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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?

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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.

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  • \$\begingroup\$ your first equation is not always true. sometimes it is off by an odd multiple of \$\pi\$ . \$\endgroup\$ Nov 13 '16 at 5:15
  • \$\begingroup\$ 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. \$\endgroup\$
    – rtclark
    Nov 13 '16 at 5:19
  • \$\begingroup\$ 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. \$\endgroup\$
    – Felix S
    Nov 13 '16 at 6:41
  • \$\begingroup\$ @Felix Exactly. The expression is always true, but you have to pay attention to what quadrant you are working in. \$\endgroup\$
    – rtclark
    Nov 13 '16 at 6:43
  • \$\begingroup\$ 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. \$\endgroup\$ Nov 14 '16 at 2:24

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