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Smith Chart

Smith Chart. Graphically solves the following bi-linear formulas. Note: works for admittance too. Just switch sign of r. Im[ r ]. Example:. 45°. | r |=1/2. Re[ r ]. Same length. Smith chart is the interior of the unit circle in the complex plane. Find Z L given r. Note:. Z eq. Z L.

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Smith Chart

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

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

  3. Find ZL given r Note: Zeq ZL Find real and Imaginary parts:

  4. Curves of constant real part: R 2.0 0.2 1.0

  5. +0.5 0.0 -0.5 Curves of constant imaginary part: X

  6. 1.3 45º |r|=1/2 1.4

  7. 45° |r|=1/2 Zeq What is Zeq at l= l/4 from the load? ZL

  8. Z0 = 100 , v= 108 m/s f = 5 MHz ZL = (70+j50)  2 m Zeq = ? Sample Problem: find Zeq Method 1:

  9. Zeq/ Z0 = (1.6+j.65) Method 2: Smith Chart ZL/ Z0 = (70+j50)/100 ZL/ Z0 = (.7+j.5) 0.104 l Move .1 l toward generator 0.204 l ZL/ Z0 = (.7+j.5)

  10. VSWR=3 Vmax ZL = ? l = 20m, Z0=100  1 m 1.7+j1.3 ZL=170+j130  Standing Wave Problem .20l .25l lmax

  11. Z0 = 100 , v= 108 m/s f = 5 MHz ZL = (140+j130)  Ys = ? l=? m ZL/Z0 = (1.4+j1.3) l/l = .06+.168=.228 YL/Y0 = (.38-j.34) Yeq/Y0 = (1.0+j1.2) Ys/Y0 = (0-j1.2) Matching

  12. Short Circuit Impedance Shunt admittance .11 l Short Circuit Admittance Ys/Y0 = (0-j1.2)

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