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I'm having a confusion with a problem given, any help will be appreciated.

For example it is given a transfer function,

$G(s)= \frac{(s+20)}{(s+1)(s+100)}$

Substitute $j\omega$ to get the frequency response,

$G(j\omega)= \frac{(j\omega+20)}{(j\omega+1)(j\omega+100)}$

The phase angle will be: $\theta(\omega) = \tan^{-1}(\frac{\omega}{20}) - \tan^{-1}(\frac{\omega}{1}) - \tan^{-1}(\frac{\omega}{100})$

What is the intuition behind the tangent function being negative when the complex is in the denominator?

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  • $\begingroup$ This follows from $\arg(a/b) = \arg(a) - \arg(b)$ $\implies \arg (1 / z) = 0 - \arg (z) = -\arg(z)\,$ and $\tan(-\alpha) = -\tan(\alpha)\,$. $\endgroup$
    – dxiv
    Commented Jun 24, 2022 at 18:11

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Let me try to give you a simpler explanation. I would assume this question is generally asked for you to analyze the transfer function, and then to sketch some plots (eg. Bode or Nyquist plots).

Let us consider one of the (denominator) terms from your transfer function $G(j\omega)$, and we name it $G_1(j\omega)$. $$G_1(j\omega) = \frac{1}{(j\omega+1)} $$ We will exploit this term by rewriting into its exponential form. $$\frac{1}{(j\omega+1)} = \frac{1}{\sqrt{\omega^2+1}e^{j\tan^{-1}{\omega/1}}} = \frac{1}{\sqrt{\omega^2+1}}e^{-j\tan^{-1}{\omega}} $$ Hence, from its exponential form, we can see that: $$|G_1(j\omega)| = \frac{1}{\sqrt{\omega^2+1}} $$ $$ \arg G_1(j\omega) = -\tan^{-1}\omega $$ Thus to find the phase angle of the overall transfer function, you will be summing the phase angles of from both the numerator and denominator terms, as written in your post.

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