Step-by-step Bode plot example. Part I

In this post we will go over the process of sketching the straight-line Bode plot approximations for a simple rational transfer-function in a step-by-step fashion. See Section 7.1 for details on the approximations. We will start with the magnitude plot and cover the phase plot in a future post.

Consider the following second-order transfer-function:

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

The first step to produce a Bode plot sketch is to factor the numerator and denominator in terms of its poles and zeros:

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

You might need the help of a numeric calculator here for large order transfer-functions. In this example all poles and zeros are real. We will consider the case of complex-poles and zeros in another post.

Once the poles and zeros have been factored, normalize the numerator and denominator:

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

This step is very important and not performing it before proceeding with the sketch is a common mistake that leads to an incorrect gain.

We proceed to identifying the pole or zero with smallest magnitude, which is this case is the pole at $s=-1$. Such a pole will show up on the magnitude Bode plot as straight-line approximation that is $0$ before $\omega=1$ and have a $-20$dB/decade slope after the pole, as shown in the figure:

After noticing that the next pole or zero happens at $s=-2$ we can proceed to draw the sketch as the red line in the next diagram, stopping right before $\omega=2$.

The value of $-6$ marked above can be calculated by using a standard $\Delta_y = \text{slope} \times \Delta_x$ argument:

$$\Delta_y = -20 \times (\log_{10}(2) -\log_{10}(1))= -20 \log_{10}(2/1) \approx -6.$$

Past $\omega=2$, it is necessary to take into account the contribution of the zero at $s=-2$, which adds to the current slope of $-20$dB/decade its own slope of $+20$dB/decade. The resulting slope of $-20+20=0$dB/decade is the straight line in the next figure:

We then proceed with the red line until right before $\omega=10$, at which point the second pole at $s=-10$ starts contributing $-20$dB/decade, as shown next:

Since there is no other pole or zero, we conclude the diagram by following the $-20$dB/decade slope line initiated at $\omega=10$:

Last but not least, we need to offset the diagram to take into account the gain of $2/10$, which corresponds to

$$20 \log_{10}(2/10)\approx-13\text{dB}$$

to obtain the final sketch:

Finally, here are some sanity checks that you should perform on your sketch:

  1. If there are no poles nor zeros at the origin then your magnitude diagram will start flat at $20 \log_{10} |G(0)|$. In this example $20 \log_{10}|2/10|\approx -13$.
  2. It will always end with a slope of $-20(n-m)$dB/decade, where $n$ is the number of poles and $m$ the number of zeros. In this example, $n=2$, $m=1$, and the plot ends with a $-20(2-1)=-20$dB/decade slope.

We will consider the sketch of the phase on a followup post.

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