How to match negative Zs with Smith Chart? ( -28.526393888 + 70.854968742i)
时间：04-05
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I am using Smith v3.10. After calculating required Zs for transistor matching, i realized that Re part is -28, and can't be put into program.
How to solve this problem? Samples from my book have no such case.
How to solve this problem? Samples from my book have no such case.
Code:
_S11=PolarToRect(0.7,-64.8)_S11 = 0.298045504 - 0.633378937i Gs=1/_S11Gs = 0.608256131 + 1.292610075i Zs=50*(1+Gs)/(1-Gs)Zs = -28.526393888 + 70.854968742i Gout=S22+(S12*S21)*Gs/(1-S11*Gs)Gout = 1.886478204 + 0.744136625i Zout=50*(1+Gout)/(1-Gout)Zout = -116.175687126 + 55.549873949i Abs(Gs)Ans = 1.428571429 Abs(Gout)Ans = 2.027939676
You can't have a negative real part.
What is the Gs value?
Sure it can have a negative real part, if it's an active network. But you can't match to it with a lossless network. You can either move Z into the RHP using series/shunt resistance, or change the transistor circuit to give a RHP Z.
Gs is Гs - source reflection coefficient
_S11 is S11 with a little smaller real part
Oh i see.
Does it mean something like changeing _S11 from (0.7, -64.8deg) to (1,-64.8deg), then real impedance part becomes zero?
I am learning an oscillator.