Phasors / and more phasors

I was reviewing some of technical papers relating to mathematical vector phasors, subject of great interest to me. Then, it occurred , something missing in all these mathematical formulas, and equations, and tutorials, perhaps. I read many papers on this subject, and it just seems to me when I look at Phasor graphs, and Phasor addition of two sinewave signals, and such, it's all illustrated at one frequency. No one really comes out and says that, in the tutorials, or inside the electrical engineering text books. So, in this paper, I had the thought of presenting Phasor Vector Analysis in a little different way, and as a function of frequency. Why ? Well, because in the real world of radio frequency communications, changing frequency is the normal operating environment, these days. I do not go into calculating the actual phase angles, I leave that as homework for others to follow up on. But, if anyone out there has to opportunity to work with any type of circular Phased Array Interferometers, this describes how the output of an I/Q discriminator works when we sweep the input of it's antenna say from 2 Ghz to 6 Ghz. I start at the top of the drawing, and work my way down. So, lets see what we can see.

First, I start at the beginning, with a 50 megacycle signal feeding a plurally resistive load. I think this part is also left out of many engineering text books. The graph on the top right attempts to show the phase angle at zero (0) degrees on the X axis. I show this for a reason.

On the second line of the drawing, shows a 50 megacycle signal feeding a series resonate circuit of 2 picofarad capacitor and 1 micro henry inductor. I chose the frequency and value of these parts that help illustrate my point. So, at this frequency, the XL of the inductor at 50 Mhz = 1.571 K ohms , and the XC of the capacitor = 1.592 K ohms. Thus, we can say the values of both the XL and the XC cancels each other, and this circuit is at resonance at 50 Mhz. The graph therefore shows the phase vector on the X axis remains at zero (0) degrees. All fine and good.

So what happens when we go up in frequency ? The third line on the drawing, we increase the source frequency to 60 Mhz. We could then calculate the new XL of the inductor as 1.885 K ohms, and the XC of the cap at 1.326 K ohms. The impedance of the inductor is higher than the cap, thus shifts the phase angle positive of the X axis, or positive on the Smith Chart.

And of course, lastly, when the source frequency is reduced to 40 Mhz. We could calculate the XL of the inductor as 1.257 K ohms, and the XC of the cap at 1.989 K ohms. Thus the resultant vector angle is shifted below the X axis, or negative on the Smith Chart.