# 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
67
The angle is:

Φ=arctan(ωc/ω)

which equals

Φ=90-arctan(ω/ωc)