1) A constant chart which is linear in distance

The Smith Chart is a superposition of two coordinate systems:

2) A constant r-x chart which is not linear in distance

The chart is set up such that finding the location of r and x for a given normalized impedance will yield the corresponding and vice-a-versa. This makes the Smith Chart a useful tool for impedance transformations

||

1.0

0.5

z = 0.3 – j1r = 0.3

x = -1

z = .3 – j1.0

The standing wave ratiois read off of the chartby noting the r value where a constant circle intersects ther axis

SWR1/SWR

1) SWR = Zmax/Z0

= zmax

= rmax

2) SWR = Z0/Zmin

= 1/zmin

= 1/rmin

g values

b values

-b values

POC PSC

To convert from zL to yl we can either:

1) Rotate around constant by /4 (180°)

2) Draw a line from zL through origin until it intersects constant

WTG = .14

WTG = .39

zL

yL

/4

We can transform z intoy by rotating z half wayaround a constant circle

Given Z = 95+j20 ona 50 line, find Y

1) Find zz = 1.9+j0.42) Draw circle

3) Draw line through origin

4) Find intersection with circle

5) Read off yy = 0.5-j0.1

6) Renormalize yY = y/Z0

= 10-j2 mSYcalc = 10.1-j2.12 mS

A 50- T-L is terminated in an impedance ofZL = 35 - j47.5. Find the position and length of the short-circuited stub to match it.

1) Normalize ZL

zL = 0.7 – j0.952) Find zL on S.C.3) Draw circle4) Convert to yL

zL

yL

5) Find g=1 circle6) Find intersection

of circle and g=1 circle (yA)

yA

7) Find distance traveled (WTG) to get to this admittance

WTG = .109

WTG = .168

8) This is dSTUB

dSTUB = (.168-.109)dSTUB = .059

A 50- T-L is terminated in an impedance ofZL = 35 - j47.5. Find the position and length of the short-circuited stub to match it.

9) Find bA

yA

bA = 1.2

10)Locate PSC

PSC

11)Set bSTUB = bA and find ySTUB = -jbSTUB

ySTUB = -1.2

12)Find distance traveled (WTG) to get from PSC to bSTUB

WTG = 0.25

WTG = 0.361

13)This is LSTUB

LSTUB = (0.361-0.25)LSTUB = .111

Our solution is to place a short-circuited stub of length .111 a distance of .059 from the load.

There is a second solution where the circle and g=1 circle intersect. This is also a solution to the problem, but requires a longer dSTUB and LSTUB so is less desireable, unless practical constraints require it.

zL

yL

yA1

WTG = .109

yA2

WTG = .332

dSTUB = (.332-.109)dSTUB = .223LSTUB = (.25+.139) LSTUB = .389