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Moderated
Re: Another off-topic about Smith chart
The Smith chart assumes that the impedances have been "normalized". Just
divide by 50 to get one in the center Bob On Wed, Mar 25, 2020, 8:05 AM vaclav_sal via Groups.Io <vaclav_sal= [email protected]> wrote: An off - topic question to any math wiz knowledgeable about Smith chart. |
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Moderated
Re: Another off-topic about Smith chart
vaclav_sal
Thanks for the reference.
Not bad first response, appreciate that very much. Cannot say much about the rest of them. In future I will limit my posts to "how much is 1 and 1 " , asking to avoid sermons is futile. Interestingly (?) on FIRST page you wrote "The chart is constructed in a rectilinear system instead of polar..." I am not going to dispute that... Will try both linear and log , but I think the linear is really easier on the eye. Many thanks. 73 AA7EJ |
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Re: NanoVNA-V2 from Tindie
Hi all,
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An S-A-A 2 device is on its way to me, and I will update nanovna-saver as soon as possible when I get it. I am sorry for having taken so long to get a new version up, but real life has interfered. :-/ -- Rune / 5Q5R On Wed, 25 Mar 2020, 20:51 hwalker, <herbwalker2476@...> wrote:
On Wed, Mar 25, 2020 at 12:21 PM, Holger M¨¹ller wrote: |
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Moderated
Re: Another off-topic about Smith chart
Thank you for the suggestion, from which I see now that
I have to correct what I wrote, as follows : for the newcomer, the whole Great Idea of that very difficult to learn Smith Chart is just of historic interest, an obsolete one, nowadays - imho, of course. |
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Moderated
Re: Another off-topic about Smith chart
I believe that the scale in question is linear... but it's measuring the "reflection coefficient", not "impedance". It has a value of -1 at the "zero ohms, short circuit" end, a value of 0 in the center of the chart, and a value of +1 at the "infinite ohms, open circuit" end.
The equations which go back and forth between reflection coefficient, and impedance in ohms, are neither a simple linear nor simple logarithmic relationship. |
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Re: NanoVNA-V2 from Tindie
On Wed, Mar 25, 2020 at 12:21 PM, Holger M¨¹ller wrote:
Subject has arrived to me today. Hope I find some time to check and test it with NanoVNA-Saver soon. ============================================ Holger, Glad to hear V2's are finally starting to make it to their destinations. I hope its performance lives up to expectation. My understanding is there is a forked version of NanoVNA-saver that you have to use with the -V2. The release at Rune's GitHub site will not work with the -V2. There is a native VNA software application for the -V2 called NanoVNA-QT that should work out of the box, after you install the appropriate driver. I don't know if you received a copy with your shipment or if you have to download it yourself. You can only perform firmware updates from the native application so you will want to install it at some point anyway. I don't have a -V2 so the above is what I have gathered from reading about it. - Herb |
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Moderated
Re: Another off-topic about Smith chart
Like a slide rule. Accurate enough to get useful results and visualize what's happening in a system.
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You can be much more accurate by using SimSmith. -- Art Greenberg WA2LLN art@... On Wed, Mar 25, 2020, at 15:10, pez@arg wrote:
Smith Chart : In this computer era, this extremely high-effort demanding, |
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Moderated
Re: Another off-topic about Smith chart
Smith Chart : In this computer era, this extremely high-effort demanding,
highly-inaccurate, old-time graphics tool became completely obsolete, because it is a totally useless one in every other aspect except that one of "just take-a-look demo" - or of fancy impression making to unsuspected, innocent, inexperienced beginners, of course |
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Moderated
Re: Another off-topic about Smith chart
Because I dropped a zero. :-)Why isn't the example = 60 + j35 ??? For example, the point 1.2 + j.7 is really 60 + j35 in a 50 Ohm system. On Wed, Mar 25, 2020 at 1:54 PM lgo51 <larryo@...> wrote: Why isn't the example = 60 + j35 ???-- Carey Fisher careyfisher@... |
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Moderated
Re: Another off-topic about Smith chart
Why isn't the example = 60 + j35 ???
At 11:24 AM 3/25/2020, Carey Fisher wrote: <snip> |
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Moderated
Re: Another off-topic about Smith chart
The Smith Chart coordinates are normalized to the impedance of whatever
system you are using. This is usually 50 Ohms so the 1.0 in the center of the chart is 50 Ohms resistance, 0 Ohms reactance. You can take any point on the Smith chart and convert to actual Ohms by multiplying by the impedance of the system (usually the main transmission line) you are making measurements in. For example, the point 1.2 + j.7 is really 6 + j35 in a 50 Ohm system. The reason for this is that you might be using different impedance transmission lines in a design and this makes it easy to move between different impedances. BTW, there are Smith Charts available with 50 Ohm markings. On Wed, Mar 25, 2020 at 11:05 AM vaclav_sal via Groups.Io <vaclav_sal= [email protected]> wrote: An off - topic question to any math wiz knowledgeable about Smith chart.-- Carey Fisher careyfisher@... |
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Off-topic: Learn Python Intro to Programming on Github
#applications
#hack
#tutorials
#training
#python
If you're interested in modifying/hacking/playing with several of the NanoVNA apps written in Python, there is an open-source area on Github with a complete intro on the subject:
Also, I had previously provided a link to a Python book - the link is now restricted but the author's webpage has a number of programming language references: ...Larry |
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Moderated
Re: Another off-topic about Smith chart
There is no scale as such. The mathematics provides the translation from the rectangular Z coordinate system to polar and the result of that math transformation yields the curves your are referencing. We and the vna actually only deal with one of the four Chart regions. The Chart actually occupies 4 quadrants.
In 2001, I wrote a paper that constructed the chart and built it in Mathcad. The paper appeared in Applied Microwave. Each piece of the chart, the real axis and the imaginary contours were assembled as individual blocks of these mapping routines. See reference that should provide a link to the paper. Alan |
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Moderated
Re: Another off-topic about Smith chart
Yes but the gentleman wants to know about the scaling of those curves which are defined by the uniquely derived? equations as shown in the URL I posted.
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Gedas, W8BYA EN70 Gallery at Light travels faster than sound.... This is why some people appear bright until you hear them speak. On 3/25/2020 11:55 AM, gary wrote:
Actually, the Smith chart is "normalized" to whatever the impedance that you are working with, usually 50 but could be 75 or 450 if open wire feed line.The operation to compute is as John described, Multiply every number on the chart by the impedance you are normalized to. |
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Moderated
Re: Another off-topic about Smith chart
Actually, the Smith chart is "normalized" to whatever the impedance that you are working with, usually 50 but could be 75 or 450 if open wire feed line.The operation to compute is as John described, Multiply every number on the chart by the impedance you are normalized to.
Gary |
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Moderated
Re: Another off-topic about Smith chart
Hi,
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The scale is neither linear nor logarithmic, and I¡¯m not sure what it is. Luckily it does not matter if you just want to use the smith chart. Generally for radio work the centre is defined as 50 ohm; that is a resistance of 50 ohms would be plotted at the centre of the chart, a resistance of 100 ohms would be plotted at the 2.0 position. The rule is divide whatever the impedance is by 50 and plot the new figure on the chart. Conversely to get a value from the chart take the reading from the chart and multiply by 50 to get the actual values. It seems complicated but a bit of practice will make it easier. Good luck 73 John M0JBA On 25 Mar 2020, at 15:05, vaclav_sal via Groups.Io <vaclav_sal@...> wrote: |
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Moderated
Re: Another off-topic about Smith chart
Not a math wiz here but one of the best ways I learned about the chart back in college was to see the mathematical calculations (and thus derivations) for the actual circles. The smith chart has unique scales based on those equations which is why they are different from things you may be used to seeing. Have a look at this simplified (but well written description) of the derivation of the chart and see if that helps you visualize why the scales are so unique and why they look the way they do. I am still in awe at the genius involved almost a century ago for someone to come up with the concept. True brain power & genius.
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Gedas, W8BYA EN70 Gallery at Light travels faster than sound.... This is why some people appear bright until you hear them speak. On 3/25/2020 11:05 AM, vaclav_sal via Groups.Io wrote:
An off - topic question to any math wiz knowledgeable about Smith chart. |
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Moderated
Another off-topic about Smith chart
vaclav_sal
An off - topic question to any math wiz knowledgeable about Smith chart.
There are oodles of stuff about Smith chart, but I have not be able to find WHAT is the actual scale of the coordinates. For example real axis ( R +j0) goes from "zero" (left) to center of the chart - MOST of the time designated as "one" . Therefore - what is the scale between 0 and 1 - linear or log? ( No need to discuss log(0)) It PROBABLY makes no difference, but IMHO log scale would "loose" some visual precision. Please, no sermons / references about what is Smith chart , how to use it etc. Sorry to post here , but I have limited choice. |
Robert Malsbury
Gary W9TD
Pablo E. Untroib
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