Antenna RF Circuit Problem

Hi all,

I am rusty in my rf circuit theory so I need some help.

Let say I have make a 1/4 lamda monopole antennna resonating at 100MHz (75cm) over an infinte ground plane using AWG14 (diameter = 0.1625cm). (I think give inductance of 1.127uH?)

I connect a 100Mhz resonat circuit at the end of the antenna. What wil be the component and value that will be required to bring the antenna back to 100Mhz resonace again?

I will very much appreciate any help.

Cheers,
Element7k

I think there also have a real part in yr antenna.
And then just choose the LC value to make the image part equal to zero.
But the most easiest way is simulated it in MWO.

practical way

In general the antenna impedance will be 35 +j0 at resonance and the j part will vary much more rapidly than the 35 part as you change the frequency.

If you can adjust the length at your fixed operating frequency, this is the best way to get j0.

In general since you have two variables R and X to change to a new set of variables R=50 X=0 for example, a two element reactive L network will work. The series element is on the side of the L with the smallest absolute value of Z facing it from the outside world and the shunt element is on the side of the L with the largest absolute value of Z facing it from the outside world.

From there it is just algebra with complex numbers to get the values of the two reactive elements. Usually one is an inductance and the other a capacitance. The selection of which set is usually done by deciding if you want extra low pass or high pass filtering. For some sources and loads they are both the same type of element. The math will tell you which.

To give you a head start, in this case the series element will be at the antenna side. Your first equation will be the admittance of the node between the two elements calculated from the left and right elements. These two should match. From the 50 side with the shunt element, you should get an admittance of 0.02 +jB1. You then calculate from the other side and get the admittance of 1/(35 -JX2) where B1 and X2 are the results of your L network elements. For the system to work X2 and B1 should be equal in magnitude and opposite in sign at the node. And the real parts of 0.02 should also exist. This should give you two equations with two unknowns.

You have many more options. One is to use a quarter wave tansmission line of 42 ohm impedance which can be done if your system is in microstrip. Another is to use an autotransformer on a ferrite core.

Hi, friends let me see if I follow..
If you shortcircuit a transmissionline, you'll have of course zero impedance as load, and infinite impedance lamda/four from the load, somewhere inbetween you'll have 50 ohms.
Shouldn't I therefore couple in my lamda/four antenna as in the picture upper left. I've connected the upper connector of the transmissionline to where I've found 50 Ohm ?
Is OK ?
If you connect as Element7K above at the node where inductor and capacitor is coupled. It is a inductor you've drawn Element7K ?
You said flatulent that its impedance is around 35Ohm,(can you give a reference ?) which I depicted as Z in the lower picture. Can't I just put a resistor of 15 Ohm in series with Z, which, forgive me I'll soon go to sleep and I am probably wrong, should therefore be in shunt with the L and the C.

Anyway Element7K, I don't think (guessing) that you can calculate the impedance of the antenna by using the formula of the inductance for a straight wire. You should measure the impedance of your antenna with a vector network analyzer.(How to couple it in ?replace the source in upper left picture with VNA ???)
Regards,
StoppTidigare
P.S. Please post your results and tommorrow I'll post my values using SmithChart, transform of 35 Ohm to 50, tomorrow.

antenna impedance

The impedance of these simple antennas were analyzed prior to 1920. The boom in the AM broadcasting put large sums available for ananysis and measurement of all parameters of vertical antennas. They even calculated and measured the combination of ground and sky wave to make sure they were equal at such a large range that the background noise was the thing that drove away listeners in those far distances. This is the origin of the 5/8 wave vertical. It was called the "antifading" antenna because of the effect described above.

The input impedance of a quarter wave line with the far end open is j0. This method approximates the reactive part of the input impedance. The radiation loss of power controls the real part of impedance.

Hi all !
How about the coupling of the lambda for antenna ? Should I bee as I drawn above, with the upper "connector" of the transmission-line attached to the antenna such a distance from its ground where we have 50 Ohm.

I mean if I couple both connector of the transmissionline at the foot of the antenn, I really short-circuit at the foot of the antenna. If I go to the other extreme, I couple one connector to the foot of the antenna and one connector at its head, the I have the opposite in impedance also. Infinity.
Somewhere in between is 50 Ohms. Am I correct about this ??? Is this the way to couple a lambda/four antenna ? Or is it as Element7K drew, putting a lambda/4 long wire at the node of the LC-tank ?

I've attached a Smithchart that shows how one transforms 35 Ohm to 50Ohm.
1. Find 35 Ohm on the Smith-chart, by doing normalization with transmissionline impedance z=r+j*x=35/50=0.7+j*0.
2. The mission is then to add series L,seriesC,shunt L or shunt C, to arrive att the midpoint 1.0.

(In upper left I've put a grey dot.Going upwards from the dot along blue circles, means adding shunt L, downward along blue circle, means adding C in shunt. I've also colored these paths in green.

If one goes upwards along red circle => add L in series,
go downwards => add C in series.)

3. I've found my startpoint 0.7 and I am going upwards along a red circle (remember, goal is to reach 1.0 ). Taking 9 steps each worh 0.05 => z_L=j*0.45. We've added this and now got normalized impedance z=0.7+j0.45.
4. Go now downwards along blue sircle, constant admittance circle. But first read the admittance value where you stopped y=g+j*b=1-j*0.65
I decide to go downward along blue circle, taking 6.5 steps each step is worth 0.1=>y_C=j0.65. Result y=1.
5. Denormalize z_L =>Z_L=z_L*50
and also y_L => Y_L=y_L/50.

Kindest regards, StoppTidigare

the Smithchart

Hi StoppTidigare,

Thank you for your comprehensive solution. I am still trying to recap on my rf theory and will update once I manage to recall some theory which I did not have to use for ages. However, I don't see the L-network that flatulent mention on your solution.

Element7k

StoppTidigare said:

the Smithchart

Click to expand...

Hi Element7K , the L-network (shaped like the letter L, can even be upside down) is really the from the Smith Chart calculated series L, shunt C that flatulent was talking about.

You have an resonant network also with an L and a C, as far as I know, if one neglect the resistance in the leads, should not alter your 50 Ohm source resistance, because the reactances cancel at resonance.

I guess that you then can consider everything to left of your LC-tank equals to 50 Ohm.(Comment wanted on this statement) I assume your source resistance is equal to caracteristic impedance of your transmission-line.

My problem though is still, the way one should connect the lambda/4 antenna to a circuit... should it bee:
i.Lambda/fourth bar connected to upper conductor of transmission-line with the other conductor grounded, or:

ii. Lambda/fourth bar connected to lower grounded conductor, with upper conductor connected at the point where impedance on the bar is 50 Ohm.

Please comment also on this one somebody!!!
Kindest regards,
StoppTidigare

The antenna really consists of the quarter wave part and a ground plane. Connect the coax center conductor to the quarter wave part and the outer coax conductor to the ground plane. If this is on a motor vehicle the ground plane is the metal body.

Hi again all,
is there an advantage having an LC-tank in resonance, before the matching section and the antenna ? Will it increase the antenna range ?

regards,
StoppTidigare

The antenna follows the same rules as one port networks. The maximum signal power you can extract from it is when it is connected to the complex conjugate load impedance. The LC and other type networks change the impedance levels so that the amplifier input impedance, after being transformed by the network, is the complex conjugate of the antenna.

This is done in situations where it is the amplifier noise that limits the reception. This is frequently above 20 MHz or so. For lower frequencies it is the atmospheric noise that limits the reception quality and in these cases it is not economically useful to do any impedance matching. Any LC networks at the input of these receivers are for removing the unwanted signals from the amplifier input so that it is less likely to be overloaded. Also to remove the image signals if it is a superhetrodyne type receiver.