# Board index » delphi » Smith Chart

## Smith Chart

2006-06-24 11:54:20 AM
delphi53
This is more of a general algorithm/math question. If you know what I
a smith chart is this will make more sense. Lets say I have an
equation for a set of circles...
(x - R/R+1)^2 + (y - 0)^2 = (1/R+1)^2
This puts the center of the circles at
((R/R+1), 0) and a Radius of 1/R+1
R is between 0 and 1.
I can map a series of these circles by sweeping R from 0 to 1 and then
plot them on the screen. My question is how can I go the other way?
Take the mouse postion and figure out what R is for the circle I am on
in the window? This problem is easy if the circle centers were all the
same. The catch is that the circle centers are a function of R as well.
Thanks.
Jim
--
www.mustangpeak.net

## Re:Smith Chart

##### Quote
This is more of a general algorithm/math question. If you know what I
a smith chart is this will make more sense. Lets say I have an
Better yet anyone know of a Smith Chart component?
Jim
--
www.mustangpeak.net

## Re:Smith Chart

"Jim" <XXXX@XXXXX.COM>writes
##### Quote
Take the mouse postion and figure out what R is for the circle I am on
in the window? This problem is easy if the circle centers were all the
same. The catch is that the circle centers are a function of R as well.
Draw a circle with center at the center of the chart and which passes
through the mouse click point of interest. In normalised units, reflection
coefficient k has magnitude of the circle radius ( 0 to 1 ), angle is 0 to
360 degrees around the chart. Now convert the complex number k to impedance
with
z = (1+k) / (1-k)
Now you have z with real (r) and imaginary (j), you can draw your r and j
circles.
Roger lascelles

## Re:Smith Chart

##### Quote
Draw a circle with center at the center of the chart and which passes
through the mouse click point of interest. In normalised units,
reflection coefficient k has magnitude of the circle radius ( 0 to 1
), angle is 0 to 360 degrees around the chart. Now convert the
complex number k to impedance with

z = (1+k) / (1-k)

Now you have z with real (r) and imaginary (j), you can draw your r
and j circles.
I knew there was an easy way and the reflection coefficent is the does
make it easy. Thanks.
Jim
--
www.mustangpeak.net

## Re:Smith Chart

##### Quote
Now you have z with real (r) and imaginary (j), you can draw your r
and j circles.
Got another one for you. How would I draw the Q curves? Where Q =
Imag(Z)/Real(Z)
Jim
--
www.mustangpeak.net