Thread: Angle of transfer function changed by rewriting equation?

1. 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"?

2. Originally Posted by Jodles
$\frac{1}{1-j\frac{w}{w_c}} = \frac{j\frac{w}{w_c}}{1+j\frac{w}{w_c}}$
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.

3. Whoops! I was a little fast there I see.

Thank you!