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Math Help - Angle of transfer function changed by rewriting equation?

  1. #1
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    Angle of transfer function changed by rewriting equation?

    Consider the transfer function \frac{1}{1-j\frac{w}{w_c}}

    Now, if I simply multiply by the conjugate I get this:
    \frac{1+j\frac{w}{w_c}}{1+\frac{w^2}{w_c^2}}

    Separating the real part and the imaginary part, and taking the angle between them: \phi = arctan(b/a) = arctan \frac{\frac{w}{w_c}}{1+\frac{w^2}{w_c^2}} / \frac{1}{1+\frac{w^2}{w_c^2}} =arctan \frac{w}{w_c}

    Now I know this should've been:  \phi = 90 - arctan \frac{w}{w_c}, by simply inspecting the same function slightly rewritten:
    \frac{1}{1-j\frac{w}{w_c}} = \frac{j\frac{w}{w_c}}{1+j\frac{w}{w_c}}
    Now , the numerator is obviously 90 degrees, and the denominator -\frac{w}{w_c}, so the answer would be  \phi = 90 - arctan \frac{w}{w_c}.

    Now why does this change by a simple rewrite and a different method? Are both equally "correct"?
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  2. #2
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    Quote Originally Posted by Jodles View Post
    \frac{1}{1-j\frac{w}{w_c}} = \frac{j\frac{w}{w_c}}{1+j\frac{w}{w_c}}
    This isn't a valid equality, you carried your algebra in the wrong way, it's supposed to be like this:
    H(j\omega) = \frac{1}{1-j\frac{w}{w_c}} = \frac{j\frac{w}{w_c}}{\frac{\omega^2}{\omega_c^2}+  j\frac{w}{w_c}}.
    Angle of H(j\omega)=90-\tan^{-1} \frac{\omega_c}{\omega} which is the same as \tan^{-1} \frac{\omega}{\omega_c} for the set of real frequencies.
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  3. #3
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    Whoops! I was a little fast there I see.

    Thank you!
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