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Thread: 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 $\displaystyle \frac{1}{1-j\frac{w}{w_c}}$

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

    Separating the real part and the imaginary part, and taking the angle between them: $\displaystyle \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: $\displaystyle \phi = 90 - arctan \frac{w}{w_c}$, by simply inspecting the same function slightly rewritten:
    $\displaystyle \frac{1}{1-j\frac{w}{w_c}} = \frac{j\frac{w}{w_c}}{1+j\frac{w}{w_c}}$
    Now , the numerator is obviously 90 degrees, and the denominator $\displaystyle -\frac{w}{w_c}$, so the answer would be $\displaystyle \phi = 90 - arctan \frac{w}{w_c}$.

    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
    $\displaystyle \frac{1}{1-j\frac{w}{w_c}} = \frac{j\frac{w}{w_c}}{1+j\frac{w}{w_c}}$
    This isn't a valid equality, you carried your algebra in the wrong way, it's supposed to be like this:
    $\displaystyle H(j\omega) = \frac{1}{1-j\frac{w}{w_c}} = \frac{j\frac{w}{w_c}}{\frac{\omega^2}{\omega_c^2}+ j\frac{w}{w_c}}$.
    Angle of $\displaystyle H(j\omega)=90-\tan^{-1} \frac{\omega_c}{\omega}$ which is the same as $\displaystyle \tan^{-1} \frac{\omega}{\omega_c}$ 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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