| View previous topic :: View next topic |
| Author |
Message |
winson Antenna Theory Regular
Joined: 12 Jul 2009 Posts: 16
|
Posted: Tue Nov 10, 2009 7:59 pm Post subject: CST Curve results problem (Parameter sweep & Direct solv |
|
|
Hi Friends,
I'm trying to simulate the phase of the single element for reflectarray.
My problem is with the same model i getting two different phase curve, one phase curve is generated from direct solving of transient time solver (just click "start" to simulate), and the second phase curve is generated by using the parameter sweep by sweeping one parameter. For S curve both technique giving the same s curve.
The two phase curve having some similarity that similar in the part from 9.3GHz to 11.5GHz, but i'm confusing why there are some part are different. Anybody can give me some suggestion?
The results is attached:
Phase curve from direct solving, arg(s)
Phase curve from parameter sweep, arg(s11)
The s curve for the single element
Actually what is the difference between arg(s) and arg(s11)?
This look like may giving different phase results.
Any suggestion will be appreciate.
Thanks in advance.  |
|
| Back to top |
|
 |
bigSteve Antenna Wizard
Joined: 14 Mar 2009 Posts: 265
|
Posted: Tue Nov 10, 2009 9:21 pm Post subject: |
|
|
Your phase curves are exactly the same.
The first curve starts at about 270 degrees, your second curve starts at -90 degrees. The difference is 360 degrees - any phase change of 360 degrees (or multiple thereof) is the same result.
The only difference in the two plots is the first one forces the phase to be between -180 and + 180, and the other one does not (and is increased by 360 degrees). |
|
| Back to top |
|
 |
winson Antenna Theory Regular
Joined: 12 Jul 2009 Posts: 16
|
Posted: Wed Nov 11, 2009 6:18 pm Post subject: |
|
|
Hi bigSteve,
Thanks for your clear explanation.
I have also try to unwrap one of the phase curve and now it show that they are exactly the same. |
|
| Back to top |
|
 |
|