- By
**zada** - Follow User

- 96 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' Control Theory' - zada

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

### ControlTheory

Bode StabilityCriterion

Other view onstability of CL

Where the PHASE of the open loop TF equals -180°(+/-n.360°), we have positive feedback.

If the AMPLITUDE RATIO at these frequencies > 0db: unstableclosed loop.

Two important measures

GAIN MARGIN

= Howmuch dB of amplitude ratio we canstilladd in the open loop before the amplitude ratio goesabove 0dB at a frequencywhere the phase crosses -180°

2. PHASE MARGIN

= ?

Phase Margin =

- How much the phase can still be increased before it reaches 0° at a frequency where the amplitude ratio is 0dB.
- Howmuch the phasecanstillbedecreasedbeforeitreaches -180° at a frequencywhere the amplitude ratio is 0dB.
- None of the above makes sense.

[Default]

[MC Any]

[MC All]

Example 1

rad/s

Given the previous Bode plot of the OPEN LOOP,

- GM = 50 dB, PM = 40°
- GM = 50 dB, PM = 90°
- GM = 30 dB, PM = 40°
- GM = 30 dB, PM = 90°
- None of the above

[Default]

[MC Any]

[MC All]

On the phase margin

The bigger the phase margin, the lessovershoot in the closed loop.

First approximation: the “damping ratio” of the closed loop = PM/100

Example:

How big do youthink the overshootwillbeif the open loop TF is

We can now state that the “disadvantage” of the I action is

- thatitincreases the OL gain at low frequencies
- thatitincreases the OL gain at high frequencies
- that it decreases the OL phase at low frequencies
- that it both decreases the OL phase and increases the OL gain at low frequencies

[Default]

[MC Any]

[MC All]

We iscanuse the stabilitycriterion to design controllers as well

- GROUP TASK 1:
- A second order processwithgain 2, damping ratio 0.5 and naturaleigenfrequency 20 rad/s is controlledwith a P controller. The time delay in the loop is 0.01s.
- What is the maximallyallowedcontrolgain
- in order for the CL to bestable
- in order for the overshoot to be smaller than 50%?

Group isTask II

Drug-inducedanasthesia

Reaction of the patient’s arterial blood pressure to a drug may vary.

Therefore a closed loop system is used. However:

Amount of drug

supplied to the patient

Desired pressure

Blood pressure

2e-sT/s

2(s+5)

Controller

Body

2/(s+2)

Sensor

- Remark: What kind of control? Why?
- Whatis the maximum time delay of the body’s response before the system willbecomeunstable?
- Determine the PM and the GM when T=0.05s? When T=0.1s?
- What is the influence of T on the step response?

Use of the Bode plot in control is

Exercise: Drug-induced anasthesia

a) Maximum T?

Zonder de dode tijd!

ωPM= 8.94 rad/s:

PM = 73.4°

Tmax = 0.1433s

Use of the Bode plot in control is

Exercise: Drug-induced anasthesia:

b) PM and GM when T=0.05s? When T=0.1s?

Zonder de dode tijd!

A- Without time delay:

PM = 73.4°

ωPM = 8.94 rad/s

B- Influence of T:

T = 0.05s: -25.6°

T = 0.1s: -51.2°

Use of the Bode plot in control is

Exercise: Drug-induced anasthesia:

c) Influence on the step response?

Download Presentation

Connecting to Server..