4.2.4. Analytical Design of a PID Compensator¶

This section presents an analytical procedure for PID compensator design. It is based on Section 7.11 of Phillips and Harbor Feedback_ Control Systems, Prentice Hall, 1988^[1].

The compensator transfer function is assumed to be

$\begin{equation}D(s)=\frac{K_Ds^2 + K_\mathrm{prop}\;s+K_I}{s}\end{equation}$

where $K_{\mathrm{prop}}$ is the proportional gain, $K_D$ is the derivative gain and $K_I$ is the integral gain. In this procedure we choose the PID gain parameters such that, given a desired location for one of the closed-loop poles $s_1$ , the equation

$\begin{equation}\left.D(s)G(s)H(s)\right|_{s=s_1}=-1\end{equation}$

is satisfied; that is we are designing a compensator that places a root of the closed-loop characteristic equation at $s=s_1$ .

The design proceeds as follows. First we express the desired closed loop pole position

$\begin{equation}s_1=|s_1|e^{j\psi}\end{equation}$

and

$\begin{equation}G(s_1)H(s_1)=\left|G(s_1)H(s_1)\right|e^{j\psi}\end{equation}$

Then the design equations (derived in Appendix B of Phillips and Harbor, 1988) are

$\begin{equation}K_{\mathrm{prop}}=-\frac{\sin(\beta - \psi)}{\left|G(s_1)H(s_1)\right|\sin\beta}-\frac{2K_I\cos\beta}{|s_1|}\end{equation}$

$\begin{equation}K_{\mathrm{prop}}=-\frac{\sin\psi}{|s_1|\left|G(s_1)H(s_1)\right|\sin\beta}-\frac{K_I}{|s_1|^2}\end{equation}$

Since there are three unknowns and only two relationships that must be satisfied, one of the gains may be chosen to satisfy a different design specification, such as choosing $K_I$ to achieve a certain steady-state response. These equations can also be used for PI and P+D controllers by setting the appropriate gain to zero. We now illustrate the design procedure with an example.

4.2.4.1. Example¶

Definitions (change these to change design)

The plant transfer function is

$G(s)=\frac{1}{(s+1)(5s+1)}$

imatlab_export_fig('print-svg')  % Static svg figures.

G = tf(1,conv([1 1],[5 1]));

The feedback transfer function is $H(s)=1$ :

H=tf(1,1);

So $G(s)H(s)$ is:

GH=G*H

GH =

  ---------------

  5 s^2 + 6 s + 1

Continuous-time transfer function.

The root locus of the uncompensated system is:

clf, sgrid(1/sqrt(2),0.25:0.25:2), hold on, rlocus(GH),hold off

From the root locus diagram, it is clear that for ideal damping the natural frequency of the closed-loop poles would be about 0.9 rad/s with a settling time of:

$T_s=\frac{4.6}{\zeta\omega_n}=\frac{4.6}{5/8}=7.36$ s

Suppose we wish to half the settling time then we need to double the natural frequency to $\omega_n = 2$ rad/s.

That is:

zeta = 1/sqrt(2); wn=2;  
s1 = -zeta*wn+j*wn*sqrt(1-zeta^2)

s1 =
  -1.4142 + 1.4142i

The steady state error of the uncompensated type 0 system is:

$\frac{1}{1+\left.G(s)H(s)\right|_{s=0}}=\frac{1}{1+\left.\frac{1}{(5s+1)(s+1)}\right|_{s=0}}=\frac{1}{2}$

For the compensated system, which is type 1:

$K_v=s\left.D(s)G(s)H(s)\right|_{s=0}=\left.\frac{s\left(K_D\,s^s+K_{\mathrm{prod}}\,s+K_I\right)}{s}\right|_{s=0}=K_I$

So if we want a steady-state _velocity _error of 20% we need

Ki=20;

4.2.4.2. Calculations¶

Having set up your problem, you shouldn?t need to change these commands

Polar form of $s_1$

m_s1=abs(s1),  p_s1 = angle(s1)*180/pi % degrees

m_s1 =
     2

p_s1 =
   135

Transfer function evaluated at $s_1$ is $G(s_1)H(s_1)$ in polar form:

[numGH,denGH] = tfdata(GH,'v');
GHs1=polyval(numGH,s1)/polyval(denGH,s1)

GHs1 =
  -0.0397 + 0.0610i

Magnitude:

mGHs1=abs(GHs1)

mGHs1 =
    0.0728

Phase²:

pGHs1=-angle(GHs1)*180/pi - 90 % degrees

pGHs1 =
 -213.0264

Hence:

beta = p_s1*pi/180; psi = pGHs1*pi/180;  % radians

From (5) and (6)

Kprop = (-sin(beta+psi))/(mGHs1*sin(beta)) - (2*Ki*cos(beta)/m_s1)

Kprop =
   33.1421

Kd = (sin(psi)/(m_s1*mGHs1*sin(beta))) + Ki/(m_s1^2)

Kd =
   10.2929

Compensator is therefore given by

D = tf([Kd, Kprop, Ki],[1, 0])

D =

  10.29 s^2 + 33.14 s + 20

  ------------------------

Continuous-time transfer function.

4.2.4.3. Evaluation of Design¶

Open loop transfer function:

Go=D*GH

Go =

  10.29 s^2 + 33.14 s + 20

  ------------------------

     5 s^3 + 6 s^2 + s

Continuous-time transfer function.

4.2.4.3.1. Root locus:¶

rlocus(Go)

4.2.4.3.2. Closed-loop transfer function:¶

DG = D*G  
Gc = feedback(DG,H)

DG =

  10.29 s^2 + 33.14 s + 20

  ------------------------

     5 s^3 + 6 s^2 + s

Continuous-time transfer function.

Gc =

      10.29 s^2 + 33.14 s + 20

  --------------------------------

  5 s^3 + 16.29 s^2 + 34.14 s + 20

Continuous-time transfer function.

4.2.4.3.3. Step response:¶

step(Gc)

4.2.4.4. Footnotes¶

[1] The proofs of the formulae given are derived in Appendix B of this text.

[2] You must be careful with angles when using packages like MATLAB, and indeed pocket calculators. It is nearly always beneficial to have a sketch so that you can correct the results. In this case a correction of $-90^\circ$ was needed.

4.2.4.5. Resources¶

An executable version of this document is available to download as the MATLAB Live Script file analrloc.mlx.

EGLM03 Modern Control Systems