Transmission Lines and E.M. Waves Lec 14

8/20/2019 Transmission Lines and E.M. Waves Lec 14

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Transmission Lines & E M. Waves

Prof. R. K. Shevgaonkar

Department of Electrical Engineering

Inian Instit!te of Technolog"# $om%a"

Lect!re '(

In this lecture, we solves some problems of transmission lines based on the theory which you are

developed so far. We will solve some problems analytically and later on we go to the solution of

the problems by graphical means that is by using the switch chart. Let us consider this problem

here; a lossless transmission line has 75 ohm characteristic impedance. The line is terminated in

a load impedance of 5 minus ! " ohms the ma#imum voltage measured on the line is "

volts. We have to find the ma#imum and minimum current and the minimum voltage on the line

also we have to find out at what distance from the load, the voltage and current are ma#imum.

This problem can be solve either analytically or graphically, we will solve here this problem

analytically.

$%efer &lide Time '()'*

&o what we are given here is that characteristic impedance which is + naught that is given as 75

ohms, the load impedance which is + l is given as 5 minus ! " ohms and magnitude of the


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voltage ma#imum on the line ma# is given as " volts. &o mod ma#imum is given as "

volts using this information then we have to find out the other parameters li-e voltage and

current minimum and distance of the voltage minimum from the load point. irst I have been all

this calculation is the calculation of the reflection coefficient. &o now, I -now the load

impedance, I -now the characteristic impedance. &o from here I can write down the reflection

coefficient at the load point gamma l which is / l minus / divided by / l plus / , if I simplify at

this comple# e#pression, I will get the gamma l that will be e0ual to 1.2)3 which an angle of

minus 125.43 degrees.

&o the magnitude of reflection coefficient mod gamma l that is e0ual to 1.2)3 and the angle of

the reflection coefficient phi l that is e0ual to minus 125.43 degrees then ne#t step could be there

once you -now the magnitude of the reflection coefficient, since noting is stored about the line

we ta-e liberty we chose the line to be lossless and therefore the magnitude of reflection

coefficient remains same on every point on the line. &o we -now that the ma#imum voltage

which you see on the line mod v ma#imum that is e0ual to mod v plus into " plus mod gamma l,

we are given this v ma# voltage which is " volts, we -now now the modulus of reflection

coefficient.

$%efer &lide Time 5(5'*


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&o we can find out the magnitude of the incident wave which is v plus modulus. &o from here,

we can substitute that is " is e0ual to mod v plus into " plus 1.2)3 inverting this relation, we

get mod v plus that is e0ual to " divided by 1.2)3 that is e0ual to 12.3 volts. &ince, we

-now the magnitude of reflection coefficient, we could calculate v ma#, we can calculate I ma#

also because once we -now the plus 0uantity, we can calculate I ma#. 6owever, we -now from

our analysis that the ma#imum current seen on the transmission line is nothing but the ma#imum

voltage divided by the characteristic impedance. &o we -now ma#, so we can divide this

0uantity by the characteristic impedance / not and we can find out directly the 0uantity which is

the ma#imum current I ma#.

&o from here, I can get the ma#imum current seen on the line I ma# that is mod ma# divided

by / naught that is e0ual to " volts divided by the characteristic impedance which is 75. &o,

that gives you the ma#imum current that is e0ual to 1".443 amperes. The I minimum which we

get on the line mod I min that if we go bac- to the e#pressions of the minimum and ma#imum

currents that will be mod v plus divided by / naught into " minus mod gamma l. We -now this

0uantity v plus which is 12.3 the + o is 75 ohms. &o this is 12.3 divided by 75, " minus

1.2)3 which is the modulus of the reflection coefficient. &o from here I can get the value of I

min that is e0ual to 1.'3 amperes.

&o the ma#imum can which you see on this line is 1".443 amperes and the minimum current will

we see on this line is 1.'3 amperes. 8nce you -now the minimum current on the line, we can

find out what is the minimum voltage and that is mod min that is nothing but / into mod I

min. &o that is e0ual to 75 into 1.'3, so that is e0ual to 1'".3 volts, so this line has a

ma#imum voltage of " volts, minimum voltage of 1'".3 volts, ma#imum current of 1".443

amperes and minimum current of 1.'3 amperes. 9ow we are also have to find out what is the

location of the voltage and current ma#imum on the line for this we have to go to the basic

relation of the ' travelling waves and when the ' travelling wave have a constructive interference

that time we have a voltage ma#imum, when the ' wave have a destructive interference, we have

the voltage minimum and it has a same location we have the current ma#imum.


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$%efer &lide Time (4'*

&o we -now that the phase different between the ' waves, the forward and the bac-ward wave

that is given as the load reflection coefficient phase minus ' beta l. &o when this 0uantity phi l

minus ' beta l, if it is even multiples of phi that time the ' wave which will have a constructive

interference and I will have voltage ma#imum, when this is all multiples of phi that time we have

destructive interference and we have a voltage minimum. &o in general when this 0uantity is

e0ual to plus minus ' m phi, where m is an integer 0uantity. We will get the voltage ma#imum.

&o the condition for voltage ma#imum is when the phase of the reflection coefficient at the load

minus ' beta l is e0ual to plus minus ' m phi that time the ' waves have constructive interference

and we get a voltage ma#imum.

&o from here we can get ' beta l that is e0ual to phi l plus ' m phi. I have chosen here the sign

positive because I want to get the lengths which are positive; we are measuring all the lengths

now towards the generator, l e0ual to at the load point. &o all the lengths which are going to

measure from the load point or the positive lengths. &o I have chosen the sign in such a way that

I get all the distances which are positive substituting now for beta that is that is ' phi by lambda


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and substituting for the phase of the reflection coefficient which you are calculated here which is

1minus 25.43 degrees, we can get now the location of the voltage ma#imum. &o let us say this is

till phi l plus ' m phi.

&o the length at which the ma#imum voltage appears, let us called that is l ma# that substituting

into this will be l ma# will be e0ual to phi l plus ' m phi into the wavelength divided by ) phi.

&ubstituting for phi l minus 125.43 degrees that is an radian:s, it is minus 1".")3 radiants. &o this

is e0ual to minus 1".")3 radiants. We can get now l ma# is e0ual to minus 1".")3 plus ' m phi

lambda divided by ) phi. 9ow the substituting different values for m, you put m e0ual to ", m

e0ual to ' and so on. ou get the l ma# or the location where the voltage is going to be ma#imum

that is 1.)"3 lambda, 1."3 lambda, 1.)"3 lambda and so on. To find out the current

ma#imum, the current ma#imum occurs where voltage is minimum that means when the phase 4

are destructive interference either we can go by that argument or we can say that the current

ma#ima and voltage ma#ima are shifted by a distance of lambda by ).

$%efer &lide Time ")("5*

&o if I add lambda by ) or 1.'53 lambda to all this value I will get the location of the current

ma#imum. &o I substituting into this I can get the location for the current ma#imum, current


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ma#imum occurs at this values plus minus 1.'53 lambda and when you are subtract 1.'53

lambda from here I still get a length positive. &o I have to choose even that value of the distance,

so the location at means the current is ma#imum that will be at 1."23 lambda, 1.223 lambda,

1".223 lambda and so on. &o you see using the basic e0uation which you are derive for a

transmission lines if you systematically proceed find the reflection coefficient, go to the

relationship of the voltages in current, write down the simple voltage e#pression in the current

e#pression on transmission line, substituting the values of the reflection coefficient, its

magnitude, its phase. We can calculate the ma#imum and minimum currents on transmission

line. We can also find the location at which the voltage or currents ma#ima or they are minima.

&o this is one of the problems which you will do you as- to solve in the real life because

normally your line is terminated in an un-nown impedance and we are supposed to find out and

where the voltage will be ma#imum so on and so on. Let us consider another problem, let us say

now we are having a transmission line which is having a characteristic impedance of 5 ohms

and this line is connected to a parallel combination of " ohm resistance and " nano farad

capacitance. In fact this is the very common problem whenever we are having a cable at higher

fre0uencies and let us say, I want to connect these cable to some circuit whose output impedance

is about " ohms is possible at that point we may have some straight capacitances or

intensically the impedance might be having a capacitance.

&o in variable you will see the output or input impedance of a circuit which will be a

combinations something li-e this. &o here the impedance in which the line is terminated is "

ohms in parallel with " nano farad capacitance, we are supposed to find out what is the &W%

on the line at a fre0uency of ' megahert/ also find the ma#imum and minimum resistance seen

on the line. &o ' things we have to find out, first thing we have to find out what is the &W%

because that is the measure of how much power is reflected on the line and also we are intersect

in finding out what is the ma#imum and minimum resistance. We will see on the line again we

proceed the same way. &o the problem we are having is a line which is connected to a resistance

of " ohms and capacitance of " nano farad, this is " nano farad, this is " ohms, the

characteristic impedance of the line + is given as 5 ohms.


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$%efer &lide Time "7(45*

&o we have to find out the 0uantities &W% rho and % ma# and % min. " is % divided by " plus ! omega % =, the fre0uency

which is given is ' megahert/.

&o the omega at the fre0uency is ' phi into ' into " to the power 2 radians or seconds. &o

omega will be 1".'523 into " to the power 7 radians per second, substituting for omega and % =

% is " ohms and = is " nano farad, we can get " divided by " plus ! 1".'5223. &o omega % =

are substitute in the value of omega and % and =, you get this 0uantity 1".'5223. &o that

impedance will be " divided by 1".23 with an angle 15".53 degrees. &o the load impedance, I

can write either in the polar form which is 12'.43 angle minus 15".53 degrees or in the real and

imaginary part which is 1'4.3 minus ! 1)5.23 ohms. 8nce I -now the load impedance now, I can


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find out the reflection coefficient gamma l which is + l minus + divided by / l plus + . &o that

is e0ual to 1'4.3 minus ! 1)5.23 minus 5 divided by 1'4.3 minus ! 1)5.23 plus 5.

$%efer &lide Time '"(*

9ow solving this, we get the reflection coefficient has 1.23 angle minus 1.43 degrees. &o the

magnitude of the reflection coefficient from here we can get and that is e0ual to 1.23. &o we get

now from here the magnitude of the reflection coefficient mod gamma l that is e0ual to 1.23,

once we get that than the &W% rho is " plus mod gamma l divided by " minus mod gamma l

that is e0ual to " plus 1.23 divided by " minus 1.23 that is e0ual to ). &o for this combination

of resistance in capacitance at ' megahert/, we will get a &W% of ). 8nce we -now the &W%

than finding out the ma#imum and minimum resistance is very straight forward we -now the

characteristic impedance of the line.

&o the ma#imum impedance will be characteristic impedance multiplied by rho and the

minimum impedance will be characteristic impedance divided by rho. &o we get % ma# on the

line which is + into rho which is 5 into )' ohms and % min, minimum resistance which will

see on the line is + divided by rho that is e0ual to 5 divided by ) that is 1"'.53.


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$%efer &lide Time ''(55*

&o in this case, we were given some combination of lambda circuit with which the line is

terminated and we were as- to find out what is ma#imum and minimum resistance on the line

and also we were as- to find out, what is the &W% which will see on the line and recall higher

the value of &W%, worse is the match. &o this by &W% is 0uite large, it is ) that means in this

case lot of power will be reflected from this terminating low. Let us now try to solve some of the

problems by using the switch chart that is the graphical means of solving the problem. ?efore,

we get into the specific problems let us first try to identify some of the special points on the

switch chart.

&o let us say, I have to identify some of the special points that is I am given the impedance 5

plus ! 75 ohms, you to mar- this on the switch chart or the impedance " plus ! ohms or an

impedance minus ! ohms or the reflection coefficient that is 1.43 angle 2 degrees or you to

mar- constant &W% circle with rho e0ual to 1'.53 or we have to find out the % min point on the

&W% circle which is having value 1".53.


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&o firstly, we ta-e the characteristic impedance of the line let us say it is 5 ohms. &o we have to

normali/e these values with 5 ohms. &o if I normali/e this impedance < which is 5 plus ! 75

this will become " plus ! 75 upon 5, so that will be 1".53. 9ow I go to the switch chart and first

I identify a circle for which the resistive part is ", we -now this is the circle which passes

through the center of the switch chart. &o this circle is % e0ual to " circle then I go to the reactive

part which is 1".53 so this circle which is a 1".53 circle.

&o the intersection point of these ' circle, this circle and this circle is the point < here which

represents and impedance 5 plus ! 75 ohms or a normali/e term " plus ! 1".53, if I ta-e a

impedance " plus ! , again we normali/e this impedance with 5 ohms. &o this will be 1.'3

plus ! , so immediately we notice that this is the resistive impedance we are tal-ing about and

the resistive impedance must lie on the hori/ontal a#is in the comple# gamma plane. &o we go to

circle which is % e0ual to 1.'3 circle and find this intersection on the hori/ontal a#is.

&o this point ? represents an impedance which is 1.'3 plus ! in normali/e terms are " ohms

plus ! in the absolute terms. The third impedance which you want to mar- on the switch chart is

minus ! ohms.


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chart here and that length is " because we -now the switch chart in comple# gamma plane has a

radius e0ual to ".

&o this length now is from center to this point is unity. 9ow we can scale it to find out what

should be the length for 1.43 and then the angle is 2 degrees which is measure from the real

a#is in the gamma plane. &o which is this line, so from here we read the angle li-e normal

convention that go in the anticloc-wise direction by 2 degrees, ta-e a radius vector whose

length will be 1.43 normali/e to this length and you will get this point A. &o the point A

represents a reflection coefficient whose magnitude is 1.43 and whose phase angle is 2

degrees. The another thing which you would li-e to get is draw a &W% circle for rho e0ual to

1'.53, how do we draw this circle. We -now the rho corresponds to the ma#imum normali/e

resistance seen on the transmission line. &o this constant &W% circle for rho e0ual to 1'.53

must pass through a point which is normali/e 1'.53 and reactance . &o we !ust find out a circle

which is 1'.53, % e0ual to 1'.53 and # e0ual to that means it is this point lie on this line.

&o this point represents here normali/e or ma#imum of 1'.53, if I draw a circle passing through

this point with the center as the center of the switch chart, this circle than represent a constant

&W% circle for rho e0ual to 1'.53. 9e#t thing we want to find out < % min seen on a constant

&W% circle of 1".53, again first we draw the constant &W% circle then on the circle you go to

a location where % will be minimum and that will give me the value of % min. &o first since the

rho e0ual to 1".53 it has be the % ma# normali/e, I find out the point which is % ma# this is 1".53

I first draw a circle passing through this point.

&o this is the constant &W% circle for rho e0ual to 1".53 then we -now the left most point on

this circle represents % min. &o if I read this point here this value corresponds to the % min. &o

the % min value from here that will be e0ual to 1.753. &o the normali/e value of the minimum

resistance on the line will be 1.753 if I multiply by a 5 ohms then I will get the minimum

impedance on the line. &o this why !ust a small e#ercise !ust to identify some points on the

switch chart when the impedances are given and the reflection coefficient is given, when the

&W% is given and then find out some other values, with this understanding now of the basic

switch chart, now we can go to a specific problem which we want to solve by using switch chart.


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Let us say I have a fifty ohm line which is terminated in a load impedance of '5 plus ! 45 ohms.

With the help of the smith chart find reflection coefficient in =artesian and polar form at the load

point reflection coefficient an impedance at a distance of 1.'3 lambda from the load end and the

&W% on the line.

&o we have to find out 4 0uantities now we are given the absolute load impedance, we are given

the characteristic impedance and we have to find out the reflection coefficient at the load at a

distance of 1.'3 lambda from the load and also the &W% on the line. To do this first we have to

normali/e of the impedance, so we can calculate the + l normali/e that is '5 plus ! 45 divided by

the characteristic impedance which is 5, so that will be 1.53 plus ! 1.73.

$%efer &lide Time 44()"*

&o the normali/e impedance which we have is 1.53 plus ! 1.73, we ta-e a switch chart and

mar- this point of 1.53 plus ! 1.73 since the reactive part is positive, this point must lie on the

upper half of the smith chart. &o first you find out a constant of circle which is 1.53, this value

and then you find out a constant reactance circle which is having a value of plus 1.73. &o this

value this % is 1.53 and this # is 1.73, this circle. &o the intersection point of these ' which is

point


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the value of the reflection coefficient, comple# reflection coefficient by measuring this distance

and normali/ing that with the radius of smith chart that will give me the magnitude of reflection

coefficient, I can measure the phase of this radius vector and that will give me the phase of the

comple# reflection coefficient.

&o if I do that I get the 0uantity of gamma l in the polar form that will be 1.5'3 angle "

degrees, you can see from here, this is almost half of this radius. &o the magnitude of the

reflection coefficient is 1.5'3, this angle is close to degree. &o its actual value is " degree,

so in polar form the reflection coefficient is 1.5'3 angle " degrees, I can either convert this

into =artesian form by using calculator or I can measure the pro!ection of these on the hori/ontal

and vertical a#is to find out the reflection coefficient in the =artesian form. &o this same thing

can be written as minus 1.3 plus ! 1.5"'3. &o these are the reflection coefficient in the polar

form these reflection coefficient in the =artesian form. The ne#t thing you have to do is to find

out the reflection coefficient at a distance of 1.'3 lambda from the load point.

9ow for this first thing we do is we draw a constant &W% circle now passing through this

point, this point


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find out the circle which is % e0ual to constant circle passing through this point and # e0ual to

constant circle passing through this point, I will get the normali/e impedance corresponding to

this point.

&o from here I can get the normali/e impedance + bar that will be 1".)3 minus ! 1".473. &o

multiplying by 5 ohms the impedance at that location will be 5 into + normali/e that is e0ual

to 7 ohms minus ! 12.53. &o the impedance at this point is 7 minus ! 12.53 ohms and the

&W% from this if I go to the right most point of this circle, this point is value, it has been the

&W%.

$%efer &lide Time )('4*

&o if we read carefully we can see this value lie between 14."3 and 14.'3, so its accurate value is

14."73 when we are doing graphical calculations that -ind of error is acceptable. &o if the point

is lying somewhere between 14."3 and 14.'3 roughly, we may say the value is 14."53, the

accurate value of this point is 14."73. &o we can read of this value and we will get a &W% rho

that is e0ual to 14."73, so in this problem we are given the load impedance we were as- to find

out the reflection coefficient at the load at a distance of 1.'3 lambda. The impedance at a distance


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of 1.'3 lambda and also the &W% on the line and see as we mention earlier that without doing

any calculation, rigorous calculation, we could estimate all these values on the transmission line.

$%efer &lide Time )"(*

&o the point which was made that or the use of switch chart ma-es the calculations of

transmission line e#tremely simple, can be seen here if we are use the analytical e#pression for

the impedance transformation we have to do with the comple# calculation. In this case !ust by

drawing now the constant &W% circle passing through this point, we can !ust read out the

values for the reflection coefficient impedance &W% and so on. Let us ta-e further problem, let

us say now I have a line which is terminated in a normali/e admittance of 1.'3 minus ! 1.53. &o

up till now we have use the switch chart li-e the impedance switch chart, let us now consider a

problem where we are dealing with the admittances, find the location of the voltage ma#imum

from the load end also find the reflection coefficient normali/ed admittance and normali/ed

impedance at a distance of 1."'3 lambda from the load.

&o in this case we are given the admittance which is directly a normali/ed admittance. &o firstly

what we are given is y which is 1."3. Let us ta-e the smith chart at the moment use the switch

chart as the admittance switch chart. &o now switch chart is the admittance one, you ta-e the


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point which is 1.'3 minus 1.53. &o find out the circle which is 1.'3 constant g circle find out

the circle which is constant b circle which is minus 1.53, the intersection point of ' gives you this

point here which is y l bar normali/e admittance at the load point.

9ow we are as- to find out the location of the voltage ma#imum from the load and also find the

reflection coefficient, normali/e admittance and normali/e impedance at a distance of 1".'3

lambda from the load. &o first thing to find out the voltage ma#imum since, we are using now the

admittance chart to voltage ma#imum does not lie on this point because if you recall the y or the

comple# gamma # is has been rotated by " degrees. &o this point now represent the voltage

ma#imum and this point represents the voltage minimum. &o if I move now towards the

generator by a distance till I reach to this point that is the length at which I will get the voltage

ma#imum.

&o first thing from this normali/e admittance, I draw as constant &W% circle move on this

circle by a distance till I reach to this point where the voltage is ma#imum. &o this distance from

here to here that gives you the distance of the voltage ma#imum from the load point. The ne#t

thing we are as- to find out is the reflection coefficient at the normali/e admittance at a distance

of 1."'3 lambda. &o I move on this constant &W% circle by a distance of 1."'3 lambda in the

cloc-wise direction because we are moving towards generator. &o various to this point which is y

bar if I read of the value here, I get the value of normali/e admittance at a distance of 1".'3

lambda from the load point, find out the constant conductance circle passing through this, a

constant acceptance circle passing through this and I get the value which is normali/e value

which is 1."3 plus ! 1.'3.

&o the admittance at a distance of 1."'3 lambda from the load is 1."3 plus ! 1.'3 to get the

reflection coefficient either I can calculate from admittance itself but we are also it have to find

out the normali/e impedance at this distance. We -now from the transmission line property that

the normali/e impedance inverse itself at a distance of lambda by ) but if we invert the

normali/e impedance that will become normali/e admittance and vice versa that means if I ta-e a

point on the constant &W% circle and rotate on this circle or move on this circle by lambda by

), I will get a value which will be inverted value of y bar which is nothing but + bar.


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&o ta-en diagonally opposite point on the constant &W% circle gives me the value of + bar and

in this point now the smith chart becomes the impedance smith chart ,to see interesting things we

started on the switch chart as a admittance switch chart mar- this point which was the load

admittance do a constant &W% circle, moved on this to find out the admittance at a distance of

1."'3 lambda and !ust ta-ing now a diagonally opposite point on this, I get the 0uantity which is

normali/e impedance and beyond this point I can start reading the switch chart li-e an

impedance switch chart.

$%efer &lide Time )7(')*

&o if I read the value for the normali/e impedance at this point that value will be 1".23 minus !

1'.23, so constant % circle which is passing through this is 1".23 the constant reactive circle

which is passing through this is minus 1'.23, so this point here is 1".23 minus ! 1'.23. 8nce I get

that then the reflection coefficient now since the impedance switch chart, the real gamma # is on

the right side I can measure now the magnitude and phase on the reflection coefficient. &o the

radius of this circle gives me the reflection coefficient magnitude, the phase measure from the

real a#is in the conventional way up to this radius vector gives me the phase of the reflection

coefficient at a distance of 1."'3 lambda. &o if I ta-e that from here I will get the reflection

coefficient which is 1.73 magnitudes and the angle will be 4' degrees.


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&o these angle is 4' degrees in this length normali/e to the radius of the switch chart that gives

you the magnitude of the reflection coefficient and that is 1.73. &o this problem clearly

demonstrate the effectiveness of the smith chart for switching over from the admittances to

impedances and vice versa. &o even if some 0uantity that given as impedance or admittance as

the need be, you may smith from the impedance to admittance !ust by ta-ing the diagonally

opposite point on a constant &W% circle and treat the switch chart as the other switch chart. &o

in this case, we started with the admittance to get diagonally opposite point here and the switch

chart becomes impedance switch chart, at some other point if you want to go bac- to the

admittance again wherever we are ta-e a diagonally opposite point you will get the normali/ed

admittances and there onwards I can start using the switch chart as the admittance switch chart

!ust to as- the 0uestion at the load point to what will be the normali/e impedance. We -now the

normali/e admittance which is given by this point.

&o this value we -now, if you want to find out now the normali/e impedance !ust simply find out

what is the diagonally opposite point on this. &o if I e#tend this radius vector, I will get a point

somewhere here and if I read off that value, I will get the normali/e impedance at the load point.

&o we can find now the use of the smith chart or one more thing and that is whatever number we

are having here, it inverts itself to get this number. &o we have you admittance become

impedance and the impedance becomes admittance but essentially what it is doing you -now, it

is inverting a comple# number that is for it is doing. &o the switch chart can be use to inverting

any comple# number.

ou ta-e the comple# number which you want to invert treat it li-e a normali/e admittance or

impedance mar- that point draw a circle pointing to that ta-e a diagonally opposite point and you

will get the inverse of the comple# number. 9ow we will solves more problem by using switch

chart li-e finding out the un-nown impedance doing the matching problem by 0uarter wave

transformer and the single step matching. &o in the ne#t lecture, we will ta-e the impedance

measurement problem and also the impedance matching problem by using the switch chart

Transmission Lines and E.M. Waves Lec 14

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Transcript of Transmission Lines and E.M. Waves Lec 14