# Impedance Matching Coaxial Cables

1. Dec 4, 2013

### DODGEVIPER13

1. The problem statement, all variables and given/known data
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?

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

3. 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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2. Dec 4, 2013

### DODGEVIPER13

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.

3. Dec 4, 2013

### sandy.bridge

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.

4. Dec 4, 2013

### DODGEVIPER13

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?

5. Dec 4, 2013

### sandy.bridge

I wasn't implying that it was wrong I was just seeing if you understood what was meant conceptually by impedance matching.