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

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: 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.

• Hi mario thank you for your detailed answer, but do you know which frequency is 73dB associated with? It is not clear to me what 20log10(48000/(0.1*100)) = 20log10(4800) = 73 dB means. Of course, without the 1/s factor, it is simply the DC gain. But with the 1/s factor, what does it mean? It is certainly not the DC gain. Dec 6 '16 at 0:54
• The 73dB are left to the point w=0.1, see the plot. Dec 6 '16 at 6:04

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.

• Yes but notice the gain is 90 dB at w = 0.1. Whereas K = 4800, and 20log10|K| = 73 dB. It doesn't tell me anything about where I should start drawing my bode magnitude plot, and nowhere can use my transfer function to tell me that it is going to be 90 dB at w = 0.1. Dec 6 '16 at 0:57
• As a rough guide, start the plot approximately a decade before the first break frequency
– Chu
Dec 6 '16 at 7:48
• As for the gain at w=0.1, the integrator contributes 20dB at this frequency and that gives a total of: 20+73=93dB. To account for the asymptotic approximation, subtract 3dB at the break frequency (w=0.1) to give overall gain of 90dB at w=0.1
– Chu
Dec 6 '16 at 7:55

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.