In a previous post we have gone step-by-step over how to sketch straight-line approximations for the magnitude diagram of the Bode plot for a simple rational transfer-function. In this post we cover the sketch of the phase diagram. See Section 7.1 for details on the approximations.

We consider the same second-order transfer-function:

$$G(s) = \frac{s+2}{s^2+11 s + 10}$$

which we normalize to:

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

As with the magnitude we start flat but, this time, the effect of the first pole at $s=-1$ will begin at $\omega=1/10=0.1$. Since this is a left-hand side pole, at that point there will be a $-45^\circ$/decade slope which will last until $\omega=10 \times 1= 10$ as shown in the next figure:

We can proceed to draw the sketch as the red line and stop right before the effect of the next zero at $s=-2$ begins at $\omega=2/10=0.2$:

The value at $\omega = 0.2$ can be calculated as

$$

\begin{aligned}

\Delta_y &= \text{slope} \times \Delta_x \\

&= -45 \times (\log_{10} (0.2) – \log_{10} (0.1)) \\

&= -45 \times \log_{10} (0.2/0.1) \approx -13.

\end{aligned}

$$

Because $s=-2$ is a zero on the left-hand side it will add a slope of $+45^\circ$/decade starting at $\omega=0.2$ and ending at $\omega=20$. Combining that with the existing slope of $-45^\circ$deg/decade we have a flat segment starting at $\omega=0.2$ and, recalling that the effect of the phase due to the pole at $s=-1$ ends at $\omega=10$, we obtain the following figure:

We can proceed to draw the red lines until $\omega=1$, when the effect of the last pole at $s=-10$ begins. At that point, since $s=-10$ is a pole on the left-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

$$-13 -45 \log_{10}(10/1) = -58^\circ$$

and, because of the residual slope of $+45$ in $[10, 20]$, the combined slope becomes flat again at that range before resuming at $-45^\circ$/decade at $\omega=20$ and then going flat at $\omega=100$ as shown below:

Note that the value of the phase at $\omega = 100$ should be

$$-58-\log_{10}(100/20) \approx -90^\circ.$$

However, it is not necessary to calculate this value in this way since, in this example where all poles and zeros are on the left-hand side, we expect that the value of the phase flattens out at

$$(m-n)\times 90^\circ=(1-2) \times 90^\circ = -90^\circ.$$

The resulting sketch is the red line:

Because the gain of $2/10$ is a positive real number, therefore with zero phase, there is no need to add or subtract any phase from the above sketch.

As with the magnitude, there are some easy sanity checks that you should perform on your phase sketch:

- The phase diagram will always start flat.
- It may have an offset of $-90^\circ$ ($+90^\circ$) per pole (or zero) at the origin and plus $180^\circ$ in case of a negative gain.
- It will always end flat at a phase equal to $$90^\circ \times (m_{-} + m_0 + n_{+} ) \, – \, 90^\circ \times (m_{+} + n_{-} + n_0 )$$ plus $180^\circ$ in case of a negative gain. In this formula $n_-$ and $m_-$ are the number of poles and zeros on the left-hand side, $n_0$ and $m_0$ are the number of poles and zeros on the imaginary axis and $n_+$ and $m_+$ are the number of poles and zeros on the right-hand side.

On a future post we will do consider more examples with poles and zeros on the right-hand side and complex poles and zeros.

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