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How to determine ABCD parameters for a TL

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satishgra

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Hi,

I am trying to find the equivalent T/PI network of a microstrip.
I found the following solution for calculating ABCD parameters for a TL.

I have few doubts in this :

1. Why is V1 = (V+) {e(jbl) + e(-jbl) } when I2 =0 {basically 2nd port is OC}
2. Why is V1 = (V+) {e(jbl) - e(-jbl) } when V2 =0 {basically 2nd port is SC}
3. If i also want to consider attenuation constant(alpha), should i replace Beta by alpha+Jbeta in the final result ??
 

Hi

First it is really important you have the correct coordinate system for this. it makes it easier to analyze. Lets say port 2 is placed at z=0, and port 1 is placed at z=-l. So the length of the line is equal to l.

First write the equation of the total voltage of a transmission line:
V(z)=(V+)e(-jbl) + (V-)e(jbl), note it is a function of z.

1.
When you have the open circuit. The reflection at at port 2 is 1. Using z=0 you get V(z=0)=(V+)+(V-)=(V+)[1+(V-)/(V+)]. We know that (V-)/(V+) is equal to 1.
Therefore V(z)=(V+)e(-jbl) + (V-)e(jbl)=(V+)[e(-jbl)+e(jbl)]
V1=V(z=-l)=(V+)[e(jbl)+e(-jbl)]=2(V+)cos(bl)

Hope this helped
 

1. Does V2 = 0 mean short circuit or terminated by matching impedence

It cant be matching impedence, bcos it is clearly stated that,
V1=(V+)e(-jbl) - (V-)e(jbl), and not V1=(V+)e(-jbl)

Therefore, it should mean that port2 is shorted and that we are also including the reflected wave

2. How does I2 turns out to be 2V+/Z0

As I2 = I1 (SC), should it not equal V1/Z0
 

Hi

Yes port 2 is shorted.

The current on the transmission line is given by
I(z)=\[\frac{(V+)}{Zo}\]e(-jbz)--\[\frac{(V-)}{Zo}\]e(jbz)

You can see that V2 is equal to 0 and if you set z=0 then I2 is equal to 2(V+)/Zo.
remember that a short gives a -1 as reflection coefficient.

When you have this then it is just a matter calculating:D
 
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hey,

Thanks for the reply...just to complete the discussion, I found a ppt on the same topic. I thought this will give even more a clear picture

Regards,
satish
 

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