Total posts : 45366
Rich,
With the new link you supplied for your new antenna plots, we may be approaching what, I hope, will be ultimate agreement. Your new data indicates that your antenna has an RF bandwidth of about 10 kHz. I, on the other hand, calculated that it is about 5 kHz. This 2/1 ratio in our results suggests that there is some common ground, and it is only necessary to find the factor that caused one of us to calculate twice the bandwidth the other did. A 10 kHz bandwidth at 1610 kHz means that the antenna Q is 161. The total antenna resistance is about 11 ohms. This means that the capacitive reactance of the antenna is 11 X 161 = 1771 ohms. The antenna capacitance must be about
1/[(1771)(2pi)(1610)(10^3)] = 56 pF.
I calculated that the antenna capacitance is 29.1 pF, which is a little more than half of 56 pF. We can now pose the question a little bit differently than before: What is the antenna capacitance? Is around 30 pF, or is it around 60 pF? I used several different formulas, and got around 30 pF with all of them. The formula I used to get 29.1 pF was taken from Antenna Engineering Handbook (Jasik). The two antenna capacitance formulas in The Low Power AM Broadcasters Handbook (which I mentioned in a previous post) both give around 30 pF. Could it be that your program has a built-in assumption that the antenna has a capacitive hat, and is adding an extra 30 pF to the antenna capacitance calculation?
The bandwidth of the antenna establishes the half-power points of the banwidth curve, which correspond to the 3dB points, and the 2.618 VSWR points. I prefer dB notation to VSWR notation because, for AM, RF bandwith is closely related to audio bandwidth, and dB notation is used for audio bandwidth curves.
