Smith Chart

------A Graphical Representation

By

Dr . Taimoor Khan

Smith Chart

Smith Chart• Smith Chart was developed in 1939 by P. Smith at Bell Telephone Laboratory USA.

• The Smith Chart has been introduced to reduce the tedious manipulations involved in calculating the characteristics of a transmission lines.

• It is graphical tool which is very much useful for solving transmission line problems.

• A Smith Chart is a conformal mapping between the normalized complex impedance plane ( z = r + j x ) and the complex reflection coefficient plane ( Γ= Γr+ jΓi).

• It is essentially a plot of the voltage reflection coefficient (Γ) in complex plane.

• It can be used to convert from voltage reflection coefficient (Γ) to normalized impedances (z=Z/Z0) and admittances (y=Y/Y0),and vice versa.

• It consists of a Polar/Cartesian plot of reflection coefficient onto which is overlaid circles of constant real and constant imaginary impedance. The standard chart is plotted for Γ≤1.

• Today, Smith Chart is a presentation medium in computer-aided design (CAD) software.

• Parameters plotted on the Smith Chart include the following:– Reflection coefficient magnitude, Γ– Reflection coefficient phase angle, θr

– length of transmission line between any two points in wavelength– VSWR

– Input Impedance, Zin

– The location of Vmax and Vmin. i.e: dmax and dmin

Parameters Plotted on SMITH CHART

(a) Typical circuit elements (b) Impedances graphed on rectangular coordinate plane (Note: ω is the angular frequency at which Z is measured.)

The Smith Chart Derivation

Reflection Coefficient in phasor form:

Example 1: A transmission line with a characteristic line impedance of Zo = 50 Ohm is terminated into the following load impedances:

a.) ZL =0 (Short Circuit)

b.) ZL =infinite (Open Circuit)

c.) ZL =50 Ohm

d.) ZL= (16.67-j16.67) Ohm

e.) ZL= (50+j150) Ohm.

Find the individual reflection coefficients and display them in the complex Γo- plane.

Solution:

Based on (1) the following reflection coefficients (Γo) can be computed:

a.) Γo=-1 for ZL =0 (Short Circuit)

b.) Γo= 1 for ZL =infinite (Open Circuit)

c.) Γo=0 for ZL =50 Ohm

d.) Γo=0.54 Angle(2210) for ZL =50 Ohm

e.) Γo=0.83 Angle (340) for ZL= (50+j150) Ohm

The different values of the reflection coefficients can be plotted in polar form:

--------------(1)

Normalized Input Impedance

• Load Reflection Coefficient:

0

00 ZZ

ZZ

L

L

+−=Γ

--------------(2)

ir j 000 Γ+Γ=Γ --------------(3)

Lie θ00 Γ=Γ --------------(4)

Contd…..

• Input Reflection Coefficient:lj

in ed β20)( −Γ=Γ=Γ

ir jd Γ+Γ=Γ )(

--------------(5)

--------------(6)

Input Impedance:

Contd…..

--------------(7)0

00 1

1

Γ−Γ+= ZZ in

Normalized Input Impedance:

)(1

)(1)( 0 d

dZdZ in Γ−

Γ+= --------------(8)

)(1

)(1)(0 d

djxrzZ

dZin

in

Γ−Γ+=+== --------------(8)

ir

irin j

j

d

djxrz

Γ−Γ−Γ+Γ+=

Γ−Γ+=+=

1

1

)(1

)(1--------------(9)

Contd…..

)1(

)1(

)1(

)1(

ir

ir

ir

irin j

j

j

jjxrz

Γ+Γ−Γ+Γ−

Γ−Γ−Γ+Γ+=+= --------------(10)

Real Part:

--------------(11)

--------------(12)

Imaginary Part:

( )[ ] 2222 11 irirr Γ−Γ−=Γ+Γ−

From (11)

----------------- (13)

22

2

1

1

1

+=Γ+

+−Γ

rr

rir ----------------- (14)

( ) ( ) 222 )( cba ir =−Γ+−Γ ----------------- (15)

Resistive Circle , r-circle

Contd….

Plot of reflection coefficient in complex plane

-1

1

Re(reflection coef.)

Im

(re

fle

ctio

n c

oe

f.)

1•j

-1•jcurves of constant r = Re(Z)

• plot Γ as a function of r– these are circles!

When: r = 1

When: 0 < r < 1

When: r > 1

When r = 0

• radius

+

0,1 r

r

r+1

1

• center

22

2

1

1

1

+=Γ+

+−Γ

rr

rir

( )2

22

1

10

=Γ+−Γ ir

22

2

11

1

11

1

+=Γ+

+−Γ ir

( )[ ] iirx Γ=Γ+Γ− 21 22

From (12)

----------------- (16)

( )22

2 111

=

−Γ+−Γ

xxir ----------------- (17)

( ) ( ) 222 )( cba ir =−Γ+−Γ ----------------- (18)

Reactance Circle, x-circle

Contd…..

Plot of Reflection Coefficient

-1

1

Re(reflection coef.)

Im

(re

fle

ctio

n c

oe

f.)

j

- j

• plot Γ as a function of x– x = ± ∞

When: x < 0

When: x > 0

When: x = 0

• from the Im part:

( )2

22

0

1

0

11 =

−Γ+−Γ ir

( )∞

=

∞±−Γ+−Γ 11

12

2ir

curves of constant x = Im(Z)

– these are also circles!

• radius

x

1,1

x

1• center

( )22

2 111

=

−Γ+−Γ

xxir

Mapping RHP to a unit circle – Smith Chart

Inductive

ResistiveLow Z

High Z

Capacitive

Admittance Transformation

Ideal Capacitors and Ideal Inductors in Smith Chart

R +L (Smith Chart)

R II L (Smith Chart)

R + C (Smith Chart)

R II C (Smith Chart)

Impedance Transformation

Ex. A Load impedance ZL=(30+j60)Ohm is connected to a 50 Ohm transmission line of 2 cm length and operated at 2 GHz. Use Smith Chart concept & find the input impedance Z in under the assumption that the phase velocity is 50% of the speed of light.

Solution

• Steps:• Normalize load impedance (ZLn=ZL/Z0)

• Locate ZLn in the Smith Chart

• Find reflection coefficient (Γo)

• Rotate reflection coefficient by twice its electrical length βd to obtain Γin(d)

• Record normalized input impedance at this spatial location d.

• De-normalize input impedance

Solution

• The normalized load impedance (ZLn):

(0.6+j1.2) Ohm

Graphical display

Smith Chart: Impedance Coordinate

Smith Chart: Admittance Coordinate

Smith Chart: Constant SWR Circles

Plotting SWR circle on the Smith Chart

– Any lossless line can be represented on the Smith chart by a circle having its origin at 1± j0 (center of the chart) and radius equal to the distance between the origin and the impedance plot.

– Therefore, SWR corresponding to any particular circle is SWR corresponding to any particular circle is equal to the value of Zequal to the value of ZLL /Z /Z00 at which the circle crosses the at which the circle crosses the

horizontal axis on the right side of the chart.horizontal axis on the right side of the chart.

If Z0 = 50 Ω and ZL = 25 + j25Ω.

Find yL, YL and SWR.

Example

Answer:

– zL is plotted on the Smith chart by locating the point where R = 0.5 arc intersects the X = 0.5 arc on the top half of the chart.

– ZL = 0.5 + j0.5 is plotted on the Figure 34 at point A and yL is plotted at point B(1- j1).

– From the Smith Chart, SWR is approximately 2.6 (point C)

PROPOSED PROPOSED SOLUTION SOLUTION

FOR FOR Example Example

SC and OC points on the Smith chartSC and OC points on the Smith chart

• The location of SC and OC points are different depending on Impedance or Admittance chart.

Points of OC and SC on Impedance ChartPoints of OC and SC on Impedance Chart

SC OC

Points of OC and SC on Admittance ChartPoints of OC and SC on Admittance Chart

OC SC

Smith Chart: Constant Impedance Magnitude Circles

• For transmission line terminated in a purely resistive load not equal to Z0 , Smith Chart analysis is very similar to the process described in the preceding section

• For example, for load impedance ZL = 37.5Ω, characteristic impedance Z0 = 75 Ω, input impedance at various distance from the load can be determined as follows– Normalized impedance, zL=0.5– zL = 0.5 is plotted on the Smith chart. A circle is drawn that

passes through point A with its centre located at the intersection of the R = 1 and X = 0 arc.

– SWR is read at the intersection of the circle and the X = 0 line on the right side, SWR = 2

Plotting Zin on the Smith Chart

SWRPoint A

– The input impedance, Zin at a distance of 0.125λ from the load is determined by extending point A to a similar position on the outside scale (point B) and moving around the scale in clockwise direction a distance of 0.125λ=0.127x720°=90°

– Rotate from point B a distance equal to the length of the transmission line ( 0.125λ at point C). Transfer this point to a similar position on the SWR=2 circle (point D)

– Normalized input impedance is located at point D (0.8 +j0.6). Actual impedance is found by multiplying the normalized impedance by the characteristic impedance of the line

– For distance of 0.3λ= 0.3x720°= 216° from the load, the normalized impedance is plotted at point E

Plotting Zin on the Smith Chart(cont.)

Ω+=+= 45j6075)6.0j8.0(inZ

Example

Determine the input impedance and SWR for a transmission line 1.25λ long with a characteristic impedance Zo = 50Ω and load impedance, ZL = 30 + j40Ω

Answer: Zin = 31.5 – j38.5, SWR = 2.9

Answer: Zin = 30 – j40, SWR = 3.0

PROPOSED PROPOSED SOLUTION SOLUTION

Example

A 30m long lossless transmission line with Zo = 50Ω operating at 2MHz is terminated by a load of ZL=60+j40Ω. If v=0.6c on the line, find:

• Γ

• VSWR

• Zin

Ω+==°∠=Γ 75.1j5.23inZ;1.2S;5635.0Answer:Answer:

Vmax versus

Vmin

Determining ZL Using Smith Chart (moving counterclockwise)

• Given that S=3 on a 50-Ω line, that the first voltage minimum occurs at 5cm from the load, that the distance between successive minima is 20cm, find the load impedance, ZL.

Answer: zAnswer: zLL=0.6-j0.8; Z=0.6-j0.8; ZLL=30-j40=30-j40ΩΩ

Example

Determine the SWR, characteristic impedance of a quarter wavelength transformer, and the distance the transformer must be placed from the load to match a 75 Ω transmission line to a load ZL = 25 - j50

Answer: 4.6; 35.2Ω; 0.1λ

EXERCISE

• An antenna, connected to a 150Ω lossless line, produces a standing wave ratio of 2.6. If measurements indicate that voltage maxima are 120cm apart and that the last maximum is 40cm from the antenna, calculate:– The operating frequency

– The antenna impedance

– The reflection coefficient. Assume v=c.

STUB MATCHING Matching a load ZL= 50-j100 to a 75Ω TL can be done by shorted

stub using Smith chart. • Steps:

1) Plot the normalized impedance, zL=0.67-j1.33 (Point A) and draw VSWR circle. Stubs are shunted across the load, thus admittances are used rather than impedances to simplify calculations. The circles and arcs on the Smith chart are now used for conductance and susceptance.

2) Normalized admittance,yL is determined by rotating the impedance plot, zL 180o(Draw a line from point A through the center to point B)

3) Rotate admittance point clockwise to a point on the impedance circle where it intersects the r = 1 circle (point C). The real component of the input impedance at this point is equal to the characteristic impedance Z0, Zin = r ± jx, where R = Z0. At point C, admittance yd = 1 + j1.7.

STUB MATCHING

• Steps (cont.):4) Distance from B to C is how far the stub must be placed, for this

example 0.09λ. The stub must have an impedance with zero resistive component and susceptance that has the opposite polarity (i.e. ys = 0 – j1.7)

5) To find the length of the stub where ys = 0 – j1.7, move around the outside circle of the Smith chart (R = 0), having a wavelength identified at point D until an admittance ys = 1.7 is found (wavelength value identified at point E).

6) If open stubs are used, rotation would begin at opposite direction (point F)

7) The distance from point D to point E is the length of the stub which is 0.084λ

PROPOSED PROPOSED SOLUTION SOLUTION

FOR FOR Stub Stub

MatchingMatching

Example

For a transmission line with a characteristic impedance Z0 = 300Ω and a load with complex impedance ZL = 450 + j600, determine:SWR, the distance a shorted stub must be placed from

the load to match the load to the linethe length of the stub.

2BdAd λ=+Note that:

In this case, we have two possible shunted stubs. To avoid confusion, we normally choose to match the shorter stub

and one at a position closer to the load lA and dA

yd.lB and dB is the

alternative stub distance and stub

length

Stub Matching Network

lx

xd

lx

xd

Smith Chart: Constant Impedance Phase Angle Circles

Few Other Applications

• Unary Operators• Squares a2• square roots • tangents tan q• cotangents cot q• inverse tangents tan-1 a• Inverse cotangents cot-1 a• Binary Operators• multiplication a • b• Division c/a• geometric mean

Please Study for your Mid Sem 1!