T
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
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.