. Two crucial aspects of this behavior are transient and steady-state responses. These concepts encapsulate how a system behaves over time, from the moment an input is applied to when the system settles into a stable state. The transient response of a system characterizes its behavior during the initial phase after a change in input. It reflects how the system reacts as it transitions from one state to another. This phase is marked by dynamic changes in the system's output as it adjusts to the new conditions imposed by the input. Characteristics of Transient Response are Time Constant, overshoot, settling time and damping. Once the transient effects have subsided, the system enters the steady-state, where its behavior becomes constant over time. In this phase, the system operates under stable conditions, and its output remains within a narrow range around the desired value, despite fluctuations in input or external disturbances. Characteristics of Steady-State Response are Steady-State Error, stability, accuracy, robustness,.

1. Transient and Steady
state Response

2. System Response
• After the engineer obtains a mathematical representation (model) of a
subsystem, the subsystem is analyzed for its transient and steady-state
responses to see if these characteristics yield the desired behavior.
• In analyzing and designing control systems, we must have a basis of
comparison of performance of various control systems. This basis
may be set up by specifying particular test input signals and by
comparing the responses of various systems to these input signals.
With these test signals, mathematical and experimental analyses of
control systems can be carried out easily, since the signals are very
simple functions of time.

3. Standard test input signals
The u(t) shows that the response is zero until t = 0

4. Standard test input signals
• An impulse; is infinite at t = 0 and zero else where, area under the
unit impulse is 1. An approximation of this type of waveform is used
to place initial energy into a system so that the response due to that
initial energy is only the transient response of a system.
• A step input; represents a constant command, such as position,
velocity, or acceleration. Typically the step input command is of the
same form as the output. For example, if the system's output is
position, the step input represents a desired position, and the output
represents the actual position. The designer uses step inputs because
both the transient response and the steady-state response are clearly
visible and can be evaluated.

5. Standard test input signals
• The ramp- input represents a linearly increasing command. For
example, if the system's output is position, the input ramp represents a
linearly increasing position, such as that found when tracking a
satellite moving across the sky with a constant speed. The response to
an input ramp test signal yields additional information about the
steady-state error.
• Parabolic inputs; is also used to evaluate a system's steady state error.
• Sinusoidal inputs- can also be used to test a physical system to arrive
at a mathematical model.

6. Which of these typical input signals to use for analyzing system
characteristics may be determined by the form of the input that the
system will be subjected to most frequently under normal operation.
If the inputs to a control system are gradually changing functions of
time, then a ramp function of time may be a good test signal. Similarly,
if a system is subjected to sudden disturbances, a step function of time
may be a good test signal and for a system subjected to shock inputs, an
impulse function may be best.
Once a control system is designed on the basis of test signals, the
performance of the system in response to actual inputs is generally
satisfactory
Standard test input signals

7. • The output response of a system is the sum of two responses:
• the forced response also called steady-state response or particular solution and
• the natural response or homogeneous solution.
• In this chapter we will discuss the response of first and second order
system. The order refers to the order of the equivalent differential
equation representing the system, the order of the denominator of the
transfer function after cancellation of common factors in the
numerator or the number of simultaneous first-order equations
required for the state-space representation.
System Response

8. Poles and Zeros of a First-Order System
The use of poles and zeros and their relationship to the time response of
a system is a technique used to derive the desired result by inspection.
• Poles of a Transfer Function
The poles of a transfer function are the values of the Laplace transform
variable, s, that cause the transfer function to become infinite. In other
words they are the roots of denominator of the transfer function.
• Zeros of a Transfer Function
The zeros of a transfer function are the values of the Laplace transform
variable, s, that cause the transfer function to become zero. In other
words they are the roots of the numerator of the transfer function.

9. Poles and Zeros of a First-Order System
• Given the transfer function, G(s) with a unit step input;
• a pole exists at s = -5 and zero exists at -2
• unit step response of the system can be found as follows;
• Thus;

10. Poles and Zeros of a First-Order System
Pole-zero plot of the system
Poles and zeros of the input and system system

11. Poles and Zeros of a First-Order System
Conclusion;
• 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.
• A pole on the real axis generates an exponential response of the form
𝑒−𝜎𝑡
where -𝜎 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.
• The zeros and poles generate the amplitudes for both the forced and
natural responses.

12. Unit step Response of first order system
• A first-order system without zeros
can be described by the transfer
function;
𝐺𝑐𝑙 𝑠 =
1
𝑇𝑠 + 1
• If the input is a unit step, where,
𝑅 𝑠 =
1
𝑠
,the Laplace transform of
the step response is;
𝐶 𝑠 = 𝑅(𝑠)𝐺 𝑠 =
1
𝑠(𝑇𝑠 + 1)
Taking the inverse transform, the step
response is given by;
𝑪 𝒕 = 𝑪𝒇 𝒕 + 𝑪𝒏 𝒕 = 𝟏 − 𝒆−𝒕
𝑻

13. Unit step Response of first order system
Equation
𝐶 𝑡 = 1 − 𝑒−𝑡
𝑇
states that initially the output 𝐶 𝑡 is zero and finally it becomes unity.
One important characteristic of such an exponential response curve 𝐶 𝑡
is that at 𝑡 = 𝑇 the value of 𝐶 𝑡 is 0.632,or the response 𝐶 𝑡 has
reached 63.2% of its total change. That is,
𝐶 𝑡 = 1 − 𝑒−1 = 0.632

14. Performance specification of first order system
for unit step input
Time constant 𝑻𝒄:
Is the time for 𝒆−𝒕
𝑻 to decay to 37% of
its initial value. Or, the time constant is
the time it takes for the step response to
rise to 63% of its final value.
At 𝑡 = 𝑇 ,𝒆−𝒕
𝑻 = 𝑒−1
= 0.37
Or 𝟏 − 𝒆−𝒕
𝑻= 0.63
the smaller the time constant 𝑇, the faster
the system response. Another important
characteristic of the exponential response
curve is that the slope of the tangent line
at 𝑡 = 0 is 1
𝑇, since
We can say pole is located at
reciprocal of time constant
.
The output would reach the final value at 𝑡 = 𝑇 if
it maintained its initial speed of response

15. Performance specification of first order
system for unit step input
Rise Time,𝑻𝒓
• is defined as the time for the
waveform to go from 0.1 to 0.9
of its final value.
𝑻𝒓 = 𝒕𝟗𝟎% − 𝒕𝟏𝟎%
𝑻𝒓 = 𝟐. 𝟑𝟏𝑻 − 𝟎. 𝟏𝟏𝑻 = 𝟐. 𝟐𝑻

16. Performance specification of first order
system for unit step input
2%, Settling time, 𝑻𝒔
• Settling time is defined as the
time for the response to reach,
and stay within, 2% of its final
value.
𝑻𝒔 = 𝟒𝑻

17. Performance specification of first order
system for unit step input
Example [page 160, Norman Nise]
A system has a transfer function, 𝐺 𝑠 =
50
𝑠+50
. Find the time constant,
𝑇𝑐 settling time,𝑇𝑠 and rise time,𝑇𝑟 of it’s step response.
Ans:
𝑇𝑐 = 0.02 s
𝑇𝑠 = 0.08 𝑠
𝑇𝑟 = 0.044 𝑠

18. Unit-Ramp Response of First-Order Systems

19. Unit-Impulse Response of First-Order
Systems

20. Second Order System
Whereas 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. Two quantities i.e. natural frequency and damping
ratio, can be used to describe the characteristics of the second-order transient
response just as time constants describe the first-order system response.
Natural Frequency, 𝝎𝒏
The natural frequency of a second-order system is the frequency of oscillation
of the system without damping.
Damping Ratio,𝜁
Is a quantity that compares the exponential decay frequency of the envelope
to the natural frequency. This ratio is constant regardless of the time scale of
the response.
Hence, 𝜁 =
𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑖𝑎𝑙 𝑑𝑒𝑐𝑎𝑦 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
𝑛𝑎𝑡𝑢𝑟𝑎𝑙 𝑓𝑟𝑒𝑞𝑢𝑛𝑐𝑦
=
𝜎
𝜔𝑛

21. Second Order System
General second order transfer function;
Consider the following servo system
Type equation here.
(a) Servo system;
(b) block diagram;
(c) simplified block
diagram.
The closed-loop transfer function of the system is
𝐶(𝑠)
𝑅(𝑠)
=
𝐾
𝐽𝑠2+𝐶𝑠+𝑘
=
𝐾
𝐽
𝑠2+𝐶
𝐽𝑠+𝐾
𝐽
In the transient-response analysis, it is convenient to
write
where 𝝈 is called the exponential decay frequency or
attenuation; 𝝎𝒏, the undamped natural frequency; and 𝜻,
the damping ratio of the system.

22. Second Order System
By definition, the natural frequency, 𝜔𝑛, is the frequency of oscillation of this
system.
𝜔𝑛 = 𝐾
𝐽 and 𝐾
𝐽 = 𝜔𝑛
2
𝜻 =
𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑖𝑎𝑙 𝑑𝑒𝑐𝑎𝑦 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
𝑛𝑎𝑡𝑢𝑟𝑎𝑙 𝑓𝑟𝑒𝑞𝑢𝑛𝑐𝑦(𝑟𝑎𝑑/𝑠𝑒𝑐)
=
𝜎
𝜔𝑛
∴ 𝐺𝑒𝑛𝑒𝑟𝑎𝑙 𝑠𝑒𝑐𝑜𝑛𝑑 𝑜𝑟𝑑𝑒𝑟 𝑠𝑦𝑠𝑡𝑒𝑚 𝑤𝑖𝑙𝑙 𝑙𝑜𝑜𝑘 𝑙𝑖𝑘𝑒;
𝑮 𝒔 =
𝝎𝒏
𝟐
𝒔𝟐 + 𝟐𝜻𝝎𝒏𝒔 + 𝝎𝒏
𝟐

23. Second Order System
• The dynamic behavior of the second-order system can then be
described in terms of two parameters 𝜻 and 𝝎𝒏. A second-order
system can display characteristics much like a first-order system or
depending on component values, display damped or pure oscillations
for its transient response.
• The poles tell us the form of the response without the tedious
calculation of the inverse Laplace transform.

24. Second order Unit-step Response
Critically damped response
• The system TF has two real and repeated poles at −𝝎𝒏
• Has a damping ratio,𝜁 = 1
• One term of the natural response is an exponential whose time constant is equal to the reciprocal of
the pole location. Another term is the product of time, t, and an exponential with time constant
equal to the reciprocal of the pole location.
𝒄 𝒕 = 𝟏 − 𝒆−𝝎𝒏𝒕 − 𝝎𝒏𝒕𝒆−𝝎𝒏𝒕,
• Is the fastest response without overshoot
poles on s-plane step response

25. Second order Unit-step Response
Under-damped second order system unit step response
• Has two complex poles at −𝜎𝑑 ± 𝑗𝜔𝑑, where, 𝜔𝑑= 𝜔𝑛 1 − 𝜁2
• Has damping ratio,0 < 𝜁 < 1
• The natural response is damped sinusoid with an exponential envelope whose time constant is
equal to the reciprocal of the pole's real part. The radian frequency of the sinusoid, the damped
frequency of oscillation, is equal to the imaginary part of the poles,
𝑪 𝒕 = 𝟏 −
𝟏
𝟏−𝜁𝟐
𝒆−𝜁𝝎𝒏𝒕
𝐬𝐢𝐧( 𝝎𝒅𝐭 + 𝝋),Where 𝝋-is phase angle
poles on s-plane step response

26. Second order Unit-step Response
Under-damped Response
approaches a steady-state value via a transient
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 in figure
on the right. This sinusoidal frequency is given
the name damped frequency of oscillation, 𝜔𝑑.
Finally, the steady-state response (unit step) was
generated by the input pole located at the origin.
Underdamped Response

27. Second order Unit-step Response
Underdamped second order system unit step response
C 𝑠 =
𝜔𝑛
2
𝑠(𝑠2+2𝜁𝜔𝑛𝑠+𝜔𝑛
2)
=
𝐴
𝑠
+
𝐵𝑠+𝐶
𝑠2+2𝜁𝜔𝑛𝑠+𝜔𝑛
2 , 𝑎𝑛𝑑 𝐴 = 1, 𝐵 = −1, 𝐶 = −2𝜁𝜔𝑛
Hence; 𝐶 𝑠 =
1
𝑠
−
(𝑠+2𝜁𝜔𝑛)
𝑠2+2𝜁𝜔𝑛+𝜔𝑛
2
Which can be written as;
C 𝑠 =
1
𝑠
−
𝑠+𝜁𝜔𝑛 +
𝜁
1−𝜁2
𝜔𝑛 1−𝜁2
𝑠+𝜁𝜔𝑛
2+𝜔𝑛
2 1−𝜁2 =
1
𝑠
−
𝑠+𝜁𝜔𝑛
𝑠+𝜁𝜔𝑛
2+𝜔𝑛
2 1−𝜁2 −
𝜁
1−𝜁2
𝜔𝑛 1−𝜁2
𝑠+𝜁𝜔𝑛
2+𝜔𝑛
2 1−𝜁2

28. Second order Unit-step Response
Under-damped second order system unit step response
Substituting 𝜔𝑑= 𝜔𝑛 1 − 𝜁2 and taking inverse Laplace transform;
and
𝐶 𝑡 = 1 − 𝑒−𝜁𝜔𝑛𝑡(cos 𝜔𝑑t +
𝜁
𝟏−𝜁𝟐
sin 𝜔𝑑t),
𝑪 𝒕 = 𝟏 −
𝟏
𝟏−𝜁𝟐
𝒆−𝜁𝝎𝒏𝒕
𝐬𝐢𝐧( 𝝎𝒅𝐭 + 𝝋), Where φ = 𝑡𝑎𝑛−1
(
1−𝜁2
𝜁
)

29. Second-order underdamped unit-step
responses for different damping ratio values
The lower the value of 𝜁, the more
oscillatory the response will be.
𝑪 𝒕 = 𝟏 −
𝟏
𝟏−𝜁𝟐
𝒆−𝜁𝝎𝒏𝒕
𝐬𝐢𝐧( 𝝎𝒅𝐭 + 𝝋),
.

30. Second order Unit-step Response
Undamped response
• Has two imaginary poles at ±𝑗𝝎𝒏
• Has a damping ratio,𝜁 = 0
• Natural response is undamped sinusoid with radian frequency equal to the
imaginary part of the poles, the absence of a real part in the pole pair
corresponds to an exponential that does not decay.
𝐂 𝐭 = 𝐀𝐬𝐢𝐧(𝝎𝒏𝐭 + 𝝋), is natural response , where A-is constant
poles on s-plane step response

31. Second order Unit-step Response
Overdamped Response
• Has real and distinct poles at , −𝜎1 𝑎𝑛𝑑 − 𝜎2
• Has a damping ratio,𝜁 > 1
• Natural response is two exponentials with time constants equal to the reciprocal of the
pole locations;
𝒄 𝒕 = 𝟏 − 𝑨𝒆−𝝈𝟏𝒕
+ 𝑩𝒆−𝝈𝟐𝒕
, where 𝐴 and 𝐵 𝑎𝑟𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠
poles on s-plane step response

32. Second order system with zero
𝐺 𝑠 =
(𝑠 + 𝑧)
𝜔𝑛
2
𝑧
𝑠2 + 2𝜁𝜔𝑛𝑠 + 𝜔𝑛
2
Which can be written as;
𝐺 𝑠 =
𝜔𝑛
2
𝑠2+2𝜁𝜔𝑛𝑠+𝜔𝑛
2 +
𝑠
𝑧
(
𝜔𝑛
2
𝑠2+2𝜁𝜔𝑛𝑠+𝜔𝑛
2)
The response will be,
𝑐𝑧 𝑡 = 𝑐 𝑡 +
1
𝑧
𝑑
𝑑𝑡
𝑐 𝑡
The effect of zero is contribute pronounced
early peaking effect to the response. The closer
the zero to the dominant pole, the more
pronounced the peaking phenomenon.
Consider a 2nd order system with
poles 𝑠12 = −1 ± 𝑗2.83

33. Performance specification of unit step
Second order under damped Response
Other parameters associated with the
underdamped response are rise time,
peak time, percent overshoot and
settling time.
• Rise time, 𝑻𝒓; The time required for
the waveform to go from 0 to 100% of
the final value.
𝑻𝒓 =
𝝅−𝝋
𝝎𝒅
𝝋 − 𝑖𝑛 𝑟𝑎𝑑𝑖𝑎𝑛
• Peak time, 𝑻𝒑; The time required to
reach the first, or maximum, peak.
𝑻𝒑 =
𝝅
𝝎𝒏 𝟏 − 𝜁𝟐

34. Performance specification of unit step
Second order under damped Response
• Percent 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.
%𝑶𝑺 = 𝒆
−(
𝜁𝝅
𝟏−𝜁𝟐
)
× 𝟏𝟎𝟎
and
𝜁 =
−𝒍𝒏(%𝑶𝑺
𝟏𝟎𝟎)
𝝅𝟐 + 𝒍𝒏𝟐(%𝑶𝑺
𝟏𝟎𝟎)

35. Performance specification of unit step
Second order under damped Response
• 𝟐% , Settling time 𝑻𝒔. The time
required for the transient's
damped oscillations to reach and
stay within 𝟐% of the steady-
state value.
𝑻𝒔 =
𝟒
𝜁𝝎𝒏

36. Exercise
Given the transfer function;
𝐺 𝑠 =
100
𝑠2 + 15𝑠 + 100
1. Find the unit step response
2. What is the nature of the response?
3. Find 𝑇𝑝, 𝑇𝑠, 𝑇𝑟 𝑎𝑛𝑑 %𝑂𝑆 and

37. Exercise-Answer
Answer: 𝐶 𝑡 = 𝐴 + 𝐵𝑒−7.5𝑡 sin 6.61𝑡 + 𝜑
𝜔𝑛 = 10, 𝜁 = 0.75, 𝑇𝑟 ≈ 0.22, 𝑇𝑝, = 0.475,
𝑇𝑠 = 0.533, %𝑂𝑆 = 2.84
The response is underdamped

38. Exercise 2
Consider the system shown in figure below, where 𝜁= 0.6 and 𝜔𝑛 =
5𝑟𝑎𝑑/𝑠𝑒𝑐.
Obtain the closed loop TF, rise time, peak time, percent overshoot, and
2% settling time, when the unity feedback system is subjected to a unit-
step input.
Plot the step response in MATLAB

39. Steady State Error
Imperfections in the system components, such as static friction, backlash,
and amplifier drift, as well as aging or deterioration, will cause errors at
steady state. In this section, however, we shall not discuss errors due to
imperfections in the system components. Rather, we shall investigate a
type of steady-state error that is caused by the incapability of a system to
follow particular types of inputs.
Control systems may be classified according to their ability to follow
step inputs, ramp inputs, parabolic inputs, and so on. This is a reasonable
classification scheme, because actual inputs may frequently be
considered combinations of such inputs. The magnitudes of the steady-
state errors due to these individual inputs are indicative of the goodness
of the system.

40. Steady State Error
Consider the unity-feedback control system with the following open-
loop transfer function G(s):
It involves the term 𝑠𝑁 in the denominator, representing a pole of
multiplicity 𝑁 at the origin. The present classification scheme is based on
the number of integrations indicated by the open-loop transfer function.
A system is called type 0, type 1, type 2, , , if N=0, N=1, N=2, , ,
respectively. Note that this classification is different from that of the
order of a system. As the type number is increased, accuracy is
improved. However, increasing the type number aggravates the stability
problem. A compromise between steady-state accuracy and relative
stability is always necessary.

41. Steady State Error
For the system shown below the closed-loop transfer function is

42. Steady State Error
Steady State Error constants
In a given system, the output may be the position, velocity, pressure,
temperature, or the like. The physical form of the output, however, is
immaterial to the present analysis. Therefore, in what follows, we shall
call the output “position,” the rate of change of the output “velocity,”
and so on.
The error constants 𝐾𝑝, 𝐾𝑣, and 𝐾𝑎 describe the ability of a unity-
feedback system to reduce or eliminate steady-state error. Therefore,
they are indicative of the steady-state performance. It is generally
desirable to increase the error constants by adding integrators to the feed
forward path, while maintaining the transient response within an
acceptable range.

43. Steady State Error
Steady State Error Constants

44. Steady State Error
This shows type 0 system is incapable of following a ramp input in the steady state.
The type 1 system with unity feedback can follow the ramp input with a finite error.
Steady State Error Constants

45. Steady State Error
Steady State Error Constants

46. Steady-State Error, System type and Error
constants

47. Example
Figure shows a mechanical vibratory system. When 9N of force (step
input) is applied to the system, the mass oscillates, as shown in Figure
below.
a) Find transfer function
b) Determine m, b, and k of the system from this response curve. The
displacement x is measured from the equilibrium position.

48. Stability
In Chapter 1 we have seen the requirements in the design of a control
system: transient response, stability, and steady-state errors. Thus far we
have covered transient response and steady state error. We are now
ready to discuss the next requirement, stability. Stability is the most
important system specification. If amoot system is unstable, transient
response and steady-state errors are points.
Physically, an unstable system whose natural response grows without
bound can cause damage to the system, to adjacent property, or to
human life. Many systems are designed with limit stops to prevent total
runaway. From the perspective of the time response plot of a physical
system, instability is displayed by transients that grow without bound
and consequently, a total response that does approach a steady-state
value or other forced response is stable.

49. Stability
There are many definitions for stability, depending upon kind of system or the
point of view. In this course we limit ourselves to linear, time-invariant
systems.
Definitions of stability for linear, time-invariant systems; using natural
response
1.A system is stable if the natural response approaches zero as time approaches infinity.
2.A system is unstable if the natural response approaches infinity as time approaches
infinity
3.A system is marginally stable if the natural response neither decays nor grows but
remains constant or oscillates.
Thus, the definition of stability implies that only the forced response remains
as the natural response approaches zero.
Bounded-input, Bounded Output (BIBO)) definition of stability; using the
total response (BIBO):
1.A system is stable if every bounded input yields a bounded output.
2.A system is unstable if bounded input yields an unbounded output.

50. Stability
Stable system Unstable system

51. Stability
If the closed-loop system poles are in the left half of the s-plane and
hence have a negative real part, the system is stable. And unstable
systems have closed-loop transfer functions with at least one pole in the
right half-plane and/or poles of multiplicity greater than one on the
imaginary axis.
Marginally stable systems have closed-loop transfer functions with only
imaginary axis poles of multiplicity 1 and poles in the left halfplane.
Unfortunately it is not always a simple matter to determine if a
feedback control system is stable due to difficulty of solving for the
roots of high order transfer function. There is, however, another
methods to test for stability without having to solve for the roots of the
denominator.

52. Routh-Hurwith Criterion
Using this method, we can tell how many closed-loop system poles are
in the left half-plane, in the right half-plane, and on the 𝑗𝜔-axis, but we
cannot find their coordinates.
The method requires two steps:
(I) Generate a data table called a Routh table and
(II) interpret the Routh table to tell how many closed-loop system poles
are in the left half-plane, the right half-plane, and on the on 𝑗𝜔 -axis.

53. Routh-Hurwith Criterion
Generating a Basic Routh Table
Consider the equivalent closed-loop transfer function;
Since we are interested in the system poles, we focus our
attention on the denominator.
We first create the Routh table shown. Begin by labeling the
rows with powers of s from the highest power of the
denominator of the closed-loop transfer function to 𝑠𝑜 .
Next start with the coefficient of the highest power of s in
the denominator and list, horizontally in the first row, every
other coefficient. In the second row list horizontally, starting
with the next highest power of s, every coefficient that was
skipped in the first row.
Initial layout for routh able

54. Routh-Hurwith Criterion
Generating a Basic Routh Table
The remaining entries are filled in as follows. Each
entry is a negative determinant of entries in the previous
two rows divided by the entry in the first column
directly above the calculated row. The left-hand column
of the determinant is always the first column of the
previous two rows, and the right-hand column is the
elements of the column above and to the right. The table
is complete when all of the rows are completed down to
𝑠𝑜.
 Routh-Hurwitz criterion declares that the number of
roots of the polynomial that are in the right half-plane
is equal to the number of sign changes in the first
column.
If the closed-loop transfer function has all poles in the
left half of the s-plane, the system is stable. Thus, a
system is stable if there are no sign changes in the first
column of the Routh table.
Completed routh able

55. Routh-Hurwith Criterion
Example
Decide whether the following system is stable or not using Routh
stability criterion
Dorm work: Decide whether the following system is stable or not using
Routh stability criterion if of the closed loop transfer function is
𝐺𝑐𝑙(s) =
1
𝑠4+8𝑠3+18𝑠2+16𝑠+5

56. Routh-Hurwith Criterion: Special cases
First case: Zero only in the first column
If the first element of a row is zero, division by zero would be required
to form the next row which will result in infinity. To avoid this
phenomenon, an epsilon, 𝜀 is assigned to replace the zero in the first
column. The value 𝜀 is then allowed to approach zero from either the
positive or the negative side, after which the signs of the entries in the
first column can be determined.
Example: Determine the stability of closed loop transfer function;
𝐺𝑐𝑙 s =
10
𝑠5 + 2𝑠4 + 3𝑠3 + 6𝑠2 + 5𝑠 + 3

57. Routh-Hurwith Criterion: Special cases
Second case: Entire Row is Zero
Sometimes while making a Routh table, we find that an entire row
consists of zeros because there is an even polynomial that is a factor
of the original polynomial with roots that are symmetrical about the
origin. This case must be handled differently from the case of a zero in
only the first column of a row.
Example: Decide whether the following system is stable or not using
Routh stability criterion if characteristics equation of the closed loop
transfer function is
𝑠5 + 𝑠4 + 4𝑠3 + 24𝑠2 + 3𝑠 + 63

58. Routh-Hurwith Criterion: Special cases
An entire row of zeros will appear in the Routh
table when a purely even or purely odd
polynomial is a factor of the original polynomial.
Even polynomials only have roots that are
symmetrical about the origin. This symmetry can
occur under three conditions of root position:
(I) The roots are symmetrical and real,(A)
(II) the roots are symmetrical and imaginary, (B)
(III) the roots are quadrantal(C).
Each of these three case or combination of these
cases will generate an even polynomial.

59. Routh-Hurwith Criterion: Special cases
Another characteristic of the Routh table is that the row previous to the row
of zeros contains the even polynomial that is a factor of the original
polynomial. Every entry in the table from the even polynomial's row to the
end of the chart applies only to the even polynomial. Therefore, the number
of sign changes from the even polynomial to the end of the table equals the
number of right-half-plane roots of the even polynomial. Because of the
symmetry of roots about the origin, the polynomial must have the same
number of left-half-plane roots as it does right-half-plane. Having for the
roots in the right and left half-planes, we know the remaining roots must be
on the 𝑗𝜔 -axis.
Every row in the Routh table from the beginning of the chart to the row
containing the even polynomial applies only to the other factor of the original
polynomial. For this factor the number of sign changes, from the beginning of
the table down to the even polynomial, equals the number of right-half-plain
roots. The remaining roots are left-half-plane roots. There can be no 𝑗𝜔 roots
contained in the other polynomial.

60. Routh-Hurwith Criterion: Special cases
Dorm work: Determine the number of right-half-plane poles in the
closed-loop transfer function;
𝐺𝑐𝑙 s =
10
𝑠5 + 7𝑠4 + 6𝑠3 + 42𝑠2 + 8𝑠 + 56