Frequency response function (polar diagram)

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Discussion Overview

The discussion revolves around the process of plotting frequency response functions on polar diagrams, specifically focusing on the mathematical steps involved in eliminating the variable ωT from the equations provided in electrical engineering notes.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • Sci-Fry presents a mathematical expression for the frequency response function and expresses confusion about the transition to a semi-circle representation in the Argand diagram.
  • Some participants suggest manipulating the equations by squaring terms and multiplying by constants to eliminate ωT, but the exact method remains unclear to them.
  • There is a back-and-forth where participants attempt to clarify the steps needed to achieve the desired form, with one participant acknowledging a mistake in their earlier reasoning.
  • Participants express frustration over the perceived simplicity of the lecturer's description of the elimination process.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the method for eliminating ωT, and there remains uncertainty about the steps required to achieve the correct form of the equation.

Contextual Notes

Participants note challenges with specific mathematical manipulations and the clarity of the lecturer's explanation, indicating potential gaps in understanding the underlying assumptions or methods.

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 Sci-Fry! :smile:

(have an omega: ω :wink:)

square y, and multiply by R2(1 + (ωT)2) …

that gives you (ωT)2/(1 + (ωT)2), which is 1 - … ? :smile:
 
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...
 
Last edited:
hmm … now i look at it, my "1 -" is missing the point :redface:

square (x - 1/2R), and multiply by R2(1 + (ωT)2) :smile:
 
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
 

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