Why does this phase calculation go to 180 instead of 90?

Question

This is all coming from the following video I am studying from http://www.youtube.com/watch?v=XSS6L42ce88

So I am working from this system $$ G(s)\,=\,\frac{4}{s^{2}+s+2}$$

and the video states the gain and phase equations for it are as follows (I agree with this bit)

$$ |G(j\omega)| = \frac{4}{\sqrt{\omega^{2} + (2-\omega^{2})^{2}}} $$ and $$\angle G(j\omega)\,=\,-\arctan\left(\frac{\omega}{2-\omega^{2}}\right)$$

Okay so I want to see what the gain and phase do when the omega tends to zero.

So obviously the gain term will end up being 2 and the phase will be 0 degrees

Now here is the bit I am struggling with The video then says let's see what the function does as omega tends to infinity.

So the gain term will end up being $4/\infty = 0$ but it says the phase will go to -180 degrees (it says that at 02:39 http://youtu.be/XSS6L42ce88?t=2m39s)

Now from how I see it, the phase term at infinity will end up being $$-\arctan\left(\frac{\infty}{2-\infty^{2}}\right) $$ Which as I see it is a very large negative number so shouldn't the phase be -90 or 90 degrees not -180 degrees

I think I have heard about this before and it's to do with being in the wrong quadrant but I am confused as hell. I have really tried with this plugging in all sorts of values.

Answer 1

Consider the breaking point, that is at $\sqrt2$. If you calculate the phase in this point, you can see that is equal to $\frac{\pi}{2}$.

When you consider $\omega$ to $0$ or $\infty$, is just a convention that you consider the first result $0$ as $0$, and the second as $-\pi$ (the minus comes from the fact that it's to the denominator)

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Let's see a generic quadratic function: $$[(\frac{j\omega}{\omega_n})^2+2\zeta\frac{j\omega}{\omega_n}+1]^{\pm 1}$$

Where $\omega_n$ is the breaking point. If we separate the imaginary part from the real, we obtain: $$[(1-\frac{\omega^2}{\omega^2_n})+(2\zeta\frac{j\omega}{\omega_n})]^{\pm 1}$$

Now, when you calculate the phase you have to start from the breaking point (because is the middle point between the result for $\omega \to 0$ and $\omega \to \infty$.

In your example you have: $$[(1-\frac{\omega^2}{(\sqrt2)^2})+(2(\frac{\sqrt2}{2})\frac{j\omega}{\sqrt2})]^{- 1}$$

The phase is: $$\phi= -\tan^{-1} \frac{2(\frac{\sqrt2}{2})\frac{j\omega}{\sqrt2}}{1-\frac{\omega^2}{(\sqrt2)^2}}$$

Thet becomes for the breaking point ($\omega=\sqrt2$): $$\phi= -\tan^{-1} \frac{\sqrt2}{1-1}= -tan^{-1} \infty = -\pi/2$$

So, in that point the phase is -$\pi/2$. Now if you consider the two extremis you obtain both $\phi=0$. You choose $0$ for $\omega \to 0$ and $-\pi$ for $\omega \to \infty$ because a convention.

Why this convention? It's possibile understand it by this:
if we plot a generic $x \mapsto 1-x^2-j 2 \zeta x$ we obtain

Where you can see that the phase converges to $-\pi$ as $x \to \infty$. And you can also see that the phase for $x=0$ (which corresponds to $1-0j$) is zero.

Answer 2

As $\omega\to \infty$, the argument ${\omega}/(2-\omega^2)$ will also go to zero, not infinity (since the bottom grows quadratically and the top only linearly). Hence 90 or -90 is not an option.