Szabályozástechnika - LaborZH, 2008. 12. 08., megoldással (angol)

← Vissza az előző oldalra – Szabályozástechnika_(info)#Labor_ZH

ebből a pdf-ből átalakítva: Control theory Exam, December 8, 2008

1. Given the following control loop:

http://i.imgur.com/tYiCFHi.png

$C(s)=0.4\cdot {\frac {1+5s}{s}}$

$P(s)={\frac {1}{(1+5s)(1+s)(1+0.2s)}}$

a./ Give the gain margin, phase margin and cut-off frequency of the system. Is the closed loop stable? (4 points)

In case of $r(t)=0$ and $y_{z}(t)=1(t)$,

2. In a sampled data system the transfer function of the process is:

$P(s)={\frac {0.4}{s^{2}-0.8s-0.2}}\mathrm {e} ^{-2s}$. The sampling time is $T_{s}=1$.

3. A continuous process is given by its state matrices:

$A=\left[{\begin{array}{rr}-1&1\\0&-8\end{array}}\right]$, $b=\left[{\begin{array}{r}1\\1\end{array}}\right]$, $c=\left[{\begin{array}{rr}2&0\end{array}}\right]$, $d=0$

4. The transfer function of a continuous plant is ${\frac {1}{(1+s)(1+10s)}}.$

In case of $u(t)=sin(0.5t)$ input signal the steady state output signal is $y(t)=Asin(0.5t+\phi )$. Give the value of A and φ. (5 points)

1.

a./

 s=zpk('s')
 C=0.4*(1+5*s)/s
 P=1/((1+5*s)*(1+s)*(1+0.2*s))
 L=C*P
 L=minreal(L)
 figure(1)
 margin(L)

--> pm=65º, wc=0.37rad/sec, Gm=23.5dB,

 gm=margin(L)

--> gm=15

 Tz=P/(1+L),
 figure(2)
 [y,t]=step(Tz)
 plot(t,y)
 grid

--> y_vég=0, ts == 20 sec

http://i.imgur.com/u332zJs.png

2.

a./

 s=tf('s');
 P=0.4/(s*s-0.8*s-0.2);
 P=zpk(P);
 %P=tf(P)%P=2/((s-1)*(5*s+1)), P=tf(P)
 Ts=1;
 Td=2;
 d=Td/Ts;
 z=zpk('z',Ts);
 G1z=c2d(P,Ts);
 Gz=G1z/(z^d)

 % Zero/pole/gain:
 %    0.27065 (z+1.302)
 % ------------------------
 % z^2 (z-0.8187) (z-2.718)

b.

Unstable, the absolute value of a pole is greater than 1.

c.

 y=step(Gz,5)
 
 % y =
 %          0
 %          0
 %          0
 %     0.2706
 %     1.5802
 %     5.6099

---

3.

 a=[-1, 1;0, -8], b=[1;1],c=[2, 0], d=0
 T0=0.2;
 kszi=0.7;
 den=[T0*T0, 2*T0*kszi,1],
 pc=roots(den)
 %   den =
 %       0.0400    0.2800    1.0000
 %  
 %   pc =
 %     -3.5000 + 3.5707i
 %     -3.5000 - 3.5707i

 k=acker(a,b,pc), G=1/dcgain(a-b*k,b,c,d)
 %   K= 2.3750 -4.3750
 %   G = 1.3889

4.

 s=zpk('s');
 P=1/((1+1*s)*(1+10*s));
 w=0.5;
 [m,f]=bode(P,w);
 A=m
 fi=f

--> A=0.1754 --> fi=-105.2551

--Haraszin Péter (vita) 2013. május 16., 13:53 (UTC)