# How to find phase angle from transfer function

From the function:

$$H(\omega) = \frac{1}{(1 + j\omega)(1 + j\omega/10)}$$

How is the phase angle obtained when it has multiple poles to get:

$$\phi = -\tan^{-1}(\omega) - \tan^{-1}(\omega/10)$$

What rule of phase angles allows you to separate the two poles into two separate inverse tangent functions?

It is just a matter of manipulating complex numbers.

$$\angle H(\omega) = \tan^{-1} \left( \frac{\Im\{H(\omega)\}}{\Re\{H(\omega)\}} \right)$$

Where $\Re \{ \cdot \}$ is the real part and $\Im \{ \cdot \}$ is the imaginary part. (NOTE: this equality is not always strictly true depending on the signs of the real and imaginary parts of $H(\omega)$. When finding the angle of an imaginary number the result may need to be adjusted depending on what quadrant the imaginary number is in.)

Expanding $H(\omega)$ gives

$$H(\omega) = \frac{1}{-\frac{\omega^2}{10}+\frac{11j\omega}{10}+1}$$

Instead of finding the real and imaginary parts of the whole expression, though you could do that, You can note that:

$$\angle H(\omega) = \angle \text{numerator of } H(\omega) - \angle \text{denominator of } H(\omega) \\ \angle H(\omega) = \tan^{-1}\left(\frac{0}{1}\right) - \tan^{-1}\left(\frac{\frac{\omega}{10}+\omega}{1-\frac{\omega^2}{10}}\right)\\$$

Using the arctangent addition,wikipedia, formula the expression can be simplified to

$$\angle H(\omega) = \phi = -\tan^{-1}(\omega)-\tan^{-1}\left(\frac{\omega}{10}\right)$$

Basically you get a phase contribution term which is the arctangent of each pole location.

• your first equation is not always true. sometimes it is off by an odd multiple of $\pi$ . – robert bristow-johnson Nov 13 '16 at 5:15
• I think that just means you have to keep track of what quadrant the answer is in. If third quadrant then yes you have to add $\pi$ for instance. – rtclark Nov 13 '16 at 5:19
• This is in the nature of the inverse tangent being calculated over a fraction. Just as an example: We want the angles of the point (1,1) in the first quadrant (45°) and (-2,-2) in the third quadrant (225°). $\phi_1 = tan^{-1}(\frac{-1}{-1})$ and $\phi_2 = tan^{-1}(\frac{-2}{-2})$ As you can see, you can simplify both expressions to $tan^{-1}(1) = 45°$ And this is why the tangent is pi periodic and you cannot distinguish opposite quadrants without thinking about the original signs in the fraction. – Felix S Nov 13 '16 at 6:41
• @Felix Exactly. The expression is always true, but you have to pay attention to what quadrant you are working in. – rtclark Nov 13 '16 at 6:43
• no @rtclark, the equation is not always true and the mod operation is on the wrong side of the = sign, and then it's still not always correct. – robert bristow-johnson Nov 14 '16 at 2:24