Step-by-step Bode plot example. Part III

We have gone step-by-step over how to sketch straight-line approximations for the magnitude and phase diagrams of the Bode plot for a simple rational transfer-function. This transfer-function had only left-hand side poles and zeros, that is it was minimum-phase (see Section 7.2). In this post we consider a non-minimum phase transfer-function with a right-hand side zero. See Section 7.1 for details on the approximations.

We consider a variation on the second-order transfer-function we addressed before:

$$G(s) = \frac{s+2}{s^2-9s -10}$$

The only difference it that the pole at $s=-10$ has been moved to the right-hand side to $s=10$. This is more evident after factoring and normalizing

$$G(s) = \frac{-2}{10} \frac{1 + s/2}{(1 + s)(1 – s/10)}.$$

Note that the above change also affects the sign of the gain, which now has become negative.

Because the sign of the gain and the poles or zeros have no effect in the magnitude of the frequency response, the magnitude straight-line approximation is the same as the one obtained before in this post:

As for the phase, we can proceed as in this post to construct the phase diagram up until $\omega =1$, at which point the effects of the pole at $s=10$ begin:

Because $s=10$ is a pole on the right-hand side, a slope of $+45^\circ$/decade is added that lasts until $\omega=100$. At $\omega=10$ the value of the phase will be

$$-13 +45 \log_{10}(10/1) = 32^\circ$$

and, because of the residual slope of $+45$ in $[10, 20]$, the combined slope becomes $+90^\circ$ at that range, which leads to a phase value of

$$32+90\log_{10}(20/10) \approx 59$$

at $\omega=20$. At that point, the slope is back at $+45^\circ$/decade and then goes flat at $\omega=100$ as shown below:

Because the gain $-2/10$ is negative, an additional phase of $-180^\circ$ is added to obtain the red line sketch as in the next figure:

Note that we expect that the phase reaches $-90$ at $\omega=100$ because we have $n_- = 1$ poles one the left-hand side, $n_+=1$ poles on the right-hand side, and $m_-=1$ zeros on the left-hand side so that

$$-180^\circ + 90^\circ \times (m_- + n_+) \, – \, 90^\circ \times (n_-) = -180^\circ + 180^\circ – 90^\circ=-90^\circ$$

See this previous post for a discussion on what phase to expect for large values of $\omega$. If you are wondering why we subtracted $180^\circ$ and not added $180^\circ$ to account for the negative gain you are not alone. We will discuss that in a future post.