ME2142 Tut 3 Soln

ME2142/ME2142E Feedback Control Systems Solutions to Tutorial Problems Set 3 March 2007

Motor 

Question 1 R(s) +

K m

Gc(s)

s (T m s  1)

-

Gc ( s )  K  p 

(a)

K i s

C(s)

. with K m  0.5 and T m  0.5 s.

Characteristic equation is 1  GH   1  3

K  p s  K i

0.5

s

s (0.5s  1)

 1

K  p s  K i 2

s ( s  2)

0

2

 s  2 s  K  p s  K i  0

Routh Array s3

1

K  p

s2

2

K i

2 K  p  K i

s1

2 s0

For stability,

(b)

2 K  p  K i 2

0

K i

and K i  0 , giving K i  0

and  2 K  p  K i .

With K  p  2 , the characteristic equation becomes 3

2

s  2 s  2s  K i  0

Or

1

K i s ( s  2 s  2)

 1

(P1.1) K i

s ( s  1   j1)( s  1   j1)

0

The open-loop transfer function has three poles (at s=0, s=-1-j1, s=-1+j1) and no zero.

For the root locus,

asymptote angles =

Real-axis intercept =

 n180

3

 60, 180 ,

n   1, 3,

 P   Z   2   0.67 .  N p  N z

Imaginary axis intercept:

3

From (a), system is limitedly stable when 2 K  p  K i , or K i  4 . Since K  p  2 .

On the imaginary axis, s   j . Substituting this into (P1.1), we have 3

2

  j   2   j 2   4  0 .

Equating either real or imaginary parts to zero gives    imaginary axis intercept.

2  1. 414 rad/s at the

Locating a test point very close to the pole at s=-1+j1, the angles of the vectors directed  o o o from the other two poles to this pole are 90 and 135 giving a total angle of 225 . Since o the sum of the angles made the vectors from all three poles to this test point must be 180 , angle of departure from the pole at s = -1+j1 is thus 180  90  135   45 The root locus plot is thus as shown RootLocus 3

2

1.414 1

60o

  s    i   x    A 0   g   a   m    I

-0.67

-1

-2

-3 -4

-3

-2

-1 RealAxis

0

1

2

Question 2

(a)

The characteristic equation for the open-loop transfer function is given by equating the denominator of G(s) to zero. Thus

2

3

2

( s  4 s  8)( s  3)  s  s  4 s  24  0

Since not all the coefficients of the equation are of the same sign, the open-loop system is unstable. (b)

The characteristic equation for the closed-loop systems is 1  G ( s)  0

This gives

or 

1

Ks ( s  2) 2

( s  4 s  8)( s  3)

0

( s 2  4 s  8)( s  3)  Ks( s  2)  0

Or  3

2

s  ( K  1) s  (2 K  4) s  24  0

(P2.1)

The Routh Array:

s

3

1

(2K -4)

s

2

(K -1)

24

s1

(2 K 2  6 K  20) ( K  1)

s

0

24

For the system to be stable, all the entries in the first column must have the same sign, or  ( K  1)  0 Or

and  (2 K 2  6 K  20)  0 K   1 and

2(K  5)( K  2)  0

Thus, for the system to be stable, K   5 .

(c)

When the system is marginally stable, K   5 and there will be a pole, or a pair of  complex conjugate poles on the imaginary axis, meaning that s  j  .

Substituting these into the characteristic equation (P2.1), we have ( j )3  4( j ) 2  j 6   24  0 Equating either real of imaginary parts to zero, we have

(d)

  

6   2.45 rad/s.

(i)

The open-loop zeros are at s  0,  2 and the poles at s   3, (2  j 2), (2  j 2) .

(ii)

On the real axis, the loci occurs in the regions (0, -2) and ( 3, )

(iii) There is only one asymptote at 180°.

Root Locus

2.5

2

1.5

1

0.5

  s    i   x    A   g   a   m    I

0

-0.5

-1

-1.5

-2

-2.5 -6

-5

-4

-3

-2 Real Axis

-1

0

1

2

Question 3

(a)

Characteristic Equation is 1 3

10( s  a )

0

s ( s  1( s  8) 2

Or 

s  9 s  18s  10a  0

Giving

1

10a 2

s ( s  9 s  18)

 1

(P3.1) 10a s( s  3)( s  6)

0

Letting K  10a , we plot the root locus for K varying from 0 to infinity. (i) There are three open-loop poles at s=0, -3, -6 and thus there are three loci. (ii) The loci occurs on the real axes in the regions (0, -3) and ( 6,  ) . (iii) The asymptotes are at   

n180

3

 60,180

  

(iii) The asymptotes intercept the real axes at

(iv) The break-away point is found from 3

Or

dK  ds

ds

 N p  N z

3

0

2

K  ( s  9 s  18 s)

From equation (P.3.1), Giving

dK 

 P   Z   3  6  3

2

 (3s  18s  18)  0

s = -1.268 or -4.732

Since the loci does exist only in the regions (0, -3) and ( 6,  ) , s=-1.268 is the break-away point. (v)

Imaginary axis crossing: Letting s  j  in the characteristic equation, we have, from Equation (P3.1), 3

2

 j  9  j18   K   0

Equating imaginary parts to zero, we have

   

18   3 2 or

  

0.

Equating real parts to zero, we have K   9 2  162 or  K   0 .

Root Locus

10

5

  s    i   x    A   g   a   m    I

0

-5

-10

-15

-10

-5

0

5

Real Axis

(b)

System is stable for  0  K   162 or  0  a  16.2 .

(c)

Using Equation (P3.1), the Routh Array is

s3

1

18

2

9

10a

1

162  10a

s s

9 s0

10a

For there to be no roots with positive real parts, all the entries in the first column must have the same sign. Thus (162  10a )  0 and Or

0  a  16.2

10a  0

Question 4 Root Locus 2

1.5

1

(a)

G ( s ) H ( s ) 

K 

0.5

s ( s  1)

  s    i   x    A

0

  g   a   m    I

-0.5

-1

-1.5

-2 -3

-2.5

-2

-1.5

-1

-0.5

0

0.5

Real Axis

Root Locus

4

3

2

1

(b)

G ( s ) H ( s ) 

K  s ( s  1)( s  2)

  s    i   x    A 0   g   a   m    I -1

-2

-3

-4

-6

-5

-4

-3

-2

-1

0

1

Real Axis

Root Locus 15

10

5

(c)

G ( s ) H ( s ) 

K ( s  4) s ( s  1)( s  2)

  s    i   x    A   g   a   m    I

0

-5

-10

-15 -4

-3.5

-3

-2.5

-2 Real Axis

-1.5

-1

-0.5

0

(d)

G ( s ) H ( s ) 

K  s ( s  1)( s  4)( s  6) Root Locus

8

6

4

2

  s    i   x    A 0   g   a   m    I -2

-4

-6

-8 -10

-5

0

5

Real Axis

(d)

G ( s ) H ( s ) 

K ( s  1) s ( s  3)( s  5) Root Locus

10

5

  s    i   x    A   g   a   m    I

0

-5

-10

-5

-4.5

-4

-3.5

-3

-2.5 Real Axis

-2

-1.5

-1

-0.5

0