# Smith chart tutorial question

I am looking at a Smith Chart tutorial on the antenna-theory.com website. Unfortunately, I'm confused by one of the examples given...I don't get the same answer, so it's safe to say I'm doing something wrong. I was hoping someone could point me in the right direction.

It's example 2 on that page. It assumes a reference impedance, Zo = 50 Ohms. It provides a reflection coefficient, Γ = -0.3 + i0.4, and calculates the corresponding load impedance, Zl = 20.27 + i*21.62 Ω using the equation below:

Zl = Zo (1+Γ) / (1-Γ)

However, I don't get the same result for Zl. My working is:

For the real component of Γ, Zl = 50Ω (1+(-0.3)) / (1-(-0.3)) = 26.9Ω

For the imaginary component: Zl = 50Ω (1+i0.4) / (1-i0.4) = 116.67 Ω

Giving Zl = 26.9 + 116.67 Ω, which is miles off their answer. I'm obviously misunderstanding something, so any help would be appreciated!

For the real component of Γ, Zl = 50Ω (1+(-0.3)) / (1-(-0.3)) = 26.9Ω

For the imaginary component: Zl = 50Ω (1+i0.4) / (1-i0.4) = 116.67 Ω

You can't split up the real and imaginary components like this. Your equation needs to be

$$\Z_l=50 \frac{1 + (-0.3+0.4j)}{1-(-0.3+0.4j)}\$$

If you then simplify this you get to the answer they provided.

One of the nice things about a Smith chart is that you can easily convert between reflection and impedance. If you had a smith chart with a grid on it, you could plot this reflection and directly read off the answer without needing to do any math.

• Thanks. I should have seen that to be honest, but I think I'm so rusty that my mind blanked out. Your pointer got me thinking in the right direction and after a bit of complex division I got the correct answer without any trouble. Commented Jan 23, 2020 at 15:16