Impedance Matching Coaxial Cables

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SUMMARY

The discussion centers on impedance matching for coaxial cables with a dielectric constant (εr = 9). The characteristic impedance (Zc) is calculated using the formula Zc=sqrt(μ/ε)((ln(b/a))/(2∏)). The participant concluded that the radius of the outer conductor of the second line must be 12 mm to achieve impedance matching. Additionally, removing the dielectric would increase the wave propagation velocity and the characteristic impedance due to a decrease in the denominator of the velocity equation (Vp=C/(sqrt(με))).

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Homework Statement


We need to connect two coax cables with the same dielectric (εr = 9) but
with different dimensions as shown. How big must be the radius of the
outer conductor of the second line in order to match two lines?

How would the velocity of the wave propagation change, if we remove
the dielectric? What’s about the characteristic impedance?


Homework Equations


Zc=sqrt(μ/ε)((ln(b/a))/(2∏))
μ=μrμ0
ε=εrε0

The Attempt at a Solution


I so far have uploaded my answer to the radius part am I right? For the last conceptual parts I am not sure and need a bit of help.
 

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For the conceptual part I am thinking that the velocity would decrease because Vp=C/(sqrt(με) so if the dielectric is removed the denominator will decrease and will lead to a faster propaagation. Furthermore the characteristic impedance is sqrt(μ/ε) so it will increase too.
 
First, what is implied when two transmission lines of different characteristic impedance are "matched"? In regards to the conceptual question, it looks like you have the right idea.
 
Zl=sqrt(μ/ε)(((ln(b/a))/(2∏))=sqrt(μ/ε)((ln(b2/a2))/(2∏)) so I can cancel 2 pi and sqrt(μ/ε) and both natural logs giving b1/a1=b2/a2 solving for b2 I get 12 mm is it still wrong I am lost?
 
I wasn't implying that it was wrong I was just seeing if you understood what was meant conceptually by impedance matching.
 

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