20 Problem A9.8 OGATA 4ed(Lead compensator, Bode plot)

Laplace.T	Root-locus	Transient response	Frecuency response

20 Problem A9.8 OGATA 4ed(Lead compensator, Bode plot)
Let us calculate a compensator that satisfies the following requirements $K_{v}=20$ , phase margin 50 and gain margin $\geq 10dB$ . The open-loop system is the following:

$\begin{displaymath}G(s)=\frac{10}{s\cdot (s+1)}\end{displaymath}$
Let's calculate K y $G_{1}(jw)$

$\begin{displaymath}\lim_{s\rightarrow 0} s\cdot G(s) \cdot G_{c}(s)=K_{c}\cdot \alpha \cdot 10=K \cdot 10=20; K=\frac{20}{10}=2\end{displaymath}$

$\begin{displaymath}G_{1}(jw)=\frac{2\cdot 10}{s\cdot (s+1)}=\frac{20}{s\cdot (s+1)}\end{displaymath}$
Let's calculate Bode plot of $G_{1}(jw)$

$\begin{displaymath}20\cdot log(20)=26.02\end{displaymath}$

Gain table:

w 1

$\frac{1}{s}$ (-20) 0 (-20)

$\frac{1}{(s+1)}$ (0) 0 (-20)

(-20) 26.02 (-40)

$\begin{displaymath}dB\vert G_{1}(j0.1)\vert=26.02-20\cdot log(0.1)=46.02\end{displaymath}$

$\begin{displaymath}dB\vert G_{1}(j10)\vert=26.02-40\cdot log(10)=-13.97\end{displaymath}$

Phase table:

w 0.1 1 10

$\frac{1}{s}$ -90 (0) -90 (0) -90 (0)

$\frac{1}{(s+1)}$ 0 (-45) -45 (-45) -90 (0)

-90 (-45) -135 (-45) -180 (0)

Bode plot by Scilab
w1=[0.1 1 10];
w2=[0.1 1 10];

gdb1=20*log10(20)
gdb(1)=gdb1-20*log10(0.1)
gdb(2)=gdb1
gdb(3)=gdb1-40*log10(10)

a(1)=-90
a(2)=-135
a(3)=-180


s=%s/(%pi*2);
g=20/(s*(s+1))
gs=syslin('c',g);
w=logspace(-1,1,100);
gsf=tf2ss(gs);
[frq1,rep] =repfreq(gsf,w);
[db,phi]=dbphi(rep);

clf;
subplot(2,1,1);
plot2d(w,db,5,logflag="ln")
plot2d(w1,gdb,2,logflag="ln")
legends(['Exact','Calculated(approximated)'],[5,2],opt=1);
xgrid;
subplot(2,1,2);
plot2d(w,phi,5,logflag="ln")
plot2d(w2,a,2,logflag="ln")
xgrid;
As you see we need a lead compensator to phase margin is 50
Let's calculate the phase margin and compensator's phase ( $\phi_{m}$ )

$\begin{displaymath}dB\vert G_{1}(w_{c})\vert=dB\vert G_{1}(j1)\vert-40\cdot log (w_{c})=0\end{displaymath}$

$\begin{displaymath}log (w_{c})=\frac{26.02}{40}\end{displaymath}$

$\begin{displaymath}w_{c}=10^{\frac{26.02}{40}}=4.47\end{displaymath}$

$\begin{displaymath}\lfloor{G_{1}(j4.47)}=-135-45*log10(w_{c})=-164.27\end{displaymath}$

$\begin{displaymath}Phasemargin=180-164.278=15.72\end{displaymath}$

$\begin{displaymath}\phi_{m}=50-15.72+5\approx 40\end{displaymath}$

We calculate the $\alpha$ of compensator

$\begin{displaymath}G_{c}(s)=K_{c}\cdot \frac{s+\frac{1}{T}}{s+\frac{1}{\alpha \cdot T}}\end{displaymath}$

$\begin{displaymath}sen(\phi_{m})=\frac{1-\alpha}{1+\alpha}\end{displaymath}$

$\begin{displaymath}\alpha=\frac{1-sen(\phi_{m})}{1+sen(\phi_{m})}=0.22\end{displaymath}$

We calculate the T of compensator

$\begin{displaymath}dB\vert G_{1}(w)\vert=-20\cdot log(\frac{1}{\sqrt{\alpha}})=-6.48\end{displaymath}$

$\begin{displaymath}26.02-40\cdot log(w)=-6.48\end{displaymath}$

$\begin{displaymath}log(w)=\frac{26.02+7.19}{40}\end{displaymath}$

$\begin{displaymath}w=6.49\end{displaymath}$

This value 6.49 is the crossover gain frequency.

$\begin{displaymath}w=\frac{1}{\sqrt{\alpha}\cdot T}\end{displaymath}$

$\begin{displaymath}T=\frac{1}{\sqrt{ \alpha}\cdot w}=\frac{1}{\sqrt{0.22}\cdot 6.49}=0.32\end{displaymath}$

$\begin{displaymath}G_{c}(s)=K_{c}\cdot \frac{s+\frac{1}{T}}{s+\frac{1}{\alpha \cdot T}}=K_{c}\cdot \frac{s+3.08}{s+13.7}\end{displaymath}$

We calculate the of compensator

$\begin{displaymath}K=K_{c}\cdot \alpha; K_{c}=\frac{K}{\alpha}=\frac{2}{0.22}=8.9\end{displaymath}$

The compensator is:

$\begin{displaymath}G_{c}(s)=8.9 \cdot \frac{s+3.08}{s+13.07}\end{displaymath}$
Calculations, checks and Bode plot by Scilab:
gdb1=20*log10(20)
gdb(1)=gdb1-20*log10(0.1)
gdb(2)=gdb1
gdb(3)=gdb1-40*log10(10)

a(1)=-90
a(2)=-135
a(3)=-180

wcg=10^(gdb1/40)
awcg=a(2)-45*log10(wcg)
mf=180+awcg
pm=50-mf+5

aux1=pm*2*%pi/360;
aux2=sin(aux1);
alp=(1-aux2)/(1+aux2)
aux3=-20*log10(1/(sqrt(alp)))
nwcg=10^((gdb1-aux3)/40)
T=1/(sqrt(alp)*nwcg)
zc=1/T
pc=1/(alp*T)
kc=2/alp


s=%s/(2*%pi);
g=10/((s+0.000001)*(s+1))
gc=kc*(s+zc)/(s+pc);
gt=g*gc;
gts=syslin('c',gt);
[mg,fcp]=g_margin(gts)
[mp,fcg]=p_margin(gts)


gc2=9.5238*(s+2.9787)/(s+14.1842);
gt2=g*gc2;
gs=syslin('c',2*g);
gcs=syslin('c',gc);
gcs2=syslin('c',gc2);
gts=syslin('c',gt);
gts2=syslin('c',gt2);
clf();
bode([gs;gcs;gts;gts2;gcs2],['Gc(jw) Book';'G(jw)*Gc(jw) Book';'G(jw)*Gc(jw)'
;'Gc(jw)';'G1(jw)' ]);
Results:
-->[mg,fcp]=g_margin(gts)
 fcp  =
 
     []
 mg  =
 
   Inf  
 
-->[mp,fcg]=p_margin(gts)
 fcg  =
 
    6.4389613  
 mp  =
 
    48.099889
As you see the phase margin is 48 when we had to give 50 should have added 12 instead of 5 to calculate the $\ phi_ (m)$ . The gain margin is infinite because the phase does not cut to -180, is always above.

Let's draw the unit-step response by Scilab

Scilab program:
s=%s;
g=10/(s*(s+1))
gc=8.9*(s+3.07)/(s+13.7);
gt=g*gc;
gc2=9.5238*(s+2.9787)/(s+14.1842);
gt2=g*gc2;
glc=g /. 1;
gtlc=gt /. 1;
gtlc2=gt2 /. 1;
gs=syslin('c',glc);
gcs=syslin('c',gtlc);
gcs2=syslin('c',gtlc2);
t=0:0.1:6;
y=csim('step',t,gs);
y1=csim('step',t,gcs);
y2=csim('step',t,gcs2);
clf();
plot(t,y,'k');
plot(t,y1,'b');
plot(t,y2,'g');
xtitle('Unit-step response','t (seg)','y(t)');
legends(['G(s)';'Gc(s)*G(s)';'Book'],[1,2,3],opt=1);
xgrid;
Let's draw the unit-ramp response by Scilab

Scilab program:
s=%s;
g=10/(s*(s+1))
gc=8.9*(s+3.07)/(s+13.7);
gt=g*gc;
gc2=9.5238*(s+2.9787)/(s+14.1842);
gt2=g*gc2;
glc=g /. 1;
gtlc=gt /. 1;
gtlc2=gt2 /. 1;
gs=syslin('c',glc);
gcs=syslin('c',gtlc);
gcs2=syslin('c',gtlc2);
t=0:0.1:3;
y=csim(t,t,gs);
y1=csim(t,t,gcs);
y2=csim(t,t,gcs2);
clf();
plot(t,t,'r');
plot(t,y,'k');
plot(t,y1,'b');
plot(t,y2,'g');
xtitle('Ramp response','t (seg)','y(t)');
legends(['Ramp';'G(s)';'Gc(s)*G(s)';'Book'],[color('red'),1,2,3],opt=1);
xgrid;