ME2142 Tut 2 Soln

ME2142/ME2142E Feedback Control Systems Solutions to Tutorial Problems Set 2 5th Sept 2012 1. G4

G3

R(s) +

G1

+

+

+

G2

C(s)

-

+

+

G5 G6

G4

R(s) +

G1

+

+

G2 1  G2G3

-

+

+

C(s)

G5 G6

G4

R(s) +

G1

+

+

G2 1  G2G3

-

+

+

G5 (1  G2G3 ) G2 G6

C(s)

R(s) +

G2 1  G2G3  G2G4

G1

G6 

C(s)

G5 (1  G2G3 ) G2

G1G2 1  G2G3  G2G4 C ( s)  R( s)   G6G2  G5 (1  G2G3 )  G1G2 1    G2   1  G2G3  G2G4  



2(a)

(i)

G1G2 (1  G2G3  G2G4 )  G1 (G6G2  G5 (1  G2G3 ))

With D(s)=0 and N(s)=0,

Gc ( s )G p ( s ) C ( s)  R( s ) 1  Gc ( s )G p ( s ) H ( s ) (ii)

With R(s)=0 and N(s)=0, the block diagram can be re-drawn as

and the closed-loop transfer function is G p (s) C (s)  D( s ) 1  G p ( s )Gc ( s ) H ( s ) (iii) With R(s)=0 and N(s)=0, the block diagram can be re-drawn as

and the closed-loop transfer function is  H ( s )Gc ( s )G p ( s )  H ( s )Gc ( s )G p ( s ) C ( s)   N ( s ) 1  G p ( s )(Gc ( s )) H ( s )) 1  G p ( s )Gc ( s ) H ( s )

[Note that for all three closed-loop transfer functions, the denominator remains the same.]

2(b)

Using the results obtained in 1(a) above, K Gc ( s )G p ( s ) K pK C ( s) s (s  1)    2 K pK R ( s ) 1  Gc ( s )G p ( s ) H ( s ) s  s  K p K 1 s (s  1) Kp

(i)

1 For a unit step, R ( s)  , s and so C ( s ) 

KpK

1 (s  s  K p K ) s 2

and

 KpK 1  c ss  lim s  2  1 s 0  (s  s  K p K ) s 

Since

r (t )  1(t ) , the steady-state error is then e ss  r ()  c()  1  c ss  1  1  0 K G p ( s) K C (s) s (s  1)  2   K pK D( s ) 1  Gc ( s )G p ( s ) H ( s ) s  s  K p K 1 s (s  1)

(ii)

For a unit step, D( s )  and so C ( s ) 

1 , s

K 1 (s  s  K p K ) s 2

and

 K 1  1 c ss  lim s  2  s 0  (s  s  K p K ) s  K p

Since

r (t )  0 , the steady-state error is then e ss  r ()  c()  0  c ss  

1 Kp

2(c)

Because the system is linear, the principle of superposition applies. Thus, C ( s) 

C (s) C ( s) R( s)  D( s ) R( s) D( s)

Using the results obtained in (b) above, c ss  1 

2(d)

1 Kp

The characteristic equation is obtained by equating the denominator of any of the closed-loop transfer functions to zero. Thus

1  Gc ( s )G p ( s ) H ( s)  0 1  (5)

Or

1 0 s(0.1s  1)

 0.1s 2  s  5  0  s 2  10s  50  s 2  2 n s   n2  0 2 n  10

giving

 n2  50 and

Therefore

 n  5 2 rad/s and  

1 2

 0.707

3.

Let the output of the Controller be M(s). 10 G 10   0.1s  1  10 K M ( s ) 1  GH 0.1s  1  10 K v v 1 0.1s  1

( s)

and

10 K p

 ( s) G    r ( s ) 1  GH



with

10 K p s (0.1s  1  10 K v )  10 K p 0.1s 2  (1  10 K v ) s  10 K p 1 s (0.1s  1  10 K v ) 100 K p

s 2  (10  100 K v ) s  100 K p

n  100 K p  10 K p

and



n2 s 2  2 n s  n2

 

10  100 K v 1  10 K v  20 K p 2 Kp

From the equations, it can be seen that increasing Kp will increase  n and thus the speed of response. Increasing Kv will increase the damping ratio and thus the damping in the system.

4 a). Pole-Zero locations:

(i)

(ii) 5

-1 0

-5

5

-1 0

Sys1

-5

Sys2 -5

-5

5

(iii)

-1 0

The pole at s=-1 will dominate the response. The response will be close to that obtained for Sys1.

-5

Sys3 -5

(iv) 0 .2

The response will be dominated by the pair of complex conjugate poles

P o le a t -1 0

-0 .5

-1 .0

Sys4 -0 .2

Step Responses:

Octave Program: sys1=zp([],-1,1); sys2=zp([],-10,10); sys3=zp([],[-1 -10],10); step(sys1); hold on; step(sys2); hold on; step(sys3);

Octave Program:

Sys4=tf(1,[0.1 1.11 1.12 0.111 0.01]); Step(Sys4)