# Homework Help: TM Lines

1. Sep 25, 2011

### Lunat1c

Hi,

I have a couple questions regarding a homework problem and would really appreciate some help.

Consider the lossless 50ohm transmission line shown below, which has an LC circuit inserted at some point along the line. If I want to find the input impedance how would I approach this problem? As far as I know, the characteristic impedance (Zo) is the same throughout the line since its independent of the length of the line. I'm thinking I should first treat the LC load as my load, take length=0.06lambda and find the input impedance as seen from the very far left of the circuit. Then I would do the same thing on the second part of the circuit and just add the two impedances, however I'm not sure if this is the correct reasoning.

Edit: I think that the two impedances found out with the above method might have to be considered in parallel to each other not in series.

http://img695.imageshack.us/img695/4592/circuitdiagram.th.png [Broken]

Last edited by a moderator: May 5, 2017
2. Sep 26, 2011

### MisterX

Start from the right. Find the input impedance for the transmission line to the right of the LC part. Then combine that impedance with the LC part to get a load impedance for the transission line on the left.

3. Sep 26, 2011

### Lunat1c

Ok so the Zin I get would be in series with j30 and in parallel with -j200?

4. Sep 26, 2011

### MisterX

no

First, let's name this "Zin2" to distinguish it from the Zin from the picture you posted.

"Zin2" would go between the circles on the rightmost dotted line.

This is in series with j30, but not in parallel with the capactor. This is basic circut theory!

5. Sep 26, 2011

### Lunat1c

sorry I misread. ok, I agree that Zin2 would be in series with j30. But wouldn't j30+Zin2 be in parallel with -j200 then?

6. Sep 26, 2011

### MisterX

Sorry, I had made a mistake writing and I just fixed it. Zin2 should be in series with j30.

Zin2 + j30 is in parallel with -j200

7. Sep 26, 2011

### Lunat1c

No problem at all. I totally agree with you. Thanks for your help.