Problem set 1 solutions

ec2204

February 15, 2013

Solution 2.

let us consider the branches of the rst transmission lines are innite length matched transmission lines.

so,

Zin = Z0ZL + Z0 tan (d)

Z0 + ZL tan (d)= Z0

if N transmission lines are conected in parallel, its eective Zin =Z0N = ZLfor rst transmiision line.now,

=ZL Z0ZL + Z0

=1N1 +N

Power reected back is given as P = 2Pin and transmitted power in each branch isPtr =(12)Pin

Nfor N=2 P = Pin/9 and Ptr = 4Pin/9for N=3 P = Pin/4 and Ptr = Pin/4for N=9 P = 0.64Pin and Ptr = 0.04Pin

Solution 4-

Z0 =

R+jLG+jC

L/C ( For high frequency)

(a) Two wire with characteristic impedance- Z0 = 300

Hence 300 = 1pi

cosh

1( d2a )

Maximum dimension is 8 mm, gives two constrains (i)2a 8mm (ii) 2a+ d 8mmFrom the gure we can easily conclude that the 2nd constrain is the eective one.

Substituting, d = 8mm 2a in the expression for characteristic impedance,300 = 1pi

cosh

1( 8mm2a2a )

Solving for a, we obtain a = 0.56mm

d = 8mm a = 6.88mm

(b) Planar Line with characteristic impedance Z0 = 15

Hence 15 =

dw (for small t and w >> d)

Constrain w=8 mm

1

Solving for d, we obtain d = 0.3185mm

(c) Coaxial Line with characteristic impedance Z0 = 72

Hence 72 =

12pi ln(

ba ) (b c)Constrain 2b=8 mm

Solving for a, we obtain a = 1.2mmand b = 4mm.

Solution 6.

First converting the current source and 100 ohm resistor combination to its Thevenin equivalent. This is a 500 Vvoltage source in series with the 100 resistor.The next step is to determine the input impedance of the 2.6 length line, terminated by the 25 resistorWe use l = 2pi 2.6 = 16.33 rad.Now

Zin = 50

[25cos(16.33) + j50sin(16.33)

50cos(16.33) + j25sin(16.33)

]= 33.7 + j24.0

The equivalent circuit now consists of the series combination of 50 V source, 100 resistor and ZinThe current in this circuit will be

I =50

100 + 3.7 + j24= 0.368 0.178

The power dissipated by the 25 resistor is the same as the power dissipated by the real part of Zin.since it is a loss lesstransmission line.

or

P25 = P33.7 =1

2I2R = 2.28W

and to nd the power dissipated by 100 resistor we need to return to the Norton conguration, with the original currentsource in parallel with the 100 resistor, and in parallel with Zin . The voltage across the 100 resistor will be the sameas that across Zin or V=IZin = (0.368 = 0.178)(33.7 + j 24.0) = 15.2 0.44. The power dissipated by the 100 ohmresistor is now

P100 =V 2

2R= 1.16W

Solution-7

Zl=73 ohm. Z0=300 ohm.

Now the quarter wave transformer needs to be matched with the characteristic impedance of the line So Z0=

Zl.Z0where Z0

is the characteristic impedance of the line. The d=2cm and frequency =200Mhz and Z0=300 ohm. Using

these 3 datas and the equations of the L, C, R and G for the 2 wire transmission line given in problem 4 we can nd the

value of a ie the radius and parameters too.

Solution-8

given,

load impedence ZL=60 + j80characteristic impedence Z0=50lenght between the stubs d0 = /8normalized load admittance yL=

YLY0

= 5060+j80 = 0.3 j0.4length of short circuited stub at A-A' is lA, it's normalised input impedence zsA = j tan (lA)then normalized admittance ysA = 1/zsA =

jtan(lA)

= jb1length of short circuited stub at B-B' is lB , it's normalised input impedence zsB = j tan (lB)then normalized admittance ysB = 1/zsB =

jtan(lB)

= jb2then at point A-A' normalised admittance is yA = ysA + yL = 0.3 j (0.4 + b1)by using impednce trnsformation relation , transform admittance at A-A' yAto B-B' as yBby using

ZB = Z0ZA + jZ0 tan (d0)

Z0 + jZA tan (d0)

normalized inpedences at A-A' and B-B' are zA = ZA/Z0, zB = ZB/Z0normalized admittances yA = YA/Y0, yB = YB/Y0 andtan (d0) = tan

(2pi

8

)= 1, expressing above equation, in terms of admittances,

2

yB =yA + j

1 + jyA

at B-B' normalized admittance yi = yB + ysB =yA+j1+jyA

jb2 here yi must be equal to 1[yi = Yi/Y0, ] substituting all thevalues

1 =0.3 j (0.4 + b1) + j

1 + j(0.3 j (0.4 + b1)) jb2

solving for b1, b2 ;b1 = 0.686tan (lA1) = 1.45lA1 = 0.346b1 = 2.114tan (lA2) = 0.473lA2 = 0.429b2 = 1.38tan (lB1) = 0.724lB1 = 0.100b2 = 3.38tan (lB2) = 0.295lB2 = 0.454

3

Answer for Question No.1

a) Given that the characteristics impedance of line is Z0 = 60

And terminated on a load consisting of

Inductance = L = 1H

Capacitance = C = 100 pF

Resistance = R = 30

The frequency of the signal transmitted along it is = = 108

So the capacitive reactance of the capacitor is = 1

100cX jj C

And the inductive reactance is = 100cL j L j

Now the equivalent circuit of the two wire transmission line can be drawn as

So the load impedance will be,

30 100 100

30

L c L

L

Z R X X j j

Z

So the reflection coefficient at the load end will be

0

0

30 60 1 1( )

30 60 3 3

LL L

L

Z Z

Z Z

And voltage standing wave ratio will be

111 3 2

111

3

L

L

L

C

R

Z0

Now the voltage at distance l from the load end can be written as

( 2 )( ) 1j l j lLV l V e e

It is clear from the above equation that the voltage will be minimum at a point where the

following condition will be satisfied

21

( 2 ) (2 1)

( 2 ) (2 1)

j le

l n

l n

nl

Where

8

8

10 1

3 10 3c

Substituting the value of and solving we get the distance of the first minimum(for n=1) from

the load is 9.423 m.

b & c.) The solution for the part b and c are same as that of part a. Only the solutions are

given below.

For part (b):-

ZL= 0.85 (42.274)

VSWR = 12.33

dmin = 3.89 m from load

for part (c):-

ZL= 0.537 (-97.14)

VSWR = 3.32

dmin = 9.08 m from load

Answer for Q. No.-3

a.) For f = 60 Hz

6

8

21.9 10

sec2 3 103p

f radv

So

6 41.9 10 80 1.5 10 1l

Hence at this frequency the line will behaves a lumped circuit and all the lumped ckt theories can

be applied. So for this case the equivalent ckt for the above line can be

So using voltage dividing rule

,

,

80 80120

12 80 12 80

104

s out in

s out

V V

V V

R1=12

R2=80 Vin

b.) For frequency 500 kHz

5

2

8

2

2 5 101.57 10

2 3 103

1.57 10 80 1.26l rad

So the transformed impedance at the input end will be

80cos(1.26) 50sin(1.26)50

50cos(1.26) 80sin(1.26)

33.17 9.57 34.5 ( 0.28)

in

in

jZ

j

Z j

So the equivalent circuit will be

So,

33.17 9.57120 120

12 12 33.17 9.57

89.5 6.46 89.7 ( 0.071)

inin

in

in

Z jV

Z j

V j

Now Vin can be written as

j l j l

in

j l j l

in L

V V e V e

V V e e

Where

0

0

80 50 3

80 50 13

LL

L

Z Z

Z Z

R1=12

Zin 120V Vin

so

1.26 1.26

89.5 6.46(42.7 100)

3

13

in

j l j l

L

j j

VV

e e

jj V

e e

Now,

,

,

3(1 ) (42.7 100) 1

13

(52.6 123) ( )

s out L

s out

V V j

V j V Ans