Antenna filter impedance
This page investigates the impendance in a 3 pole antenna matching filter.
If you follow the text below you will understand the match of an antenna to a transmitter. I have used lot of jw-calculation,
and if you don't know this method you can still follow the text and see the result in figures and diagram.
All contribution to this page are most welcome
Background
The reason I am doing this page is because I didn't really understod the output filter from a class-C transmitter to the antenna.
Most often there is a coil and two capacitor. To really understand this I went back to my old school-books and picked up the
jw (J-Omega) method. This method substitut inductances and capacitances with reactances. The resistors is still the same.
jw substitut
The nice thing with jw is that when you have subsituted the capacitor and the inductor you can see them as resistors and then you
can use the same method to calculate resistans as in a simple DC measurment. It sound complicated, but look at the figure below and I will explain.
Figure 1. shows an output filter connected to an antenna. This filter is supposed to be matched for an 50 ohm antenna at 50MHz. Ztot is the impedance the transmitter will see when it looks into this filter (is connected to this filter). This far you are still
with me. If we look into figure 2 we can see that the antenna is substituted with a 50 ohm resistor to ground. The antenna acts as a
resistor and in this case the antenna was 50 ohm so there is nothing strange here.
If figure 3 I have substituted all elements to impedances (Z). Remember I told you it was possible and I will tell you later how. Z4 is for example the 50 ohm resistor and
Z2 is the coil. Don't mind about any values right now, we comes to that later.
Since all element is impedances I can calculate on this schematic as if it would be resistors. I can now use all the rules one use in DC
calculation. For example Z3 and Z4 are parallel couppled, so In figure 4 have replaced them with Zant.
Don't forget it is the Ztot I want to calculate, so in the frame you can see the formula for the Ztot.
Ztot is equal to Z1 parallel with (Z2 + Zant). If you look in figure 4 you can also see this.
How to transform an inductor and capacitor to an impedance
I use a metod called jw. It is a mathematic way where you use complex values. Real and Imaginary numbers. I will not explain more
about real and imaginay numbers, because you will have to learn it from basic in some math-book. If you know a little you can easy follow this.
What one does is to substitut the capacitor value to XC wich is 1/jwC
The inductor is substituted with XL wich is jwL
the w (Omega) is equal to 2*pi*frequency w=2*pi*f
Setting up the equation
I now know how to substitut the L and C to impedances and I know how to calculate the Ztot. What I will do now is
to put the XC = (1/jwC) and XL = (jwL) expression in formula and find the final formula for Ztot.
In the first 2 expressions (red boxes) I calculate Z3 parallel with Z4 and the next red box is Z2+ Zant. See figure 4 at the top.
The rest is for calculating Ztot and some cleaning up and finally in the big red box you can find the formula for this filter.
If I put the components value in this formula for R, L and C and 2*pi*50MHz for the w-calculation I will get the impedance from this
filter (Ztot).
I have made impedance calculation for different frequency and plotted them in a diagram (next picture). Please look at it and you
will understand what this formula has calculated.
Impedance diagram
To see what the impedance curve would look like, I started to calculate the impendace at 30MHz and increased 1MHz untill
66MHz. The values are plotted in a diagram.
The blue line represent the Ztot for different frequency. If we look at the actual frequency wich is 50MHz we
will find that the impedance is 50ohm, wich we wanted from the begining.
How did I get the components value from the begining?
I used a filterprogram for this and we can now be sure that the filterprogram calculated it right.
Frequency respons
One main reason to use a filter is to filter away overtones from the carrier.
I have made a simulation of this filter and below you can se the frequency respons from this filter.
The Y-axle show the attenuation in dB at the antenna side.
At 50Mhz there is no attenuation but you can see the higher frequency the more attenuation. Remember that the scale is
logarithmic soo the attenuation is quit good.
Final word
I started with an output filter of descrete components. Then I transformed the antenna and the rest of the components to impedances. Then I
transfomed the inductances and capacitances to reactances with jw-method and put it all in one big equation. (Puhhh...)
After some confusion I got the final formula (in the red rectangle) for impedance.
I know the values of the components and I calculated the impedance for different frequency and found out that the impedance
at 50MHz actually is 50ohm.
This was a small introduction to jw calculation on filter and I hope you understod some of it.
Antenna filter impedance
How to transform an inductor and capacitor to an impedance
