Steps to hand draw Bode magnitude plot for a particular transfer function

Question

Can anyone guide me through how to hand draw a correct Bode magnitude plot for

$$G(s) = \dfrac{48000}{s(s+0.1)(s+100)}$$

• Don't even need phase plot, just need correct magnitude plot
• As few calculation needed as possible, in other words, just use rules for poles and zeros (pole -> -20dB/Dec, zero -> +20dB/Dec) instead of calculating raw numbers

The MATLAB produced Bode plot can be seen here:

Answer 1

First check the transfer function for poles and zeros. In this case the numerator is a constant, so there are no zeros. The poles are the zeros of the denominator. By inspection we see that they are at s=0, s=-0.1, s=-100. For each pole we get additional -20dB/dec and additional -90 degrees phase shift.

The gain starts to change at the corresponding positive value of the pole. E.g. for s=-100 we will see a change at w=100. The phase changes from about one decade before this frequency to about one decade above this frequency by -90 degrees. Exactly at the corresponding frequency the change because of this pole is -45 degrees.

Before we can draw the diagram we need to find a starting point. This is a little bit tricky because of the 1/s term.

We can rewrite the equation and deal with the parts separately $$ G(s) = \dfrac{48000}{s(s+0.1)(s+100)} = \frac 1s \cdot \dfrac{48000}{(s+0.1)(s+100)} $$ Now we have a 1/s-term that is infinity at zero and 1 at w=1. It will decrease by 20dB per decade. Now we look at the other part $$ \dfrac{48000}{(s+0.1)(s+100)} $$ For s=0 the gain is 48000/(0.1*100) = 4800. At w=0.1 it will start to decrease by 20dB/dec, at w=100 it will start to decrease by 40dB/dec.

Now we know everything about the components and can construct the Bode plot.

The result looks as shown below. The 1/s term is red. The other term is blue and the sum of these two is yellow.

Starting at w=1, we have 0dB for the 1/s term.

The other term has 20*log10(4800) ~ 73dB left to the corner frequency and at w=1 it is 20dB below that value, so we have 53dB.

This is our first point! At this point we have a slope of -40dB/dec. This slope will extend one decade to the left (so we have 93dB there) and then continue with -20dB/dec. To the right it will extend up to w=100 and then continue with -60dB/decade. At w=100 the magnitude is 53dB - 2decades*40dB/dec = -27dB.

Answer 2

Write it in the form \$\dfrac{K}{s(1+s/\omega_1)(1+s/{\omega_2})}\$, then you have the breakpoints and the correct gain term.

Answer 3

As few calculation needed as possible, in other words, just use rules for poles and zeros

• Start with setting s = 0 then clearly (at 0 Hz) the gain is infinite
• At s = infinity the response is clearly zero

There are two more to calculate by setting s = -0.1 (this results in a pole along the sigma axis) and s= -100, resulting in another pole along the sigma axis.