Finding the Angle of a Complex Number: Tips and Tricks

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Homework Help Overview

The discussion revolves around finding the angle of a transfer function represented as a complex number. Participants explore methods to determine the phase without breaking the function into real and imaginary components.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Some participants suggest using the arctangent function to find the angle based on the real and imaginary parts, while others question the necessity of this method and seek a faster approach.

Discussion Status

The conversation is ongoing, with participants sharing hints and questioning the logic behind certain suggestions. There is no explicit consensus on the best approach, and multiple interpretations of the problem are being explored.

Contextual Notes

Participants note that the problem is not strictly a homework question, and there is a discussion about the rules or methods for determining the angle of complex numbers.

asi123
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Homework Statement



Hey guys.
I have the next transfer function

http://img195.imageshack.us/img195/7924/scan0002l.jpg

And I want to find the angle of it.
I know I can break it into REAL and IMAGINARY but I'm looking for a faster way, is there?

Thanks.

Homework Equations





The Attempt at a Solution

 
Last edited by a moderator:
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asi123 said:
I have the next transfer function

And I want to find the angle of it.

Hi asi123! :smile:

Hint: if w R and C are all real, then the angle (phase) of 1 + jwRC is arctan(wRC) :wink:
 
asi123 said:
I know I can break it into REAL and IMAGINARY but I'm looking for a faster way, is there?

Not really; finding the real and imaginary parts here is fairly quick. Just multiply both the numerator and denominator by the conjugate of the denominator, then [tex]\phi=\arctan\left(\frac{\text{Im}[H]}{\text{Re}[H]}\right)[/tex].
 
gabbagabbahey said:
Not really; finding the real and imaginary parts here is fairly quick. Just multiply both the numerator and denominator by the conjugate of the denominator, then [tex]\phi=\arctan\left(\frac{\text{Im}[H]}{\text{Re}[H]}\right)[/tex].

For this on, but I'm asking in general.

Thanks.
 
tiny-tim said:
Hi asi123! :smile:

Hint: if w R and C are all real, then the angle (phase) of 1 + jwRC is arctan(wRC) :wink:

Ok, I don't see any logic in that.
This is not a homework question, I know the answer, why would you give me a hint?
Is there a rule how to get to the angle?

Thanks.
 

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