Problem 4.1: Design of a
position control system using the root locus method
(a)
State the functions of the proportional,
integral and derivative parts of a PID controller.
(b)
Fig. 1 shows a
position control system.
G(s) = 60
+ + G(s)
a s
Fig.1
Determine the value of a in the velocity feedback loop so that the damping ratio x of the closed loop poles is 0.5.Use the root locus
method.
Solution:
The characteristic polynomial of the system with velocity feedback is
F(s) = s2+(60a+2)s+60
= ( s2+2s+60)+60as
Draw the root locus for
s2+2s+60
and determine a for x=0.5
X
a for desired root location is .0958 (Answer)
Problem 4.2: Design of a
phase lag compensator using the root locus method
(a)
Write down the steps necessary for the design of a phase-lag compensating
network for a system using the root locus method.
(b)The
uncompensated loop transfer function for a system is
GH(s) = K/[s(s+2)].
(i) Use root locus
method to design a compensator so that the damping ratio of the dominant complex
roots is 0.45 while the system velocity constant, Kv
is equal to 20.
(ii)
Find the overall gain setting
Solution:
(a) Theory
(b) Uncompensated root locus is a
vertical line at s =-1 and results in a root on the ξ line at s = -1 +
j2 as shown below.
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ξ=.45 |
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j2
jw
j1
-2 s 0
Measuring the gain at this root, we have K = (2.24)2 = 5
Kv of uncompensated system = K/2 =5/2 =2.5
Ratio of zero to pole of compensator is │z/p│ = α = Kvcomp/Kv uncomp =20/2.5 = 8
Examining the following Fig, we
find that we might set z =-0.1 and then p = -0.1/8. The difference of the angles from p and
z at the desired root is approx 1 0,
and therefore s =-1+j2 is still the location of the dominant roots.
ξ=.45
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j1
-z -p
Sketch of the compensated root locus is shown above.
Compensated system transfer function is
Gc(s)GH(s) = 6(s+0.1)/[s(s+2)(s+0.0125)],
where
(K/α) 5 or K=40 in order to account for the attenuation of the lag
network.
Problem 4.3: The root locus procedure
i
Plot the root locus
for the characteristic equation of a
system when
1+ K/[s(s+2) (s+2+j2) (s+2-j2)] = 0
as K varies from zero to infinity.
Solution: i
-6 -4 -2 -1 0
(ii) separate loci =4
(iii) Angles of asymptotes= 45,135,225,315 degs.
Centre of asymptotes= (-2-2-2)/4=-1.5
(iv)
Charac.eqn.
s(s+2) (s2+4s+8) +K=0
Apply RH criterion for stability
and obtain K crit.
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S4 |
1 |
12 |
K |
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S3 |
6 |
8 |
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S2 |
32/3 |
K |
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S1 |
8-(9K/16) |
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S0 |
K |
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(v) Kcrit =
14.2 and roots of the auxiliary eqn are
10.66s2+14.2 = 10.66(s2+ 1.33) = 10.66(s+j1.153) (s-j1.153)
The locus crosses the im axis at j1.153 and –j1.153
Problem 4.4: To determine the value of the gain
at the limit of
stability using root locus
The
open loop transfer function of a unity feedback system is:
s(s+1)(s+2)
i.
Draw the root locus
and the asymptotes on the plot
ii.
Calculate the value
for the asymptote centroid.
iii.
Calculate the number
of asymptotes
iv.
Calculate the value
of w at which the locus crosses
the imaginary axis, and find the corresponding value of K at the limit of
stability
v.
Find the value of K
required to place a real root at s =
-5.
For this value of K, find the location of the complex roots.
Solution:
(i)
2
1
-3 –2.5 -2 -1.5 -1 -.5 0 0.5 1
Real Axis
(ii)sP= -3/3 = -1
(iii) 3 asymptotes @ -600, 600, -1800
(iv) w=Ö2
K= 6 @ stability limit
(v) K= 60,
Roots are @ s= -5,1-j3.32, 1+j3.32
Problem 4.5 : Breakaway point and the limiting value of gain for
stability
Construct the root locus for the
control system shown in Fig.2. Follow
the following six
steps:
(i)
Locate the poles on the s-plane.
(ii)
Draw the segment of the root locus on the real axis.
(iii)
Draw the three separate loci.
(iv)
Determine the angles and the center of the asymptotes.
(v)
Evaluate the breakaway point for one of the loci.
(vi)
Determine the limiting value of the gain K for stability using the Routh-Hurwitz criterion.
K s(s+4)(s+6)
Fig.2
Solution:
-Characteristic eqn. s(s+4)(s+6)+K=0.
Poles are at s= 0, -4, -6. There are no zeros.
On the real axis, there are loci between the origin and –4 and from –6 to –infinity.
Asymptotes intersect the real axis at
s =[0+(-4) +(-6)]/3 =-3.333
Angles of asymptotes are +/- 60 deg. and 180deg.
-dK/ds = 3s2+20s+24= 0
Therefore, the breakaway point is on the real axis at s=-1.57.
0
K= 240
The application of Routh’s criterion to the characteristic equation
s(s+4)(s+6)+K =0 gives the following array:
s3 1 24 0
s2 10 K 0
s1 240-K/10 0
s0 K
The value of K which makes the rows1 vanish is K=240. This is the limiting value of K for stability.
Problem 4.6:
Breakaway point and the limiting value of gain for stability
Plot
the complete root locus of the following characteristic equation of a system,
when K varies from zero to infinity.
1
+ K = 0.
Follow
the following seven steps:
i.
Locate the poles on the s-plane
ii.
Draw the segment of the root locus on the real axis
iii.
Draw the four separate loci
iv.
Determine the angles and the center of the asymptotes.
v.
Evaluate the breakaway point for one of the four loci.
vi.
Determine the limiting value of the gain K for stability using the Routh-Hurwitz criterion.
vii.
Determine the angles of departure at the complex poles.
Solution:
-6 -5 -4 -3 -2 -1
· -four separate loci
f =(2q+1)/4, q =0,1,2,3
f =45,135,225,315 deg.
asymptote center = ( -4-4-4)/4 =-3
· Breakaway point is obtained from
K =p(s) =-s(s+4) (s+4+j4)(s+4-j4) between s= -4 and s=0
Search s for a maximum value of p( s) by trial and error . It is approximately s=-1.5
· Write the characteristic equation and Routh array. Limiting value for stability is K=570
· Angle of departure at complex poles is obtained from the angle criterion.
Angles are 90,90,135 and 225 deg.
Problem 4.7: Root locus analysis when G(s) contains a zero
(a)
Suggest a possible
approach to the experimental determination of the frequency response of an
unstable linear element within the loop of a closed-loop control system.
(b)
Fig.3 shows a system
with an unstable feed-forward transfer function.
R(s)
C(s)
10(s+1) s (s-3)
-
Fig.3
Sketch the root locus plot and
locate the closed loop poles. Show that although the closed-loop poles lie on
the negative real axis and the system is not oscillatory, the unit-step
response curve will exhibit overshoot.
Solution:
(a)
Measure the frequency response of the unstable element by using it as part of a stable system. Consider a unity feedback system whose OLTF is G1*G2.. Suppose that G1 is unstable.. The complete system can be made stable by choosing asuiatble linear element G2. Apply a sinusoidal signal at the input.. Meausure the error signal e(t), the input to the unstable element, and x(t) the output of the unstable element. By changing the frequency and repeating the process , it is possible to obtain the frequency rsponse of the unstable element.
(b)
-2
Closed loop zero
Closed loop transfe function is
C(s)/R(s) =10(s+1)
C(t) = 1+1.666exp(-2t) –2.666exp(-5t)
C(t) Rough sketch-Actual values may be
plotted.
Although the system is not oscillatory, the response exhibits overshoot. This is due to the prsence of azero at s=-1.
Problem 4.8:
Root locus plot giving the number of asymptotes,centroid,
and the frequency at which the locus crosses the imaginary axis
(a) Draw
the root locus plot for a unity feedback system with
G(s) = K
.
s(s+5) 2
Calculate and record values for the asymptote centroid, number of asymptotes, and value of the frequency,
w at which the locus crosses the
imaginary axis. Also, show the asymptotes on the plot.
(b) How would you obtain qualitative information
concerning the stability and performance of the system from the root locus
plot?
Solution:
(a) Poles at s=0, two at s=-5. Root locus crosses imaginary axis at w=5
Centroid at –10/3=-3.33
Three asymptotes at (=/-) 60deg., -180deg
K= 250 at stability limit
(b) Descriptive
Problem 4.9:
Design the value of K in the velocity feedback loop to limit the overshoot for a step input
using root locus
625 s2
Ks
Fig.4
(b)
Determine the step response for the
designed value of K.
Solution:
System without velocity feedback:
625
s2 +625 roots s=-/+ j25
Open loop Bode diagram magnitude curve is a straight line with slope of –40dB/decade, crossing 0dB at w=25
With velocity feedback added the root locus is as shown with K as the gain.
OLTF = 625/s2 = 625
1+(625Ks/s2) s(s+625K)
27 K
22.5 ..
Root at –12.5 +j21.65
wn= 25
By inspection, any desired value of x is available. Choosing x =0.5 to get Mpt < 1.25.
Mpt= 1+ exp(-PI x/sqrt(1-x2)) =1.163 for x =0.5
K= .04 and roots are at s= -12.5+ j 21.65
Step response:
C(t) = 1- exp(-xwn T) sin( wn sqrt(1-x2)+ tan-1(sqrt(1-x2)/ x) )
t
Problem 4.10: To determine the s
co-ordinate when
the imaginary co-ordinate at some point, w on the root locus is given
(a)
Give the value of w where the root locus of
G(s) = K
(s+1) (s+1-j)(s+1+j)
crosses the imaginary axis.
(b)
Give the values of s
where the root locus of
G(s) = K(s+1)(s+5)
s2(s+2)
leaves
and re-enters the real axis.
(c)
Draw the root locus
for
G(s)= K(s+2)
s(s+1)
(s+5) (s+8)
What is the s co-ordinate when the imaginary co-ordinate at some point is w = 1.5?
Solution:
(i) The
root locus in
the s-plane follows a circular path when there are
(a) two
open loop zeros and one open loop pole
(b) two
open loop poles and one open loop zero
( c ) two open loop poles and two open loop
zeros
Ans: (b )
(ii) The algebraic sum of the
angles of the vectors from all poles and zeros to a point on any root locus segment is
(a) 1800
(b) 1800 or odd
multiple thereof
(c )1800
or even multiple thereof
Ans: ( b)
(iii)
The algebraic sum of the angles of the vectors from all poles and zeros to the
point on any root-locus segment is
(a) 1800
(b) 1800
or odd multiple thereof.
(c ) 1800 or even multiple thereof
(d) 900
or odd multiple thereof.
Ans: (
d)
(iv) Whenever there are more
poles than zeros
in the open loop transfer function G(s), the number of root locus segments is
(a)
Equal to the number
of zeros
(b)
Equal to the number
of poles
(c)
Equal to the
difference between the number of poles and
number of zeroes
(d)
Equal to the sum of
the number of poles and
number of zeroes.
Ans: (b)
(v)
If the number of
poles is M and the number of zeroes is N, then the number of root-locus
segments approaching infinity is equal to
(a)
M+N
(b)
M/N
(c)
MN
(d)
M-N
Ans:( d )
(vi ) If the system poles lie on
the imaginary axis of the s-plane, then the system is
(a)
stable
(b)
unstable
(c)
conditionally stable
(d)
marginally stable
Ans:( d )
( vii ) The Routh array for a system
is shown below:
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s3 |
1 |
100 |
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s2 |
4 |
500 |
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s1 |
25 |
0 |
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s0 |
500 |
0 |
How many roots are in the right-half plane?
(a)
One
(b)
Two
(c)
Three
(d)
Four
Ans:( b )