1. UNIVERSITI MALAYSIA PERLIS SCHOOL OF ELECTRICAL SYSTEM ENGINEERING
CHAPTER 3
Time
Response
EET302 : CONTROL SYSTEMS ENGINEERING
Dr. Norkharziana Mohd Nayan

2. Time Response Outlines Poles & Zero System Response Undamped
1st Order System
2nd Order System
Undamped
Under damped
Over damped
Dominant Pole
Stability : Routh Hurwitz Stability Criteria
Steady state
Unit step
Unit ramp
Unit parabola
CO2: Ability to analyze system’s response in time or frequency domain and
analyze control system problems by utilizing control system graphical
tools such as root locus or bode plot.

3. 3.1 Introduction TIME RESPONSE OF A CONTROL SYSTEM Transfer function
Defines as the output of the closed loop system as the function of time
Obtained by :
solving the diff. equation governing the system
Transfer function of the system
Transfer function
=mathematically modelling of a control system
=ratio of the output system to the input system
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How 6-Axis Industrial Robots Work

4. 3.1 Introduction
Output response of a system is the sum of 2 responses: forced & natural responses.
Poles:
Values of the Laplace transfom variable, s, that cause the TF to infinite.
Any roots of the denominator of the TF that are common to roots of the numerator.
Zeros
Values of the Laplace transfom variable, s, that cause the TF to zero.
*TF= Transfer Functions
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How 6-Axis Industrial Robots Work

5. Robotic ARM BLOCK DIAGRAM
Reconfigurable Robot- Arm. 
Robotic ARM BLOCK DIAGRAM
Ref:

6. The block diagram below shows 3 link robotic arm, it consists of 3 DC motor, quadrature encoder for each link. Each link of the simplified robotic arm can be controller seperately. The PID based control application block diagram for the control of a single link is shown bellow.

7. ELEVATOR CONTROL SYSTEM
Elevator Control System is the system responsible for coordinating all aspects of elevator service such as travel, speed, and accelerating, decelerating, door opening speed and delay, leveling and hall lantern signals. It accepts inputs (e.g. button signals) and produces outputs (elevator cars moving, doors opening, etc.).

8. 3.2 Poles, Zeros and System Response.
Tulis Eq. 4.1 kat whiteboard
A pole of the input function generates the form of the forced response (that is, the pole at the origin generated a step function at the output).
A pole of the transfer function generates the form of the natural response (that is, the pole at -5 generated e-5t).
A pole on the real axis generates an response of the form e-αt, where -α is the pole location on the real axis.
Thus, the farther to the left a pole is on the negative real axis, the faster the exponential transient response will decay to zero (again, the pole at 5 generated e5t; see Figure 4.2 for the general case).
The zeros and poles generate the amplitudes for both the forced and natural esponses (this can be seen from the calculation of A and B in Eq. (4.1)).
Continue Ex.4.1
G=Gain
R= Reference input
C=controlled variable
Figure 3.1: (a) System showing input and output; (b) Pole-zero plot of the system; (c) Evolution of a system response.

9. 3.3 First Order System. The transfer function,
For a unit step of; 1/s,
Its response,
A, B and B’ are constant. For K=1 and t=1/a then,
Time constant (1/a), is defined as the time for e-at to decay to 37%of its initial value or the time it takes for step response to reach 63% of its final value.
Time constant can be considered as a transient response spec for a 1st order syst.
Since its related to the speed at which the system response to a step input.

10. Figure 3.2: (a) First Order Response to a Unit Step.
3.3 First Order System.
Step response,
For a 1st order system with G(s)= K/(s+a), the step response is…
The output in time domain will be..
Figure 3.2: (a) First Order Response to a Unit Step.

11. 3.3 First Order System.
The pole of the transfer function is at –a, the farther the pole from the imaginary axis, the faster the transient response.
Rise time (Tr), the time the response to go from the 0.1 to 0.9 of its final value. Tr=2.2/a.
Settling time (Ts), time range when the response to reach and stay within 2% of its final value. Let c(t) = 0.98 then the Ts=4/a.

12. Example 3.3: Assume unit response given as input in figure below.
Find Tc, Tr, Ts, gain K and transfer function G(s) of the system.
98%
90%
63% Ts
10% Tr Tc

14. 3.4 Second Order System. The transfer function, For Impulse response,
Where,
Standard Form,
Where K is the dc gain,
 is the damping ratio, n is the undamped natural frequency.
Where

15. Figure 3.3: Second Order System, pole plots and Step Response.
To become familiar with the wide range of responses, we take a look at numerical examples of the second order system responses shown in Figure 4.7.
All examples are derived from Figure 4.7(a), the general case, which has two finite poles and no zeros.
The term in the numerator is simply a scale or input multiplying factor that can take on any value without affecting the form of the derived results.
By assigning appropriate values to parameters a and b, we can show all possible second-order transient responses.
The unit step response then can be found using C(s)=R(s)G(s), where R(s)=1/s, followed by a partial-fraction expansion and the inverse Laplace transform.

16. Figure 3.3: Second Order System, pole plots and Step Response.
Poles tell us the form of the response without the tedious calculation of the inverse laplace transform

17. General 2nd order TF
In this section, we define two physically meaningful specifications for second-order systems.
These quantities can be used to describe the characteristics of the second-order transient response just as time constants describe the first-order system response.
The two quantities are called natural frequency and damping ratio.
Compare G(s)

18. Plot for an underdamped σ
2nd Order System.
Step Response for 2nd Order System Damping Cases.
Relationship of ζ and ωn with pole location

19. Figure 3.5: Second Order Response as a Function of Damping Ratio.
We see that the various cases of second-order response area function of zeta; they are summarized in Figure

20. 3.4.1 Over Damped Response. The transfer function,
Poles at the origin from the unit step and two real poles from the system.
Constant force response and natural responses.

21. Example 3.5: Over Damped Response.
Find the step response of the system.
Solution:
Expand the partial fraction.
Take the inverse Laplace Transform.

22. 3.4.2 Under Damped Response. Under Damped transfer function,
When 0 < < 1 The transfer function is,
The Pole position is,

23. Figure 3.6: Second Order Response as a Function of Damping Ratio.

24. 3.4.2 Under Damped Response. Peak Time Settling Time
Peak Time
Settling Time
Rise time,Tr = Time required for the waveform to go from 0.1 to 0.9 of the final value.
Peak time, Tp= Time required to reach the first or maximum peak
Settling time, Ts= Time required for the transient’s damped oscillations to reach and stay within ±2% of the steady state value

25. Overshoot
% Overshoot, %OS = The amount that the waveform overshoots the steady state, or final, value at the peak time, expressed as a percentage of the steady-state value

26. Performance Measures. Poles position
Figure 3.8: Lines of constant peak time, Tp , settling time, Ts , and percent overshoot, %OS Note: Ts2 < Ts1 ; Tp2 < Tp1; %OS1 < %OS2

27. Pole Placement.
Figure 3.9: Step responses of second-order underdamped systems as poles move:
with constant real part;
with constant imaginary part;
with constant damping ratio

28. 3.4.3 Critically Damped.
The transfer function,

29. Example 3.6: Critically Damped Response.
Find the step response of the system.
Solution:
Expand the partial fraction.

30. Dominant Pole.
The formula that describing %OS, Ts, Tp were derived only for system with two complex poles and no zeros.
A system with more than two poles or zeros can be approximated as a second order system that has just two complex dominant poles. c

31. Dominant Pole.
Will approach second order system
Cannot be represented as second order system
Figure 3.11: Component responses of a three-pole system: (a) pole plot; (b) component responses: non-dominant pole is near dominant second-order pair (Case I), far from the pair (Case II), and at infinity (Case III).

32. Effect of adding a zero to a two-pole system
The closer is the zero to dominant poles, the greater its effect on transient response.
As the zero move away from dominant poles, the response approaches that of the two pole system.
Starting at poles 1±j2.828, then consecutively add zeros at -3, -5, -10.

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Kinestatic
Spacial/ Visual
Musical
Interpersonal
Intrapersonal
Intelligence

34. 3.5 Stability. (i) Stable system. Natural response approaches zero.
Poles in LHP.
(ii) Unstable system.
Natural response grows.
Poles in RHP.
(iii) Marginally stable system.
Natural response neither grows/approaches zero.
Poles on j axis.
Figure 3.12: Closed-loop poles and response: a. stable system; b. unstable system

35. 3.6 Routh-Hurwitz Stability Criteria.
What is Routh-Hurwitz Criterion (RHC)?
Through the RHC method we can tell how many close-loop system poles are in the left half plane, in the right half-plane and on the jw-axis. We can find the number of poles in each section of the s-plane, but cannot find their coordinate.
The number of roots of the polynomial that are in the right half-plane is equal to the number of changes in the first column.
The RHC method requires two steps;
(1) Generate the data table called Routh table.
(2) Interpret the Routh table to tell number of close loop system poles in the left half plane, in the right half-plane and on the jw-axis.

36. The Close-Loop Transfer
function.
Initial layout for the
Routh-Hurwitz Table.
Completed Routh Table.

37. Example 3.7: Routh-Hurwitz.
Make a Routh table from the system shown below.
Solution:
Find the equivalent close loop system.
Figure (b) above.
Interpretation:
There are two sign changes in the first column.
1  -72  103
The system is unstable, two poles exist in the right half plane.

38. Example 3.7: Routh-Hurwitz.
The number of RHP poles = The number of SIGN CHANGES of COL 1
TWO sign changes:
RHP Poles =2

39. Example 3.8: Routh-Hurwitz.
Solution:
Two sign changes: 2 RHP (UNSTABLE)
Poles: 2 LHP and 2 RHP

40. Example 3.9: Routh-Hurwitz.
Assume  is small POSITIVE : TWO sign changes
Poles: 2 RHP, 3 LHP
Solution:

41. Example 3.10: Routh-Hurwitz.
Solution:
Assume e is small positive: Two sign changes
Poles: 2 RHP, 3 LHP

42. Example 3.11: Routh-Hurwitz.
Solution:
NO sign changes: No RHP (STABLE)
Row of ZEROS indicate existence of complex poles & Symmetric Equations
Poles: 1 LHP and 4 on jw axis

43. Entire row is zero
When a purely even or odd polynomial is a factor of the original polynomial.
Even polynomial only have roots that are symmetrical about the origin.
The symmetry can occur under 3 conditions:
The roots are symmetrical and real.
The roots are symmetrical and imaginary.
The roots are quadrantal. j 2 3 1 1  3 2

44. Example 3.12: Routh-Hurwitz.
Solution:
2 sign changes: 2 RHP (symmetric)
Poles: 2 RHP, 4 LHP and 2 on j axis

45. Use of Routh Hurwitz Criteria
Main use is to determine the position of the poles, which in turns can determine the stability of the response.
Example
A closed-loop transfer function is given by
Determine the range for K for the system to be always stable and its oscillating frequency before it becomes unstable.

46. Solution: Characteristic equation is: Expand the equation
Form the Routh’s array s4 1 3 K s3 2 s2 s1 s0

47. Solution: For no sign change Referring to row 4 which gives,
and row 5,
Hence its range,
Oscillating frequency,

48. Steady state From the diagram Consider And
R(s)
E(s) + -
B(s)
From the diagram
Consider
And
Use the final value theorem and define steady state error, ess that is given by

49. Unit step Unit step input, From Steady state error,
We define step error coefficient,
Thus, the steady state error is
By knowing the type of open-loop transfer function,
we can know step error coefficient and thus the steady state error

50. Unit step SYSTEM TYPE= Value of n in the denominator
= no. of pure integrations in the fwd path
n=0 → Type 0 system
n=1 → Type 1 system
n=2 → Type 2 system
For open-loop transfer function of type 0:
For open-loop transfer function of type 1:
For open-loop transfer function of type 2:
Y(s)
R(s)
E(s) + -

51. Unit step
Example:
A first order plant with time constant of 9 sec and dc gain of 5 is negatively feedback with unity gain,
determine the steady state error for a unit step input and the final value of the output.
Solution:
The block diagram of the system is
As we are looking for a steady state error for a step input, we need to know,
Knowing the open-loop transfer function, then
And steady state error of,
Its final value is,
Y(s)
R(s) + -

53. Unit Ramp , while its Laplace form is
As in the above section, we know that
Thus, its steady state error is
Define ramp error coefficient,
Which the steady state error as
Just like for the unit step input we can conclude the steady state error for a unit ramp through the type
of the open-loop transfer function of the system.
For open-loop transfer function of type 0:
For open-loop transfer function of type 1:
For open-loop transfer function of type 2:

54. Unit Ramp - + Example: A missile positioning system is shown.
(i) Find its closed-loop transfer function
(ii) Determine its undamped natural frequency and its damping ratio if
(iii) Determine the steady state error, if the input is a unit ramp.
(iv) Cadangkan satu kaedah bagi menghapuskan ralat keadaan mantap untuk (iii).
Compensator DC motor + -

56. Unit Ramp
Solution:
(a) By Mason rule, the closed-loop transfer function is ,
(b) If
Comparing with a standard second order transfer function

57. Unit Ramp Comparing Thus undamped natural frequency rad.s-1 and
damping ratio of
(c) To determine the ramp error coefficient, we must obtain its open-loop transfer function
As it is a type 1, the system will have a finite ramp error coefficient, putting
Hence steady state error of

58. Unit Parabola Its time function ,while its Laplace
,thus its steady state error is
Define parabolic error coefficient,
Similarly we can determine its steady state error by knowing the type of the open-loop transfer function
For open-loop transfer function of type 0:
For open-loop transfer function of type 1:
For open-loop transfer function of type 2:

59.  Unit step Unit ramp Unit parabolic Type 0 Type 1 Type 2
In summary we can make a table of the steady state error for the above input
Unit step
Unit ramp
Unit parabolic
Type 0 
Type 1
Type 2

60. Quiz 2 (2%) ONE time answer ONLY 10 Q
Fill in correct Name, , Matrix & Programme