Given a bode diagram

enter image description here

From the figure, I see that when $\omega = 100 \text{ rad/s}$ the magnitude response starting to go down and when $\omega = 1000 \text{ rad/s}$ the slope become higher. Then

$$H(s) = \frac{1}{\left(\frac{s}{100} + 1 \right)\left(\frac{s}{1000} + 1 \right)}$$

Then I tried to plot it on MatLab and here's what I got

enter image description here

The plot shape is close but both Y-axis are different. Where did I go wrong?


1 Answer 1


The slope of the lowpass is 20dB/decade or 6dB/octave, that means it's a simple first order lowpass filter. At the corner frequency the gain is -3dB.

Same of the phase. It goes from 0 to -90, so it's a first order filter. At the corner frequency, it's 45 degrees.

Looking at both phase and level we conclude that it's a first order lowpass with $f_c = 1000 rad/s$, hence the whole thing is simply

$$H(s) = \frac{1}{\frac{s}{1000}+1}$$


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