Introduction¶

Basic Ideas¶

Besic Concepts

被控变量(Controlled Variable, CV)
设定值(Setpoint, SP)
操纵变量(Manipulated Variable, MV). Directed controlled by actuator and influence CV.
扰动变量(Disturbance Variables, DV). Variables that also influence CV and is not influenced by MV.

Dynamic Characteristics of Process¶

Typical Process¶

自衡过程(self-balanced process)

When input changes, process can converge to another equilibrium. This process includes 纯滞后过程, 单容过程 and 多容过程.

纯滞后过程、单容过程 and 多容过程

纯滞后过程

\[ G(s)=e^{-\tau s} \]

单容过程

\[ G(s)=\frac{K}{1+T s}e^{-\tau s} \]

多容过程

we use one order or two order model to approximate high order object

\[ G(s)=\frac{K}{(1+T_1 s)(1+T_2 s)}e^{-\tau s} \]

非自衡过程(Non-self-balanced process)

积分过程, 开环不稳定

积分过程

we change valve into 计量泵, then

\[ A\frac{dH}{dt}= q_i-q_o \]

making Laplace transformation and we get

\[ \frac{H}{Q_i}=\frac{1}{As} \]

开环不稳定

\[ G(s)=\frac{|K|}{|T|s-1} \]

Mechanisim Modelling¶

There are typical steps.

List Differential equations by physical mechanisim.
find its static point.
find dynamic relationship at this static point. Linearize the model if necessary.

Q1. Model dynamic process of liquid tank.

Answer

we know that

\[ A\frac{dh}{dt}=Q_i-Q_o \]

with \(Q_O=k\sqrt{h}\), where \(A\) is the cross area of the tank, \(k\) is the coefficient of the pipe. So

\[ A\frac{dh}{dt}=Q_i-k\sqrt{h} \]

Now we have to linearize it.

at static point \(t=0\), we have \(dh/dt=0\), so \(Q_{i0}=Q_{o0}\). Denote

\[ \begin{align*} \Delta h&=h-h_0 \\ \Delta Q_i&=Q_i-Q_{i0}\\ \Delta Q_o&=Q_o-Q_{o0} \end{align*} \]

So

\[ \begin{align} A\frac{d\Delta h}{dt}=\Delta Q_i-\Delta Q_o \label{eqh} \end{align} \]

Use Taylor extansion to get the linear part.

\[ \begin{align*} \Delta Q_o&=k\sqrt{h} - Q_{o0}\\ &\approx \frac{d Q_o}{dt}\Bigg|_{h=h_0} \Delta h \\ &= \frac{k}{2\sqrt{h_0}}\Delta h \end{align*} \]

So back in euqation \(\ref{eqh}\), we have

\[ A\frac{d\Delta h}{dt}=\Delta Q_i- \frac{k}{2\sqrt{h_0}}\Delta h \]

Make Laplace Transiformation and we get

\[ AsH(s)=Q_i(s)-\frac{k}{2\sqrt{h_0}}H(s) \]

If we denote \(R=\frac{2\sqrt{h_0}}{k}\), we get

\[ G(s)=\frac{H(s)}{Q_i(s)}=\frac{R}{RAs+1} \]

Detection Part¶

This Detection part can usually be described as an one order model

\[ G_m(s) = \frac{K_m}{T_m s+1}e^{-\tau_m s} \]

Valve Part¶

气动阀的气开、气关选择

气开、气关阀

气开阀: 有气则开，无气则关
气关阀: 无气则开，有气则关

Criteria: Without signal, should the valve be closed or open.

Characteristics

A valve is composed by two parts in series.

\[ G(s)=G_{v1}(s)\times G_{v2}(s) \]

where \(G_{v1}(s)\) denotes the relationship between the input signal and the displacement of valve core, which is usually one order model and \(G_{v2}(s)\) denotes the relationship between the displacement and the output flow, which is usually non-linear and can be used to compensate some non-linear part in the control system.

Here we talk about some details about the latter relationship. Firstly, we denote the flow characteristic \(f\) to be

\[ \frac{Q}{Q_{max}}=f\left(\frac{l}{L}\right) \]

where \(\frac{Q}{Q_{max}}\) denotes the relative flow and \(\frac{l}{L}\) denotes the relative displacement of the valve core. And we define \(R=\frac{Q_{max}}{Q_{min}}=30\). The following graph is acquired when the pressure difference between the front and the back of the vavle stays the same.

Typical Valves

(i) Linear valve

\[ \frac{Q}{Q_{max}}=K\frac{l}{L}+C \]

with \(K=\frac{R-1}{R}\) and \(C=\frac{1}{R}\).

(ii) Equipercent valve

\[ \frac{Q}{Q_{max}}=R^{\left( \frac{l}{L}-1 \right)} \]

which is deduced by ODE

\[ \frac{d(Q/Q_{max})}{d(l/L)}=K\frac{Q}{Q_{max}} \]

(iii) Quick-Opening valve

\[ \frac{Q}{Q_{max}}=\frac{1}{R}\left[1+(R^2-1)\frac{l}{L}\right]^{1/2} \]

which is deduced by

\[ \frac{d(Q/Q_{max})}{d(l/L)}=K\left(\frac{Q}{Q_{max}}\right)^{-1} \]

(iv) ParaBola valve

\[ \frac{Q}{Q_{max}}=\frac{1}{R}\left[1+(\sqrt{R}-1)\frac{l}{L}\right]^{2} \]

which is deduced by

\[ \frac{d(Q/Q_{max})}{d(l/L)}=K\left(\frac{Q}{Q_{max}}\right)^{1/2} \]

When we install a valve, the pressure would not be constant, cause the resistence of the pipe. Usually the curve would bend left up, which is determined by Pressure-drop ratio

\[ s=\frac{\Delta P|_{l=L}}{P} \]

if \(s=1\), then the curve does not change.

Define the gain of valve to be

\[ K_v=\frac{\Delta Q/Q_{max}}{\Delta l/L} \]

which is the tangent of the above graph.

General Object and its modelling¶

Test Method

For one order object with pure delay, we can use a phase step as an input and use the response to get its corresponding model.

Method for One order object with pure time delay

We use Two-Point Method usually.

(i) Gain

\[ K=\frac{O_f-O_i}{I_f-I_i} \]

with units.

(ii) time constant \(T\) and time daley constant \(\tau\)

choose \(y_{283}(t_1)\) and \(y_{632}(t_2)\) and then

\[ T=1.5\times (t_2-t_1),\quad \tau=t_2-t_0-T \]

where \(t_0\) denotes the start time of input step function.

PID Parameters Determination¶

After getting a one order model of the system, i.e. \(K\), \(T\), \(\tau\), then we can use them to get PID controller parameters.

响应曲线法¶

ZN method: only applies to \(0<\tau<T\)

控制器类型	K_C	T_i	T_d
P	\(T/(\tau K)\)	\(\infty\)	\(0\)
PI	\(0.9T/(\tau K)\)	\(3.33\tau\)	\(0\)
PID	\(1.2T/(\tau K)\)	\(2.0\tau\)	\(0.5\tau\)

\(\lambda\) method: all application

控制器类型	K_C	T_i	T_d
P	\(T/(\tau K)\)	\(\infty\)	\(0\)
PI	\(T/(2\tau K)\)	\(T\)	\(0\)
PID	\(T/(1.2\tau K)\)	\(T\)	\(0.5\tau\)

Both method are same for \(T_d\) and P controller.

Others¶

经验法
临界比例度法

在闭环下，先使用纯比例控制，从小到大调节\(K_C\)，对于给定\(K_C\).使用小幅度阶跃输入，使得等幅振荡下，记录周期\(P_u\)和此时的比例增益\(K_{Cmax}\)，求出对应参数。

注: 该方法以得到4：1衰减比，合适超调为目标。

衰减曲线法: 针对纯比例闭环下得不到等幅振荡的情况

注：和经验法结合使用