20220725 compensator in automatic control principle

Bode Definition of amplitude and frequency of graph

Amplitude is defined as $$20\log |G(j{\omega }_i)|$$

therefore ,0 dB Refers to the constant amplitude .

in addition , The starting point , namely ${\omega }_i=0$ when , You can use $s=0$ Calculation dB Count .

tail 1 form , Help regular memory

Write a tail 1 Forms are easier to remember rules .

First order system $G(s)=\frac{1}{\frac{1}{a}s+1}$,$a$ It's the cut-off frequency

The starting point ：$G(s)=1$, namely 0 dB.
Guess what ： When ${\omega }_i$ smaller , The starting point is dominant ; When ${\omega }_i$ more , Degenerate to $G(s)=\frac{1}{s}$, That is, the integral link , The amplitude of the integration link is a slope -20 dB The slash of , lagging 90°.

When $a=5$ rad/sec,Bode The graph is as follows

When $a=10$ rad/sec,Bode The graph is as follows

notes 1： The higher the cut-off frequency , The fidelity of high-frequency signals will be better , That is to say, the reaction ability of the system is faster . Give a step , The shorter the adjustment time .

notes 2： On the left is from 0 dB At the beginning , In line with the previous conjecture .

PD System $G(s)=\frac{1}{a}s+1$

The starting point ：$G(s)=1$, namely 0 dB.
Guess what ： When ${\omega }_i$ smaller , The starting point is dominant ; When ${\omega }_i$ more , Degenerate to $G(s)=s$, That is, the differential link , The amplitude of the differential link is a slope 20 dB The slash of , leading 90°.

When $a=5$ rad/sec,Bode The graph is as follows

Lead compensator $G(s)=\frac{s+5}{s+10}$

Write the tail first 1 form , namely $$G(s)=\frac{5}{10}(\frac{1}{5}s+1)\frac{1}{\frac{1}{10}s+1}$$ This is pure proportion +PD+ First order system

Guess link ： When ${\omega }_i$ Very hour , Pure proportion works , That is, the amplitude is $20\log \frac{5}{10}$; When ${\omega }_i=5$ and ${\omega }_i=10$,PD And the first-order system respectively .

$G(s)=\frac{5}{10}$ Of Bode The graph is as follows ：

$G(s)=\frac{1}{5}s+1$ Of Bode The graph is as follows ：

$G(s)=\frac{1}{10}s+1$ Of Bode The graph is as follows ：

$G(s)=\frac{s+5}{s+10}$ Of Bode Figure is the superposition of the above three figures , as follows ：

Hysteresis compensator $G(s)=\frac{s+10}{s+5}$

formula ： Write a tail 1 form , Find the key frequency , Denominator amplitude down , The molecular amplitude is up .
skill ： For minimum phase systems ,Phase It probably shows Magnitude The slope of .