Smith Chart Graphically solves the following bi-linear formulas Note: works for admittance too. Just switch sign of 

Smith chart is the interior of the unit circle in the complex plane 45° |  |=1/2 Example: Same length Re[  ] Im[  ]

Find Z L given  Note: ZLZL Z eq Find real and Imaginary parts:

Curves of constant real part: R

Curves of constant imaginary part: X

45º |  |=1/2

What is Z eq at l= /4 from the load? 45° |  |=1/2 ZLZL Z eq

2 m Z L = (70+j50)  f = 5 MHz Z 0 = 100  v= 10 8 m/s Z eq = ? Method 1: Sample Problem: find Z eq

Z L / Z 0 = (70+j50)/100 Z L / Z 0 = (.7+j.5) Move.1 toward generator Z eq / Z 0 = (1.6+j.65) Method 2: Smith Chart

Standing Wave Problem 1 m Z L = ? V max  = 20m, Z 0 =100  VSWR= j1.3 Z L =170+j130  l max

l=? m Z L = (140+j130)  f = 5 MHz Z 0 = 100  v= 10 8 m/s Y s = ? Z L /Z 0 = (1.4+j1.3) Y L /Y 0 = (.38-j.34) Y eq /Y 0 = (1.0+j1.2) Y s /Y 0 = (0-j1.2) l/ = =.228 Matching

Shunt admittance Y s /Y 0 = (0-j1.2) Short Circuit Impedance Short Circuit Admittance.11