Frequency response function (polar diagram)

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Sci-Fry
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This isn't really a homework question but I felt this is the most relevant section for this. I therefore apologize for not following the standard post template.

I was going through my electrical engineering notes on frequency response functions. It was explaining how to plot frequency response functions on polar diagrams and had the following two steps, which I didn't quite understand. It got to this:

[tex]G(j\omega) = x + jy = \frac{1}{R}\frac{1-j\omega T}{1+(\omega T)^2}[/tex]

It is easy to write:

[tex]x=\frac{1}{R}\frac{1}{1+(\omega T)^2} ; y=\frac{1}{R}\frac{-\omega T}{1+(\omega T)^2}[/tex]

That's fine, but I didn't understand how they jumped to the next step:

From which it is straightforward to eliminate \omegaT to give

[tex](x-\frac{1}{2R})^2 + y^2 = (\frac{1}{4R})^2[/tex]

The point of writing it in this form is that the frequency response can now be plotted as a semi-circle on the Argand diagram. However, I don't understand how they did this step so easily. Is there something I'm not spotting? What is the method from getting from the previous step to this one?

Many thanks,
Sci-Fry
 
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Hi tiny-tim, thanks for the response (and the omega :P)!

Hmmm... I'm probably being incredibly daft, but I see what you're doing...

[note: couldn't get some of the latex to work so I'm using mimtex off another server]

{1+%28%20\omega%20T%29^2}%20=%201%20-%20\frac{1}{1%20+%20%28%20\omega%20T%29^2}%20=%201%20-%20xR.gif


Is that what you were getting at? I'm still not sure what exactly to do after that. Playing around with the numbers gets me to the wrong answer:

mimetex.gif


mimetex.gif


That gets rid of the omega, but it's not what I'm looking for. I know I'm being stupid, just wish I knew where...
 
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Oh that makes sense now, got it. Thanks.

Still difficult to spot though. I don't get how the lecturer can call that a straightforward elimination :P