ee 370l controls laboratory laboratory exercise #7 root...
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EE 370L
Controls Laboratory
Laboratory Exercise #7
Root Locus
Department of Electrical and Computer Engineering
University of Nevada, at Las Vegas
1. Learning Objectives
To demonstrate the concept of error and feedback
To demonstrate how to draw and analyze the root locus of a given system.
2. Equipment Usage
In this laboratory exercise, you will be exposed to the FEEDBACK control system. You will
use MATLAB and SIMULINK software to design a feedback control system.
3. Error and Feedback
Open Loop Systems: An open-loop controller, also called a non-feedback controller, is a type
of controller which computes its input into a system using only the current state and its model of
the system. A characteristic of the open-loop controller is that it does not use feedback to
determine if its input has achieved the desired goal. This means that the system does not observe
the output of the processes that it is controlling. Consequently, a true open-loop system cannot
engage in machine learning and also cannot correct any errors that it could make. It also may not
compensate for disturbances in the system.
Closed Loop Systems: A closed-loop or feedback control system is one in which an input forcing
function is determined in part by the system response. The measured response of a physical system
is compared with a desired response. The difference between these two responses initiates actions
that will result in the actual response of the system to approach the desired response. This in turn
drives the difference signal toward zero.
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Unity Gain Feedback System Response:
Transfer function of the system above can be defined as:
Assuming G = N/D, transfer function reduces to
If terms in numerator and denominator are not comprise, no cancellation of terms exists and
poles of transfer function can be restated as the roots of .
System stability:
For a feedback system to be stable all the poles of transfer function shall be located at the left
half plane. A simple technique to locate stable poles is root locus systems. A root locus
corresponding to G traces all the poles of H, as K varies from 0 to +. Thus, root locus is a family
of curves of the roots of .
A common technique in control system design is to plot the root locus corresponding to a plant
in order to discover where the closed-loop poles can be placed using a gain compensator. A typical
root locus is depicted below.
X O
p1 = 0 z1 = -0.5
X
p2 = -1
s = --1.45
X
p3 = -2
s = --1.25
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Second Order Systems Response:
In this Experiment, you are given G(s) having the form
( )
( )
( )
System is critically damped when
, is over-damped when
and under-damped
when
For under-damped case, systems response is as follows, where
Tr = rise time
Tp = peak time
Ts = settling time
Mp = Peak value
Figure 1. Step Response of an Under-damped Systems
Ideally, we would like the motor output to follow the input. Therefore, the design goal is to
have Tr and Ts as small value and Mp as close to unity.
Explicit formulas can be found for these parameters. First, note that the unit step response for t >
0 is given by
( ) { ( )
} [
( ) ]
{[ (
)
]
(
)}
From the above relationship, we can derive the following relationships:
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(
√ )
√ ( )
√ ( )
(
[
( ) ]
)
√ ( )
Steady state error to a step response is as follows:
| ( ) |
More often, we are also interested in steady state error for ramp response. A typical ramp
response is shown below:
Figure 2. Ramp Response of a closed loop Systems
ess1
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Steady state error to a ramp response is as follows:
| ( ) |
Using final value theorem, we get the following formula:
|
[ ( ) ]|
In addition to time response, we are also interested in frequency response parameters.
Frequency response of a closed loop systems is as follows:
| ( )|
√[ ( ) ] ( )
Figure below depicts |H(jω)| (with ω plotted on a logarithmic axis) for an under-damped
systems:
Figure 3. Frequency Response
Frequency Response
where,
m
b = bandwidth
Mm = Peak at resonance
From the above relationship of H(jω), we can derive the following relationships:
√ (
)
√ (
)
√(
)
( )
√ ( )
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Given the target specification of Tr, Tp, Ts, Mp, ess1, ωm, ωb, and Mm, the above formulas can
be used to determine whether a choice of K exists such that the specification goals are satisfied
and, if so, what range of values of K achieves the specifications.
4. Root Locus
Steps for sketching Root Locus:
1. Root locus begins at pole and ends at zero. The number of separate loci is equal to the number
of poles. m loci ends at zeros and n-m loci ends on infinities.
2. Root loci are symmetrical with respect to the horizontal real axis.
3. On the real axis, a point is on the root locus if the sum of poles and zeros on its right ride is an
odd number.
4. Asymptotes centroid is on the real axis as
∑ ∑
The angle of the asymptotes with the respect to the real axis is
( )
5. Points that locus crosses the imaginary axis can be derived by:
a. Routh-Hurwitz criterion.
b. Characteristic equation: ( ) ( ) (For low order equations only)
6. Breakaway points on the real axis:
( )
or
( ) ( ) ( ) ( )
7. The tangents to the loci at the breakaway point are equally spaced over .
8. The angle of locus departure from a pole:
∑( )
∑( )
( )
Matlab Commands
Plotting root locus:
First set the gain range.
Example. K = -50:.2:50; (this creates a vector varying from -50 to 50 with increment of 0.2)
Type r = rlocus(num, den,k), where num and den are the plant numerator and denominator.
To plot the root locus, type
plot(r)
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To find the gain and frequency corresponding to a point on the locus, type
k=rlocfind(num,den)
In this experiment we will calculate closed-loop response from open-loop data. Before
calculating either step response or Bode plots, the closed-loop transfer function must be found by
typing the following commands
nc=k0*num;
dc=den+k0*[0 0 num];
where “k0” is a specific gain. Note that two extra zeros are padded to ensure that “den” and “[0 0
num]” have the same length for addition.
The closed-loop unit step response can be generated by typing
step(nc,dc)
If you need more control over the time interval in which the computation is carried out, define
a vector “t” and type
step(nc,dc,t)
The magnitude of the closed-loop transfer function |H(jω)| versus ω can be plotted with the
commands
[mc,pc,wc]=bode(nc,dc);
semilogx(wc,mc)
Root Locus Example:
Consider the control system
with
s
sKsK P 2)(
,
1
1)(
ssG ,
4
4)(
ssH
Then
41
2
41
24)()()(
sss
sk
sss
sKsHsGsK P where the parameter PKk 4
Now draw the Root Locus:
R(s) C(s)
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1. There are 3 branches.
2. The real axis between –1<s<0 and –4<s<-2 is on the root locus because these areas are to the
left of an odd number of poles and zeros.
3. The Root Loci start (K 0 ) at the poles located at –4, -1 and 0 then end at the zero located at -2
as well as (2) zeros at s .
4. For large values of s, the Root Loci are asymptotic to asymptotes with angles,
OOO
a
k270,90
13
18012
5. The intersection of the asymptotes lies on the real axis at
5.1
2
3
2
25
1-3
2410
as
6. A Breakaway point occurs along the root locus 01 s at a relative maximum value of k.
Calculate 2
41
)(
)(
s
sss
sN
sDk and tabulate versus “s”.
s 2
41
)(
)(
s
sss
sN
sDk
-0.25
402.0
75.1
75.375.025.0
225.0
425.0125.025.0
k
-0.50
583.050.1
50.350.050.0
2
41
)(
)(
s
sss
sN
sDk
-0.75
488.025.1
25.325.075.0
2
41
)(
)(
s
sss
sN
sDk
-0.625
575.0375.1
375.3375.0625.0
2
41
)(
)(
s
sss
sN
sDk
-0.6
583.04.1
4.34.06.0
2
41
)(
)(
s
sss
sN
sDk
-0.55
589.045.1
45.345.055.0
2
41
)(
)(
s
sss
sN
sDk max
From the table it can be seen the maximum k is acquired around -0.55 which is the breakaway
point.
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s
-1 -4
-1.5
-2
707.0
js 65.065.0
k 1.5447 - 0.0531j 1.5
Breakaway @
s = -0.55
Root Locus can be calibrated to find gain at important points. Here
38.04
5.1
4
kKP
The Matlab commands to generate the root locus are:
>> Num=[1 2];Den=conv([1 0],conv([1 1],[1 4]));
>> GH=tf(Num,Den)
Transfer function:
s + 2
-----------------
s^3 + 5 s^2 + 4 s
>> rlocus(GH)
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More Root Locus Examples:
MatLab Program:
num = [1]
poles = [0;-1+j;-1-j;-3]
den = poly(poles)
rlocus(num,den)
Notes:
n=4 poles, m=0 zeros
asymptotes at ±45,±135
asymptote cg @ (-3-1-1+0)/4 = -1.25
angle of departure, d
( d + arctan(0.5) + 90 + 135) = 180
d = - 71.6 deg
Root Locus
Real Axis
Imagin
ary
Axis
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0-5
-4
-3
-2
-1
0
1
2
3
4
5
System: GH
Gain: 1.48
Pole: -0.625 + 0.63i
Damping: 0.704
Overshoot (%): 4.43
Frequency (rad/sec): 0.888
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MatLab Program:
zeros = [-1+3*j;-1-3*j]
num = poly(zeros)
poles = [0;-1+2*j;-1-2*j;-2]
den = poly(poles)
rlocus(num,den)
Notes:
n = 4 poles, m = 2 zeros
asymptotes at ±90 degrees
asymptote cg @ [(-2-1-1+0)-(-1-1)]/2 = -1.0
angle of departure, d
[( d + arctan(2)+(180-arctan(2))+ 90) -
(-90+90) ] = 180
d = -90 deg
Angle of Arrival, a
[(90+arctan(3)+(180-arctan(3))+90)-(-( a
+90)]=180
a = 90 deg
MatLab Program:
num = [1]
poles=[0;-1+4*j;-1-4*j;-2]
den = poly(poles)
rlocus(num,den)
Notes:
n = 4 poles, m = 0 zeros
asymptotes at ±45,±135
asymptote cg @ (-2-1-1+0)/4 = -1.0
angle of departure, d
( d + arctan(4) + (180-arctan(4)) + 90) = 180
d = -90 deg
5. Prelab
1. The loop transfer function of a single-loop negative feedback system is
( )
( )
a. Based on the second-order model and equations given above, find the value of K for
critical damping.
b. Using formulas given above, find the range of K for which the closed-loop system satisfies
the following specifications:
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Tr ≤ 100ms
Tp ≤ 120ms
Mp ≤ 1.1
Ts ≤ 150ms
ess1 ≤ .045
ωm ≥ 10r/s
ωb ≥ 50r/s
Mm ≤ 1.1.
2. The loop transfer function of a single-loop negative feedback system is
( ) ( )
( )( )
a. Analyze the system, do the hand calculation and plot root loci of the system step by step.
b. Using MATLAB, plot the root loci again and compare will your hand sketch.
c. Determine the range of feedback gain K such that the closed-loop system is stable. For
each value of K where instability begins, find the corresponding frequency of oscillation.
6. Experiment and Postlab
1. The loop transfer function of a single-loop negative feedback system is
( )
( )
a. Construct a Simulink model. You should plot the step input and the response of the system
on a single scope using the MUX block.
b. Verify the system is oscillatory for any K value in the corresponding interval by printing
the output of the scope block for this value of K.
c. Verify the system is stable for any K value in the corresponding interval by printing the
output of the scope block for this value of K.
d. Suppose we want the final value of our response to be within 0.01% of the final value of
our input (step) signal. That is, the steady state error is 0.01%. Further, let us assume the final
value is achieved for t ≥ 10 seconds. Using your Simulink block, determine for what values of K is
the steady state error 0.01% (for t ≥ 10 seconds)?
Notice that in designing a control system we first analyzed the stability of our open loop plant.
If our open loop plant is unstable, we use feedback to stabilize the system. Then we pick the
values for the parameters in our control law so our control objectives are achieved. Satisfying all
the design requirements is the goal of control theory.
2. The loop transfer function of a single-loop negative feedback system is
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( ) ( )
( )
a. Analyze the system, do the hand calculation and plot root loci of the system step by step.
b. Using MATLAB, plot the root loci again and compare will your hand sketch.
c. Find each value of K where system is critically damped.
d. Find each value of K where instability begins, find the corresponding frequency of
oscillation.
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