Angle of transfer function changed by rewriting equation?

**Jodles** · November 21st 2010, 01:23 AM

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"?

**rebghb** · November 21st 2010, 02:38 AM

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.

**Jodles** · November 21st 2010, 03:19 AM

Whoops! I was a little fast there I see.

Thank you!

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