Control Systems Analysis: Transfer Functions and Feedback Effects

This report analyzes first and second-order control systems, focusing on determining their transfer functions from experimental data and assessing the impact of feedback. The analysis includes the derivation of transfer functions for both open-loop and closed-loop systems, considering unity and negative feedback configurations. MATLAB simulations are used to generate time-response graphs and data, which are then used to evaluate system performance characteristics such as rise time, settling time, overshoot, and steady-state values. The report compares the performance of open-loop and closed-loop systems, highlighting the effects of feedback on system stability, transient response, and steady-state accuracy. The findings demonstrate how feedback affects the gain and time constants of first-order systems and influences the transient response of both first and second-order systems, including changes in rise time, settling time, and overshoot. The report also explores the design considerations for second-order systems, and the necessity for software simulations and potentially PID controllers to enhance precision. The MATLAB code used to generate the results is also included in the appendices.

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.

Contents
First Order System.....................................................................................................................................1
a) Determine the first order Transfer Function..................................................................................1
b) Closed loop transfer function (unity negative feedback, H(s) = 1)...............................................2
c) Closed loop transfer function (negative feed back, H(s) = 0.5).....................................................4
d) A comparison of open loop and closed system.............................................................................6
Second Order System.................................................................................................................................7
a) Determine the second order Transfer Function.............................................................................7
b) Closed loop transfer function (unity negative feedback, H(s) = 1)...............................................8
c) Closed loop transfer function (negative feedback, H(s) = 0.5)...................................................10
d) Performance of the open loop system compared with the closed loop systems..........................12
References................................................................................................................................................13
Appendices..............................................................................................................................................14
MATLAB Code.......................................................................................................................................14
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.
A first order system is modelled by the general transfer function shown below:
G( s)= k
1+τs ...... (i)
Where:
k = dc gain, and
τ = time constant.
By inspection, the response of the first order can be fixed to fit the model thereby
evaluating, k that 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) .
From the response, the system reaches a steady state value of 1 and the time it takes
the response to get to 63.2% (0.632) is about 0.5 seconds. The system seems to step at
0 seconds. The time constant τ=0.5-0=0.5 seconds. Also, the dc gain from the
graph=1.
Substituting in equation 1 yields
G( s)= 1
0.5 s +1 ...... (ii)

2
Figure 1: Open loop transfer function response
System information for step input from 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 feedback, H(s) = 1)
For a negative feedback system, according to (Nise, 2015) :
T (s )= G(s )
1+G(s) H (s) ...... (iii)
With G(s) as the open loop transfer function. From the previous section
G( s)= 1
0.5 s +1

3
For a unity feedback system
H(s) = 1
Substituting in the general equation, we have
1+ ( G( s) ×1 ) = ( 1+ 0.5 s+1
0.5 s+1 )= 0.5 s+ 2
0.5 s +1 ...... (iv_a)
Therefore
G(s)
1+ G(s)= ( 1
0.5 s+ 1 )× ( 0.5 s+1
0.5 s +2 )= 1
0.5 s+2 ...... (iv_b)
Hence
T (s )= 1
0.5 s+ 2 ...... (v_a) OR
T (s )= 0.5
0.25 s+1 ...... (v_b)
The transfer function in its canonical form becomes
From which
k = 0.5
τ = 0.25

4
Figure 2: Closed loop transfer function response (Unity feedback)
System information for step input from 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 (iii), T (s )= G(s )
1+G( s) H (s) ...... (vi)
G(s) as obtained from part a 1
0.5 s+ 1
Using a similar analogy as part (b),

5
G( s) H (s)=( 1
0.5 s +1 )×0.5= 0.5
0.5 s+1 ...... (vii_a)
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 ...... (vii_b)
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 ...... (vii_c)
Therefore, closed loop transfer function is:
T (s )= 1
0.5 s+1.5 ...... (viii_a)
Changing to its canonical form
T (s )= 0.67
0.33 s+1 ...... (8b)
From which
k = 0.667
τ = 0.33

6
Figure 3: Closed loop transfer function response (H(s) = 0.5)
System information for step input from MATLAB:
RiseTime: 0.7323
SettlingTime: 1.3040
SettlingMin: 0.6030
SettlingMax: 0.6666
Overshoot: 0
Undershoot: 0
Peak: 0.6666
PeakTime: 3.5153

7
d) A comparison of open loop and closed system
Table 1: First order system response summary from MATLAB response
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
From the system response characteristics, percentage overshoot all the systems is
at 0%. This support the fact that a first order system has no overshoot. A feedback
path however influence the value of dc gain k and time constant, τ. The results shows
that applying a unity feedback reduces the values of the two parameters; dc gain k and
time constant, τ by ½. In order to restore the desired steady state response involves
doubling of the gain so as to get the desired steady state value. For a H(s) = 0.5
feedback path, both the gain and time constant are reduced by a 1/3. This suggest that
application path in a first order system affect time constant and dc gain in equal
proportions.
The findings also reveal interesting information on the effects of a feedback path
on the transient response of a first order system. The rise time and settling time seems
shows a significant improvement when a feedback path is added. A feedback system
therefore improve system transient characteristics by ensuring that the control system
takes the least possible time to settle and to rise. Depending on application, this might
not be desirable in some cases. 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 variations in transient response characteristics are consistent with the fact that a
feedback system changes the τ, affection the location of poles and zeroes (not
applicable in 1st order system) thus influencing the desired response.

8
Second Order System
a) Determine the second order Transfer Function.
A second order system can be generally be modelled by the equation shown below.
G( s)= kss ωn
2
s2 +2 ζ ωn s +ωn
2 ...... (ix)
Settling time is given by:
T s= 4
ζ ωn
...... (x_a)
Peak time is given by:
T p= π
ωn √ 1−ζ2 ...... (x_b)
From (Xu & Ding, 2017),
From (x_a) ωn= 4
T s
= 4
4 =1=¿ ζ ωn =1 ...... (xi_a)
Tp = 0.5, therefore 3 9.478=ωn
2 (1−ζ2 ) ...... (xi_b)
From (11a), it can be expressed as ζ as: ζ = 1
ωn
...... (xi_c)
Substituting (xi_c) in (xi_b):
3 9.478=ωn
2 (1− 1
ωn
2 )=ωn
2 −1
Hence, ωn
2=40.478=¿ ωn= √ 40.478=6.36 ...... (xi_d)
Also, ζ = 1
ωn
= 1
6.36 =0.157 ...... (xi_e)
By inspection, from the response, final value = 2.5 = kss
Hence, ω = 6.36 , ζ = 0.157 and kss = 2.5

9
Substituting the values in (xi):
G(s)= 101.20
s2 +2 s+40.48 ...... (xii)
Figure 4: Open loop transfer function response
System information for step input from MATLAB:
RiseTime: 0.1857
SettlingTime: 3.6395
SettlingMin: 1.5835
SettlingMax: 4.0150
Overshoot: 60.6008
Undershoot: 0
Peak: 4.0150
PeakTime: 0.5066

10
b) Closed loop transfer function (unity negative feedback, H(s) = 1)
From equation (iii), T (s )= G(s )
1+G( s) H (s) ...... (iii)
H(s) = 1. Substituting and solving for T(s) using similar steps in section 1(b):
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
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
Therefore, (s)= 101.20
s2 +2 s+141.68 ...... (xiii)