开云体育

Hello

I appeal to your wisdom. It's on the sidelines of nanoVNA but some participants in this forum seem to me to know a lot of things, which is not the case everywhere

With a circuit as attached 2F-01.png, I adapt to 50 ? an antenna on two separate frequencies. In the example the same antenna shows

Frequency = 7.1 MHz Impedance = 50.01 - J 0 ohms
Frequency = 3.65 MHz Impedance = 121.1 + J 41.1 ohms

In the case of a real and not simulated construction, it is my nanaoVNA which indicates the values of Z to me. This is how I set up my 80-40-20m antenna with 4 traps

To calculate the circuit, I separately calculate
1/ a 3.650 MHz high pass
2/ a 7.100 MHz low pass
I cascade the two adapters. The best combination shows:

Frequency = 3.65 MHz Impedance = 50.03 + J 0 ohms
Frequency = 7.1 MHz Impedance = 48.38 - J 24.62 ohms

It's obviously not perfect; there is still reactive on 7.1 MHz

Putting the complete circuit into an equation and calculating the roots is not simple at all. There are various calculators on the Net for 'T' or 'pi' adapters but I don't see my double 'L' circuit

Have a way to calculate, in one go, the L and C components so as to obtain a result of 50 +j0 on two frequencies.

To illustrate I put the EZnec file corresponding to my example and the .s1p résult calculated
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F1AMM
Fran?ois

Hi Fran?ois,

When you do matching HP and LP calculations, you get perfect results at targeted frequencies. That's ok, but don't forget that at 7,1 MHz, you are not in the flat transfer part for HP network. So you get small mismatch and loss at this frequency. Demonstration below with SimSmith.

You don't give us details about your two matching cells, but schematic is probably as drawn in picture HP_LP.

First report : your LP cell is absolutely useless, simply as at 7,1 MHz the complex impedance is already 50 Ohms (50.01 +0j). Values for L1 and C1 are also not realistic at these frequencies. So keep only the HP cell, as drawn in picture HP-only.

Second report : No change for SWR value (1,64) at 7,1 MHz without the LP cell (normal behavior). This value is also comparable to what can be found in your s1p file at 7,1 MHz (SWR = 1,62). The Bode plot (BodePlot picture) highlights loss (0,3 dB) and mismatch (SWR = 1,64) values.

73 - Jean-Roger

Ah, a French :)

Hello

I advanced a little on the subject. To simplify, I will already give the almost satisfactory response. The base of the antenna is :

- Frequency = 3.647 MHz Impedance = 120.79 + J 41.13 Ohms
- Frequency = 7.1 MHz Impedance = 201.6 - J 212.3 Ohms

I cascaded one behind the other, starting from the antenna to the TX
- A low pass adapter (granted on 7.1 MHz)
- A high pass adapter (granted on 3.65 MHz)

Regarding the adapter passes high, the closest to the TX, I fix that at 3.65 MHz I want it to bring 50 ?. It will therefore bring 50 ?. Obviously at 7.1 MHz, it brings back an impedance which is not 50 ?

In the simplified (and inaccurate) calculation I imposed that the "low pass addator, that on the side of the antenna brings back to 7.1 MHz 50 ?

To optimize I made sure that the impedances brought back by this low pass serve from the entrance to the top pass adapter.

The objective was, on the TX side that we see 50 ? for 3.65 MHz and 7.1 MHz. In fact is to adjust the intermediate impadance between the two adapters (impadance that I have arbitrarily fixed at 50 ?) and here is the result
ROS 3,650: 1.00
ROS 7,100: 1.02

The adapter diagram is attached (Ok.png).

Here is where I am.
- Either someone has already treated this case and there is a calculation module
- Either this module does not have (or it cannot be found) and I will write code to resolve the network by sweeping the values of this intermediate impedance.

In the example it is 62 + j32

As the Excel file is small, I have it, attached. The problem is that it is not an operational calculation sheet (armored) but the calculations are from. 62+j32 are in the Q5 and Q6 cells of the sheet "Ajusté"

73
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F1AMM
Fran?ois