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Smith Chart question

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Tom Bruhns

Jan 1, 1970
0
Hi Rick,

Yes, I derived it myself. I did it much as you describe, but more
specifically as follows. First, realize that the Smith chart is
simply a linear plot of reflection coefficient with respect to the
chart's normalization impedance. Then the circle around (but not
centered on) the transmission line impedance will have its center at a
real value, given the real normalization and line impedances. And the
circle will intersect the real axis when the line's reflection
coefficient is real, once at +r and once at -r (your quarter wave
length change). (Reminder: r = magnitude of reflection coefficient of
the line's load with respect to the line impedance.) You can find
those impedances where the circle crosses the real axis from
Zline*(1+r)/(1-r) and Zline*(1-r)/1+r). You can then express those as
reflection coefficients to the Smith-normalization impedance. They
define the end points of a diameter of the circle. Since we're now in
the linear space we want, just average those two to find the center.

As I was trying to generalize it to arbitrary complex impedances
(which I think I see, but need to totally convince myself before
posting it), I realized that the whole formulation becomes simpler
still if you get rid of the line impedance by representing it as a
reflection coefficient with respect to the chart normalization
impedance. So, as before, define r=magnitude of the reflection
coefficient representing the circle around the transmission line (that
is, with respect to Zline), and add p = the reflection coefficient of
Zline with respect to Z0 of the chart. Note that p is real valued
(because of our restrictions on Z0 and Zline), but can be positive or
negative--it's not just the magnitude of rho. Then, in the chart's
linear reflection coefficient domain, the circle center will be at:

center = p*(1-r^2)/[1-(p*r)^2]

Dat's it. And it does check out as equivalent to the earlier one I
posted. If I didn't screw up the math, the radius comes out (in the
linear refl.coeff. units)

radius = r*(1+p-p^2-p^3)/[1+p*r-(p*r)^2-(p*r)^3]

And I'm wanting to say that if you put things in polar coordinates,
it's "obvious that" you can just rotate things so they look just the
same as that simple formulation. But I got a bit tangled up in my
polar math, and am not happy with that yet.

Now mind you, someone will come in here and say that "all this is
obvious from thus and such property of the coordinate transformation."
That's wonderful, but I've never been through learning all those
properties before, so I work it out and by so doing, I learn it and
internalize it.

Again, my thanks to you for getting me to think about this deeper than
I had before!

Cheers,
Tom

Rick said:
Tom Bruhns said:
Rick said:
Try also putting in a section of transmission line: note that the
point moves in a circle centered on the line's characteristic
impedance.

Only if the line's characteristic impedance is the same as the
normalising impedance of the chart. Otherwise, the equation for
the centre of the circle is a bit of a bitch.

Having not actually gone through that manipulation before, I thought
it might be worthwhile. _IF_ the line impedance is real, and of
course the Smith chart normalization impedance is real, then it's not
too bad. (In fact, properly formulated it shouldn't be bad even for
complex impedances). If r is the magnitude of the reflection
coefficient with respect to the line, whose impedance is Z1 (a real,
not complex, value) and Z0 is the chart normalization impedance, then
the circle will be centered at a Z0 reflection coefficient given by

(Z1^2-Z0^2)/[Z1^2+Z0^2+2*Z1*Z0*(1+r^2)/(1-r^2)]

Ahh. Nice - did you derive that expression yourself? When I went
through this a few months ago I couldn't find anything in any of
my numerous books on RF/microwave...they all seem to think that
you're only interested in tlines of the same impedance as that
used to normalise your chart.
I clearly chose the most convoluted method possible to derive
my form of the same equation, in the parametric u-v coordinates:

u = (Ztl*Ztl-Z0*Z0)*(m->real)/(Ztl*Ztl+Ztl*Ztl*(m->real)+Z0*Z0*(m->real)+Z0*Z0*(m->imag)*(m->imag)+Z0*Z0*(m->real)*(m->real));
v = 0 ;

I took the impedance as one point on the circle, transformed
it through a quartwave to another impedance, converted both
of these to u-v coordindates, and that, coupled with the
observation that the circle must lie on the v=0 line is
enough to derive the equation of the circle. The equation
is now nestling somewhere in the 5,000 lines of code that
I've so far written towards my Smith Chart program.
 
T

Terry Given

Jan 1, 1970
0
Active8 said:
IIRC, you can use the scales on the perimeter of the chart and/or
the bottom scales to design your microstrip stubs *as well as* the
matching sections. Pretty cool chart, really. I wish I could find a
comprehensive online reference that covers it all as opposed to bits
and pieces.

didn't Smith write a book ?
 
R

Rick

Jan 1, 1970
0
Tom Bruhns said:
Hi Rick,

Yes, I derived it myself. I did it much as you describe, but more

[major snip]

Thanks, Tom. I'll digest that little lot at leisure.
Now mind you, someone will come in here and say that "all this is
obvious from thus and such property of the coordinate transformation."

That was my initial worry when I saw your post - that it was already
well-known and I'd wasted time puzzling it out for myself!
 
R

Rick

Jan 1, 1970
0
Active8 said:
On 01 May 2004 15:01:40 GMT, Rick wrote:

LOL. That's way more lines than I've written, maybe :) IIRC, right
now I basically have classes to convert between Z, Y, and S params
and do the math on those complex (or polar) numbers. I think you can
multiply say a Z param with an S param. It's a bunch of function
overloading wrapped up in the param classes which derive from an
abstract base class, Param.

I managed to get a "Series C" button working that starts with the
last plotted point and restricts the cursor movement to the constant
R circle as you move the mouse. Get to where you're going and click
to plot the point and draw the arc. The status bar shows the cursor

This is the single thing I hate most about all Smith Chart applications,
whereby the cursor is hijacked and constrained to a cirlce or arc
until you click. I've implemented the constraint in a slightly different
way - the mouse cursor is still visible, and the software computes where
the line between the mouse and the centre of the constraint crosses the
constraint circle. That way, the mouse still moves in the normal way,
and you can hover near the centre of the constraint to really whizz
the marker round, or go far outside the constraint to effectively
create sub-pixel accuracy on the actual marker.
 
A

Active8

Jan 1, 1970
0
This is the single thing I hate most about all Smith Chart applications,
whereby the cursor is hijacked and constrained to a cirlce or arc
until you click. I've implemented the constraint in a slightly different
way - the mouse cursor is still visible, and the software computes where
the line between the mouse and the centre of the constraint crosses the
constraint circle. That way, the mouse still moves in the normal way,
and you can hover near the centre of the constraint to really whizz
the marker round, or go far outside the constraint to effectively
create sub-pixel accuracy on the actual marker.

I think I know what you mean.

IIRC, my app does not hide the mouse pointer. I can still activate
another window, I think. The app is somewhere in a project
directory. I'll someday get it out and play some more. I think I
just match the x y movements of the mouse, but I like your idea.
 
D

Dick

Jan 1, 1970
0
Terry Given said:
well said.

Is it any good?
Excellent book, very comprehensive, well written and with plenty of
practical examples, I've a copy on the shelf behind me I've had for
years. If you get a copy make sure it's still got the set of transparent
overlay charts inside the back cover pocket.
 
T

Terry Given

Jan 1, 1970
0
Dick said:
Excellent book, very comprehensive, well written and with plenty of
practical examples, I've a copy on the shelf behind me I've had for
years. If you get a copy make sure it's still got the set of transparent
overlay charts inside the back cover pocket.

Thanks Dick. Its on my list, if I ever see it...

Terry
 
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