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