How do you design a PID controller for MIMO systems

Status
Not open for further replies.
R

RITESH KAKKAR

Banned
Joined
Jul 6, 2010
Messages
263
Helped
7
Reputation
14
Reaction score
7
Trophy points
1,298
Location
INDIA
Activity points
0
Hello,
What is MIMO multi input multi out put application where it is used?
How do you design a PID controller for MIMO systems
 

can you design analog integrator??
if Vin is 200mv what will be the output?/
 

Attachments

  • output.jpg
    88.4 KB · Views: 50
  • op-amp-integrator-equation-02.gif
    1.3 KB · Views: 88

I presume you answered the question yourself with the simulation circuit.

You can also calculate the output by referring to integral calculus rules in a math handbook.

u(t) = ∫sin ωt dt = ?

Consider that term "c" (initial value) in the equation is undetermined. Furthermore OP offset voltage causes the integrator output to slowly drift away. A real integrator needs additional means to reset the output. In case of a PID controller, the feedback loop will achieve this.
 
I presume you answered the question yourself with the simulation circuit.
Click to expand...

Sin= - Cos +C

but the value magnitude is less than input it should be more as per formula.

- - - Updated - - -

A real integrator needs additional means to reset the output.
Click to expand...

What this mean?
 

but the value magnitude is less than input it should be more as per formula
Click to expand...
No, there is factor ω in the sine integral.
 

how to understand it in simple why out put is not correlated to formula?
 

RITESH KAKKAR said:
how to understand it in simple why out put is not correlated to formula?
Click to expand...
What is not correlated to formula?

It works perfect.

Vout(t)=Initial voltage in capacitor + (0.2/Π)·[cos(100·Π·t)-1]

- - - Updated - - -

The mathematical plot is the same as your output given by LTspice.

 
RITESH KAKKAR said:
How you get the formula?
Click to expand...

First: Set the polarities. Vcapacitor=-Vout

Second: write differential equation: Vcapacitor=1/RC * ∫Vin(t) dt
NOTE: In this case is an indefinite integral which means it will give me a constant value after integration

Third: Find that constant value resulted after integration by applying the Initial conditions (IC) i.e. Vcapacitor(t=0) = Initial voltage in capacitor.

Fourth: Vout=- Vcapacitor solved above
 
First: Set the polarities. Vcapacitor=-Vout
Click to expand...
How to do in simulation?

Third: Find that constant value resulted after integration by applying the Initial conditions (IC) i.e. Vcapacitor(t=0) = Initial voltage in capacitor.
Click to expand...

When i applying skip initial condition it get transient response.



Vin 200mV
freq 50hz

Vout = -200mV / 1k*10uF=-20 what this mean?
 

Attachments

  • aaa.jpg
    376.8 KB · Views: 51

Last edited:

Well you tell me because you have calculated that.. Voltage divided by seconds are volts/s.

I presume comes from the integral?
Click to expand...

ok, volts/sec

what then 20v/sec = 20msec
 

Attachments

  • aaa.jpg
    376.6 KB · Views: 48

Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…