Smith Chart question

Paul Burridge · Apr 29, 2004

Hi all,

Firstly, apologies for raising a matter of relevance to electronics on
this group.
I have posted a .gif image to a.b.s.e under the same subject heading
in which a Smith chart is displayed with four impedance
transformations shown marked in black, blue, red and yellow. The black
arrow shows the effect of adding series capacitance; the red arrow
shows the addition of shunt inductance. What do the blue and yellow
arrows indicate, though? I'm particularly interested in the effect of
the yellow arrow, which starts off in the direction of increasing
series inductance, peaks and then starts heading south again towards
increasing C. Is that really what it's showing and would such a
transformation that passes through a zenith be used in real life? This
has got me completely baffled for some reason. The answer should be
obvious but for some reason I have a complete mental block on seeing
straight. Any informed advice welcome.

<puzzled>

Active8 · Apr 29, 2004

On Thu, 29 Apr 2004 16:53:31 +0100, Paul Burridge wrote:

Hi all,

Firstly, apologies for raising a matter of relevance to electronics on
this group.

Shit like that will get you killfiled.

I have posted a .gif image to a.b.s.e under the same subject heading
in which a Smith chart is displayed with four impedance
transformations shown marked in black, blue, red and yellow.

It's orange, Paul.

The black
arrow shows the effect of adding series capacitance; the red arrow
shows the addition of shunt inductance. What do the blue and yellow
arrows indicate, though?

Series L

I'm particularly interested in the effect of
the yellow arrow, which starts off in the direction of increasing
series inductance, peaks and then starts heading south again towards
increasing C.

It's *not* heading toward series C or even decreasing series L. It's
moving in a positive direction along a constant reactance circle.
Positive reactance being inductive. It's heading for infinity.

Is that really what it's showing and would such a
transformation that passes through a zenith

you have a zenith? good TV.

be used in real life?

Maybe, the actual impedance it's heading for is unknown since we
don't know how the chart's been normalized. In real life, you'd want
to head from this end point to the center of the chart normalized to
the characteristic impedance of a transmission line or the input
immpedance of a filter maybe. OTOH, you could be feeding a high Zin
FET, so IDKWTF.

This
has got me completely baffled for some reason. The answer should be
obvious but for some reason I have a complete mental block on seeing
straight. Any informed advice welcome.

A pshrink and an optometrist? A pshrink and an optometrist were out
golfing one day...

puzzled

--
Best Regards,
Mike

Paul Burridge · Apr 29, 2004

On Thu, 29 Apr 2004 12:18:31 -0400, Active8 <reply2group@ndbbm.net>
wrote:

On Thu, 29 Apr 2004 16:53:31 +0100, Paul Burridge wrote:

Hi all,

Firstly, apologies for raising a matter of relevance to electronics on
this group.

Shit like that will get you killfiled.

I know. Jim Thompson in particular would be furious at me.

I have posted a .gif image to a.b.s.e under the same subject heading
in which a Smith chart is displayed with four impedance
transformations shown marked in black, blue, red and yellow.

It's orange, Paul.

It looks orange through this scanner. The 'black' line is actually
dark green!

The black
arrow shows the effect of adding series capacitance; the red arrow
shows the addition of shunt inductance. What do the blue and yellow
arrows indicate, though?

Series L

Eh?? One's in the inductive region, the other's in the capacitive. And
"series L" by itself doesn't really tell me anything.

It's *not* heading toward series C or even decreasing series L. It's
moving in a positive direction along a constant reactance circle.
Positive reactance being inductive. It's heading for infinity.

Er, no. All the lines I've marked are on constant *resistance* circles
which is I assume what you meant to say. I'm still none the wiser
though. :-|

<equally baffled>

Tim Wescott · Apr 29, 2004

Paul Burridge wrote:

On Thu, 29 Apr 2004 12:18:31 -0400, Active8 <reply2group@ndbbm.net
wrote:

On Thu, 29 Apr 2004 16:53:31 +0100, Paul Burridge wrote:

Hi all,

Firstly, apologies for raising a matter of relevance to electronics on
this group.

Shit like that will get you killfiled.

I know. Jim Thompson in particular would be furious at me.

Really, if you don't feel qualified to comment about Bush you could at
least snipe about Blair.

I have posted a .gif image to a.b.s.e under the same subject heading
in which a Smith chart is displayed with four impedance
transformations shown marked in black, blue, red and yellow.

It's orange, Paul.

It looks orange through this scanner. The 'black' line is actually
dark green!

The black
arrow shows the effect of adding series capacitance; the red arrow
shows the addition of shunt inductance. What do the blue and yellow
arrows indicate, though?

Series L

Eh?? One's in the inductive region, the other's in the capacitive. And
"series L" by itself doesn't really tell me anything.

You don't care if it's in the "inductive" region, you only care where
it's going. Series L will add inductance with constant resistance,
shunt L will add inductance with constant conductance, similar for
capacitive series & shunt. The blue line is entirely in the capacitive
region, but it will take you all the way over to inductive if you keep
it up.

It's *not* heading toward series C or even decreasing series L. It's
moving in a positive direction along a constant reactance circle.
Positive reactance being inductive. It's heading for infinity.

Er, no. All the lines I've marked are on constant *resistance* circles
which is I assume what you meant to say. I'm still none the wiser
though. :-|

equally baffled

Only two of the lines are on constant resistance circles; the other two
are on constant conductance circles. The circles centered on the right
edge are usually taken as resistive; the circles centered on the left
are usually taken as conductive (you can do it the other way if you feel
like it).

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Active8 · Apr 29, 2004

On Thu, 29 Apr 2004 17:50:58 +0100, Paul Burridge wrote:

On Thu, 29 Apr 2004 12:18:31 -0400, Active8 <reply2group@ndbbm.net
wrote:

On Thu, 29 Apr 2004 16:53:31 +0100, Paul Burridge wrote:

Hi all,

Firstly, apologies for raising a matter of relevance to electronics on
this group.

Shit like that will get you killfiled.

I know. Jim Thompson in particular would be furious at me.

I have posted a .gif image to a.b.s.e under the same subject heading
in which a Smith chart is displayed with four impedance
transformations shown marked in black, blue, red and yellow.

It's orange, Paul.

It looks orange through this scanner. The 'black' line is actually
dark green!

The black
arrow shows the effect of adding series capacitance; the red arrow
shows the addition of shunt inductance. What do the blue and yellow
arrows indicate, though?

Series L

Eh?? One's in the inductive region, the other's in the capacitive. And
"series L" by itself doesn't really tell me anything.

effect of adding series inductance.

It's *not* heading toward series C or even decreasing series L. It's
moving in a positive direction along a constant reactance circle.
Positive reactance being inductive. It's heading for infinity.

The top one is. The bottom is headed for 0 net reactance, but if it
doesn't change it's way, it'll keep on heading from net capacitive
thru 0 to net inductive.

Er, no. All the lines I've marked are on constant *resistance* circles
which is I assume what you meant to say. I'm still none the wiser
though. :-|

right. They're on const R circles. They're both traversing the
reactance circles in the same positive direction, in this case
clockwise. I can't get any clearer without drawing pictures or using
many more words than necessary. Opfen der book.

equally baffled

--
Best Regards,
Mike

Active8 · Apr 29, 2004

On Thu, 29 Apr 2004 10:13:24 -0700, Tim Wescott wrote:

Paul Burridge wrote:

On Thu, 29 Apr 2004 12:18:31 -0400, Active8 <reply2group@ndbbm.net
wrote:

On Thu, 29 Apr 2004 16:53:31 +0100, Paul Burridge wrote:

Hi all,

Firstly, apologies for raising a matter of relevance to electronics on
this group.

Shit like that will get you killfiled.

I know. Jim Thompson in particular would be furious at me.

Really, if you don't feel qualified to comment about Bush you could at
least snipe about Blair.

I have posted a .gif image to a.b.s.e under the same subject heading
in which a Smith chart is displayed with four impedance
transformations shown marked in black, blue, red and yellow.

It's orange, Paul.

It looks orange through this scanner. The 'black' line is actually
dark green!

The black
arrow shows the effect of adding series capacitance; the red arrow
shows the addition of shunt inductance. What do the blue and yellow
arrows indicate, though?

Series L

Eh?? One's in the inductive region, the other's in the capacitive. And
"series L" by itself doesn't really tell me anything.

You don't care if it's in the "inductive" region, you only care where
it's going. Series L will add inductance with constant resistance,
shunt L will add inductance with constant conductance, similar for
capacitive series & shunt. The blue line is entirely in the capacitive
region, but it will take you all the way over to inductive if you keep
it up.

It's *not* heading toward series C or even decreasing series L. It's
moving in a positive direction along a constant reactance circle.
Positive reactance being inductive. It's heading for infinity.

Er, no. All the lines I've marked are on constant *resistance* circles
which is I assume what you meant to say. I'm still none the wiser
though. :-|

equally baffled

Only two of the lines are on constant resistance circles; the other two
are on constant conductance circles.

Good observation.

The circles centered on the right
edge are usually taken as resistive; the circles centered on the left
are usually taken as conductive (you can do it the other way if you feel
like it).

Which brings me to what I forgot to mention. That looks like a real
good friggin' chart like the ones from AD. All on has to do is look
at the printing to see if they're headed in the negative or positive
resistance, conductance, reactance or susceptance direction. Then
it's a matter of knowing that negative net reactance is capacitive
as is net positive susceptance.

--
Best Regards,
Mike

Jim Thompson · Apr 29, 2004

On Thu, 29 Apr 2004 10:13:24 -0700, Tim Wescott
<tim@wescottnospamdesign.com> wrote:

Paul Burridge wrote:

[snip]
Hi all,

Firstly, apologies for raising a matter of relevance to electronics on
this group.

Shit like that will get you killfiled.

I know. Jim Thompson in particular would be furious at me.

Really, if you don't feel qualified to comment about Bush you could at
least snipe about Blair.

[snip]

480)460-2350 | |
| E-mail Address at Website Fax

Tom Bruhns · Apr 29, 2004

Getcherself a Smith chart program, and experiment with it. In fact,
you probably already have one in the form of RFSim99. Shunt L's and
C's result in arcs that follow a constant admittance curve. For
example, set up a port1 source, and a load consisting of parallel 150
ohm resistor and capacitor of, say, 0.01pF. Simulate. Set the plot
to smith x/y. Set the frequency range to 100MHz to 101MHz. Note that
the point is on the first larger admittance circle from the center of
the plot. Now "tune" the capacitor up in steps and see where the
point goes. Stop when you get to 14.8pF. Now add in a series
inductor between the source and the parallel RC. Start with it very
tiny, and adjust it up. Note that you are moving along a constant
resistance line. Note that when you get to about 112nH, you're at the
center of the chart: 50 ohms in this case. You just designed a
matching network to match from 150 ohms resistive to 50 ohms resistive
at 100MHz. Note that if you swap to a shunt inductor and series
capacitor, you can do the same thing, on the upper half of the chart.
Works for any sort of ladder network you want to put together: pi,
L-pi, T, several sections...

RFSim99's Smith chart, used this way, isn't as nice as something like
WinSmith, but it's servicable.

Try also putting in a section of transmission line: note that the
point moves in a circle centered on the line's characteristic
impedance. In a line with loss, it will be a spiral inward as you
lengthen the line.

Shunt L and shunt C don't change the admittance but do change the
suseptance: the two are orthogonal. Series L and series C don't
change the resistance, but do change the reactance. Again the two are
orthogonal. Inductors add suseptance or reactance; capacitors
subtract (in that their reactances and suseptances are negative).

Cheers,
Tom

(I'm quite sure I've seen a good Smith chart tutorial on the web. Do
a google search...)

Paul Burridge <pb@notthisbit.osiris1.co.uk> wrote in message news:<ap82909knm8ti9gfe40d3sf0c49uhg9f1u@4ax.com>...

Hi all,

Firstly, apologies for raising a matter of relevance to electronics on
this group.
I have posted a .gif image to a.b.s.e under the same subject heading
in which a Smith chart is displayed with four impedance
transformations shown marked in black, blue, red and yellow. The black
arrow shows the effect of adding series capacitance; the red arrow
shows the addition of shunt inductance. What do the blue and yellow
arrows indicate, though? I'm particularly interested in the effect of
the yellow arrow, which starts off in the direction of increasing
series inductance, peaks and then starts heading south again towards
increasing C. Is that really what it's showing and would such a
transformation that passes through a zenith be used in real life? This
has got me completely baffled for some reason. The answer should be
obvious but for some reason I have a complete mental block on seeing
straight. Any informed advice welcome.

puzzled

Paul Burridge · Apr 29, 2004

On Thu, 29 Apr 2004 10:13:24 -0700, Tim Wescott
<tim@wescottnospamdesign.com> wrote:

Really, if you don't feel qualified to comment about Bush you could at
least snipe about Blair.

That would profit me not a jot. However, I've a much more practical
solution to our lying PoS Prime Minister: I'm leaving the country -
and mostly because of him and his 'government, too.
Enough's enough.

You don't care if it's in the "inductive" region, you only care where
it's going. Series L will add inductance with constant resistance,
shunt L will add inductance with constant conductance, similar for
capacitive series & shunt. The blue line is entirely in the capacitive
region, but it will take you all the way over to inductive if you keep
it up.

It's *not* heading toward series C or even decreasing series L. It's
moving in a positive direction along a constant reactance circle.
Positive reactance being inductive. It's heading for infinity.

Er, no. All the lines I've marked are on constant *resistance* circles
which is I assume what you meant to say. I'm still none the wiser
though. :-|

equally baffled

Only two of the lines are on constant resistance circles; the other two
are on constant conductance circles. The circles centered on the right
edge are usually taken as resistive; the circles centered on the left
are usually taken as conductive (you can do it the other way if you feel
like it).

I find it better personally to simply convert all the susceptances to
reactances by taking the inverse of each before I "un-normalise" so
don't worry about that aspect.
Sorry to say that notwithstanding both yours and Mike's efforts so far
I'm still pretty much in the dark and my original question hasn't been
answered.

Paul Burridge · Apr 29, 2004

On Thu, 29 Apr 2004 13:20:27 -0400, Active8 <reply2group@ndbbm.net>
wrote:

right. They're on const R circles. They're both traversing the
reactance circles in the same positive direction, in this case
clockwise. I can't get any clearer without drawing pictures or using
many more words than necessary. Opfen der book.

I have geoffneted der buch but all the examples in it are really
easy-peasy ones that use combinations of tracks along those constant
resistance circles where those tracks are all near vertical - like the
black and red ones on the chart I posted. I understand those alright.
What I'm having difficulty with is when a track/trace/arrow - call it
what you will - follows a near *horizontal* path. It appears to be
moving sideways and neither towards the inductive region or the
capacitive region.
Are you saying - I *think* this is what you are saying - that the
orange line, for example, solely represents an increase in series L
right along its length, notwithstanding that it starts to drop back
down towards the capacitive region after about the halfway point?
Similarly, are you saying that even though the blue line is nearly
horizontal that it actually, in fact, represents an increase in series
L, too?
If that's the case then surely it's misleading for publishers to show
the top hemisphere of the chart as being inductive territory and the
bottom hemisphere as being the capacitive region?

<befuddled>

Tim Wescott · Apr 29, 2004

Paul Burridge wrote:

On Thu, 29 Apr 2004 10:13:24 -0700, Tim Wescott
tim@wescottnospamdesign.com> wrote:

Really, if you don't feel qualified to comment about Bush you could at
least snipe about Blair.

That would profit me not a jot. However, I've a much more practical
solution to our lying PoS Prime Minister: I'm leaving the country -
and mostly because of him and his 'government, too.
Enough's enough.

You don't care if it's in the "inductive" region, you only care where
it's going. Series L will add inductance with constant resistance,
shunt L will add inductance with constant conductance, similar for
capacitive series & shunt. The blue line is entirely in the capacitive
region, but it will take you all the way over to inductive if you keep
it up.

It's *not* heading toward series C or even decreasing series L. It's
moving in a positive direction along a constant reactance circle.
Positive reactance being inductive. It's heading for infinity.

Er, no. All the lines I've marked are on constant *resistance* circles
which is I assume what you meant to say. I'm still none the wiser
though. :-|

equally baffled

Only two of the lines are on constant resistance circles; the other two
are on constant conductance circles. The circles centered on the right
edge are usually taken as resistive; the circles centered on the left
are usually taken as conductive (you can do it the other way if you feel
like it).

I find it better personally to simply convert all the susceptances to
reactances by taking the inverse of each before I "un-normalise" so
don't worry about that aspect.
Sorry to say that notwithstanding both yours and Mike's efforts so far
I'm still pretty much in the dark and my original question hasn't been
answered.

Get a copy of the ARRL's UHF Experimenter's handbook, and a good book on
microwave amplifiers (or a book on transmission lines). They'll give
you more info on Smith chart's than you can handle.

Isn't there info to google on for this?

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Reg Edwards · Apr 30, 2004

The Smith Chart went out of date with the invention of the personal
computer.

As an approximation it suffers from serious misleading errors at low
frequencies.

It was adopted in the late 1930's from simplification of similar charts
which had been in use since the Victorian age to cater for the splurge in
the use of HF coaxial lines particularly during WW2.

Old timers now use it only for sentimental, recreational purposes.
----
Reg, G4FGQ

Ric · Apr 30, 2004

Tom Bruhns <k7itm@aol.com> wrote:

Getcherself a Smith chart program, and experiment with it. In fact,
you probably already have one in the form of RFSim99. Shunt L's and
C's result in arcs that follow a constant admittance curve. For

no - they follow a curve of constant CONDUCTANCE. No such thing as
a curve of constant admittance.

example, set up a port1 source, and a load consisting of parallel 150
ohm resistor and capacitor of, say, 0.01pF. Simulate. Set the plot
to smith x/y. Set the frequency range to 100MHz to 101MHz. Note that
the point is on the first larger admittance circle from the center of
^^^^^^^^^^^^^^^^^

conductance circle

the plot. Now "tune" the capacitor up in steps and see where the
point goes. Stop when you get to 14.8pF. Now add in a series
inductor between the source and the parallel RC. Start with it very
tiny, and adjust it up. Note that you are moving along a constant
resistance line. Note that when you get to about 112nH, you're at the
center of the chart: 50 ohms in this case. You just designed a
matching network to match from 150 ohms resistive to 50 ohms resistive
at 100MHz. Note that if you swap to a shunt inductor and series
capacitor, you can do the same thing, on the upper half of the chart.
Works for any sort of ladder network you want to put together: pi,
L-pi, T, several sections...

RFSim99's Smith chart, used this way, isn't as nice as something like
WinSmith, but it's servicable.

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.

--
Rick

Active8 · Apr 30, 2004

On Thu, 29 Apr 2004 17:08:19 -0700, Tim Wescott wrote:

Paul Burridge wrote:

On Thu, 29 Apr 2004 10:13:24 -0700, Tim Wescott
tim@wescottnospamdesign.com> wrote:

Really, if you don't feel qualified to comment about Bush you could at
least snipe about Blair.

That would profit me not a jot. However, I've a much more practical
solution to our lying PoS Prime Minister: I'm leaving the country -
and mostly because of him and his 'government, too.
Enough's enough.

You don't care if it's in the "inductive" region, you only care where
it's going. Series L will add inductance with constant resistance,
shunt L will add inductance with constant conductance, similar for
capacitive series & shunt. The blue line is entirely in the capacitive
region, but it will take you all the way over to inductive if you keep
it up.

It's *not* heading toward series C or even decreasing series L. It's
moving in a positive direction along a constant reactance circle.
Positive reactance being inductive. It's heading for infinity.

Er, no. All the lines I've marked are on constant *resistance* circles
which is I assume what you meant to say. I'm still none the wiser
though. :-|

equally baffled

Only two of the lines are on constant resistance circles; the other two
are on constant conductance circles. The circles centered on the right
edge are usually taken as resistive; the circles centered on the left
are usually taken as conductive (you can do it the other way if you feel
like it).

I find it better personally to simply convert all the susceptances to
reactances by taking the inverse of each before I "un-normalise" so
don't worry about that aspect.
Sorry to say that notwithstanding both yours and Mike's efforts so far
I'm still pretty much in the dark and my original question hasn't been
answered.

Get a copy of the ARRL's UHF Experimenter's handbook, and a good book on
microwave amplifiers (or a book on transmission lines). They'll give
you more info on Smith chart's than you can handle.

Isn't there info to google on for this?

Piles. And the UHF book isn't nearly as good at explaining the Smith
Chart as Chris Bowick's little book, which Paul has. I'm suspecting
either a lack of fundamentals, a lack of an intuitive feel for
electronics in general, or something else yet to be revealed.
--
Best Regards,
Mike

Active8 · Apr 30, 2004

On Fri, 30 Apr 2004 00:07:16 +0100, Paul Burridge wrote:

On Thu, 29 Apr 2004 13:20:27 -0400, Active8 <reply2group@ndbbm.net
wrote:

right. They're on const R circles. They're both traversing the
reactance circles in the same positive direction, in this case
clockwise. I can't get any clearer without drawing pictures or using
many more words than necessary. Opfen der book.

I have geoffneted der buch but all the examples in it are really
easy-peasy ones that use combinations of tracks along those constant
resistance circles where those tracks are all near vertical - like the
black and red ones on the chart I posted. I understand those alright.
What I'm having difficulty with is when a track/trace/arrow - call it
what you will - follows a near *horizontal* path. It appears to be
moving sideways and neither towards the inductive region or the
capacitive region.
Are you saying - I *think* this is what you are saying - that the
orange line, for example, solely represents an increase in series L
right along its length, notwithstanding that it starts to drop back
down towards the capacitive region after about the halfway point?
Similarly, are you saying that even though the blue line is nearly
horizontal that it actually, in fact, represents an increase in series
L, too?

Yeah. Look at the *whole* const R circle that the line is on and
realize that clockwise is moving in a positive net reactance
direction. positive net reactance is inductive, so you're moving in
an increasing inductance direction.

If that's the case then surely it's misleading for publishers to show
the top hemisphere of the chart as being inductive territory and the
bottom hemisphere as being the capacitive region?

I know what picture yer referring to, but I didn't have a problem
with that because the fundamentals of reactance have been in my
beaner since I was a teen. I could also see that the lines were arcs
drawn on a circle and no one said I had to start or stop at any
particular point on the circle. He did draw them with an arrow.
Remember rays? the keep going in one direction.

befuddled

--
Best Regards,
Mike

Paul Burridge · Apr 30, 2004

On Fri, 30 Apr 2004 08:02:58 GMT, Rick<rik_nntp@dsl.pipex.com> wrote:

Tom Bruhns <k7itm@aol.com> wrote:
Getcherself a Smith chart program, and experiment with it. In fact,
you probably already have one in the form of RFSim99. Shunt L's and
C's result in arcs that follow a constant admittance curve. For

no - they follow a curve of constant CONDUCTANCE. No such thing as
a curve of constant admittance.

This is another problem one faces in trying to get to grips with this
aspect of the science. Admittances, susceptances, reactance,
conductance, resistance, impedances and so on. Quite a lot of similar
sounding terms to differentiate between. The difficulty being
compounded when someone makes a mistake like the one above. And
earlier Mike used the word "reactance circle" when he meant
*resistance* circle. Thank god I'd done enough homework to be able to
spot those errors for myself and the only outstanding query was the
one I originally posted about - which Mike has answered to my
satisfaction now.
Thanks, all.
--

The BBC: licenced at public expense to spread lies.

Paul Burridge · Apr 30, 2004

On Fri, 30 Apr 2004 05:05:10 -0400, Active8 <reply2group@ndbbm.net>
wrote:

Isn't there info to google on for this?

Piles. And the UHF book isn't nearly as good at explaining the Smith
Chart as Chris Bowick's little book, which Paul has. I'm suspecting
either a lack of fundamentals, a lack of an intuitive feel for
electronics in general, or something else yet to be revealed.

I did - at the outset a few weeks ago- come across a rather nifty
little interactive site for Smith chart manipulation. Analogue
Electronics, IIRC. Very useful for getting a feel for it. My original
questiion though was not fathomable from it as the scale of the
virtual chart was very small. But in all other respects it was very
helpful. As for Bowick's book, again, no specific info on what happens
when an arc is moving through what *appears* to be neutral points. I
reailse now that it's not; just the necessarily distored nature of the
chart making it appear so at first sight.
So yeah, if the yellow arrow were to continue all the way down to the
horizon at infinity, it still wouldn't be going any more capacitive or
less inductive; it's just the distortion of the chart making it appear
as if it is.
Thanks.
HTF I've got the above right or I *am* in trouble!

--

The BBC: licenced at public expense to spread lies.

Paul Burridge · Apr 30, 2004

On Fri, 30 Apr 2004 03:33:23 +0000 (UTC), "Reg Edwards"
<g4fgq.regp@ZZZbtinternet.com> wrote:

The Smith Chart went out of date with the invention of the personal
computer.

As an approximation it suffers from serious misleading errors at low
frequencies.

It was adopted in the late 1930's from simplification of similar charts
which had been in use since the Victorian age to cater for the splurge in
the use of HF coaxial lines particularly during WW2.

Old timers now use it only for sentimental, recreational purposes.

As someone who's probably older than the Chart, Reg, I'm surprised to
hear you say that. I believe a lot of folks would disagree with you...
--

The BBC: licenced at public expense to spread lies.

Active8 · Apr 30, 2004

On Fri, 30 Apr 2004 13:34:50 +0100, Paul Burridge wrote:

On Fri, 30 Apr 2004 05:05:10 -0400, Active8 <reply2group@ndbbm.net
wrote:

Isn't there info to google on for this?

Piles. And the UHF book isn't nearly as good at explaining the Smith
Chart as Chris Bowick's little book, which Paul has. I'm suspecting
either a lack of fundamentals, a lack of an intuitive feel for
electronics in general, or something else yet to be revealed.

I did - at the outset a few weeks ago- come across a rather nifty
little interactive site for Smith chart manipulation. Analogue
Electronics, IIRC. Very useful for getting a feel for it. My original
questiion though was not fathomable from it as the scale of the
virtual chart was very small. But in all other respects it was very
helpful. As for Bowick's book, again, no specific info on what happens
when an arc is moving through what *appears* to be neutral points. I
reailse now that it's not; just the necessarily distored nature of the
chart making it appear so at first sight.
So yeah, if the yellow arrow were to continue all the way down to the
horizon at infinity, it still wouldn't be going any more capacitive or
less inductive; it's just the distortion of the chart making it appear
as if it is.
Thanks.
HTF I've got the above right or I *am* in trouble!

Yeah. Sounds like you got it. When I wrote my (unfinished) smith
chart app's GUI, I had to write the equations for the circles,
calculate the intersections, plot them, and then I said "f it" and
captured different color schemes to use for the app. The whole time
I thought the process sucked because of what the chart really is. A
distorted cartesian plane.

Think of resistance as being the y axis with all horizontal grid
lines representing const R, and the others, const X. remember the
basic triangle that desribes a phasor? the base is the R, the height
is the net X, and the hypotenuse is the Z = R + jX = sqrt(R^2 + X^2)
....

if you take the vertical lines at infinity ( 00 ) , pinch the ends
together and bend them back around to the same point ( 00 ), you get
the R circles. With the horizontal lines, you pinch one end together
and spread the other ends out. Actually, you're pinching both ends,
but bending the top and bottom back around in their respective
directions (leaving the x axis alone) to make 2 families of circles
but the 0 R cicle that bounds the chart makes it look otherwise.
--
Best Regards,
Mike

Paul Burridge · Apr 30, 2004

On Fri, 30 Apr 2004 10:50:12 -0400, Active8 <reply2group@ndbbm.net>
wrote:

Yeah. Sounds like you got it. When I wrote my (unfinished) smith
chart app's GUI, I had to write the equations for the circles,
calculate the intersections, plot them, and then I said "f it" and
captured different color schemes to use for the app. The whole time
I thought the process sucked because of what the chart really is. A
distorted cartesian plane.

Think of resistance as being the y axis with all horizontal grid
lines representing const R, and the others, const X. remember the
basic triangle that desribes a phasor? the base is the R, the height
is the net X, and the hypotenuse is the Z = R + jX = sqrt(R^2 + X^2)
...

if you take the vertical lines at infinity ( 00 ) , pinch the ends
together and bend them back around to the same point ( 00 ), you get
the R circles. With the horizontal lines, you pinch one end together
and spread the other ends out. Actually, you're pinching both ends,
but bending the top and bottom back around in their respective
directions (leaving the x axis alone) to make 2 families of circles
but the 0 R cicle that bounds the chart makes it look otherwise.

It's a pretty horrifying sight for anyone who hasn't seen it before. I
imagine Mr. Smith was some sort of tortured genius who prolly shot
himself in the head shortly after devising it.

Failing that, maybe
someone else should have. ;-)

--

The BBC: licenced at public expense to spread lies.

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