# The simulation of a microwave antenna made of nanofilm Ti3C2 (Mxene)

• Alekseykolesnik
In summary: This is why I would like to know the effective Er of the dipole element, in the simulation; It may be different from 3.1.Also, as you are modelling a layer of Ti3C2, with Er = 1.00, I am wondering how that layer changes the effective Er of the dipole. I have not had time to read through the article yet, to see how the layer is supposed to work.But this simulation is not bad. You should be able to improve the simulation by tweaking the geometry.
Alekseykolesnik
Hello! I am a student of the radio engineering faculty and now I am engaged in master's work. The purpose of my work is to find application for Ti3C2 in antenna technology. I found a lot of information from article http://advances.sciencemag.org/content/4/9/eaau0920

But I can't understand how to simulate the Ti3C2 dipole in CST Microwave Studio. I have been trying for five months to get the same results as in article (S11 values), but without the result. Apparently, I do not understand something in terms of building a model. Could you help me with it, please?
The fact is that I modeled the dipole in the CST Studio program with the parameters that I managed to get from the article. The remaining parameters were set based on pure common sense and logic…

Parameters:
The thickness of the nanofilm-1.4 µm
The length of the dipole – 0,0625 m
Frequency-2.4 GHz
The width of the dipole was set to 0.005 m

I considered a 100-µm-thick layer of lossy substrate as the PET sheet. The permittivity value is 3.1, the el. conductivity = 2,07e-3 S/m.

For the simulation the Ti3C2-layer was used a surface impedance model instead.

A discrete 50-ohm port is used as a source

The simulation was carried out in the time domain with an accuracy of -60 dB. As a result, the simulation S11 came out to -8,5dB, which does not correspond to the data obtained in the article (-36dB).

The version of CST Studio is 2017.

If you will not complicate, prompt, please, that I do not correctly. I had no previous experience with modeling nanomaterials. Maybe I need to improve the accuracy or reduce the step of the meshing. On the other hand, maybe I create a surface impedance model of a Ti3C2-layer incorrectly.

Screenshots and model are attached

P.S. Sorry for my English!

@Alekseykolesnik Welcome to PF.
Sorry for the delay. Your English is very good.

Unfortunately I do not have CST Microwave Studio.
I cannot open your pack.rar on this system, maybe use a screenshots.pdf file?

Dielectric constant of substrate is 3.1 nearby, but is 1.00 far away = free space.
I wonder what is the effective dielectric constant and the velocity factor?

Can you do a frequency sweep to measure the resonant frequency of your model dipole?

berkeman
Ok, I attached the screenshots in pdf.

I didn't fully understand what you said about the dielectric constant of the substrate. After all, the dipole is applied to the substrate, respectively, it is in close contact with the substrate. That is, the value 3.1 should be used, right?

I have a graph of S11 versus frequency, isn't that the frequency sweep?

#### Attachments

• Screenshots.pdf
1.2 MB · Views: 257
Last edited:
I do NOT see any big problems with your model.
The far-field radiation pattern has deep nulls on the Y-axis. That is correct for a dipole.
The gain is a maximum of 2.23 dBi which is good, I would expect 2.15 dBi for a dipole.

Do you model the dielectric as polyester or as PET ?

Alekseykolesnik said:
I have a graph of S11 versus frequency, isn't that the frequency sweep?
That is correct. I can now see it in screenshots.pdf; Your result S11 shows the reflected energy is a minimum at a frequency of 2.35 GHz.

Alekseykolesnik said:
The simulation was carried out in the time domain with an accuracy of -60 dB. As a result, the simulation S11 came out to -8,5dB, which does not correspond to the data obtained in the article (-36dB).
The difference could be due to several things. Their signal generator may be better matched to their dipole, or their dipole may have higher resistive losses than yours.

To better match the dipole to the generator, plot the complex impedance of the dipole at the required frequency. Adjust the length of the dipole to reduce the reactance to zero. Adjust the width of the dipole to bring the resistance closer to 50 ohm. Repeat the process until S11 gets closer to their figure of -36dB.

That will also eliminate estimates of dielectric constant Er, and VF.
Alekseykolesnik said:
I didn't fully understand what you said about the dielectric constant of the substrate. After all, the dipole is applied to the substrate, respectively, it is in close contact with the substrate. That is, the value 3.1 should be used, right?
When a dipole is implemented as a resonator on a PCB with a ground plane, almost all the energy propagates in the dielectric between the ground plane and the dipole element. The velocity of that wave is less than c, by a velocity factor of vf = 1/√( Er ). It will not radiate efficiently with a ground plane.
However, you have NO ground plane, so the energy propagates around the element, mostly through free space with Er = 1.00, but also through the thin dielectric layer with Er = 3.1; Since the dielectric is thin, and only on one side, it plays only a small part. I would guess that in your model the combined Er of space and the PE is going to be less than Er = 1.1 Modelling capacitors or transmission lines with asymmetric dielectrics is difficult without FEM, but we can calculate an estimate for the effective Er.

We know the length of the half-wave dipole is 62.5 mm.
299.792458 / ( 2 * 0.0625 ) = 2398.34 MHz
We know centre frequency (from S11 minimum) is at 2.35 GHz.
Frequency ratio is 2.39834 / 2.35 = 1.02057
The square of 1.02057 is the effective Er = 1.04156

If you reduce the length of the dipole to 62.5 / 1.02057 = 61.24 mm the centre frequency will move back up to 2.4 GHz. That will improve the match and lower S11.

Notice that the presence of a dielectric near the dipole can have a significant effect on the velocity factor. Changes in design need to be monitored. An insulated wire dipole will be affected by the thickness and material of the insulation.

Baluncore said:
The gain is a maximum of 2.23 dBi which is good, I would expect 2.15 dBi for a dipole.
The authors of the invention told me in correspondence that the gain of a half-wave dipole can never be greater than 2.15 dBi. It is an axiom. That's what confuses me.

Baluncore said:
Do you model the dielectric as polyester or as PET ?
It seems to be the same in Russia.
It says here that PET belongs to the polyester family.
https://en.wikipedia.org/wiki/Polyethylene_terephthalate
Thank you! Now I am optimizing the size of the dipole to obtain the required S11 (-36dB). I hope that I will succeed.

I thought that there are some features in the modeling of this type of Ti3C2 antenna as opposed to conventional metal antennas...
Maybe I need to improve the accuracy or reduce the step of the meshing, or I should use other boundary conditions.

This file contains sheet resistance for nanofilms of a certain thickness. Depending on the thickness of the nanofilm (and its resistance), the parameter S11 varies greatly. Maybe I need to set the ti3c2 material in the program as ohmic sheet with pure active resistance?

#### Attachments

• aau0920_SM.pdf
1.9 MB · Views: 238
I ran a parameterization of the length of the dipole in the CST programme and the value of S11 is less than -13dB failed to obtain.

Alekseykolesnik said:
The authors of the invention told me in correspondence that the gain of a half-wave dipole can never be greater than 2.15 dBi. It is an axiom. That's what confuses me.
I agree. But if the dipole is longer than half wavelength it can have gain greater than a dipole. Also, I expect the error margin of FEM to be maybe 0.08 dB.
Try changing mesh to have more elements, does 2.23 dB change towards 2.15 dB.
Try changing dipole length to see if 2.23 dB changes towards 2.15 dB.

Alekseykolesnik said:
I ran a parameterization of the length of the dipole in the CST programme and the value of S11 is less than -13dB failed to obtain.
Impedance match must be correct resistive and correct reactive.
A dipole usually has Zo ≈ 70 ohms. But you drive it with 50 ohms.
Try changing the dipole width from x = 5 mm to x = 3 mm, or 8 mm, does impedance match improve and S11 become less ?

Baluncore said:
Impedance match must be correct resistive and correct reactive.

I have only the sheet resistance for Ti3C2-layer. The article does not specify the reactance, that's why I don't know how to set properties of Ti3C2-material in CST Studio rightly.

Changing the width of the dipole in the larger and smaller side leads to a shift in resonance at a frequency of 2.25 GHz, S11 does not change. The increase in impedance of the port to 70 Ω also leads to a shift of resonance frequency of 2.25 GHz and reduce S11 to -23dB.

Alekseykolesnik said:
Changing the width of the dipole in the larger and smaller side leads to a shift in resonance at a frequency of 2.25 GHz, S11 does not change.
You must fix excitation frequency to 2.4GHz then match excitation impedance by changing the Mxene dipole dimension of larger and smaller sides.

I notice your substrate is same size as dipole. Maybe make the substrate bigger so you can keep it fixed during experiment. Here is a possible construction of model.

Make substrate sheet, 0.1 mm thick, below zero. Substrate; Polyester; Brick.
Xmin=-15e-3; Xmax=+15e-3; Ymin=-50e-3; Ymax=+50e-3; Zmin=-0.1e-3; Zmax=0;

Then print dipole on surface of substrate above zero. Dipole; Mxene; Brick.
Xmin = -5e-3; Xmax = +5e-3; Ymin = -31e-3; Ymax = +31e-3; Zmin = 0; Zmax = 1.4e-6;

Cut a 2 mm slot, through Mxene dipole for excitation feedpoint.
Do not cut the substrate; Vacuum; Brick.
Xmin = -15e-3; Xmax = +15e-3; Ymin = -1e-3; Ymax = +1e-3; Zmin = 0; Zmax = 1e-3;

Define excitation at the cut in Mxene dipole.
You must fix excitation at 2.4GHz.

Keep dimensions of substrate and vacuum cut fixed.
Adjust x and y dimension of Mxene to get (R+jX) ≈ 50 + j 0.

That will match dipole to 50 ohm excitation, S11 <= -36 dB, SWR = 1.0x?
Experiment with z dimension and properties of Mxene.

I did everything you said. As a result, I got the same S11 value as before (-13dB). I found the impedance graph (I attached it) but I don't know what exactly it shows. Maybe this graph shows the input impedance of the dipole?
Last Friday, I set the value of this schedule at a frequency of 2.4 GHz in the settings of the impedance of the port (~110 Ohm). So I was able to get the value S11=-37dB, which was necessary. Today I spent many hours, but I could not get the value S11=-36dB, but only S11=-26dB. I can't figure out if I made a mistake on Friday or if I'm doing something wrong now. I will try again to reach the value of S11=-36dB. Thank you for helping me! Maybe you know what this graph shows?

I believe the “discrete port impedance, (magnitude)” is the vector sum of the real and imaginary complex impedance components.
Z = ( R + j X ); Magnitude = √ ( R2 + X2 )

You need to plot port impedance as separate complex components, resistance R ohms, and reactance X ohms.
Then change the dipole ±x and ±y size to get Z = ( 50 + j 0 ) ohms at 2.4 GHz.

As substrate and vacuum gap are now much bigger than the dipole, you may ignore them while you;
Adjust dipole ±y to significantly change the reactance; imaginary = zero.
Adjust dipole ±x to significantly change the resistance; real = 50 ohm.
You can probably use component sweep to find optimum values of x and y.

Good afternoon. Thank you very much!
I spent a lot of time experimenting with the model again. I did sweep in two ways: w(0.001...0.007) and l (0.028...0.033), changing them at the same time. As a result, the minimum S11 was about -13 dB, and with changes in the sizes of w and l, the resonance was shifted towards 2.3 GHz.
(Xmin=-w; Xmax=w; Ymin=-l; Ymax=l)

The sweep on active resistance of port (from 40 to 120 Ohms), at l=0.028, w =0.003:

I tried to change the size of the dipole in a wider range (l=0.025...0.037, w=0.001...0.004):

At some sizes, the imaginary part of the dipole impedance approached zero, and the real part in all cases ranged from 50 to 230 Ohms. When the real part had a value of about 50 Ohms, the imaginary part had a value of -40 Ohms.

The real part:

The imaginary part:

Therefore, to achieve the input impedance of the dipole is 50+j0 Ohms out. If the real part is close to 50 Ohms, the imaginary part at this time is a value far different from 0 Ohms and S11 with a very large (-13dB).

It was possible to determine several points at which the reactive component of the dipole impedance tends to zero, with the active equal to 90.4 Ohms/ 104.2 Ohms / 119.6 Ohms. In this case, if you set the impedance of the port to 90.4 Ohms, then there is an excellent agreement (S11=-55 dB), but this contradicts the article, which says that the port had a resistance of 50 Ohms and S11=-36dB.

Alekseykolesnik said:
Good afternoon. Thank you very much!
Your afternoon is 1:34 AM here in Australia!
Thanks for the report. It is always good to see a dip in the response of a dipole. You now better understand the degrees of freedom available with a dipole, and how to model the R and X of a complex impedance.

Minimising reflected energy, S11 can be achieved in a number of ways. Some good and some bad. There is loss resistance in the dipole that heats the dipole, and there is radiation resistance that relates the ratio of voltage and current at the driven point. The S11 of a 50 ohm carbon resistor looks good, but radiates in the IR band which is not what you want. You need to know why S11 changes, and make sure it is because of increased RF radiation. Keep an eye on the far-field of the dipole response to make sure you are improving things in the right direction.

Are you chasing a false target? Did the original reported dipole film have a higher surface resistance and so lower reflected energy? Compare S11 and far-field radiation in dBi. Also measure the model surface temperature of the film. If the published radiation was less than yours, you can increase the resistance of your film to approach their model.

Check exactly how the published example dipole was connected to the virtual generator.

A theoretical thin dipole has a radiation resistance of about 70 ohms, not the 50 ohms typical of RF solid dielectric transmission lines. It is therefore necessary to transform the impedance of the line or the generator to the dipole. That requires a matching network which can be as simple as a quarter wavelength of line having an impedance that is the geometric mean of the two impedances being matched. Z = Sqrt( Z1 * Z2 ). If you need to apply your dipole you will need to match the impedances involved.

I do not see a direct question, so what direction will you go from here.

Baluncore said:
Your afternoon is 1:34 AM here in Australia!
Hahaha)

The main target was to make sure that the program simulates my model correctly. I think, that I have achived it, haven't I?

Now, my task is simulating the other antennas with Ti3C2 nanofilm, for example a dish antenna.

The main problem was the difference in results of S11- parameter between my model and their model. It was confusing and I thought I was doing something wrong, like setting the material or the boundary conditions.

Baluncore said:
Check exactly how the published example dipole was connected to the virtual generator.
The authors of the article do not provide information about this. It is known only that they installed 50 Ohm port.

Maybe they really set a matching line between the dipole and port and therefore my results differ from their results.

Baluncore said:
The S11 of a 50 ohm carbon resistor looks good, but radiates in the IR band which is not what you want. You need to know why S11 changes, and make sure it is because of increased RF radiation.
Please excuse me, but I don't understand what you're saying...

Alekseykolesnik said:
Please excuse me, but I don't understand what you're saying...
A perfectly matched load does not reflect energy. Energy not radiated as RF will heat the dipole and dielectric, or be reflected. A 50 ohm resistor can be perfectly matched to a 50 ohm line, so S11 = –90dB. The resistor does not radiate RF, so energy must be converted to heat.

The lesson here is that S11 = –40dB is not always good. You can get better S11 by having more resistance and less radiation. Make the film thinner, with higher resistance, less RF radiation and more heat, which will be closer to the published dipole specification.

I'm sorry. I was able to achieve the value of the parameter S11 as the authors of the article. S11=-38.7 dB at 2.4 GHz! At the same time R=48.96+0.48 Ohm. But I was confused by the geometrical dimensions of the dipole, they are wonderful: l=0,0242 m, w=0.01 m.
The authors of the article in a personal letter to me said that the width of the dipole is 5 mm, but I have a width of 10 mm. The length of the dipole I have much less.
Because of the increase in width, it turns out that I increased the band of the dipole, and with the length it is not clear at all.

I also wondered what the distance between the arms of the dipole affects. I would also like to know why you showed me the distance between the arms is 2 mm.

A very important point is that these results were achieved only if the port is set to the extreme points of the dipole, as shown in the figure.

You probably know if I can set the port this way?

Prior to this, the port was installed between the edges of the dipole, as shown below. This did not allow me to achieve the desired result.

Excuse me please and thank you very-very much!

You are able to design and model antennas. Do not be sorry, just be happy.

My 2 mm gap was a guess. A dipole is a resonator. The gap in the middle, where the port current feeds the element, should be as small as practical, so currents flowing in the two sections are not phase shifted by the feed.
You must consider and model the connection of a real port to the dipole. Here you have the same series current flowing in the line as in the dipole. That is very tight coupling and will be broad-band. If you make the dipole continuous, with zero gap, then you can feed it at two points with a port voltage, and maybe see a higher Q resonator.

I cannot say why the port position in the gap is so important. I would expect if the port was symmetrical to see lower resistance. The corner points will have higher resistance = greater RF losses, with implications to S11.
Take care that it is the real world you are modelling, and not an artefact of the numerical model.
Maybe you need a high conductivity area of surface near the gap where the port is attached, silver paint.
Do not attach the port to a corner, or an edge. Attach the port to an area on the surface away from the film edge and see what changes.
You must learn to think like an EM disturbance on a transmission line resonator, or a wave spreading out over a structure from the port.

Alekseykolesnik said:
Because of the increase in width, it turns out that I increased the band of the dipole, and with the length it is not clear at all.
Maybe if you search the degrees of freedom space by changing the film resistance you will see a change in width.
The best indicator of tuned wavelength is not the amplitude of the current, or the far-field strength, (a cosine function of λ), but is the polarity sign of the reactance, (a sine function of λ).

Baluncore said:
That requires a matching network which can be as simple as a quarter wavelength of line having an impedance that is the geometric mean of the two impedances being matched. Z = Sqrt( Z1 * Z2 ). If you need to apply your dipole you will need to match the impedances involved.

A matching network must be copper (for example) or ti3c2 and the thickness and width must be the same as the real conductor or the same as the Ti3C2 dipole? I suppose there should be a matching network with a short circuit at the end?

Where you use a transmission line element as a coupler it can be open or short. In both cases it will reflect energy, but a closed end will conduct DC bias current to electronics while an open circuit line will not. To change between an open or shorted line, change the length of the line by one quarter of a wavelength.

When you build an Mxene dipole, it must couple to the electronics.
That will decide how the antenna matching may be done.

https://en.wikipedia.org/wiki/Quarter-wave_impedance_transformer

Sorry, could you guess how the dipole was connected to the port in CST MWS based on the image of the dipole structure? Head boils... I even tried to connect a quarter-wave matching network, but did not get any results.

When you implement such an antenna you will need to connect to the film. That might be a patch of silver paint so you can solder the connection, as in picture B above.

In your model, try crossing the ends of the dipole at the gap with 1mm wide strips of conductive silver. Make it the same as the dipole width, 10 mm long for now. Think of the silver as the terminal of the feedline or the electronics.

See if, and by how much, that changes the sensitivity of the complex impedance to the port connection point.

You mean the thickness of the silver ink to make equal to 1.4 e-6 m?Like the picture?

If so, then it is necessary to set the silver ink not as a metal, but as a material with surface impedance, and I do not know the resistance of silver ink for a given thickness.

Or do you mean the design below?

## 1. What is a microwave antenna made of nanofilm Ti3C2 (Mxene)?

A microwave antenna made of nanofilm Ti3C2 (Mxene) is a type of antenna that uses a thin film of titanium carbide (Ti3C2) known as Mxene to transmit and receive microwave signals. Mxene is a two-dimensional material that is highly conductive and can be manipulated to create antennas with unique properties.

## 2. How does a nanofilm Ti3C2 (Mxene) antenna work?

A nanofilm Ti3C2 (Mxene) antenna works by converting electrical energy into electromagnetic waves, which can then be transmitted or received. The thin Mxene film acts as a conductor and is able to efficiently transmit and receive microwave signals due to its unique properties.

## 3. What are the advantages of using nanofilm Ti3C2 (Mxene) for microwave antennas?

There are several advantages to using nanofilm Ti3C2 (Mxene) for microwave antennas. These include its high conductivity, lightweight and flexible nature, and its ability to be easily manipulated to create antennas with specific properties. Additionally, Mxene is resistant to corrosion and can operate at high temperatures, making it suitable for a variety of applications.

## 4. What are the potential applications of a nanofilm Ti3C2 (Mxene) antenna?

A nanofilm Ti3C2 (Mxene) antenna has a wide range of potential applications, including in wireless communication systems, radar technology, and satellite communication. It can also be used in medical devices, such as wireless implants, and in the development of wearable technology.

## 5. What are the challenges in simulating a microwave antenna made of nanofilm Ti3C2 (Mxene)?

Simulating a microwave antenna made of nanofilm Ti3C2 (Mxene) can be challenging due to the complex nature of the material and the need for specialized software and equipment. Additionally, accurately predicting the behavior of the antenna in different environments and under different conditions can be difficult. Further research and development are needed to improve the accuracy and efficiency of simulations for Mxene antennas.

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