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linear feedback controller analysis - 18.17

Given the system elements (you should assume negative feedback),

G( s) =

K

H( s) = 1

s-------------------------2 + 3s + 2

 

 

First, find the characteristic equation,. and an equation for the roots,

 

 

 

 

 

K

 

 

 

 

 

 

1 +

-------------------------

( 1) = 0

 

 

 

 

 

 

s

2

 

 

 

 

 

 

 

 

 

 

+ 3s + 2

 

 

 

 

 

 

 

s2 + 3s + 2 + K = 0

 

 

 

 

 

roots

 

=

– 3 ± 9 – 4( 2 + K)

= – 1.5 ±

1 – 4K

 

 

-----------------------------------------------

 

2

 

-------------------

 

 

 

 

 

 

 

 

 

 

2

 

Next, find values for the roots and plot the values,

 

K

 

 

 

root

 

 

 

jω

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

-1

 

 

 

 

 

 

σ

2

 

 

 

-2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

 

 

 

-3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

******CALCULATE AND PUT IN NUMBERS

18.3.1Approximate Plotting Techniques

The basic procedure for creating root locus plots is,

1. write the characteristic equation. This includes writing the poles and zeros of the

linear feedback controller analysis - 18.18

equation.

 

 

 

 

1 + G( s) H( s)

( s + z1) ( s + z

2)…

( s + zm)

 

---------------------------------------------------------------

= 0

= 1 + K( s + p ) ( s + p )…

( s + p )

 

1

2

n

 

2.count the number of poles and zeros. The difference (n-m) will indicate how many root loci lines end at infinity (used later).

3.plot the root loci that lie on the real axis. Points will be on a root locus line if they have an odd number of poles and zeros to the right. Draw these lines in.

4.determine the asymptotes for the loci that go to infinity using the formula below. Next, determine where the asymptotes intersect the real axis using the second formula. Finally, draw the asymptotes on the graph.

β( k)

=

± 180° ( 2k + 1)

 

 

k [ 0, n m – 1]

-----------------------------------

 

 

 

 

n m

 

 

 

σ =

( p1 + p2 + + pn) ( z1

+ z2

+

+ zm)

-------------------------------------------------------------------------------------------

 

 

n m

 

 

 

5.the breakaway and breakin points are found next. Breakaway points exist between two poles on the real axis. Breakin points exist between zeros. to calculate these the following polynomial must be solved. The resulting roots are the breakin/breakout points.

A = ( s + p1) ( s + p2)…

( s + pn)

B = ( s + z1) ( s + z2)… ( s + zm)

d

 

d

 

0

 

-----A B A

----B =

 

ds

 

ds

 

 

 

6. Find the points where the loci lines intersect the imaginary axis. To do this substitute the phasor for the laplace variable, and solve for the frequencies. Plot the asymptotic curves to pass through the imaginary axis at this point.

( jω + z1) ( jω + z2)…

( jω + zm)

1 + K(--------------------------------------------------------------------------jω + p1) ( jω + p2)…

( jω + pn) = 0

• Consider the example in the previous section,

linear feedback controller analysis - 18.19

Given the system elements (you should assume negative feedback),

G( s) =

K

H( s) = 1

2----+-----3---s----+-----2-

s

 

Step 1: (put equation in standard form)

 

 

 

 

K

 

 

1

1 + G( s) H( s) = 1 +

-------------------------

 

 

 

( 1)

--------------------------------

s

2

+ 3s + 2

= 1 + K( s + 1) ( s + 2)

 

 

 

 

 

Step 2: (find loci ending at infinity)

 

 

 

 

m = 0

n = 2

(from the poles and zeros of the previous step)

n m =

2

(loci end at infinity)

Step 3: (plot roots)

 

 

 

 

 

 

 

 

 

 

 

 

jω

 

σ

-2 -1

Step 4: (find asymptotes angles and real axis intersection)

β( k)

=

180° ( 2k + 1)

 

k I[ 0, 1]

 

jω

-------------------------------

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

β( 0)

 

180° ( 2( 0)

+ 1)

 

90°

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

------------------------------------

=

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

σ

 

 

 

 

 

 

 

 

 

 

 

β( 1)

 

180° ( 2( 1)

+ 1)

 

270°

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

------------------------------------

=

 

 

 

 

 

 

 

 

 

2

 

 

-2

-1

 

 

 

σ =

( 0) ( – 1 – 2)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

----------------------------

= 0

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

asymptotes

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

linear feedback controller analysis - 18.20

Step 5: (find the breakout points for the roots)

 

A = 1

B = s2 + 3s + 2

jω

d

 

0

 

d

----A =

 

----B = 2s + 3

ds

 

 

ds

A

d

 

d

 

-----B B

----A = 0

 

ds

 

ds

 

σ

-2

-1

1( 2s + 3) ( s2 + 3s + 2) ( 0) = 0

-1.5

 

 

 

 

 

 

2s + 3 = 0

s = –1.5

Note: because the loci do not intersect the imaginary axis, we know the system will be stable, so step 6 is not necessary, but we it will be done for illustrative purposes.

Step 6: (find the imaginary intercepts)

1 + G( s) H( s) = 0

1

 

1 + Ks-------------------------2 + 3s + 2

= 0

s2 + 3s + 2 + K = 0

( jω ) 2 + 3( jω ) + 2 + K = 0

ω 2 + 3jω + 2 + K = 0

ω 2 + ω ( –3j) + ( – 2 – K) = 0

 

 

 

 

ω

=

3j ± ( –3j) 2 – 4( – 2 – K)

=

3j ± – 9 + 8 + 4K

=

3j ± 4K – 1

--------------------------------------------------------------2

---------------------------------------------

-------------------------------

 

 

 

2

 

2

In this case the frequency has an imaginary value. This means that there will be no frequency that will intercept the imaginary axis.

• Plot the root locus diagram for the function below,