Author Topic: 75 Ohm why?  (Read 3080 times)

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Online vk6zgo

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Re: 75 Ohm why?
« Reply #25 on: May 17, 2022, 02:18:52 am »
That graph gets copied everywhere, but for some reason the assumptions behind its derivation never come along for the ride. Guess: teflon dielectric, copper conductor, 100MHz.

Probably air dielectric, back in 1929.
By the way, there is such a thing as 75 Ohm twin line!
 

Online vk6zgo

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Re: 75 Ohm why?
« Reply #26 on: May 17, 2022, 02:24:30 am »
Guess: teflon dielectric, copper conductor, 100MHz.
I doubt Teflon was around in 1929 :)

but nylon was...
Nylon sucks as a dielectric.
Air dielectric coax cables used various spacers, including Perspex, & porcelain.
The former was used up to the 1950s in some coax I have seen.
 

Online vk6zgo

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Re: 75 Ohm why?
« Reply #27 on: May 17, 2022, 02:34:36 am »
Medium wave broadcasting used a plethora of impedances, from 600 \$\Omega\$, to 200 \$\Omega\$, depending on the open wire line used.
One place I worked at, used  a "6 wire" cable, consisting of two paralleled conductors mounted on a standoff insulator, with four conductors at earth potential, mounted on a square steel frame surrounding them.---sort of "ersatz" coax! ;D
 

Offline E Kafeman

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Re: 75 Ohm why?
« Reply #28 on: May 18, 2022, 07:15:38 pm »
Quote from: E Kafeman
An ideal a 1/2 wave center-feed dipole antenna in free space have an impedance around 83 Ohm.
No, ideal half-wave dipole with center feed and wire thickness 0.001 wavelength placed in a free space has Z = 73.1 + j*42.5 Ω.

There is no NO except No, You is very misunderstanding, it is all too basic knowledge if you know anything about antenna math. If you want to accuse someone having wrong, do you need to have own knowledge and in this case at least know anything about wave math.

I'll try to explain at a low level:

Circle chart with 2π circumference and it related math is where it often start when learning wave theory at school,  how waves behaves and related math expressions such as: 32+4i2=52. i is a way to tell it is an imaginary value.
Even more basic can it be written (3X,4Y).
3 and 4 express a position in a 2D coordinate system. It is an rectangular expression for the actual value which have a length of 5. It is same relations for an dipole antenna impedance. Assume 72+j42 as complex impedance. Impedance value is then 83.3 Ohm (722x422=83.32).

73.1 + j42.5 Ω is a expression in rectangular form, in a coordinate system.  It describes in this case an impedance with a vector value of 84 Ohm.
Ideal impedance value with added exemptions for air dielectric for this dipole is around 83 Ohm expressed as an vector length.
Using your rectangular numbers "73.1 + j42.5 Ω", then is "84.6 Ω ∠30.2o" same value expressed as impedance with an specified angle. Many calculators have inbuilt translators between these math forms so it is not complicated to calculate.

These angles are expressed in degrees but can also be expressed in radians which often is to prefer when doing antenna calculations or any wave related calculations. A circle with circumference 2π radians is perhaps remembered?
Typical half-wave dipole antenna impedance with air as dielectric is an absolute vector length with a value around 83 Ohm and with an angle of "π/6", which can be expressed in this form "83Ω ∠ π/6".
Radian is of very basic and usable angle-value when calculating impedance in any form related to antennas.

Another kind of impedance is characteristic impedance in free space which is approximated to 120π Ohm. No need to specify angle for that value as it already is stated in its expression but it can be written including angle.
« Last Edit: May 18, 2022, 10:22:31 pm by E Kafeman »
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Offline radiolistener

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Re: 75 Ohm why?
« Reply #29 on: May 19, 2022, 02:43:40 pm »
You is very misunderstanding, it is all too basic knowledge if you know anything about antenna math. If you want to accuse someone having wrong, do you need to have own knowledge and in this case at least know anything about wave math.

What do you mean when you're talking about "misunderstanding"?

I calculated radiation resistance value according to the math taken from old-school antenna modelling books, here is the formula which I used for a center-feed half-wavelength dipole:

Za = Ra + j*Xa

where

Ra = (Zenv / (2*pi)) * ((EulerGamma + log(2 * k * l) - Ci(2 * k * l) + cos(2 * k * l) / 2 * (EulerGamma + log(k * l) + Ci(4 * k * l) - 2 * Ci(2 * k * l)) + sin(2 * k * l) / 2 * (Si(4 * k * l) - 2 * Si(2 * k * l))));

Xa = (Zenv / (2 * pi)) * (Si(2 * k * l) + sin(2 * k * l) / 2 * (EulerGamma + log(k * l) + Ci(4 * k * l) - 2 * Ci(2 * k * l) - 2 * log(l / r)) + cos(2 * k * l) / 2 * (2 * Si(2 * k * l) - Si(4 * k * l)));

Zenv = sqrt( (μ0*μ) / (ε0*ε) );

k = 2 * pi / lambda

Za - antenna impedance
Ra - active part (due to antenna radiation)
Xa - reactive part (due to reflection from environment)
Zenv - environment impedance (around antenna)
k - wave number
l - dipole arm length
r - dipole arm conductor radius
lambda - wave length
μ0 - vacuum permeability constant
ε0 - vacuum permittivity constant
μ - environment relative permeability
ε - environment relative permittivity
EulerGamma - Euler-Mascheroni constant
Si(x) - sine integral function
Ci(x) - cosine integral function

For vacuum μ=1 and ε=1, so Zenv = 376.73 Ω.
For air μ=1 and ε=1.0006, so Zenv = 376.62 Ω.
As you can see air environment impedance is almost the same as for vacuum (free space).

For a half-wave dipole I got result as Za = 73.1 + j*42.5 Ω.
If you don't believe, you can check my calculations.

To calculate Ci(x) and Si(x) you can use Casio online calculator: https://keisan.casio.com/exec/system/1180573427

For example, if you open the book "ANTENNA THEORY ANALYSIS AND DESIGN, Constantine A. Balanis, Wiley 2005", you can find exactly the same radiation impedance value for a center-feed half-wave dipole, see chapter 4.6 HALF-WAVELENGTH DIPOLE, page 184. See picture in attachment. You can find the same value using different approach, include induced EMF method.

Exactly the same value 73 + j*43 Ω for a half-wave dipole is mentioned in different sources, so I think my calculations are correct. Isn't it?

Of course that math doesn't include thermal loss in the antenna radiator conductor, but since it is small enough we can ignore thermal losses due to radiator heating in that discussion.

73.1 + j42.5 Ω is a expression in rectangular form, in a coordinate system.  It describes in this case an impedance with a vector value of 84 Ohm.

Za = 73.1 + j*42.5 Ω is antenna radiation resistance of the half-wave dipole.
Active part Ra=73.1 Ω here describes resistance due to energy loss for antenna radiation.
Reactive part Xa=42.5 Ω here describes resistance due to energy reflection from environment.

This is what I'm talking about when I said about 73.1 + j*42.5 Ω. You're needs to match coax feeder impedance with that 73.1 + j*42.5 Ω value. Not with 84 Ω.

I'm not sure what is the reason to mention about impedance modulus 84 Ω in context of antenna matching with coax feeder. Can you explain please?

Circle chart with 2π circumference and it related math is where it often start when learning wave theory at school,  how waves behaves and related math expressions such as: 32+4i2=52. i is a way to tell it is an imaginary value.
Even more basic can it be written (3X,4Y).
3 and 4 express a position in a 2D coordinate system. It is an rectangular expression for the actual value which have a length of 5. It is same relations for an dipole antenna impedance. Assume 72+j42 as complex impedance. Impedance value is then 83.3 Ohm (722x422=83.32).

We all know about complex numbers and Polar/Cartesian coordinate system. But if you use scalar length 83 of a complex impedance vector 73+j43, you will lose important things (related to wave flow direction) which is very mandatory for impedance matching. Because 73 Ω and 43 Ω here have different meaning. 73 Ω here is resistance due to energy loss for antenna radiation. While 43 Ω is reactance due to reflected wave, note - this is not energy loss. This reflected wave flows back from antenna to the source. With proper impedance matching you can catch reflected wave energy and push it back into antenna again.

Impedance matching is not just about resistance transformation. It also needs to conjugate reactive part of impedance in order to keep reflected wave inside antenna+matching circuit system.

This is why vector antenna analyzer is much more useful for impedance measurement than scalar one.
« Last Edit: May 19, 2022, 07:23:31 pm by radiolistener »
 

Offline E Kafeman

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Re: 75 Ohm why?
« Reply #30 on: May 23, 2022, 05:41:47 pm »
What do you mean when you're talking about "misunderstanding"?
I calculated radiation resistance value according to the math taken from old-school antenna modelling books, here is the formula which I used for a center-feed half-wavelength dipole:

Nice write up and still didn't you understand why you not seems to understand why a dipole impedance with advantage can be expressed as a vector.

You're needs to match coax feeder impedance with that 73.1 + j*42.5 Ω value. Not with 84 Ω.
Sorry. You is wrong and do not seem to understand. Try a suiting good book. Balanis book is not for beginner, it is a bible only useful if you understand its content before you opening it.

In this case is 84 Ω impedance magnitude but an angle is needed as well to create an vector which in this case describes delay between a current and a voltage wave (phasor). Check previous text i wrote about angle format.
Impedance is an magnitude with an certain angle There are a lot of wave gif-animations at internet that shows how this vector varies over time as it is a wave function.
It can be described using resistance and reactance value or a number of other ways.

It is perfectly usable numeric format for impedance matching and especially for designing filters.When you becomes skilled enough will you understand that it often are advantages to use this numeric format.For my part, when designing filters and matching circuits above 10 GHz is it so much simpler to think most of the math in angles and wavelengths as discrete components anyway are of less use.
At lower frequencies is it easier with rectangular numbers. To conjugate match a reactive load, just add its opposite reactance value in serial, in this case a capacitor.

At least one additional reactive component is needed if it should also be a resistive match, often 50 Ohm.A common problem is that matching seldom is done at antenna feeding point and , say at 10 GHz, will 5 mm transmission line from antenna feeding point change actual impedance a lot.
That distance problem can be used as an advantage by matching using just PCB traces.Using this kind of delay as a part of impedance matching is simple. One special length is even more simple, quarterwave transmission line transformer, which is well known and with simple math.
In this case, add an open stub to take care of reactive part.

Lets take it a step even simpler, below are a few lines from  Wikipedia about "Electrical Impedance" : https://en.wikipedia.org/wiki/Electrical_impedance
_____________________________


The impedance of a two-terminal circuit element is represented as a complex quantity  ZZ. The polar form conveniently captures both magnitude and phase characteristics as
  _____________________________


Impedance Z is a bit more formal defined as the ratio of phasor voltage to phasor current.
If you want to learn a bit about this is it some real simple math examples at hamradioschool. https://www.hamradioschool.com/post/complex-impedance-part-3-putting-it-all-together
Diagram below is borrowed from that linked page.

Z represents impedance magnitude with a shown angle.
__________________


    Quote
I'm not sure what is the reason to mention about impedance modulus 84 Ω in context of antenna matching with coax feeder. Can you explain please?

    No. I can't learn you that.

 You have not yet got any understanding for how vector math in many cases are much simpler then rectangular math, adding/subtracting and paralleling vectors as tuning and filter element.

From there to learn about how transmission lines are used as simple transformers, by just rotating the angle in easy calculated degrees or radian seem to be a bit ower you current knowledge.
another example is how an stub with a certain length can correct reactive part is simple when you got the right math skills. Stub lengths are often measured in wavelengths for a certain angle adjustment.

Maybe the most describing tool for an impedance curve over frequency is a Smith chart which is a polar plot of \Gamma .
Magnitude of \Gamma is given directly by the distance of a point to the center . Its evolution along a transmission line is likewise described by a rotation around the chart's center where one full circle corresponds to a half wavelength.
Rectangular format is less convenient for a Smith-chart.
Quote
This is why vector antenna analyzer is much more useful for impedance measurement than scalar one.

 ??? ? Scalar can not be used to read any angle, we know that. So what? Wrong tool for a certain job have never been any useful.
Vector analyzer says it by its name, it is a vector measurement tool.

A vector is a magnitude with a certain angle.
« Last Edit: May 23, 2022, 05:49:34 pm by E Kafeman »
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Offline emece67

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Re: 75 Ohm why?
« Reply #31 on: May 23, 2022, 07:22:24 pm »
.
« Last Edit: August 19, 2022, 05:30:24 pm by emece67 »
 

Offline E Kafeman

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Re: 75 Ohm why?
« Reply #32 on: May 23, 2022, 09:52:05 pm »

>
Quote
An ideal a 1/2 wave center-feed dipole antenna in free space have an impedance around 83 Ohm.

It is correct but is lacking an angle if we want to call it an impedance vector. Without angle is it only a magnitude of the impedance. A scalar value is if no angle exist or at least not is measured. In this case do an angle exist.


Quote
Quote from: E Kafeman on Today at 05:41:47 pm
??? ? Scalar can not be used to read any angle, we know that. So what? Wrong tool for a certain job have never been any useful.

Err, you were the one that gave a scalar value to the impedance of a dipole. Here:
Quote from: E Kafeman on May 15, 2022, 07:23:11 am
An ideal a 1/2 wave center-feed dipole antenna in free space have an impedance around 83 Ohm

Scalar was never mentioned. A few messages before that did I include its vector for a more complete expression several times, such as "83Ω ∠30o".and I explained alternative format for the angle and included simple math examples, and you is then finding it uncler in my next reply.
Can not do that for each reply, explain what impedance is. Someone must know it without misunderstanding what I am writing and without me starting from first page in schoolbook each time I write an impedance value.

It is a bit too repeating to add that an angle is related to an impedance vector in every sentence but ok point is taken, you did find it unclear.
Just in case someone still not understand how a vector is related to corresponding rectangular math format:
 
 83Ω ∠30
 83 x cos30 = 71Ω
 83 x sin30 = 41Ω
 71+j41 Ω
 
 No need for cleverness needed to understand that an angel is needed to describe a vector direction.
Vector impedance magnitude is 83Ω and it is not a scalar value, to avoid any confusion.

« Last Edit: May 23, 2022, 10:01:31 pm by E Kafeman »
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Offline radiolistener

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Re: 75 Ohm why?
« Reply #33 on: May 23, 2022, 11:15:43 pm »
You have not yet got any understanding for how vector math in many cases are much simpler then rectangular math, adding/subtracting and paralleling vectors as tuning and filter element.

I well understand both representation of a complex values and understand how to transform between them. I'm using both representation in my code (digital signal processing and other things), it's just depends on context. From my point of view operating with complex impedance in rectangular form is more simple. But probably this is just a matter of habit and depends more on what calculations you have to deal with more often. Isn't it?

From there to learn about how transmission lines are used as simple transformers, by just rotating the angle in easy calculated degrees or radian seem to be a bit ower you current knowledge.

I know how impedance transformation is performed on a transmission line. Some years ago I made some software which needs to perform some calculations for transmission lines and I made all calculations in rectangular representation of a complex impedance. It was pretty easy and simple. If I understand correctly, you're talking that it's better to use Polar representation. I can assume that for some tasks it is really can be more convenient. But for my needs it was more easy to perform calculations in rectangular form.

For another example, some time ago I was playing with my custom software and firmware for a vector network analyzers. All math the same was performed in rectangular form.

What I don't understand is why you think that the Polar representation is better than rectangular one?


another example is how an stub with a certain length can correct reactive part is simple when you got the right math skills. Stub lengths are often measured in wavelengths for a certain angle adjustment.

When you know a stub length and stub properties you can calculate a stub input impedance, after that you can just easy combine two impedance in rectangular form in order to get resulting impedance. To get reactive part all what you need is just to use sum of two values, it's more easy than deal with angle. Isn't it? :)

May be our vision is different because I see it from software point of view and you see it from manual calculation point of view. Usually such calculations in software are performed with complex data type which represents rectangular form, so for me it looks more easy to perform calculations in rectangular form.

You're needs to match coax feeder impedance with that 73.1 + j*42.5 Ω value. Not with 84 Ω.
Sorry. You is wrong and do not seem to understand. Try a suiting good book. Balanis book is not for beginner, it is a bible only useful if you understand its content before you opening it.

In this case is 84 Ω impedance magnitude but an angle is needed as well to create an vector which in this case describes delay between a current and a voltage wave (phasor).

If it's more easy for you to understand values in Polar representation, you can easy convert from rectangular to Polar representation.

For 73.1 + j*42.5 Ω:

Vector length: sqrt(73.1^2 + 42.5^2) = 84.56
Vector angle: atan2(42.5, 73.1) = 30.17°

So, the half wave dipole impedance in Polar representation is 84.56 ∠+30.17° Ω. But not 84.56 Ω.

84.56 Ω and 84.56 ∠+30.17° Ω are different impedance's. Because first one 84.56 Ω is a scalar value which means 84.56 ∠+0° Ω in Polar form or 84.56 + j*0 Ω in rectangular form.

In other words you're needs to match half wave dipole with 73.1 + j*42.5 Ω or 84.56 ∠+30.17° Ω, but not 84.56 Ω.

If you ask why 84.56 Ω is different? I can explain it on example. Let's assume that we have two transmission lines with impedance Z0 and Z1. Let's see what VSWR we will have at the point between these transmission lines:

Γ = (z1 - z0) / (z1 + z0)
VSWR = (1 + abs(Γ)) / (1 - abs(Γ))

Now just apply Z0=84.56 Ω = 84.56+j*0 Ω  and Z1=84.56 ∠+30.17° Ω = 73.1 + j*42.5 Ω => VSWR = 1.738

As you can see 84.56 Ω is not matched with 84.56 ∠+30.17° Ω. So, if you try to match half-wave dipole with 84.56 Ω you will fail, because it leads to VSWR = 1.738 :)



By the way, you're the first people that I see who like to talk about antenna complex impedance in Polar form instead of rectangular one. Usually all talking about antenna complex impedance is in rectangular form. For example you can open any book about antennas and almost all talking about impedance in rectangular form  :)

Try a suiting good book. Balanis book is not for beginner, it is a bible only useful if you understand its content before you opening it.

Yes, you're needs some basic physics and RF knowledge for any book about antenna. But I don't agree that Balanis book content is complicated. He provided pretty clean math and explanation. Believe me, many other books about antennas have very complicated math with a lot of formulas on hundred pages with something like multi spherical integrals and other stuff with almost no explanation what is going on :) So, I think Balanis book is very good, because it has a good text descriptions.  :)

Balanis explains some question very short, but you can find almost all basic things about antennas in his book.
« Last Edit: May 24, 2022, 12:34:24 am by radiolistener »
 

Offline E Kafeman

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Re: 75 Ohm why?
« Reply #34 on: June 01, 2022, 08:20:03 am »
From my point of view operating with complex impedance in rectangular form is more simple. But probably this is just a matter of habit and depends more on what calculations you have to deal with more often. Isn't it?
Take two any complex circuits and parallel connect them. If ever doing such impedance math do you know why polar math is faster and simpler in these situations.
Rectangular math can sometimes be preferred and sometimes not but as RF technician must you be able to know what math to perform and when.
Feeding a PC  for doing calculation, then doesnt it matter but at lab work must often much of the math be solved on the fly and basic knowledge is needed.
What kind of antenna impedance matching, frequency or wideband and if it should be done with or without a Smith chart and pen or just using brain does often make math selection easy.
Frequency is affecting type of math because for certain frequencies do not some parameters and other are added.

In general:
-30 MHz and nominal value printed on big capacitor and inductors, Farad and Henry, is used as reliable parameters.

Below 100-3000 MHz and nominal value is of limited interest. It is not ideal environment and components are not ideal.
For impedance matching must VAN be used and matching network reactive components must be calculated using its measured Touchstone files.

Above 10 GHz and all is about angles and wavelengths at a PCB with no typical reactive components. Stub filters rules, bandpaass lowpass or as matching network between to reactive loads. Pure resistive loads do not exist above 30 MHz why it often is complex impedanses over wide frequencies at both ends of a transmission line.

More ideal math used below 30 MHz have limited use at higher frequencies as well as the opposite.

What math to select have never been a complicated choice. You chose same math language as you colleagues and customers are using if you want to be understood.

Especially if for a frequency range is not always any angels given, just assumed. Signal transformers, ac relays, speakers, have you ever seen its impedance given as a rectangular value? It is often just magnitude given.
Speaker and microphone impedance, as it vary a lot over frequency is often an average value given called nominal impedance and no angle included. https://en.wikipedia.org/wiki/Electrical_characteristics_of_dynamic_loudspeakers

If dealing with signal processing do I see it more natural to describe it as a dynamic varying signal, an amplitude with an rotating angle.
If describing a modulated RF signal do we often have similar situation, an amplitude with a angle that varies over time and over covered bandwidth. Is often described as a carrier wave and with an additional math describing modulating signal effect on carrier signal.

A unmodulated signal can be written: V(t)=A cos(2 x pi x f ) but it is getting more complicated to fully describe a signal behavior for  for modulation type FM   https://en.wikipedia.org/wiki/Frequency_modulation
PSK is a modulation that even include phase in its name. Simplest PSK can be described as two alternating carriers, both same frequency but with opposite phase but many more phases can be added.

Quote
What I don't understand is why you think that the Polar representation is better than rectangular one?
To understand that must be understood what impedance describes.
Impedance is a vector with an angle which is repeated regularly with frequency, as result of phase difference between ac voltage and current.
That is pretty much whole idea with the impedance unit, a vector and a angle.
 
Resistance is not directly measurable as you probably know, it is a fictive unit and indirect result of measuring applied DC voltage and resulting current values and relative phase.

Impedance is not needed to be translated to rectangular resistance and reactance, it is just one of several ways to do sub-definitions.

If you tries to measure impedance using directly measurable properties, can it be done in a number of ways.
Power factor and voltage are two measurable units which can be directly measured and can be used to  calculate impedance value.
Most common is to measure ac voltage and current magnitudes and its phase difference from which impedance can be calculated as magnitude and angle.

Quote
You're the first people that I see who like to talk about antenna complex impedance in Polar form instead of rectangular one.

From a bit more educating perspective: "We will often find it convenient to express this value in polar form." https://eng.libretexts.org/Bookshelves/Electrical_Engineering/Electronics/Book%3A_AC_Electrical_Circuit_Analysis%3A_A_Practical_Approach_(Fiore)/02%3A_Series_RLC_Circuits/2.3%3A_Series_Impedance

I am not alone. We are a lot of people that are skilled enough to understand why and how it it is less brain-consuming to calculate vectors in polar format.
You have already got several links where you can learn about this.

Quote
When you know a stub length and stub properties you can calculate a stub input impedance, after that you can just easy combine two impedance in rectangular form in order to get resulting impedance. Isn't it? :)


"When you know" is the part you must know how to calculate. "After that combine" - there is not much to combine in a simple stub-matching case. If you assume you is skilled enough to teach me what math to use, you must have at least beginners skills in this field.

In school do they usually show how to do a stub matching using a Smith chart printed at a paper for drawing of arcs using compass.
It is not for no reason a Smith chart is called an "RF engineers analogue computer ".

Just to show how simple it can be for an impedance matching using stubs will I describe an example and include detailed description how I solve this. It should make it more understandable how simple this actually is.

To make this example somewhat realistic, a bit above a school example will we assume a real unknown FR4 PCB, 1mm single side, and a VNA to do measurements with.
No complex filters just a single frequency matching will be done.
We can take as an example our ideal dipole and say that we want to stub match it to a 50 Ohm transmission line and antenna resonant frequency is 5 GHz.
To solve this impedance matching is it so simple that just a virtual Smith chart is needed, the one you have in your memory.

If using a virtual Smith chart and a virtual clock do I not need any real Smith chart at paper nor a pen as this is very simple math.
Call it a rough estimations but result is often just as good as PC-calculated result.
I am not clever or special skilled in math why I always tries to simplify needed calculations and I will try to explain that as well.
It saves me a lot of time in the RF lab to be able to calculate simpler things in my head.

Can you shut your eyes and see a virtual clock? I use a clock mainly because I finds it simple to estimate angles using hour hands.
Can you shut your eyes and do the same for a Smith chart? See a virtual diagram and even assume your written arcs?
If using real paper charts long enough, you will learn this too. No skill, just practicing.
I do not use a calculator for calculation at this level. I will try to show how I instead solves needed math.

We can follow a standard formula often learnt in school starting with transforming real impedance part using a quarter wave transformer.
Normalizing is of limited value this time. Finally, her starts the calculation to impedance match an ideal dipole to 50 Ohm characteristic impedance:

5 GHz wavelength is around 60 mm in length in free space but we must take in account the FR4 PCB dielectric so length is shorter then 60 mm.
A 15 mm long thin strip cu tape is placed on PCB and VNA is connected in one end of this strip using bottom side CU layer as ground. Now is the tape cut with an exacto until 0.25 wavelength is found. Lets for simple numbers assume length was found to be 10 mm.
Transformer impedance is calculated: (50=0.5x100) (72x0.5) =36 which happens to be square of 6 and needed transformer impedance is then 60 Ohm.
Soldering a smd 60 Ohm resistor in opposite end of the thin cu line and connect to ground and adjust width for 60 Ohm using VNA. After that width is found can resistor be removed.

We now know where we are at Smith chart as we have described a half circle around 60 Ohm at resistive center line and the arc is rotating is counter-vise from starting antenna impedance and have exactly reached the admittance line that is crossing 50 Ohm impedance line (and also crosses 0 Ohm). There are two such important admittance circles in a Smith chart and each have two alternative ways to find 50 Ohm.
It is easy to calculate impedance for this point as it is a special case due to the quarter wave transforming but we have no use for it for the moment.

It is now only remaining to add a pure reactive serial load. I select a serial shortcut stub and select a tape-width corresponding to 50 Ohm characteristic impedance. We know what width it is for actual PCB as we have previously decided trace width for impedance corresponding to 60 Ohm.
50 Ohm is 40% wider then measured 60 Ohm width given by this quote: (60x60)/(50x50) => 36/25 => 14/10 => 40%.
Remember the inverse square relation, it is useful to remember.
Actual angle to reach impedance for 50 Ohm do I estimate to be two hours, from 5 to 3 o'clock which is 1/6 wavelength. 40/6 = 20/3 or 6 mm.
I cut the cu tape length slightly longer and add a 50 Ohm resistor as load at end of resulting matching net.
Now can I check using my still connected VNA if the stub needs to be adjusted or else just save actual result.
No calculator was harmed during this process.
No rectangular math was needed, it is as most about angels and vectors lengths  in wavelength and if using real compass and Smith chart is calculation not much more then two connected arcs written at the paper.

If you prefer rectangular math, practice by calculate same impedance match circuit using head calculation and show how it went to calculate this stub matching. By trying will you learn that rectangular math do sometimes sucks big.
I can do that kind of math as well but it is a it would harm my brain.
Many other stub types can be selected to achieve same final matching result and there is at least 4 admittance lines that we can use to find 50 Ohm, but I think this example is the simplest alternative to understand.

Maybe was above description a bit hard to follow . It is as a practical job much faster then what I can describe it.If assumed that I oversimplified, it is not so. It is even simpler if using a real Smith chart but it takes longer time using a real pen instead of a virtual one.

 A very similar impedance matching lesson is shown in linked relative short two-part video, a pretty basic 5 minutes lesson anyone that need basic mastering of impedance matching will have use of.
Here is also a less virtual Smith chart used and a lot of clutter is done around Smith char, which I recommend. Much recommended video as well as several others i same series. Notice that math is still very simple, no calculator needed.
https://youtu.be/GdSV2-6sYqk

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84.56 Ω and 84.56 ∠+30.17° Ω are different impedance's. Because first one 84.56 Ω is a scalar value.
What I originally wrote was that an ideal dipole have a feeding impedance around 83 Ohm.
That is fully correct as a technical definition. It is a magnitude and not a coordinate. A value that can have a specific direction but it is not required.
It is no need to express angle for an impedance and is neither always used.
 
Compare sqrt(32x42)=5. If 3 and 4 are ac voltage and current, what is 5? It is impedance magnitude expressed in Ohm.
It is not a angle-less value even if its actual angle not is expressed. If doing more complex impedance matching have you maybe heard about effective impedance or modulus impedance which is similar to absolute impedance.
About absolute relative modulus: http://www.mathwords.com/a/absolute_value_of_a_complex_number.htm

No angle exist for these expressions, can not be expressed in rectangular format, neither polar format as it only is a magnitude. Impedance expressed for wideband antennas, speakers and transformer can not be given including angle but can still be very useful values.
Some single frequency transformers do even have its impedance given in percent.


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For another example, some time ago I was playing with my custom software and firmware for a vector network analyzers. All math the same was performed in rectangular form.
I have also designed a number of software for the RF industry. Mainly for measurement chambers active measurements, especially EMI related, as well a for lab-bench simpler antenna design and impedance matching software matching using lossy discrete components.This later software must calculate optimal impedance topology using real world component impedances and do needed math as fast as possible to make result appear "live" at PC screen. I have now sold that software since 20 years. A bit beautified now as first versions was written in assembler in DOS and had a commandoline as only GUI, but it was back then good enough to develop first embedded cellphones antennas and its matching network.
For this software,  I do mostly do as when doing head-calculations, most math is done using simplified integers and if possible is shift operations used instead of conventional math to keep calculation speed up as much as possible. As similar calculations are repeated for each sweep, maybe 10 times/second if VNA is sweeping fast enough, is not precision calculations needed for for each sweep. It is enough to compensate if there is something changed in data stream from VNA and each sweep have often just minor changes compared to previous sweep.That is why calculation for this software not seems to take any time at all, optimized wideband or multiband component selection and optimized topology takes a fraction of time relative VNA sweep time.
Numeric presentation is in whatever format actual user prefer and most IEEE-488 compatible VNAs are supported.
Here is a presentation, showing me doing a retuning of an existing antenna in real time. Develop and verify a discrete matching network in less then 10 minutes.
Most people that is using this software do not need any instructions but not all knows about Wheelers cap, which is shown in this video.
https://youtu.be/RyMFun_KhAc

This software, AnTune, is else mainly focused as an aid for optimized embedded antenna design.
« Last Edit: June 01, 2022, 08:29:16 am by E Kafeman »
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