Question E9G08:
How is a Smith chart normalized?  

Choices:  
A: Reassign the reactance axis with resistance values  
B: Reassign the resistance axis with reactance values  
C: Reassign the prime center’s impedance value  
D: Reassign the prime center to the reactance axis    

Correct Answer is  C: Reassign the prime center’s impedance value  

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Normalization on a Smith chart means you choose a reference impedance, $Z_0$, and interpret the chart’s center (the “prime center”) as that value. In other words, you “reassign the prime center’s impedance value” so that the center corresponds to $Z = Z_0$ (i.e., $z = 1 + j0$ in normalized units).

Why the correct answer is correct:
- Normalizing is done by dividing all impedances by a chosen $Z_0$:  
  $$z = \frac{Z}{Z_0}$$  
  This makes the prime center represent $z = 1 + j0$, which is $Z = Z_0$ in real ohms. Choosing a different $Z_0$ “reassigns” the impedance value of the chart’s center.

Why the other choices are incorrect:
- A: Reassign the reactance axis with resistance values  
  - The chart’s geometry (resistance circles and reactance arcs) does not change during normalization; you don’t swap what axes represent.
- B: Reassign the resistance axis with reactance values  
  - Same issue: normalization doesn’t relabel axes; it rescales values by $Z_0$.
- D: Reassign the prime center to the reactance axis  
  - The prime center lies on the resistive axis at $X=0$. Moving it to the reactance axis makes no sense and is not part of normalization.

Useful formulas and tips:
- Normalize impedance and admittance:  
  $$z = \frac{Z}{Z_0}, \quad y = \frac{Y}{Y_0} = Y Z_0$$
- Reflection coefficient in terms of normalized impedance or admittance:  
  $$\Gamma = \frac{z - 1}{z + 1} = \frac{Z - Z_0}{Z + Z_0}$$  
  $$\Gamma = \frac{1 - y}{1 + y} = \frac{Z_0 - Z_L}{Z_0 + Z_L}$$
- De-normalize to get real ohms:  
  $$Z = z \, Z_0$$
- Rules of thumb:  
  - Prime center: match, $Z = Z_0$, $\Gamma = 0$, VSWR = 1.  
  - Right edge: open-circuit ($|\Gamma| = 1$, $Z \to \infty$).  
  - Left edge: short-circuit ($|\Gamma| = 1$, $Z = 0$).  
  - Top half: inductive (+X), bottom half: capacitive (−X).

Example: With $Z_0 = 75\,\Omega$, the prime center is $75\,\Omega$. A $150\,\Omega$ load plots at $z = 2 + j0$ on the resistive axis to the right of center.