1 the system the actual roll angle is

1

Executive Summary:

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There are two main types of control systems, an open loop and a closed loop system. (Gene F
Franklin 2015) .

As the output signal, which in this experiment is being adjusted as well as being used in the control
computation it is known as a closed loop or fee dback control system, where the output is the
command roll angle.

There are six main components in a feedback control system; the desired value, the controller, the
compactor, a final control element, the measurement inducer and the communication paths. (Gene F
Franklin 2015) . These components are used to ensure that the actual value in the system is attuned
to the desired value. This system is a rocket attitude -control system, where the set point is the
command roll angle. Where the actual command roll angle is compared with th e desired roll angle
and through the system the actual roll angle is adjusted to match the desired command roll angle.

The root locus diagram was invented by Evans and it can be used as a guide to understand the
design’s system (Gene F Franklin 2015) .
To start, a tra nsfer function must be written for the closed loop and the characteristic equation is
written where the roots are the poles of the transfer function. (Gene F Franklin 2015)
To be able to study the roots, the equation must be written in polynomial form with a K value. The
root lo cus is the plot of all the root values where the k values range from zero to infinity (Gene F
Franklin 2015) .

The main objective of control design is system stability. In order to achieve this, it is necessary to
ensure that the poles lie in the left half of the s -plane. In t he rocket attitude control system in our
problem, the open loop poles are located at 0, -2 and -5 and since one pole is located at 0 the system
marginally stable. To address this marginal stability, the closed loop poles were relocated by
increasing the controller gain K. The gain was adjusted to make the relative damping ratio of the
closed loop system equal to 0.45. This moved the location of all three poles to the left half plane
thereby stabilising the system.

The gain margin of this system is 31.2 d B and the phase margin 82.3 °. These values suggest that
the system designed is robust .

If the gain is increased further, the gain and phase margins would reduce, increasing the risk of
instability in an operating environment. When K = 35.2 dB, there is no damping and any
oscillations would continue indefinitely.

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Contents

Executive Summary: ………………………….. ………………………….. ………………………….. ………………….. 1
Part (i) ………………………….. ………………………….. ………………………….. ………………………….. …………. 3
Part (ii) ………………………….. ………………………….. ………………………….. ………………………….. ………… 4
Part (iii) ………………………….. ………………………….. ………………………….. ………………………….. ………. 4
Part ( iv) ………………………….. ………………………….. ………………………….. ………………………….. ……….. 6
Part (v ) ………………………….. ………………………….. ………………………….. ………………………….. ………… 9
Conclusion ………………………….. ………………………….. ………………………….. ………………………….. …… 9
References ………………………….. ………………………….. ………………………….. ………………………….. …. 10
Appendix ………………………….. ………………………….. ………………………….. ………………………….. …… 11

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Part (i) Root L ocus D iagram

Root locus diagram was constructed using Matlab command rlocus(sys) and the damping ratios are
overlaid on the plot using sgrid(z,wn) command.

Figure 1: Root Locus Diagram

The root locus diagram above shows a zero at -9 and poles at -2, -5 and 0. When the gain (K
value) is kept below 35.2 the system is stable.

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Part (ii) Domi nant roots

For a stable system it is a requirement that the po les should lie on the left half plane. (Jacqueline
Wilkie 2002) . Dominant roots are also known as dominant poles. (Jacqueline Wilkie 2002) . The
output of a transfer function is very much dependent on the poles near the origin . These poles near
the origin of the s plane are called dominant poles. (Jacqueline Wilkie 2002) . For a stable system,
the dominant p oles are usually those with the smallest negative real parts. (Jacqueline Wilkie 2002) .
The system step responses become faster as the dominant poles move away from the vertical axis of
the s -plane (Jacqueline Wilkie 2002) . As the dominant poles travel further away from the imaginary
axis or complex axis in the s plane. Th is means th e dominant rootes with m ove closer towards the
left hand plane . (Jacqueline Wilkie 2002) . W hen designing a controller, it is important to consider
the points of the poles on the left hand plane so that the performance of the system can be improved
and the system becomes more stable . (Jacqueline Wilkie 2002) .

Those poles far from the origin has little effect on the output. H ence the effect of the non -dominant
poles can be neglected in the analysis (Gene F Franklin 2015) .

Part (iii ) K value

In this part, the K value that makes the relative damping ratio of the closed loop system equal to
0.45 was found.

On the root locus diagram this is when the locus of the dominant roots intersects K = 0.45 line. By
zooming into the root locus diagram shown in Figure 1 above, the value of K is found to be between
1.7 and 2.04 giving damping ratios (zeta) of 0.488 and 0.434 respectively. This is shown in Figure
2.

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Figure 2 Close u p of root locus diagram showing the variation of K close to 0.45

To obtain an exact value of K at 0.45, the Matlab function damp(sys) is applied to the feedback as
shown in the Matlab program in the Appendix. The resulting output is shown below.
K wn zeta
1.70 1.69 0.489
1.71 1.70 0.487
1.72 1.70 0.485
1.73 1.71 0.483
1.74 1.71 0.481
1.75 1.71 0.480
1.76 1.72 0.478
1.77 1.72 0.476
1.78 1.73 0.474
1.79 1.73 0.473
1.80 1.74 0.471
1.81 1.74 0.4 69
1.82 1.75 0.468
1.83 1.75 0.466
1.84 1.76 0.464
1.85 1.76 0.463
1.86 1.77 0.461
1.87 1.77 0.459

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1.88 1.77 0.458
1.89 1.78 0.456
1.90 1.78 0.455
1.91 1.79 0.453
1.92 1.79 0.452
1.93 1.80 0.450

Hence the K value that makes the relative damping ratio equal to 0.45 is 1.93.

The pole zero map shown below after the varying the controller gain K = 1.93 is shown below. All
three poles are now on the left half plane indicating a stable system. The pair of compl ex conjugate
poles are on the left half plane produce exponentially damped sinusoidal step responses (Jacqueline
Wilkie 2002) . These are considered to be dominant poles as the real pole is located a t distance
further than 5 times the distance from the vertical axis in comparison to the distance to this pair of
poles.
Figure 3 Pole -zero map of the system when the controller gain K = 1.93

The system shown in Figure 3 is a good design as the real system has to be robust to take into
account of unexpected variatio ns that can occur in the operating environment. This is done by
studying the phase and gain margins as described below.

Part ( iv) Phase and Gain Margins

System models are not expected to be absolutely accurate. The behaviour of a model is not always
obtained in a test environment of a real system. (Jacqueline Wilkie 2002) . The real system has to be
robust to consider of unexpected variations that can occur in the operating environment. (Jacqueline
Wilkie 2002) . Hence a margin of error is included in design specifications. (Jacqueline Wilkie
2002) . The more uncertainty there i s on these variations, the more margin that has to be built into
the system.

A system becomes unstable if at least one pole is on the right -hand plane. (Jacqueline Wilkie 2002) .

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A gain margin and a phase margin ensure that a resulting system are not performing near the bounds
of their stability (Jacqueline Wilkie 2002) .

Gain Margin
Gain is the proportion of the magnitude of the output to the magnitude of the input at steady state.
(Jacqueline Wilkie 2002) .

The gain margin is the factor by which the gain can be increased (or decreased) before instability
arises . The gain margin represents how much gain that can be added (in dB) to the gain plot before
the closed loop system would be unstable (Jacqueline Wilkie 2002) .

To calculate the gain margin the phase cross over point must be determined . (Jacqueline Wilkie
2002) . This is the point o n the Bode diagram where the phase intersects the horizonal axis at -180 °.

At this point by looking at the gain the gain must be calculated. (Jacqueline Wilkie 2002) . To
calcula te the gain margin zero must be subtracted f rom the above gain calculated. (Jacqueline
Wilkie 2002) .

If the gain margin of the system is negative then the system is unstable, a nd it is ideal f or the gain
margin t o be greater than six to eight db. (Jacqueline Wilkie 2002) . If the sys tem does not me et
these requirements then by redesigning the controller the gain margin and stability of the system
can be improved. (Jacqueline Wilkie 2002) .

When the gain crosses one or zero d B on the horizontal axis it is desirabl e that there is a phase
margin between forty and sixty degrees. (Jacqueline Wilkie 2002) . Simil arly , when the phase
crosses ov er at negative 180 ° it is ideal that the gain margin is below the zero dB line . (Jacqueline
Wilkie 2002) .

Phase margin
Phase is the shift of the signal measured as an angle or fraction of a wave length. (Jacqueline
Wilkie 2002) . The phase margin r epresents how much phase that can be lost before instability
occurs (Jacqueline Wilkie 2002) .
To calculate the phase margin , initia lly the gain crossover m ust be found. (Jacqueline Wilkie 2002) .
The ga in crossover is where the gain curve c rosses the horizontal ax is at zero d b. The next step is to
using the phase measured in degrees versus frequency plot , d etermine the phase at t hat point.
(Jacqueline Wilkie 2002) . The final step is to calculate the phase margin which is equal to the phase
point subtract negative 180 °. (Jacqueline Wilkie 2002) .
From t he calculation of the phase margin we can determine whether the system is unstable
(Jacqueline Wilkie 2002) . Thi s will occur whe n t he phase margin calculated is below forty five
degrees. If the system is u nstable , it might be ideal to alter the system to make it stable. (Jacqueline
Wilkie 2002) . This can be done by altering the controller which will consequen tly change the phase
margin value , which will impr ove the phase margin making the system stable . (Jacqueline Wilkie
2002) .

The gain margin and phase margin can be read directly from a Bode plot shown in Figure 4 are 31.2
dB at 3.16 rad/s and 82 ° at 0.192 rad /s respectively .

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Figure 4: Bode diagram

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Part (v ) Time Respo nse
The time response of the closed loop system to a unit step is investigated in this part.

Figure 5 Step response

The time response graph indicates the system is underdamped with a moderate overshoot. The
system reaches a steady state level at a reasonable time. The peak time is just over 2 seconds and a
settling time is around 5 seconds.

Conclusion
In this assignment, a rocket altitude control system that was marginally stable was stabilised by
ad ding a gain controller. The roo t-locus diagram wa s used to adjust the gain by rel ocating the
closed loop poles . The gain was adjusted to make the relative damping ratio of the closed loop
system equal to 0.45. This moved the location of all three poles to the left half plane thereby
stabilising the system .

The Bode plot shows that the gain margin and phase margins are 31.2 dB and 82 ° respectively .
The se are more than the 6 – 8 dB and 40° – 60° margins recommended (Jacqueline Wilkie 2002) for
a stable syst em and confirms that the control choice is appropriate.

This is established by the step response that sh ows a steady state to be reached in 5 seconds.

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References

Gene F Franklin, J David Powell, Abbas Emami -Naeini. 2015. Feedback Control of Dynamic
Systems. New Jersey : Pearson.

Jacqueline Wilkie, Michael Johnson, Reza Katebi. 2002. Control engineering An introductory
course. New York: PALGRAVE.

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Appendix

Matlab script

%Linear Systems ; Control – Assignment2.m
clear all
clc
s=tf( ‘s’ );

g = 1/(s*(s^2+7*s+10));
h = (9+s);
G = g*h; %equivalent to Matlab series command

rlocus(G) %Part (i) of Assignment 2
zeta = 0,0.15, 0.3,0.45, 0.75,1;
wn = 0:10:100;
sgrid(zeta,wn); % from this plot k is between 1.7 and 2.04

fprintf( ‘%4s %4s %5s
‘ , ‘k’ ,’wn’ ,’zeta’ ); %print heading for values
below
for k = 1.7:0.01:2.04
T = feedback(k*g,h);
wn,zeta = damp( T);
%k, wn, zeta
fprintf( ‘%4.2f %4.2f %5.3f
‘ , k,wn(1),zeta(1)); %print values
if abs(zeta(1) – 0.45)

x

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