Why does a pole with increasing phase start at -180°?

[jegues]

Homework Statement

Understanding the bode plot for,

$$G(s) = \frac{1}{2s-1}$$

Homework Equations

The Attempt at a Solution

Usually when I am drawing bode plots for either real poles or zeros, I simply solve for the break frequency knowing that a zero/pole will give me a +/-20dB/decade asymptote after the break frequency on my magnitude plot, and a +/-90° phase shift across two decades, with the center (i.e. the +/-45° point) at my break frequency on my phase plot.

This has always worked when I have,

$$(s+a) \quad a>0$$

but it seems to change when,

$$(s - a) \quad a>0 \quad \text{or, } \quad (-s+a) \quad a>0$$

I can't wrap my head around why a pole has an increasing phase that starts at -180°. I was expecting a phase decrease from 0° to -90°.

Can someone explain what I am misunderstanding?

 

[donpacino]

You did solve the bode plot correctly.

When a pole is in the right half plane (greater than zero) the phase will increase during the pole.
It is important to note that this means the system is unstable.

 

[rude man]

There is no such thoing as a Bode plot for poles in the right-hand plane.

Zeros, OK.

 

[milesyoung]

There is no such thoing as a Bode plot for poles in the right-hand plane.

A Bode plot is usually just defined as a magnitude and phase plot of $H(s)\rvert_{s = j\omega}$ versus $\omega$, where $H(s)$ is the transfer function of the system. So strictly speaking, regardless of the implications of having poles in the right-half plane, the Bode plot for the system is still well-defined.

I can't wrap my head around why a pole has an increasing phase that starts at -180°. I was expecting a phase decrease from 0° to -90°.

I'm not sure why you would expect that. When in doubt, go back to basics:
$$\left. H(s) \right|_{s = j\omega} = \left. \frac{1}{2s - 1} \right|_{s = j\omega} = \frac{1}{2j\omega - 1} = \frac{1}{-1 + j2\omega} \Rightarrow\\ \arg(H(j\omega)) = \arg\left(\frac{1}{-1 + j2\omega}\right) = \arg(1) - \arg(-1 + j2\omega) = 0 - (\arctan(-2\omega) + \pi) = -\arctan(-2\omega) - \pi$$
Try evaluating that for $\omega \ll 1$ and $\omega \gg 1$.

 

[rude man]

A Bode plot is usually just defined as a magnitude and phase plot of $H(s)\rvert_{s = j\omega}$ versus $\omega$, where $H(s)$ is the transfer function of the system. So strictly speaking, regardless of the implications of having poles in the right-half plane, the Bode plot for the system is still well-defined.

No. The gain is undefined. The output approaches infinity for any finite input. So a "magnitude" of gain does not exist, neither does a "phase".
.

 

[milesyoung]

No. The gain is undefined. The output approaches infinity for any finite input. So a "magnitude" of gain does not exist, neither does a "phase".

It's true that the Bode plot for an unstable system does not give you information about its steady-state gain and phase shift (since it has no steady state to speak of), but that doesn't mean the Bode plot itself is undefined, which, if you read my post carefully, was what I was commenting on.

$H(s)\rvert_{s = j\omega}$ evaluates fine for $0 < \omega < \infty$, be it for a stable or unstable $H(s)$.