Introduction to Smith Charts - Educated India...

23
Introduction to Smith Charts Dr. Russell P. Jedlicka Klipsch School of Electrical and Computer Engineering New Mexico State University Las Cruces, NM 88003 September 2002

Transcript of Introduction to Smith Charts - Educated India...

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Introduction to Smith Charts

Dr. Russell P. JedlickaKlipsch School of Electrical and Computer

EngineeringNew Mexico State University

Las Cruces, NM 88003

September 2002

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EE521 Lecture 3 08/22/02

Page 1 of 22

zZZ

R j XZL

L

o

L L

o= =

+

Γ Γ Γ ΓL L L Le jL

r i= = +∠Γ

Γ LL o

L o

Z ZZ Z

=−+

Smith Chart Summary

A Smith Chart is a conformal mapping between the normalized complex impedance plane ( z =r + j x ) and the complex reflection coefficient plane. First, the normalized impedance is givenas

Consider the right-hand portion of the normalized complex impedance plane. All values ofimpedance such that R $ 0 are represented by points in the plane. The impedance of all passivedevices will be represented by points in the right-half plane.

The complexreflection coefficient may be written as a magnitude and a phase or as real and imaginary parts.

Remember that and the p 'L is measured with respect to the positive real 'r axis. 0 1≤ ≤Γ L

The reflection coefficient in terms of the load, ZL, terminating a line, Zo is defined as

Rearranging the above equation we get

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EE521 Lecture 3 08/22/02

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Z ZL oL

L=

+−

11

ΓΓ

z r j xL L LL

L= + =

+−

11

ΓΓ

( )Γ ΓLL

LL

Lr i

rr r

−+

+ − =

+

10

11

22

2

( )Γ ΓL LL L

r i x x− + −

=

1

1 122 2

rr

L

L10

+

,

11+ rL

11

,xL

So we get the conformal mapping by dividing through by Zo (remember 'L is complex).

Substituting in the complex expression for 'L and equating real and imaginary parts we find thetwo equations which represent circles in the complex reflection coefficient plane.

The first circle is centered at

whose location is always inside the unit circle in the complex reflection coefficient plane. Thecorresponding radius is

So it is observed that this circle will always be fully contained within the unit circle. The radiuscan never by greater than unity. The second circle is centered at

whose location is always outside the unit circle in the complex reflection coefficient plane. Notethat the center of these circles will always be to the right of the unit circle. The corresponding

radius is . The radius can vary between 0 and infinity.1xL

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EE521 Lecture 3 08/22/02

Page 3 of 22

VSWRVV

L

L

= =+

−max

min

1

1

Γ

Γ

The first circles, those centered on the real axis, represent lines of constant real part of the loadimpedance (rL is constant, xL varies). The circles whose centers reside outside the unit circlerepresent lines of constant imaginary part of the load impedance (xL is constant, rL varies).

Circles centered at the match point ( ZL = Zo, or 'L = 0 ) are equidistance from the origin (| 'L | =constant) and are called constant VSWR circles. This related quantity is defined as

Note that the VSWR can vary between .1 ≤ ≤ ∞VSWR

The phase of the reflection coefficient is given by the angle from the right-hand horizontal axis. We have -180° # p'L # +180°. Above the horizontal axis is positive, below negative.

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EE521 Lecture 3 08/22/02

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When we place these other circles on the REFLECTION COEFFICIENT PLANE we call it aSMITH CHART. It can also be referred to as an:

impedance chart

admittance chart

immittance chart (impedance and admittance)

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INSERT SMITH CHART GRAPHIC HERE

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EE521 Lecture 3 08/22/02

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( )zj

j ensionlessL =+

= +100 100

502 2 dim

Normalized Impedance

The transmission line of characteristic impedance, Zo, is terminated in a load impedance, ZL.

If the characteristic impedance is Zo = 50 + j 0 S and the load impedance is ZL = 100 + j 100 S,the normalized load impedance is

Now enter this on the Smith Chart as shown in below.

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INSERT SMITH CHART GRAPHIC HERE

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EE521 Lecture 3 08/22/02

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VSWR = 4 3.

( )Γ =

−+

=+ −+ +

Z ZZ Z

jj

L o

L o

100 100 50100 100 50( )

VSWR =+−

=11

4 265ΓΓ

.

VSWR Circles

To construct the VSWR circle for a lossless line, place the compass center at the center of theSmith Chart ( ZL = Zo, or 'L = 0 ) and construct a circle through zL.

To read off the VSWR on the line, note the intercept of the VSWR circle with the right-half zeroreactance line (it is the horizontal line that bisects the top and bottom of the chart.

For the previous example, this is shown in Figure 2 where we read off the VSWR as

Let’s check this against the calculated VSWR. First, compute the reflection coefficient.

Taking the magnitude we find . Then the VSWR is Γ = 0 62.

This tells us that the graphically obtained value (VSWR = 4.3) is accurate to about onesignificant figure.

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INSERT SMITH CHART GRAPHIC HERE

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EE521 Lecture 3 08/22/02

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YZL

L=

1

Yj

j SiemensL =+

= −1

100 1000 005 0 005. .

( )yYY

jj ensionlessL

L

o= =

+= −

1100 100

1500

0 25 0 25. . dim

Load Admittance

The load admittance is the reciprocal of the load impedance.

For the previous example, we find

Normalized Load Admittance

The normalized load admittance is the reciprocal of the normalized load impedance which can beexpressed as

We can find this more easily by noting that yL is on the constant VSWR circle at a pointdiametrically opposed from zL. This is shown in Figure 3.

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INSERT SMITH CHART GRAPHIC HERE

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EE521 Lecture 3 08/22/02

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Z ZZ jZ

Z jZin o

L o

o L

=+

+

tan

tan

2

2

πλπλ

l

l

( )

( )Z

j j

j j

jj

jin =+ +

+ +=

+= −100

100 100 1004

100 100 1004

1001 2

1200 100

tan

tan

π

π Ω

Input Impedance

The input impedance to the line of characteristic impedance, Zo, a length R from the load, ZL, isfound via the equation.

Where 8 is the wavelength on the transmission line.

Consider the following example with ZL = 100 + j 100 S.

Now consider this process on the Smith Chart.

1. Enter the normalized load impedance, zL = 1 + j 12. Construct the constant VSWR circle. 3. Travel clockwise (transform towards the generator) on the circle by a distance equivalent

to the line length, starting at zL . Remember the distance on the Smith Chart is in termsof wavelength. Specifically, wavelength on the transmission line, which is notnecessarily the free space wavelength. One full revolution is 8/2. The outer scale iscalibrated in wavelengths.

4. Read off zin. In this case, zin = 2 - j 1.5. Compute Zin = zin * Zo = ( 2 - j 1 ) * 100 = 200 - j 100 S.

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INSERT SMITH CHART GRAPHIC HERE

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EE521 Lecture 3 08/22/02

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YZ j

j Siemensinin

= =−

= −1 1

200 1000 004 0 002. .

Input Admittance

Use the Smith Chart to find the input admittance. Consider the example above.

1. Enter the normalized load admittance on the VSWR circle. This is accomplished byentering the normalized load impedance, drawing the VSWR circle and marking the pointdiametrically opposed.

2. Travel clockwise (towards the generator) from yL on the VSWR circle by a distanceequivalent to the line length.

3. Read off yin = 0.44 + j 0.2

4. Find Yin = yin * Yo = ( 0.44 + j 0.2 ) / (100) = 0.004 + j 0.002 Seimens.

Let’s compare the result from the Smith Chart on the next page to the analytical result. From theexample above we found Zin = 200 - j 100 S.

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INSERT SMITH CHART GRAPHIC HERE

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EE521 Lecture 3 08/22/02

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Z ZZ jZ

Z jZin o

L o

o L

=+

+

tan

tan

2

2

πλπλ

l

l

Drill Problem # 1

Consider a load impedance, ZL = 100 - j 25 S, connected to a transmission line of characteristicimpedance Zo = 50 S. The line is R = 3/8 8 long.

Find the following quantities using the Smith Chart on the next page.a) zL = ______________

b) VSWR = ______________

c) yL = ______________

d) zin = ______________

e) Zin = ______________

f) yin = ______________

g) Yin = ______________

h) Check the answer in part e) using the analytical result.\

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INSERT SMITH CHART GRAPHIC HERE

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EE521 Lecture 3 08/22/02

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Z ZZ jZ

Z jZin o

L o

o L

=+

+

tan

tan

2

2

πλπλ

l

l

Drill Problem # 2

Consider a load impedance, ZL = 0 - j 0 S, connected to a transmission line of characteristicimpedance Zo = 50 S. The line is R = 1/4 8 long.

Find the following quantities using the Smith Chart on the next page.a) zL = ______________

b) VSWR = ______________

c) yL = ______________

d) zin = ______________

e) Zin = ______________

f) yin = ______________

g) Yin = ______________

h) Check the answer in part e) using the analytical result.

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INSERT SMITH CHART GRAPHIC HERE

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EE521 Lecture 3 08/22/02

Page 20 of 22

Drill Problem #3

Use the Smith Chart on the next page to find the input impedance to the parallel connected linesZo1 = Zo2 = 70.7 S. Furthermore, ZL1 = ZL2 = 50 + j 0 S. The line lengths are R1 = R2 = 8/4. The characteristic impedance of the main line is Zo = 50 S to which the two parallel lines areconnected, what is the VSWR on this main line?

Drill Problem #4

Use the Smith Chart to find the input impedance to the parallel connected lines Zo1 = 50 S andZo2 = 70.7 S. Furthermore, ZL1 = 50 + j 0 S and ZL2 = 100 + j 0 S. The line lengths are R1 = 8/2and R2 = 8/4. The characteristic impedance of the main line is Zo = 50 S to which the twoparallel lines are connected, what is the VSWR on this main line?

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INSERT SMITH CHART GRAPHIC HERE

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INSERT SMITH CHART GRAPHIC HERE