Phase of transfer function changes by multiplying with conjugate?

Discussion in 'Math' started by Jodles, Nov 20, 2010.

  1. Jodles

    Thread Starter New Member

    Nov 20, 2010
    1
    0
    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"?
     
  2. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    The angle is:

    Φ=arctan(ωc/ω)

    which equals

    Φ=90-arctan(ω/ωc)
     
Loading...