ENG540 Control Systems Analysis: First and Second Order Systems Report

This report analyzes first and second-order control systems, focusing on determining transfer functions from experimental data and examining the impact of feedback on system performance. The analysis includes open-loop and closed-loop systems with unity and non-unity feedback configurations. The report uses MATLAB simulations to generate time-response graphs and system information, such as rise time, settling time, overshoot, and peak time. The study explores how feedback affects gain, time constants, and overall system behavior, including rise time, settling time, and overshoot in both first and second-order systems. The report compares the performance of open-loop and closed-loop systems, highlighting the trade-offs between different feedback configurations. The second-order system analysis reveals the effects of feedback on overshoot, peak time, and final value. Overall, the report provides a detailed analysis of control systems, demonstrating the importance of feedback in shaping system responses and achieving desired performance characteristics.

ENG540 CONTROL SYSTEMS ANALYSIS
Registration Number:
Coursework: First and second order systems, identification and analysis
Aim: To determine the transfer function of a first and a second order system
obtained from experimental data and carry out analysis on the effect of feedback
on the systems.

Table of Content
First Order System.....................................................................................................................................1
a) Determine the first order Transfer Function..................................................................................1
b) Closed loop transfer function (unity negative feed back, H(s) = 1)..............................................2
c) Closed loop transfer function (negative feed back, H(s) = 0.5)....................................................3
d) Performance of the open loop system compared with the closed loop systems...........................5
Second Order System.................................................................................................................................6
a) Determine the second order Transfer Function.............................................................................6
b) Closed loop transfer function (unity negative feed back, H(s) = 1)..............................................7
c) Closed loop transfer function (negative feed back, H(s) = 0.5)....................................................8
d) Performance of the open loop system compared with the closed loop systems...........................9
References................................................................................................................................................11
Appendices...............................................................................................................................................12
Appendix 1: MATLAB Source code...............................................................................................12
List of Figures
Figure 1 : Open loop transfer function response............................................................................................1
Figure 2 : Closed loop transfer function response (Unity feedback).............................................................3
Figure 3 : Closed loop transfer function response (H(s) = 0.5).....................................................................4
Figure 4 : Open loop transfer function response............................................................................................7
Figure 5 : Closed loop transfer function response (Unity feedback).............................................................8
Figure 6 : Closed loop transfer function response (H(s) = 0.5).....................................................................9
List of Tables
Table 1 : First order system response summary(from MATLAB response).................................................5
Table 2 : Second order system response summary (from MATLAB response)..........................................10

1
First Order System
a) Determine the first order Transfer Function.
The general form a first order transfer function is:
G(s)= k
1+τs ...... (1)
where:
k = gain, and
τ = time constant.
Given a step response, k represents the steady state value of the response while τ
represents the value of time (in seconds) when the response gets to 63.2% (1 - e-1) of
its final (steady state) value (Nise, 2019) (Dorf & Bishop, 2017) . This approach is
used to obtain the numerical transfer function of the system whose response is given
in this exercise.
From the given step response, the final value is 1 and the time it takes the response to
get to 63.2% (0.632) is approximately 0.5 seconds. Hence, the open loop transfer
function of the system is given by:
G(s)= 1
1+0.5 s OR
G(s)= 1
0.5 s +1 ...... (2)
Figure 1: Open loop transfer function response

2
System information (as generated by MATLAB):
RiseTime: 1.0985
SettlingTime: 1.9560
SettlingMin: 0.9045
SettlingMax: 1.0000
Overshoot: 0
Undershoot: 0
Peak: 1.0000
PeakTime: 5.2729
b) Closed loop transfer function (unity negative feed back, H(s) = 1)
The transfer function of a closed loop control system with a general negative feedback
is given by (BELL, 1981) :
T (s )= G(s )
1+G( s) H (s) ...... (3)
Where G(s) is the open loop transfer function. From section (a), G( s)= 1
0.5 s +1 and
H(s) = 1 for unity feedback system.
Substituting G(s) in equation (3),
1+G( s)=1+ 1
0.5 s +1 =1+0.5 s+1
0.5 s +1 = 0.5 s +2
0.5 s +1 ...... (4a)
G(s)
1+ G(s)= 1
0.5 s +1 × 0.5 s+1
0.5 s+2 = 1
0.5 s +2 ...... (4b)
Hence, closed loop transfer function is:
T (s )= 1
0.5 s+ 2 ...... (5a) OR
T (s )= 0.5
0.25 s+1 ...... (5b)

3
k = 0.5
τ = 0.25
Figure 2: Closed loop transfer function response (Unity feedback)
System information (as generated by MATLAB):
RiseTime: 0.5493
SettlingTime: 0.9780
SettlingMin: 0.4523
SettlingMax: 0.5000
Overshoot: 0
Undershoot: 0
Peak: 0.5000
PeakTime: 2.6365
c) Closed loop transfer function (negative feed back, H(s) = 0.5)
From equation (3), T (s )= G(s )
1+G( s) H (s) ...... (6)
G(s) remains 1
0.5 s+ 1
From the same analogy in part (b),
G(s) H (s)= 1
0.5 s+1 × 0.5= 0.5
0.5 s+1 ...... (7a)

4
1+G( s)H (s)=1+ 0.5
0.5 s+1 = 0.5+0.5 s+1
0.5 s+1 = 0.5 s+1.5
0.5 s +1 ...... (7b)
G(s)
1+ G(s) H (s) = 1
0.5 s+1 × 0.5 s+1
0.5 s+ 1.5 = 1
0.5 s+1.5 ...... (7c)
Hence, closed loop transfer function is:
T (s )= 1
0.5 s+1.5 ...... (8a) OR
T (s )= 0.67
0.33 s+1 ...... (8b)
k = 0.667
τ = 0.33
Figure 3: Closed loop transfer function response (H(s) = 0.5)
System information (as generated by MATLAB):
RiseTime: 0.7323
SettlingTime: 1.3040
SettlingMin: 0.6030
SettlingMax: 0.6666
Overshoot: 0
Undershoot: 0
Peak: 0.6666
PeakTime: 3.5153

5

6
d) Performance of the open loop system compared with the closed loop systems
Rise time % overshoot Settling time Final value
Open Loop 1.15 sec 0% 1.96 sec 1.0
Unity feedback 0.549 sec 0% 0.978 sec 0.50
H(s) = 0.5 0.732 sec 0% 1.30 sec 0.667
Table 1: First order system response summary(from MATLAB response)
One notable observation is that the percentage overshoot for all the three system
is 0%. Since this is a first order system, no overshoot is observed in any setup. Adding
a feedback path, however, affects the value of gain, k, and time constant, τ. For
instance, in unity feedback system, both the values of gain and time constant values
are reduced by half. To restore the initial (desired) steady state response requires
doubling of the gain so as to get the desired steady state value. For a feed back path of
H(s) = 0.5, both the gain and time constant are reduced by a third. It therefore seems
that a feedback path in first order control system affects the gain and time constant in
an equal proportion.
Both rise time and settling time seems to have improved when a feed back path is
included. Improvement in this case is interpreted as the ability of the system to take
the least time possible to rise and settle. This may not be always the case, depending
on the application. The rise time and settling improves from 1.15 seconds and 1.96
seconds for open loop system to 0.732 seconds and 1.3 seconds for H(s) = 0.5 and
0.549 seconds and 0.978 seconds for unity feedback systems, respectively. The
observed trend is expected since different feedback paths change the values of τ,
altering the poles and zeros (if present) locations, and hence the performance of
system to get the desired output.

7
Second Order System
a) Determine the second order Transfer Function.
The general form on an open loop second order transfer function is:
G(s)= kss ωn
2
s2 +2 ζ ωn s +ωn
2 ...... (9)
The settling time is given by: T s= 4
ζ ωn
...... (10a)
Peak time is given by: T p= π
ωn √ 1− ζ2 ...... (10b)
From (Xu & Ding, 2017),
Given that Ts = 4s, then ζ ωn= 4
T s
= 4
4 =1=¿ ζ ωn =1 ...... (11a)
Tp = 0.5, then 3 9.478=ωn
2 (1 − ζ2) ...... (11b)
From (11a), we can express ζ as: ζ = 1
ωn
...... (11c)
Substituting (11c) in (11b):
3 9.478=ωn
2 (1 − 1
ωn
2 )=ωn
2 − 1
Hence, ωn
2=40.478=¿ ωn= √ 40.478=6.36 ...... (11d)
Also, ζ = 1
ωn
= 1
6.36 =0.157 ...... (11e)
From the response plot given, final (steady state) value = 2.5 = kss
Therefore, ω = 6.36 , ζ = 0.157 and kss = 2.5
Substituting the values in (9):
G(s)= 101.20
s2 +2 s+40.48 ...... (12)

8
Figure 4: Open loop transfer function response
System information (as generated by MATLAB):
RiseTime: 0.1857
SettlingTime: 3.6395
SettlingMin: 1.5835
SettlingMax: 4.0150
Overshoot: 60.6008
Undershoot: 0
Peak: 4.0150
PeakTime: 0.5066
b) Closed loop transfer function (unity negative feed back, H(s) = 1)
From equation (3), T (s )= G(s )
1+G( s) H (s) ...... (3)
and H(s) = 1. Substituting and solving for T(s) following the steps in section 1(b), we
have: G( s) H (s)= 101.20
s2 +2 s+40.48
1+G(s)H (s)=1+ 101.20
s2 +2 s +40.48 = s2 +2 s +141.68
s2 +2 s+ 40.48

9
G( s)
1+ G(s) H (s)= 101.20
s2+2 s+ 40.48 × s2+ 2 s+ 40.48
s2 +2 s+ 141.68 = 101.20
s2 +2 s +141.68
Hence, T (s )= 101.20
s2+2 s+141.68 ...... (13)
Figure 5: Closed loop transfer function response (Unity feedback)
System information (as generated by MATLAB):
RiseTime: 0.0935
SettlingTime: 3.7596
SettlingMin: 0.2938
SettlingMax: 1.2623
Overshoot: 76.7258
Undershoot: 0
Peak: 1.2623
PeakTime: 0.2639
c) Closed loop transfer function (negative feed back, H(s) = 0.5)
From equation (3), T (s )= G(s )
1+G( s) H (s) ...... (3)
and H(s) = 0.5. Substituting and solving for T(s) following the steps in section 1(b),

10
we have: G(s) H (s)= 101.20
s2 +2 s+40.48 ×0.5= 50.60
s2+2 s+40.48
1+G(s)H (s)=1+ 50.60
s2 +2 s +40.48 = s2 +2 s +91.08
s2 +2 s +40.48
G( s)
1+ G(s) H (s)= 101.20
s2+2 s+ 40.48 × s2 +2 s+ 40.48
s2 +2 s+ 91.08 = 101.20
s2+ 2 s+91.08
Hence, T (s )= 101.20
s2+2 s+91.08 ...... (14)
Figure 6: Closed loop transfer function response (H(s) = 0.5)
System information (as generated by MATLAB):
RiseTime: 0.1186
SettlingTime: 3.7161
SettlingMin: 0.5383
SettlingMax: 1.9090
Overshoot: 71.8092
Undershoot: 0
Peak: 1.9090
PeakTime: 0.3292