 30j jkz

E x e y e

= +

Since is defined as the phase of yE relative to that

of xE ,

o30 , =

1 o0

30tan 71.56 ,

10

= =

( ) o0tan 2 tan 2 cos 0.65 or 73.5 , = = =

( ) o0sin 2 sin 2 sin 0.40 or 8.73 , = = =

Since 0, < the wave is right-hand elliptically polarized.

Q.14 Explanation :

( ) ( )1 2cos 2sinE x t kz y t kz = +

( ) ( ) 2cos 2cos / 2x t kz y t kz = +

/ 21 2 2

jkz jkz jE x e y e e = +

1 1 o0 tan tan 1 45 ,

ay

ax

= = =

/ 2 =

Hence, wave 1E is RHC

Q.15 Explanation :

/ 22 2 2


Wave 2E has the same magnitude and phases as wave

1E except that its direction is along z instead of z+ .

H e n c e , t h e l o c u s o f r o t a t i o n o f E will match the left hand

instead of the right hand. Thus, wave 2E is LHC

Q.16 Explanation :

( ) ( )3 2cos 2sin ,E x t kz y t kz =

/ 23 2 2


Wave 3E is LHC

Q.17 Explanation :

/ 2 2 2jkz jkz juE x e y e e= +

Reversal of direction of propagation (relative to wave 3E

) makes wave 4E RHC


6/12


EX Page No. 6 of 12


Q.18 Explanation :

( ) ( ) sin 2cosE x t kz y t kz = + + +

Wave direction is .At 0z z = ,

sin 2cosE x t y t = +

Tip of E rotates in accordance with right hand ( with

thumb pointing along z ). Hence, wave state is RHE

z 1-1

-2

2

y t= /2

t= 0

Figure : Locus of versus timeE

x

Propagation in a Lossy Medium

Q.1 Explanation :

The phase angle by which the magnetic field leads the

electric field is where is the phase angle of c

8 9

0.1 362

2 10 10 9

= =

Hence, quasi-conductor

1/ 21/ 2

0

" 1201 1

' 'c

rr

j j

= =

= ( )1/ 2

125.67 1 2 71.49 44.18j j = +

o84.04 31.72=

Thereforeo31.72 =

Since ( ) o1/ , leads by ,or by 31.72cH k E H E =

. In other words, H lags E by o31.72 .

Q.2 Explanation :

( ) 0 10 ,z zE z E e e = =

44

9 9

5 10 363.32 10

2 3 10 10 9

= =

.

Hence, medium is a low-loss dielectric

( )4120 5 10 120

. 0.032 Np/m2 2 2 9r

= = = =

3 0.032 410 10 , ln10 0.032 ,ze z = =

287.82mz =

Q.3 Explanation :

for a good conductor,

s

1, and for any material .

= =

Hence,

9 62 2 2 5 10 3 10p sf

f

= = = =

= ( )49.42 10 m/s

Q.4 Explanation :

Since the phase angle of ois 45c , the material is a good

conductor. Hence,

( )o45 o o1 28.1 28.1cos 45 28.1sin 45jc j e j

= + = = + ,

oro28.1cos45 19.87

= =

Since 1/ 1/ 2 0.5 Np/m,s = = =

20.5 2.52 10 S/m19.87 19.87

= = =

Q.5 Explanation :

Since = for a good conductor, and 0.5 = , it

follows that 0.5 = .

Therefore,

2 24 12.57 m

0.5

= = = =


7/12

EX Page No. 7 of 8



6 710 12.57 1.26 10 m/spu f= = =

Q.7 Explanation :

From the given expression forE,

( )92 10 rad/s , =

( )30 Np/m , =

( )40 rad/m =

22 2 2 2 ' '

0 0 2' ,r r

c

= = =

22 "

22 " r

c

= = .

Using the above values for , and , we obtain the

following :

' 1.6r =

" 5.47r =

1/ 2"

1' '

c j

=

=

1/ 2 1/ 2"0

''

377 5.47

1 1 1.61.6

r

rrj j

=

( )o36.85157.9 je=

30 4025 ,x j xE z e e =

30 40

o36.85

1 1 25

157.9

x j x

jc

H k E x z e e

e

= =

=o30 40 36.850.16 x x jy e e e

{ }Re j tH He= =

( ) ( )30 9 o0.16 cos 2 10 40 36.85 A/mxy e t x

Q.8 Explanation :

A material is characterized as an imperfect dielectric if for

a specific frequency 0, 1

( ) ( )

( )( )( )1210 . 10

Hz2 9 8.854 10

f

>

200Hzf>

Q.9 Explanation :

( ) ( ) v, cos mz

xo xE z t E e t z a =

and rad3 m =

2 2m

3

= =

Q.10 Explanation :

Propagation constant j = +

= attenuation constant

= phase constant

3j = +

and intrinsic impedance

( )72 103

ojj

j

= =

+

o =

( )( )7 72 10 4 10

3

j

j

=

+

( ) ( )0.8 3 7.54 2.51j j = + = +


8/12


EX Page No. 8 of 12


Q.11 Explanation :

( )j j = +

= 6 72 100 10 4 10j

( )6 120.1 2 100 10 9 8.854 10j + Q.12 Explanation :

For a good conductor = , and for any material skin

depth1

= , Hence phasre velocity

Pv

= =

9 62 2 5 10 2 10Pv f = =

= 46.28 10 m/s

Q.13 Explanation :

For low loss dielectric medium propagation constant

p j = +

2p j

= +

2 2

o r

o r

= =

41 1 5 10 120120

2 2 2 16r

= = =

Np0 024

m =

Q.14 Explanation :

Loss tangent tan

=

tan =

6 122 850 10 8 854 10 9 0 15 =

3 S63 8 10m

=

EM Power Density

Q.1 Explanation :

( ) ( )750sin 2 10 mA/mH x t ky=

( ) ( )70 0 50sin 2 10 mV/mE y H z t ky = =

( )( )

( )2

20 6 650 120 10 50 102 2

avS z x y = =

( )20.48 W/my=

Q.2 Explanation :

( )0120

409r

= =

The wave is traveling in the negative x-direction

( )2 2

23 2 13

0.05 W/m2 2 40

avS x x x

+ = = =

Q.3, 4, 5 Explanation :

(3) Since 0.2 = = , the medium is a good conductor

( ) ( ) ( )0.2

1 1 1 0.054

c j j j

= + = + = +

( )o450.0707 je=

we have

20 2 0.4 o25 cos cos 45

2 2 0.0707

z zav

c

ES z e z e

= =

= ( )0.4 2125 W/mzz e

(4) ( )8.68 8.68 0.2 1.74 dBA z z z= = =

(5) 40 dB is equivalent to 4

10

. Hence

( )4 2 0.4 410 , ln 10 0.4z ze e z = = =

or z= 23.03 m


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EX Page No. 9 of 8



0 120 60 188.54r

= = = =

[ ]2 2 60 0188.5

9 16 10

2 2

av y xS x H H x = + = +

( )22.36 mW/m=

( )32.36 10 20 47.13 mWavP S A= = =

Q.7 Explanation :

2 20 0

,2 2

avav

E ES

S

= =

( )2

24.56125.67

2 2.4 = =

But

20 377 377, 9

125.67r

r r

= = = =

Hence,

883 10 1 10 m/s

3p

r

cu

= = =

Q.8 Explanation :

( )

( )2

2 3 2 2

0

, 1 mW/cm 10 W/cm 10 W/m2

av

E RS

= = =

3 4

2

3 10 1 1.2 1010

2 120R R

= =

1/ 24

1.2 1034.64m

10R

= =

Q.9 Explanation :

The real and imaginary parts of the intrinsic impedance

are equal hence the medium is a good conductor.

Q.10 Explanation :

( )2

21, cos2

zoavg

EP t z e

=

( )2

1,0 cos

2

oavg

EP t

=

& ( )( )

22 101

,10 cos2

oavg

EP t e

=

&( )

( )

( )2 10,10

,0 1

avg

avg

P t e

P t

=

2010

100e =

( )1 log 0.120

e =

10.115m =

Q.11 Explanation :

For good conductor,

20.115

=

54.6m =Q.12 Explanation :

j

=

j =

n

j

=

2

nf

j

=

( )( )7

35 35 0.115 0.115

2 4 10

j jf

j + +

=

o =

1.02 MHzf =


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EX Page No. 10 of 12

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Q.13 Explanation :

The conductivity of the medium is calculated with

( )

( )

0.115 1Re Re

35 1

j

j

+ = = +

3 S3.29 10m

=

Q.14 Explanation :

0.4 =

2 25m

0.4

= = =

772 10 5 10 m/s

0.4Pv

= = =

1 1Po o r r

v = =

83 10, 1P r

r r

v

= =

nonmagnetic material

83 10P

r

v

=

2 28 8

7

3 10 3 1036

5 10r

Pv

= = =

Q.15 Explanation :

The electric field and magnetic field are 4 out of

phase, hence, the medium is a good conductor, thus

54 10 = =

and6

5

2 25 10 m

4 10

= = =

Q.16 Explanation :

( )2

21, cos

2

zoav

Ep t z e

=

( )2

1at 0, ,0 cos

2

oav

Ez p t

= =

( )2

21at , , cos2

doav

Ez d p t d e

= =

( )

( ) 2,0 1

,

av

dav

p t

p t d e =

( )

( )2,0

,

av d

av

p te

p t d

=

( ) ( )1

at , , ,02

av avz d p t d p t= =

22de =

1

log 22

e d

=

5

1

log 22 4 10ed =

0 2758md =

Q.17 Explanation :

&2

= =

2

=

5

4 10 20 0888577

=

7 S2 10m

=


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EX Page No. 11 of 8

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LOGIC GATE ElectronicsAndCommunicationEngineering


12/12


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Exp. EMF & Antenna Practice Paper 1.pdf