3.5. Analytical Root-Locus Design of Phase-Lead Compensators¶

This MATLAB Live Script presents an analytical procedure for phase-lead design. It is based on Section 7.8 of Phillips and Harbor Feedback Control Systems, Prentice Hall, 1988^[1]. For the procedure it is convenient to write the compensator transfer function as

$\begin{equation}D(s)=\frac{a_1 s + a_0}{b_1s + 1} \end{equation}$

In this procedure we choose $a_1$ , $a_0$ , and $b_1$ such that given $s_1$ , the equation

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

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

In equation (2) we have four unknowns, including $K$ , and only two relationships (magnitude and phase) that must be satisfied. Hence, we can arbitrarily assign values to two of the unknowns. $K$ is easily eliminated since

$KD(s)=\frac{K a_1 s + Ka_0}{b_1 + 1}$

so if we assume that $K=1$ for the design procedure we eliminate one of the unknowns. The other unknown that can be eliminated is $a_0$ which can be seen to be the DC gain of the compensator. Its value can therefore be chosen to satisfy the steady-state error requirements of the design and we need only to determine values for $a_1$ and $b_1$ .

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

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

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

$\begin{split}\begin{eqnarray} {a_1} &=& \frac{\sin \beta + {a_0}\left| G(s_1)H(s_1) \right|\sin (\beta - \psi )}{ \left| s_1 \right|\left| G(s_1)H(s_1)\right|\sin \psi}\\ {b_1} &=& \frac{\sin (\beta - \psi ) + a_0\left| G(s_1)H(s_1) \right|\sin \psi }{ - \left| s_1 \right|\sin \psi } \end{eqnarray}\end{split}$

Given $a_0$ , $G(s)H(s)$ , and the desired closed-loop pole location $s_1$ , (5) and (6) give the remaining compensator coefficients. This procedure places a closed-loop pole at $s=s_1$ ; however, the locations of the remaining poles are unknown and may be unsatisfactory. In fact, some may be unstable!

For the case that $\psi$ is either $0^\circ$ or $180^\circ$ , equations (5) must be modified to give the single equation

$\begin{equation}a_1|s_1|\cos\beta \pm \frac{b_1|s_1|}{|G(s_1)H(s_1)|} \pm \frac{1}{|G(s_1)H(s_1)|}+a_0=0\end{equation}$

where the plus sign applies to the case $\psi = 0^\circ$ and the minus sign applies to $\psi=180^\circ$ . For this case, the value of either $a_1$ or $b_1$ can also be assigned. An example is now given to illustrate the procedure.

3.5.1. Example¶

An executable version of this document is available as a MATLAB Live Script analrloc.mlx. You can use it to design a Lead Compensator for other systems by downoading that script and changing the set-up parameters.

3.5.1.1. Definitions (change these to change design)¶

The plant transfer function is :

G = tf(1,[1 0 0]);

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

H = tf(1,1);

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

GH=series(G,H)

GH =

---

s^2

Continuous-time transfer function.

The desired closed-loop poles are:

s1 = -2 + 2j;

Now the DC gain of this type 2 system will be:

$\begin{split}\begin{eqnarray*}K_a &=& \left. s^2D(s)G(s)H(s) \right|_{s = 0}\\ &=& \left. s^2\frac{a_1 s + a_0}{b_1 + 1} \times \frac{1}{s^2} \right|_{s = 0}\\ &=& a_0.\end{eqnarray*}\end{split}$

For the purpose of illustration let us arbitrarily take a value of $a_0=8/3$ :

a0 = 8/3;

3.5.1.2. Calculations¶

(You shouldn’t need to change these commands)

Polar form of $s_1$

m_s1=abs(s1),  p_s1 = (angle(s1)*180/pi + 90) % degrees

m_s1 =
    2.8284

p_s1 =
   225

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

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

GHs1 =
   0.0000 + 0.1250i

Magnitude:

mGHs1=abs(GHs1)

mGHs1 =
    0.1250

Phase:

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

pGHs1 =
   -90

Hence angles are:

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

beta =
    3.9270

psi =
   -1.5708

From (5)

a1 = (sin(beta) + a0*mGHs1*sin(beta - psi))/(m_s1*mGHs1*sin(psi))  
b1 = (sin(beta + psi) + a0*mGHs1*sin(beta))/(-(m_s1)*sin(psi))

a1 =
    2.6667

b1 =
    0.1667

Compensator is therefore given by

numD = [a1, a0], denD = [b1, 1]

numD =
    2.6667    2.6667

denD =
    0.1667    1.0000

which in normal form:

$D(s)=K_c\left(\frac{s+z_1}{s+p_1}\right)$

has

Kc = a1/b1, z0 = a0/a1, p0 = 1/b1

Kc =
   16.0000

z0 =
     1

p0 =
    6.0000

Now make a transfer function

D = tf(Kc*[1, z0],[1, p0])

D =

  16 s + 16

  ---------

    s + 6

Continuous-time transfer function.

3.5.1.3. Evaluation of Design¶

Open loop transfer function:

Go = series(D,GH)

Go =

   16 s + 16

  -----------

  s^3 + 6 s^2

Continuous-time transfer function.

Root locus:

rlocus(Go)

Closed-loop transfer function:

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

DG =

   16 s + 16

  -----------

  s^3 + 6 s^2

Continuous-time transfer function.

Gc =

         16 s + 16

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

  s^3 + 6 s^2 + 16 s + 16

Continuous-time transfer function.

Step response:

step(Gc)

As an exercise, you should examine the effect of designing for a range of DC gains in the range $0\le K_a\le10$ .

3.5.2. Footnotes¶

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

EGLM03 Modern Control Systems