ME-314 Introduction to Control Engineering is a course taught to Mechanical Engineering senior undergrads. The course is taught by Dr. Bilal Siddiqui at DHA Suffa University. This lecture is about time response of systems derived by inspection of poles and zeros. First and second order systems are considered, along with higher order and nonminimum phase systems

1. ME-314 Control Engineering
Dr. Bilal A. Siddiqui
Mechanical Engineering
DHA Suffa University

2. Time Response of Systems through
Poles and Zeros
• With transfer functions, it is easy to represent interconnected
components of a system graphically.
• Each transfer function is represented as a block, with inputs and
outputs.
• One way to use transfer functions is to multiply it with the input
𝑌 𝑠 = 𝐺 𝑠 𝑅 𝑠 and take the inverse Laplace 𝑦 𝑡 = 𝕷−1 𝑌 𝑠
• But, this is tedious and time consuming.
• We seek a technique which yields results in minimum time.
• This can be done easily by just looking at roots of transfer function
numerator and denominator,
R(s) Y(s)

3. System Output (Nise)
• Response of a system is solution of the differential equations
describing it.
• This response is composed of two parts:
– Particular solution – also known as Forced Response
– Homogenous solution – also known as Natural response
• Let the transfer function be 𝐺 𝑠 = 𝑁(𝑠)/𝐷 𝑠
• Roots of denominator D(s) are called poles of G(s)
• Roots of numerator N(s) are called zeros of G(s)
• For the transfer function G 𝑠 =
𝐶 𝑠
𝑅 𝑠
=
𝑠+2
𝑠+5
, unit step response
is found as 𝐶 𝑠 = 𝐺 𝑠 R s =
s+2
s s+5

4. Poles and Zeros and system response
• On the s-plane, poles
are plotted as ‘x’ and
zeros as ‘o’.
• Recall 𝑠 = 𝜎 + 𝑗𝜔
• 𝜎 determines
exponential decay/
growth
• 𝜔 determines
frequency of
oscillation

5. Recall -Imaginary Frequency Domain

6. Recall - Complex Frequency Domain
Therefore, Laplace transform is a better
representation for general functions.
S-domain is the general complex plain.

7. Some observations on P’s and Z’s
• A pole of input function generates the form of forced response
(i.e., pole at origin generated a step function at output)
• A pole of transfer function generates the form of natural
response (the pole at -5 generated e-5t).
• A pole on real axis generates an exponential response of form e-αt
where α is pole location on real axis. Thus, the farther to the left
a pole is on the negative real axis, the faster the transient
response will decay to zero (again, the pole at 5 generated e-5t ).
• Zeros and poles generate amplitudes for both forced and natural
responses
• We will learn to write response by just inspecting poles and zeros

8. An Example (Nise)
• Write the output c(t). Specify the forced and
natural parts of the solution

9. Class Quiz 02
• You have 2 minutes to solve this.

10. First Order Systems
• 1st Order Systems: A simple pole in the denominator
• If the input is a unit step, where R(s)=1/s
 Prove this

11. Time Constant
• We call TC=1/a the “time
constant” of the response.
• The time constant can be
described as the time for the step
response to rise to 63% of its
final value.
• Since derivative of e-at is -a,
when t=0, a is the initial rate of
change of the exponential at t=0.

12. Significance of Time Constant
• Time constant can be considered a transient
response specification for a 1st order system,
since it is related to the speed at which the
system responds to a step input.
• Since the pole of the transfer function is at -a,
we can say the pole is located at the reciprocal
of the time constant, and the farther the pole
from the imaginary axis, the faster the transient
response (T=1/a)

13. Rise and Settling Times

14. System Identification of 1st Order
Systems
• There are many 1st order systems around us
– DC Motor
– Chemical reactors in chemical plants
• Often it is not possible or practical to obtain a
system’s transfer function analytically.
• Perhaps the system is closed, and the component parts
are not easily identifiable.
• With a step input, we can measure the time constant
and the steady-state value, from which the transfer
function can be calculated.
The motor transfer
function can simply be
written as
𝝎 𝒎 𝒔
𝑬 𝒂 𝒔
=
𝑲
𝒔 + 𝜶

15. Example of Sys ID of 1st Order Sys
• We determine that it has 1st order
characteristics (No overshoot
and nonzero initial slope).
• From the response, we measure the time
constant, that is,
the time for the amplitude to reach 63% of
its final value.
• Since the final value is about 0.72, and the
curve reaches 0.63x0.72 =
0.45, or about 0.13 second. Hence,
a=1/0.13= 7.7.
To find K, we realize that the
forced response reaches a steady
state value of K/a= 0.72.
Substituting the value of a, we
find K = 5.54.
𝐺 𝑠 =
5.54
𝑠 + 7.7

16. Another Example

17. Second Order Systems
• Compared to the simplicity of a first-order
system, a second-order system exhibits a wide
range of responses that must be analyzed and
described.
Varying a first-order system’s parameter
simply changes the speed of the response,
changes in the parameters of a second-order
system can change the form of the response.
• Let us look at some examples

18. Step Responses of 2nd Order Systems

19. Step Responses of 2nd Order Systems

20. Under-damped Response
• The transient response consists of an exponentially decaying amplitude
generated by the real part of the system pole times a sinusoidal waveform
generated by the imaginary part of the system pole.
• The time constant of the exponential decay is equal to the reciprocal of the
real part of the system pole.
• The value of the imaginary part is the actual frequency of the sinusoid, as
depicted. This sinusoidal frequency is given the name damped frequency of
oscillation, 𝜔 𝑑.

21. Example
• By inspection, write the form of the step response
• First, we find the form of the natural response.
• Factoring denominator of transfer function, poles at 𝑠 = −5 ± 𝑗13.23
• The real part, -5, is the exponential rate of decay of transient response
• It is also the reciprocal of time constant of decay of oscillations.
• The imaginary part, 13.23, is the frequency (rad/sec) for the sinusoidal
oscillations.
• See the previous slide. It is obvious the response should be like
• This is an exponential plus a damped sinusoid.

22. Undamped Response
• This function has a pole at the origin that comes from the unit step
input and two imaginary poles that come from the system.
• The input pole at the origin generates the constant forced response,
and the two system poles on the imaginary axis generate a sinusoidal
natural response whose frequency is equal to the location of the
imaginary poles.
• Note that the absence of a real part in the pole pair corresponds to an
exponential that does not decay.
• In general, 𝑐 𝑡 = 𝐾1 + 𝐾4 cos(𝜔𝑡 − 𝜙)

23. Critically Damped System
• This function has a pole at the origin that comes from the unit step
input and two multiple real poles that come from the system.
• The input pole at the origin generates the constant forced response,
and the two poles on the real axis generate a natural response
consisting of an exponential and an exponential multiplied by time
• Hence, the output can be estimated as 𝑐 𝑡 = 𝐾1 + 𝐾2 𝑒−𝜎𝑡
+ 𝐾3 𝑡𝑒−𝜎𝑡
• Critically damped responses are the fastest possible without the
overshoot that is characteristic of the under-damped response.

24. Summary of 2nd Order System Step Responses

25. Summary of 2nd Order System Step
Responses

26. Assignment 4a

27. The General 2nd Order System
• 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.

28. Natural Frequency, 𝜔 𝑛
Damping Ratio, 𝜁
• The natural frequency of a 2nd -order system is the
frequency of oscillation of the system without
damping. For example, the frequency of oscillation of
a spring-mass sys is 𝑘/𝑚.
• Exponential frequency: Since the time constant
(a=1/TC) has the units of 1/sec, we call it the
‘exponential frequency’. For a second order system,
this is the real part of the complex pole.
• Damping Ratio is ratio of frequency of exponentially
decaying oscillations to natural frequency

29. Coming back to our cart example

30. Standard Form of Second Order System
• Transfer function of this 2nd order system 𝐺 𝑠 =
1
𝑚𝑠2+𝑏𝑠+𝑘
can be rewritten as
𝑮 𝒔 =
𝝎 𝒏
𝟐
𝒔 𝟐 + 𝟐𝜻𝝎 𝒏 𝒔 + 𝝎 𝒏
𝟐
• It is obvious that 𝜔 𝑛 =
𝑘
𝑚
is the undamped natural frequency
of the system if the damper is removed from the system.
• The damping ratio can be shown to be 𝜁 =
𝑏
2 𝑘𝑚
• The transfer function in bold above is called the standard form
of 2nd order systems. It is very useful.
Solving for poles of transfer function yields
𝑠1,2 = −𝜁𝜔 𝑛 ± 𝑗𝜔 𝑛 1 − 𝜁2
𝜔 𝑑 = 𝜔 𝑛 1 − 𝜁2 is also
called damped natural
frequency

32. Assignment 4b
• For each system, find value of 𝜁, 𝜔 𝑛 and report type of response
Hint

33. General Underdamped 2nd Order Systems
• Our first objective is to define transient specifications
associated with underdamped responses.
• We relate these specifications to pole location.
• Finally, we relate pole location to system parameters.
• Therefore, from user specification for response, we will find
appropriate values of system poles (i.e. damping and nat. freq).
• Step response of a gen 2nd order sys (𝜁 < 1 for underdamped)
• Taking the Inverse Laplace Transform
Remember: poles of
transfer function were
𝑠1,2
= −𝜁𝜔 𝑛 ± 𝑗𝜔 𝑛 1 − 𝜁2

34. Relationship between Damping Ratio
and Type of Response
• Lower the 𝜁, the more oscillatory the response

35. Second Order Underdamped Response
Specifications
• Rise time, Tr. Time required for
waveform to go from 0.1 of the final
value to 0.9 of the final value.
• Peak time, TP. Time required to reach
the first, or maximum, peak.
• Percent overshoot, %OS. The amount
that the waveform overshoots the
steadystate, or final, value at the peak
time, expressed as a percentage of the
steady-state value.
• Settling time, Ts. Time required for
damped oscillations to reach and stay
within 2% of the steady-state value

36. Expressions for Tr, TP, TS and %OS
• Without formal derivation (they are similar to those
for 1st order systems), we state that
Normalized rise time, 𝜔 𝑛 𝑇𝑟

37. Rise Time and %OS

38. Example: Find Specs from TF
Normalized rise time, 𝜔 𝑛 𝑇𝑟

39. Pole Locations and 2nd Order System
Parameters
• We know 𝜔 𝑑 = 𝜔 𝑛 1 − 𝜁2
• This damped natural
frequency is equal to the
imaginary part of the poles
• The exponential frequency,
which is the real part of the
poles is given as 𝜎 𝑑 = 𝜁𝜔 𝑛
• From the figure, it is clear
that cos 𝜃 = 𝜁
• The more the angle, the
higher the frequency of
oscillation (𝜔 𝑑) and the less
the actual damping (𝜎 𝑑)

40. System Poles and Specifications
• Since 𝑇𝑠 =
𝜋
𝜔 𝑑
it is vertical lines and 𝑇𝑃 =
𝜋
𝜎 𝑑
are
horizontal lines on the s-plane.
• Since 𝜁 = cos 𝜃, radial
lines are lines of constant
𝜁.
• Since percent overshoot is
only a function of 𝜁,
radial lines are lines of
constant percent
overshoot, %OS.

42. An Example (Nise)

43. Component Design

44. Assignment 4c
• Questions 4,5,6,7 and 14 of Chapter 4 of Nise.
• Problem 29 of Chapter 4 of Nise

45. Higher Order Systems (extra poles)
• So far, we analyzed systems with one or two poles.
• Formulas for overshoot, settling time, peak time derived only for
system with two complex poles and no zeros.
• If a system has more than two poles or has zeros, we cannot use the
formulas.
• Under certain conditions, a system with more than two poles or with
zeros can be approximated as a 2nd -order system that has just two
complex dominant poles. If this holds, we can use 2nd order formulars
• Consider an underdamped second order system with complex poles at
− 𝜁𝜔 𝑛 ± 𝑗𝜔 𝑛 1 − 𝜁2 and a real pole at −𝛼 𝑟. Step response:

46. If the real pole is five times farther to the left than the dominant
poles, we assume that the system is represented by its dominant
second-order pair of poles. Otherwise, this approx does not work

47. Example
c2, with its third pole at -10 and farther from the
dominant poles, is the better approximation of c1, the
pure 2nd -order system response; c3, with a third pole
close to the dominant poles, yields the most error.

48. System Response with Zeros
• We saw that zeros of a
response affect the
amplitude but do not
affect the shape of the
response.
• We add a real-axis zero
to a two-pole system.
• The zero will be added
first in left half-plane
and then in right half-
plane and its effects
noted and analyzed.

49. Adding zeros
• If we add a zero to the transfer function, yielding (s+a)T(s), the
Laplace transform of response will be
𝑠 + 𝑎 𝐶 𝑠 = 𝑠𝐶 𝑠 + 𝑎𝐶 𝑠
• Thus, the response consists of two parts:
– derivative of the original response, sC(s)
– scaled version of the original response, aC(s)
• If ‘a’ is very large, the response is approximately aC(s)
• As ‘a’ becomes smaller, derivative term has more effect.
• For step responses, the derivative is typically positive at
the start of a step response.
• Thus, for small values of a, we can expect more overshoot
than in 2nd order systems.

50. Example (zero in left half plane)
• Starting with a two-pole system with poles at
− 1 ± 𝑗2.828 , we add zeros at -3,-5, and -10.

51. Adding zeros in Right Half Plane
(Non-minimum Phase Systems)
• An interesting phenomenon occurs if ‘a’ is negative, placing
the zero in the right half-plane.
• Since 𝑠 + 𝑎 𝐶 𝑠 = 𝑠𝐶 𝑠 + 𝑎𝐶 𝑠 , we see that the
derivative term, SC(s) which is positive initially, will be of
opposite sign from the scaled response term, aC(s).
• Thus, if the derivative term, sC(s), is larger than the scaled
response, aC(s), the response will initially follow the
derivative in the opposite direction from the scaled response.
• A system that exhibits this phenomenon is known as a
nonminimum-phase system.

52. Real world examples of
Nonmimimum Phase Systems
• Imagine parallel parking where the output is the distance
from the kerb. When you first start to move backwards
while turning the steering wheel, the driver moves away
from the kerb before he/she moves closer. This is the reason
why parallel parking is hard and is a part of the driving test.
• Imagine roasting a chicken on a coal hearth. If the
temperature is too low, you add more coal to heat the
furnace up. But at first step, you actually achieve the
opposite, the temperature is reduced, because added coal
dumps the fire. In the second step, fire gets more power and
the temperature begins to rise.
• I took these examples from .