在 LabVIEW中模拟 PID控制器 英文版

Topics：

How the PID controller works

Purpose:

By controlling the PID controller, make the water level of Q1 and Q2 is at a steady state

Description:

Q a tributary of the feed water pump

Q1 is the outflow of water from the upper reservoir (outlet to the bottom)

𝑄1(𝑡)=39ℎ1(𝑡)𝑄2(𝑡)=36ℎ2(𝑡)

0,44

Q2 is the outflow of water from the lower reservoir

0,46

V1 is the volume of liquid in the tank which surface is S1

V2 is the volume of liquid in the bottom tank bottom which area depends on the water column S2(h2)

Initial Value : 40 Final Value : 80 Step Time : 200

Involving formulas:

Diagram and Panel:

appropriate fields enter the values:

Initial Value : 40 Final Value : 80 Step Time : 200

From the library Simulation / Signal Generation drag the window block diagram Step Signal. In the

This block will be responsible for control of the pump - Control [cm3/s].

Using sum block --- Simulation / Signal Arithmetic / Summation and integral element (integrator) from the library Simulation / ContinuousLinearSystem, build differential equation:

Fig.1

in which the flow Q1 (t) will be calculated on the basis of formulas 𝑄1(𝑡)

=39ℎ1(𝑡)0,44

In the same loop Please modeled tank Center described differential equation which after substitution

Fig.2

in which the flow Q2 should be realized in accordance with the formula

0,46

𝑄2(𝑡)=36ℎ2(𝑡)

Connect the two tanks by connectingQ1 (outflow  suply). changing from from eg. 40 to 80 cm3/s at the time t = 200s.

In this exercise will be create a PID controller which is given by transmittance

Check the operation of the model, watching the liquid levels h1 and h2 at the pump capacity Q

Fig.3

Run the application in continuous operation mode. Change slider setting Select PID controller setting

Fig.4

Conclusion

1 PID parameter debugging is to adjust the control parameters (proportional gain, integral gain / time, differential gain / time) for the system to achieve the best control effect.

2 It can be seen from the formula and fig.4 that when using the P

controller, the gain of the example is large, and the output of the controller is larger under the same error. However, if the proportional gain is too large, the system will be unstable. Conversely, if the proportional gain is small, the controller will be less sensitive if its output is smaller for the same amount of error. If the proportional gain is too small, when there is interference, the control signal may not be large enough to correct the interference.

3 Integral control accelerates the system approach setpoint and

eliminates the steady-state error that can occur with a pure proportional controller. The larger the integral gain, the faster the approach to the setpoint. However, since the integral control will accumulate all past errors, it may overshoot the feedback value.

4 Differential control considers the future error, calculates the first

derivative of the error, and multiplies it by a positive constant Kd. The control of this derivative will respond to changes in the system. The larger the derivative result, the control system responds more quickly to the output. Differential control can improve setting time and system stability, but may also result in high-frequency gain and noise, so the differential control is usually coupled with a low-pass filter to limit it.