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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?

Null
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calvinjarrod
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1 Answers1

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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.

rtclark
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    your first equation is not always true. sometimes it is off by an odd multiple of \$\pi\$ . – robert bristow-johnson Nov 13 '16 at 05:15
  • 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 05:19
  • 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 06:41
  • @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 06:43
  • 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 02:24
  • Ok, you are correct. – rtclark Nov 14 '16 at 02:42
  • I felt that the complexity was getting a little much and was taking away from the actual point of the question. Thank you for all your input. – rtclark Nov 14 '16 at 04:44
  • Albert Einstein is quoted as saying: *"Explanations of things should be as simple as possible, but no simpler."* – robert bristow-johnson Nov 14 '16 at 04:48
  • @robertbristow-johnson Please elaborate why you think the equation isn't correct in modulo pi. Apparently you've never had to do with modular arithmetic if you can't tell a modulo operation from a congruence (https://en.wikipedia.org/wiki/Modular_arithmetic). – Felix S Nov 14 '16 at 14:49
  • I admit I could have used a congruence sign instead of equality but I was assuming the context was clear. Proof that the equation is just plain wrong now, taking H(w) from the OP: arg(H(100)) = -173.7°; atan(imag(H(100))/real(H(100))) = 6.3° != -173.7°. Both results are congruent modulo pi though, as expected. – Felix S Nov 14 '16 at 15:39
  • okay, the angle does have a range of length \$2 \pi \$ (most often from \$-\pi \$ to \$ +\pi \$ and the \$ \tan^{-1}(\cdot) \$ has a range of \$ -\frac{\pi}{2} \le \tan^{-1}(x) \le +\frac{\pi}{2} \$ and, if the angle is expressed as a natural sum of the angles of zeros minus the sum of angles of the poles, then the unwrapped phase will exceed that \$ \pm \frac{\pi}{2} \$ limit. so the unwrapped principle value of the angle (which is between \$ -\pi \$ and \$ +\pi \$) is: – robert bristow-johnson Nov 15 '16 at 00:22
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    $$ \operatorname{Arg} \{H(\omega) \} = \begin{cases} \tan^{-1}\left(\frac{\Im \{H(\omega) \}}{\Re \{H(\omega) \}}\right) &\text{if } \Re \{H(\omega) \} > 0, \\ \frac{\pi}{2} - \tan^{-1}\left(\frac{\Re \{H(\omega) \}}{\Im \{H(\omega) \}}\right) &\text{if } \Im \{H(\omega) \} > 0, \\ -\frac{\pi}{2} - \tan^{-1}\left(\frac{\Re \{H(\omega) \}}{\Im \{H(\omega) \}}\right) &\text{if } \Im \{H(\omega) \} < 0, \\ \tan^{-1}\left(\frac{\Im \{H(\omega) \}}{\Re \{H(\omega) \}}\right) \pm \pi &\text{if } \Re \{H(\omega) \} < 0, \\ \text{undefined} &\text{if } H(\omega) = 0. \end{cases} $$ – robert bristow-johnson Nov 15 '16 at 00:30
  • now the natural wrapped angle \$ \angle H(\omega) \$ is, for a causal system, something that usually starts at zero and goes negative from there (if there is no polarity reversal at DC). this wrapped phase angle is related to the unwrapped principal value Arg function as $$ -\pi < \angle H(\omega) + 2 n \pi = \operatorname{Arg} \{H(\omega) \} \le +\pi $$ where \$ n \in \mathbb{Z} \$ is whatever integer is necessary to bring the Arg function to be within \$ |\operatorname{Arg} \{H(\omega) \}| \le \pi \$ . – robert bristow-johnson Nov 15 '16 at 00:46
  • If the terms in parentheses are phasors, then can't we just do the following? $$ \angle H(\omega) = \angle numerator\ of H(\omega) - \angle denominator\ of H(\omega) $$ $$= arctan(\frac{0}{1}) - \angle \{\ (1 + jw)(1 + jw/10)\ \} $$ $$= -\angle (1 + jw)-\angle (1 + jw/10) $$ $$ = -arctan(w) - arctan(w/10)$$ – Ralph Apr 10 '19 at 01:02