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Hi Kell,
The unit you purchased is the latest version of the 2.8 inch Nanovna.
The H4 version has the same RF and operating system design but with a 4inch display and a uP with larger flash memory for about $15 more. The F is different in terms of its uP and operating system and its twice the price but has a full metal enclosure.?
There is a recent thread on the pros and cons of each in the forum.?
Personally, as a starting point, I would get the H4 version as it is useful up to about 1.3GHz. If you want cutting edge (hobbiest) technology, you should go for the V2 that goes up to 3GHz - BUT - the current version only has a 2.8inch display.?
Your call.?



On Wed, 25 Mar 2020 at 7:53 PM, Kell Bodholt<Kbodholt@...> wrote: Wish I would have subscribed to this nanovna@... site before buying.? Live and learn.? I just ordered and received today a 【Upgraded】AURSINC Vector Network Analyzer 10KHz -1.5GHz HF VHF UHF Antenna Analyzer Measuring S Parameters, Voltage Standing Wave Ratio, Phase, Delay, Smith Chart(Latest Version REV3.4) from Amazon.? On the back is a sticker NanoVNA-H, H/W version 3.4, 50k-1.5GHz. So based upon what I reading here what I bought is not the latest (-F or -H4) is my assumption accurate, excluding the V2 and/orSAA2 I see posted above ?? Should I return and get the -F or -H4?

The display is rather small, assuming 2.3" and I can barely read the SWR values with my reading glasses.? If I was to return this and get the 4" screen -F or -H4 is the actual font size larger?

Wish I would have subscribed to this nanovna@... site before buying. Live and learn. I just ordered and received today a 【Upgraded】AURSINC Vector Network Analyzer 10KHz -1.5GHz HF VHF UHF Antenna Analyzer Measuring S Parameters, Voltage Standing Wave Ratio, Phase, Delay, Smith Chart(Latest Version REV3.4) from Amazon. On the back is a sticker NanoVNA-H, H/W version 3.4, 50k-1.5GHz. So based upon what I reading here what I bought is not the latest (-F or -H4) is my assumption accurate, excluding the V2 and/orSAA2 I see posted above ? Should I return and get the -F or -H4?

The display is rather small, assuming 2.3" and I can barely read the SWR values with my reading glasses. If I was to return this and get the 4" screen -F or -H4 is the actual font size larger?

I also just received notice from USPS that the S-A-A 2 is in transit. With any luck I should receive it within a few days. I placed my order on the 8th of March, so given the current state of affairs with COVID-19 I can't really complain about the transit time. Other members who ordered around the 8th will probably also start receiving postal delivery notices.

I look forward to putting the device through its paces when it arrives.

- Herb

On Wed, Mar 25, 2020 at 02:22 PM, Rune Broberg wrote:

Hi all,
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, Good to hear from you. Oristo has also been curiously absent of late. I'm hoping real life means your NanoVNA-saver efforts resulted in a commercial opportunity you couldn't turn down. I am also waiting for delivery of an S-A-A 2, as is Kurt and a few other members. Time will tell when delivery occurs.

Glad to hear you anticipate adding support of the S-A-A 2 to NanoVNA-saver. I haven't investigated how many of the NanoVNA console commands are duplicated, or how increased measurement points via USB transfer are performed (maybe like the NanoVNA raw scan command?).

On an un-related topic, were you ever able to figure out the coding for the NanoVNA-F "capture" command? My own efforts at a python script using the "capture" command have proved fruitless. The NanoVNA-H and NanoVNA-H4 follow ch045's original script but the NanoVNA-F seems to need some extra handling.

Regards,

- Herb

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.
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.

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

Hi all,
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

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.

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.

On Wed, Mar 25, 2020 at 12:21 PM, Holger Müller 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

Like a slide rule. Accurate enough to get useful results and visualize what's happening in a system.

You can be much more accurate by using SimSmith.

--
Art Greenberg
WA2LLN
art@...

Hi,

$Subject has arrived to me today. Hope I find some time to check
and test it with NanoVNA-Saver soon.

ciao
Holger (DG5DBH)

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

Why isn't the example = 60 + j35 ???

Because I dropped a zero. :-)

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 ???

At 11:24 AM 3/25/2020, Carey Fisher wrote:

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.

<snip>

--
Carey Fisher
careyfisher@...

Why isn't the example = 60 + j35 ???

At 11:24 AM 3/25/2020, Carey Fisher wrote:

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.

<snip>

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.
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.

--
Carey Fisher
careyfisher@...

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

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

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.

Gedas, W8BYA EN70

Gallery at
Light travels faster than sound....
This is why some people appear bright until you hear them speak.

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