The document covers control engineering topics, specifically focusing on the stability of dynamical systems, including static and dynamic stability, system response, and the importance of poles in determining stability. It discusses feedback influences on system stability, the Routh-Hurwitz criterion for analyzing closed-loop stability, and steady-state errors resulting from system configuration and input types. It emphasizes the relationship between system configuration and steady-state errors, highlighting the role of gain and integrators in achieving desired performance.

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

2. Stability of Dynamical Systems
• Recall there are two types of
system stabilities:
– Static Stability: Initial tendency of
a system to come back to
equilibrium after disturbance. A
statically stable system is not
necessary stable. A statically
unstable system is not necessarily
unstable (e.g. nonminimum phase
systems).
– Dynamic Stability: A system may
tend initially towards equilibrium,
but may overshoot the mark and
this overshoot may keep increasing
with time. Such a system is
necessarily unstable. Dynamic
stability is a necessary and
sufficient condition of stability

3. Stability Requirements from
Control Systems
• System response consists of two parts
𝑐 𝑡 = 𝑐𝑓𝑜𝑟𝑐𝑒𝑑 𝑡 + 𝑐 𝑛𝑎𝑡𝑢𝑟𝑎𝑙(𝑡)
• For a control system to be useful, the natural response must
eventually approach zero, thus leaving only the forced response.
• Control systems must be designed to be stable, 𝑐 𝑛𝑎𝑡𝑢𝑟𝑎𝑙 ∞ → 0
• In many systems transient response you see on a time response
plot can be directly related to the natural response.
• If the natural response decays to zero as time approaches infinity,
the transient response will also die out, leaving only the forced
response.
• If the system is stable, the proper transient response and steady-
state error characteristics can be designed

4. System Stability in terms of Poles
• Shape of response is determined only by poles. Zeros
determine only the magnitude of response
• Poles in the left half of s-plane yield either pure
exponential decay (𝑒 𝜎𝑡) or damped sinusoidal
(𝑒 𝜎𝑡cos𝜔 𝑑 𝑡) natural responses.
• These natural responses decay to zero as time
approaches infinity.
• Stable systems have closed-loop transfer functions
with poles only in the left half-plane, i.e. real part
of all poles is negative (𝝈 < 𝟎)
• Unstable systems have at least one pole in right hand
plane; real part is positive (𝝈 > 𝟎)

5. Another type of instability
• A system is marginally stable if there are complex
pair of poles on imaginary axis (𝜎 = 0)
• Another type of instability can occur if the system has
repeated poles on the imaginary axis. This will cause
natural response to be of the type 𝐴𝑡 𝑛 cos 𝜔 𝑑 𝑡, where
n+1 is the number of repeated roots at imaginary axis.
• Unstable systems have closed loop transfer functions
with at least one pole in the right half-plane and/or
repeated poles on the imaginary axis.

6. Determining Stability of Systems when
Loop is Closed
• Feedback is generally useful. Too much feedback
can also destabilize a system originally stable
without feedback.
• Consider the system 𝐺 𝑠 =
3
𝑠 𝑠+1 𝑠+2
• We know this system is stable in open loop with
non-oscillatory response.
• Let us now close the loop. Is the system still stable
with feedback?

7. Stability of Close Loop Systems
• The closed loop transfer function is
𝐶 𝑠
𝑅 𝑠
=
𝐺
1 + 𝐺
=
3
𝑠 𝑠 + 1 𝑠 + 2 + 3
• So now, we have to find the poles of above
• Turns out the non-oscillatory open loos
response has now become oscillatory

8. Destabilizing Effect of Feedback
• Consider the system 𝐺 𝑠 =
7
𝑠 𝑠+1 𝑠+2
• We know this system is stable in open loop. We
also know the response is non-oscillatory
• Let us now close the loop. Is the system still stable
with feedback?
• Turns out, feedback is destabilizing in this case

9. An Important Observation
• If all closed loop transfer function poles are in left half-
plane, then they are of type (𝑠 + 𝛼𝑖) or (𝑠 + 𝜎𝑖 ∓ 𝑗𝜔 𝑑 𝑖
),
where 𝛼𝑖, 𝜎𝑖 > 0.
• No term in the polynomial can be missing, otherwise
there will be poles on imaginary axis or some positive
and negative coefficients got cancelled.
• Thus, a sufficient condition for a system to be stable is
that all signs of the coefficients of the denominator of
the closed-loop transfer function are the same. If
powers of s are missing, the system is either unstable
or, at best, marginally stable.

10. Routh Hurwitz Criterion – Reading
Assignment
• It’s a method that yields stability information
without the need to solve for the closed-loop
system poles.
• 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
imaginary-axis.
• The method is archaic and of practically no
use. I am not going to waste a whole lecture on
this, and so this is part of your reading
assignment. Watch these lectures [1, 2, 3]

11. Assignment 5a
• Watch these lectures [1, 2, 3] on Routh
Hurwitz method by Brian Douglas.
• Solve the following problems from Nise Ch 6
• Questions 42, 55, 57

12. Steady State Errors
• Control systems analysis and design focus on
three specifications:
– transient response (𝑇𝑠, 𝑇𝑝, 𝑇𝑟, %𝑂𝑆, 𝜁, 𝜔 𝑛)
– stability
– steady-state errors
• Steady-state error is the difference between the
input and the output for a prescribed test input
as 𝑡 → ∞.

13. Test Inputs
• Different test inputs are used depending on
which type of inputs need to be tracked.

14. Types of Test Inputs
• Step inputs represent constant position and thus are
useful in determining the ability
of the control system to position itself with respect
to a stationary target, such as a
satellite in geostationary orbit. An antenna position
control is an example of a system that can be tested
for accuracy using step inputs.
• A position control system that tracks a satellite that
moves across the sky at a constant angular velocity,
would be
tested with a ramp input to evaluate the steady-state
error between the satellite’s
angular position and that of the control system.
• Parabolas, whose second derivatives are constant,
represent constant acceleration inputs to position
control systems and can be used to represent
accelerating targets, such as a missile to determine
the steady-state
error performance.

15. Quantizing Steady State Errors
• In Fig (a), steady stay errors to step input are shown. Output 1
has zero steady-state error, and output 2 has a finite steady-state
error, 𝑒2 ∞ .
• In Fig (a), steady stay errors to ramp input are shown. Output 1
has zero steady-state error, Output 2 has a finite steady-state
error, 𝑒2 ∞ , and Output3 has infinite steady state error

16. Sources of Steady State Errors
• Steady state errors arise from the configuration of the system itself
and the type of applied input.
• Consider the following system
• Here E(s)=R(s)-C(s), so if 𝑒 𝑡 → 0, then 𝑐 𝑡 → 𝑟 𝑡 . But, 𝑐 𝑡 =
𝐾𝑒 𝑡 , so if 𝑐 𝑡 ≠ 0 then e(t) is also not zero!
• Due to system configuration (a pure gain K in the forward path), an
error must exist. 𝑐 ∞ = 𝐾𝑒 ∞ ⇒ 𝑒 ∞ =
1
𝐾
𝑐 ∞
• Larger the value of K, smaller the error!
• Another way to look at this is the final value theorem.
• From figure,
𝐶 𝑠
𝑅 𝑠
=
𝐾
1+𝐾
, then 𝐸 𝑠 = 𝑅 𝑠 1 −
𝐾
1+𝐾
=
𝑅 𝑠
1+𝐾
• For unit step input, 𝐸 𝑠 =
1
𝑠
1
1+𝐾
• Apply final value theorem, 𝑒 ∞ =
1
1+𝐾
, which proves steady state
error reduces with increasing gain.

17. Sources of Steady State Errors
• Consider now a system with integrator in forward path.
• Due to system configuration (integrator 1/s in the forward path), the
error goes down to zero
• As c(t) increases, e(t) will decrease, since e(t)=r(t)-c(t).
• This decrease will continue until there is zero error, but there will still
be a value for c(t) since an integrator can have a constant output
without any input.
• Since
𝐶 𝑠
𝑅 𝑠
=
𝐾
𝑠+𝐾
, then 𝐸 𝑠 = 𝑅 𝑠 1 −
𝐾
𝑠+𝐾
=
𝑠𝑅 𝑠
𝑠+𝐾
• For unit step input, 𝐸 𝑠 =
1
𝑠
𝑠
𝑠+𝐾
=
1
𝑠+𝐾
• Apply final value theorem, 𝑒 ∞ = lim
𝑠→0
𝑠
𝑠+𝐾
= 0, which proves steady
state goes to zero, regardless of value of gain.
• This proves steady state error is a function of system configuration
and type of test input.

18. Generalizing the Steady State Error for
Unity Feedback Systems
• Consider G(s) in forward loop with unity feedback
• Close loop tf is then 𝑇 𝑠 =
𝐺 𝑠
1+𝐺(𝑠)
• Since 𝐸 𝑠 = 𝑅 𝑠 − 𝐶 𝑠 = 𝑅 𝑠 − 𝑅 𝑠 𝑇 𝑠 = 𝑅 𝑠 1 − 𝑇 𝑠
• Apply the final value theorem
• Thus, 𝒆 ∞ = 𝐥𝐢𝐦
𝒔→∞
𝒔𝑹 𝒔 𝟏 − 𝑻(𝒔
• Similarly, C s = G s 𝐸 𝑠 ⇒ 𝐸 𝑠 =
𝑅 𝑠
1+𝐺 𝑠
• Thus, 𝒆 ∞ = 𝒍𝒊𝒎
𝒔→∞
𝒔
𝑹 𝒔
𝟏+𝑮 𝒔

19. Example
Verify this on Simulink!

20. Another Example
Verify this on Simulink!

21. Class Quiz 3
Hint:

22. Static Error Constants

23. Example
• Evaluate the static error constants and find the expected error for the
standard step, ramp, and parabolic inputs.
Verify this on Simulink!

24. System Type
• The values of the static error constants, again, depend upon the form of G(s),
especially the number of pure integrations (1/s) in the forward path.
• Since steady-state errors are dependent upon the number of integrations in the
forward path, we define system type to be the number of pure integrations in
the forward path.
• Therefore, a system with n = 0 is a Type 0 system. If n = 1 or n = 2, the
a Type 1 or Type 2 system, respectively.

25. Example

26. Assignment 5b