Bode Plots Little Tutorial – Part 1

12/11/2011 by Daniele Salatti | 0 comments

While working as a consultant I’m struggling trying to finish my studies in Computer Engineering. The Bachelor degree seems so far away…
Anyway, I’m preparing for an exam right now and I would like to share with you a bit of what I learn along the road, so I’m going to write down a little tutorial on how to draw a bode plot. The transfer function comes from an actual exercise I found in a book.

Given the following transfer function:

$G(s) = \frac{50(s+100)}{s(s+1)(s^2+10s+100)}$

Plot the Bode diagram and determine the gain and phase margins.

Step 1 – Extract the constituent parts of the transfer function

That’s easy. In practice we are going to decompose the transfer function into its constituent parts. To simplify that process, we are going to rewrite the transfer function in the Bode form.

$G(s) = \frac{50 \cdot 100(1+\frac{s}{100})}{s(1+s)100(1+\frac{1}{10}s+\frac{s^2}{100})}=50 \frac{(1+\frac{s}{100})}{s(1+s)(1+\frac{1}{10}s+\frac{s^2}{100})}$

As I said that’s an easy step. Now we need to extract the following parts from the transfer function:

A constant (K)
Poles at the origin
Zeros at the origin
Real Poles
Real Zeros
Complex conjugate Poles
Complex conjugate Zeros

1 – A constant (K)

A no-brainer. The value of K is 50.

$K = 50$

Write it down and let’s move forward.

2 – Poles at the origin

For a definition of what a pole is, please refer to the page on wikipedia. We need to find out if there are poles in zero, and of course there’s one. It’s easy to spot it at a glance: just give a look at the denominator:

$s(1+s)(1+\frac{1}{10}s+\frac{s^2}{100})$

So, we got a pole in 0, let’s write it down and move to the next step.

$P = \left\{{0}\right\}$

3 - Zeros at the origin

Again, for a definition of what a zero is, please refer to the page on wikipedia. This time we have no luck: it’s easy to see that there aren’t zeroes in 0.

Nothing to write this time…just move to the next step.

4 – Real Poles

We got one! Give a look at the denominator again:

$s(1+s)(1+\frac{1}{10}s+\frac{s^2}{100})$

We have a pole in -1.

$P = \left\{{0, -1}\right\}$

5 - Real Zeros

Looking at our numerator it is easy to spot: there’s a zero in -100.

$(1+\frac{s}{100})$

Write it down and…yeah, you guessed it: move to the next step.

$P = \left\{{-100}\right\}$

6 - Complex conjugate Poles

Finally, the fun begins! But you have to wait for this part…

To be continued…

Author: Daniele Salatti

Developer and software analyst with experience leading small development teams to build Object Oriented Internet/Intranet applications using C#, ASP.NET and MS SQL Server.

Daniele Salatti

The unhandled exception