1. Design via Root Locus
CEN455: Dr. Nassim Ammour

2. Root Locus Techniques
Root locus is a graphical presentation of the closed-loop poles as a system parameter k is varied.
The graph of all possible roots of this equation (K is the variable parameter) is called the root locus.
The root locus gives information about the stability and transient response of feedback control systems.
Closed-loop system
Equivalent Transfer Function
Zeros of T(s) are zeros of G(s) and poles of H(s).
Poles of T(s) depends on gain K
CLCF is a function of K.
Root Locus graphically shows poles of T(s) as K varies
CEN455: Dr. Nassim Ammour

3. Magnitude and phase Of F(s) at S
Evaluation of a Complex Function via Vectors
Any complex number, 𝜎+𝑗𝜔, described in Cartesian coordinates can be graphically represented by a vector,
𝑀= 𝑧𝑒𝑟𝑜 𝑙𝑒𝑛𝑔𝑡ℎ𝑠 𝑝𝑜𝑙𝑒 𝑙𝑒𝑛𝑔𝑡ℎ𝑠 = 𝑖=1 𝑚 (𝑠+ 𝑧 𝑖 ) 𝑗=1 𝑛 (𝑠+ 𝑝 𝑗 )
Magnitude and phase Of F(s) at S
If 𝐹 𝑠 = 𝑖=1 𝑚 (𝑠+ 𝑧 𝑖 ) 𝑗=1 𝑛 (𝑠+ 𝑝 𝑗 )
𝜃= 𝑧𝑒𝑟𝑜 𝑎𝑛𝑔𝑙𝑒𝑠− 𝑝𝑜𝑙𝑒 𝑎𝑛𝑔𝑙𝑒𝑠= 𝑖=1 𝑚 ≺ 𝑠+ 𝑧 𝑖 − 𝑗=1 𝑛 ≺ 𝑠+ 𝑝 𝑗
Problem:
Given
Find F(s) at the point 𝑠=−3+𝑗4
Solution:
Any complex number can be represented by a vector
For zero ( point 𝑠 1 =−1) the vector is:
𝑠− 𝑠 1 =𝑠+1= −3+𝑗4 − −1 =−2+𝑗4= − 𝑡𝑎𝑛 −1 4 −2 =
For pole at 0:
For pole at -2:
Magnitude and phase
(polar form)
Vector magnitude,
Vector angle,
Vector
CEN455: Dr. Nassim Ammour

4. Poles location for different values of K
Defining the Root Locus
Poles location for different values of K
CEN455: Dr. Nassim Ammour

5. Defining the Root Locus (contd.)
Gain less than 25, over-damped.
Gain = 25, critically damped.
Gain over 25, under-damped.
Stable system, as no pole on right-hand plane.
During underdamped, real parts are same; so settling time (which is related to real part) remains the same.
Damping frequency (imaginary part) increases with gain, resulting in reduction of peak time.
CEN455: Dr. Nassim Ammour

6. Properties of Root Locus
The closed-loop transfer function 𝑇 𝑠 = 𝐾𝐺(𝑠) 1+𝐾𝐺 𝑠 𝐻(𝑠)
Module condition
Angle condition
𝑠 0 is a pole if 1+𝐾𝐺 𝑠 0 𝐻 𝑠 0 =0 ⟹𝐾𝐺 𝑠 0 𝐻 𝑠 0 =−1=1≺ 2𝑘 𝑘=0,±1,±2,…
𝐾𝐺 𝑠 𝐻 𝑠 =−1⟹ 𝐾𝐺 𝑠 𝐻 𝑠 =1⟹𝐾= 1 𝐺 𝑠 𝐻 𝑠
Find if the point -2+j3 is on root locus for some value of gain, K:
From the angle condition
For the point −2+j( ) which is on root locus, the gain K is:
Σ zero angle - Σ pole angle
𝐾= 𝑝𝑜𝑙𝑒 𝑙𝑒𝑛𝑔𝑡ℎ𝑠 𝑧𝑒𝑟𝑜 𝑙𝑒𝑛𝑔𝑡ℎ𝑠 = 𝐿 3 𝐿 4 𝐿 1 𝐿 2 = (1.22) (2.12)(1.22) =0.33
Not a multiple of So, −2+j3 is not in the root locus
(can not be a pole for some value of K).
CEN455: Dr. Nassim Ammour

7. Sketching the Root Locus
Number of branches: Equals the number of closed loop poles.
Symmetry: Symmetrical about the real axis (conjugate pairs of poles, real coefficients of the characteristic equation polynomial).
Real axis segments: For K > 0, root locus exists to the left of an odd number real axis poles and/or zeros (angle condition).
Real-axis segments of the root locus
Start and end points: The root locus begins at finite and infinite poles of 𝐺 𝑠 𝐻 𝑠 and ends at finite and infinite zeros of 𝐺 𝑠 𝐻 𝑠 .
Asymptotes: The root locus approaches straight lines as asymptotes as the locus approaches infinity. the equation of the asymptotes is given by:
Intersection with Real axis
𝜎 𝑎 = 𝑓𝑖𝑛𝑖𝑡𝑒 𝑝𝑜𝑙𝑒𝑠− 𝑓𝑖𝑛𝑖𝑡𝑒 𝑧𝑒𝑟𝑜𝑠 ≠𝑓𝑖𝑛𝑖𝑡𝑒 𝑝𝑜𝑙𝑒𝑠−≠𝑓𝑖𝑛𝑖𝑡𝑒 𝑧𝑒𝑟𝑜𝑠
Angle in radian of the asymptote with real axis
Complete root locus for the system
𝜃 𝑎 = (2𝑘+1)𝜋 ≠𝑓𝑖𝑛𝑖𝑡𝑒 𝑝𝑜𝑙𝑒𝑠−≠𝑓𝑖𝑛𝑖𝑡𝑒 𝑧𝑒𝑟𝑜𝑠 𝑓𝑜𝑟 𝑘=0,±1,±2,…
CEN455: Dr. Nassim Ammour

8. Sketching the Root Locus: Example
Problem: Sketch the Root Locus for the system shown in the following figure.
Solution: Calculate asymptotes to find real axis intercept:
The angles of the lines that intersect at −4 3 is given by:
For higher values of k, the angles would begin to repeat.
CEN455: Dr. Nassim Ammour

9. Sketching the Root Locus: Example
There are four poles and one finite zero. Root locus begins at poles and ends at zeros.
Where are the other zeros? At infinity.
Asymptote
Three zeros at infinity are at the ends of the asymptotes.
CEN455: Dr. Nassim Ammour

10. Refining the Sketch Real-Axis Breakaway and Break-in Points:
σ1: Breakaway point (leave the real axis);
σ2: Break-in point (return to the real axis ); .
Breakaway point: at maximum gain on the real axis between -2 and -1.
Break-in point: at minimum gain (increases when moving towards a zero) between +3 and +5.
CEN455: Dr. Nassim Ammour

11. Differentiating K with respect to 𝜎 (max and min)
Finding Breakaway and Break-in Points
One method to find break-in and break-away points is a variation on the transition method. Breakaway and break-in points satisfy
the relationship:
-1.45
3.82
Zi and pi are negative of zero and pole values, respectively, of G(s)H(s).
Example: From figure, we get
Method 2: Differentiation
𝐾𝐺 𝑠 𝐻 𝑠 = 𝐾(𝑠−3)(𝑠−5) (𝑠+1)(𝑠+2) = 𝐾( 𝑠 2 −8𝑠+15) ( 𝑠 2 −3𝑠+2)
Along the real axis ( 𝑠=𝜎) 𝑎𝑛𝑑 𝐾𝐺 𝑠 𝐻 𝑠 =−1
𝐾( 𝑠 2 −8𝑠+15) ( 𝑠 2 −3𝑠+2) =−1
Solving for K
𝐾= −( 𝜎 2 −3𝜎+2) ( 𝜎 2 −8𝜎+15)
𝑑𝐾 𝑑𝜎 = (11 𝜎 2 −26𝜎−61) ( 𝜎 2 −8𝜎+15) 2 =0
Differentiating K with respect to 𝜎 (max and min)
Solving for 𝜎,
𝜎=− 𝑎𝑛𝑑 𝜎=3.82
CEN455: Dr. Nassim Ammour

12. Imaginary-Axis Crossing
Stability: the system's poles are in the left half-plane up to a particular value of gain K.
PROBLEM: For the system, find the frequency and gain, K, for which the root locus crosses the imaginary axis. For what range of K is the system stable?
SOLUTION: The closed-loop transfer function
Characteristic Eq.:
We get Routh table as follows:
For K > 0, only 𝑠 1 row can be zero.
Forming the even polynomial by using the 𝑠 2 row (above) with K= 9.65,
Gives
Thus, the root locus crosses the imaginary-axis at ±j1.59 at a gain of 9.65 So, the system is stable for 0 ≤ K < 9.65
A complete row of zeros yields the possibility for imaginary-axis roots.
CEN455: Dr. Nassim Ammour

13. Angles of Departure and Arrival
Departure: from complex poles.
Arrival: to complex zeros.
Angle of departure:
Zero: -2
Poles:-3, -1+j, -1- j
we calculate the sum of angles drawn to a point 𝜀 close to the
complex pole, -1 + j
𝜃 3
𝜃 4
CEN455: Dr. Nassim Ammour

14. Improving System Response
Desired transient response
(percent overshoot and settling time)
Obtained transient response
(gain adjustment)
Improving Transient Response
We need to speed up the response at A to that B, without affecting the percent overshoot (compensation by adding poles and zeros in order to move the root locus and put to the desired pole on the root locus for some value of gain).
If the system gain is adjusted to meet the transient response specification, steady-state error performance deteriorated (The higher the gain, the smaller the steady-state error, but the larger the percent overshoot).
Dynamic compensators (function of s), can be designed to meet transient and steady-state error specifications simultaneously.
Transient response is improved with the addition of differentiation, and steady-state error is improved with the addition of integration in the forward path.
a. Sample root locus, showing possible design point via gain adjustment (A)
and desired design point that cannot be met via simple gain adjustment (B);
Improving Steady-State Error
b. Responses from poles at A and B
CEN455: Dr. Nassim Ammour

15. Compensators
Ideal compensators, pure integration PI for improving steady-state error or pure differentiation PD for improving transient response are implemented using active amplifiers. Compensators implemented with passive elements Lead and Lag are not ideal compensators (steady-state error is not driven to zero).
If a satisfactory design cannot be obtained by adjustment of the proportional gain alone, then dynamic compensator is used.
Compensator transfer function :
lead compensation if z < p and lag compensation if z > p.
Configurations
Both methods change the open-loop poles and zeros, thereby creating a new root locus that goes through the desired closed-loop pole location.
Cascade compensation
Feedback compensation
CEN455: Dr. Nassim Ammour

16. Ideal Integral Compensation (PI) Improving Steady-State Error
Steady-state error can be improved (without appreciably affecting the transient response) by placing an open-loop pole at the origin, because this increases the system type by one.
To solve the problem, we also add a zero close to the pole at the origin, the angular contribution of the compensator zero and compensator pole cancel out, point A is still on the root locus, and the system type has been increased.
If we add a pole at the origin to increase the system type,
the angular contribution of the open-loop poles at point A
is no longer 180°, and the root locus no longer goes through point A,
System operating with a desirable transient
response generated by the closed-loop poles at A
we have improved the steady-state error without appreciably affecting the transient response
Pole at A is:
on the root locus without compensator;
not on the root locus with compensator pole added;
approximately on the root locus with compensator pole and zero added
CEN455: Dr. Nassim Ammour

17. Example1
Given the system of Figure (a), operating with a damping ratio of 0.174, show that the addition of the ideal integral compensator shown in Figure (b) reduces the steady-state error to zero for a step input without appreciably affecting transient response.
Closed-loop system
a. before compensation;
b. after ideal integral
compensation
SOLUTION
We first analyze the uncompensated system to determine the location
of the dominant, second-order poles
The root locus for the uncompensated system is shown in Figure. A damping ratio of is represented by a radial line drawn on the s-plane at °.
Searching along this line we find that the dominant poles are ± j3.926 for a gain, K, of for the third pole on the root locus beyond —10 on the real axis.
for the same gain as that of the dominant pair, K = 164.6, we find that the third pole is approximately at
This gain yields Position constant 𝐾 𝑝 = lim 𝑠→0 𝐺(𝑠) = =8.23.
Hence, the steady-state error is:
CEN455: Dr. Nassim Ammour

18. Example 1-Conted
Adding an ideal integral compensator with a zero at - 0.1, we obtain the root locus shown in Figure.
Another section of the compensated root locus is between the origin and
Searching this region for the same gain at the dominant pair, K = 158.2, the fourth closed-loop pole is found at
Same transient response
Poles and gain are approximately the same
Damping Ratio unchanged (with K = 158:2).
Steady State Error is ZERO!.
Desired response
CEN455: Dr. Nassim Ammour

19. Implementation of an ideal integral compensator
The value of the zero can be adjusted by varying 𝐾 2 𝐾 1
The error and the integral of the error are fed forward to the plant G(s).
Made steady-state error zero.
However, it is expensive to implement as the integrator requires active elements.
We may want to use the solution presented next: Lag
Compensation.
PI controller implementation
CEN455: Dr. Nassim Ammour

20. Improving Steady-State Error
Lag Compensation
Improving Steady-State Error
Similar to the Ideal Integrator, however it has a pole not on the origin but close to the origin (fig c) due to the passive networks.
Steady State Improvement:
Before compensation: The static error constant,
KVo, for the system is:
After compensation:
The effect on the transient response is negligible:
If the lag compensator pole and zero are close together, the angular contribution of the compensator to point P is approximately zero degrees. point P is still at approximately the same location on the compensated root locus.
CEN455: Dr. Nassim Ammour
Root locus: a. before lag compensation; b. after lag compensation

21. Example2
Compensate the system of Figure (a), whose root locus is shown in Figure (b), to improve the steady-state error by a factor of 10 if the system is operating with a damping ratio of
SOLUTION
The uncompensated system error from Example 1 was with Kp = A tenfold improvement means a steady-state error of:
𝑒 ∞ = =0.0108, 𝑠𝑖𝑛𝑐𝑒 𝑒 ∞ = 1 1+ 𝐾 𝑝 ⇒ 𝐾 𝑝 =91.59
The improvement in Kp from the uncompensated system to the compensated
system is the required ratio of the compensator zero to the compensator pole:
𝑧 𝑐 𝑝 𝑐 = 𝐾 𝑝𝑁 𝐾 𝑃𝑜 = =11.13
Arbitrarily selecting
𝑝 𝑐 =0.01
𝑧 𝑐 = 𝑝 𝑐 ≈0.111
The compensated system
CEN455: Dr. Nassim Ammour

22. Example2-Conted
Comparison of the Lag-Compensated and the Uncompensated Systems
On the ζ= line
the second-order dominant poles are at —0.678 ±j3.836 with gain K=158.1
The third and fourth closed-loop poles are at and (on real axis)
The fourth pole of the compensated system cancels its zero.
Root locus for compensated system
CEN455: Dr. Nassim Ammour

23. Example2-Conted
The transient response of both systems is approximately the same with reduced steady state error
Less Steady State Error
the transient responses of the uncompensated
and lag-compensated systems are the same
CEN455: Dr. Nassim Ammour

24. Ideal Derivative Compensation (PD) Improving Transient Response
The objective is to design a response that has a desirable percent overshoot and a shorter settling time than the uncompensated system. (two approaches).
1. Ideal derivative compensation (Proportional-plus-Derivative (PD) active elements ): a pure differentiator is added to the forward path of the feedback control system.
sensitive to high frequency noise.
2. Lead Compensation: (not pure differentiation) approximates differentiation with a passive network by adding
a zero and a more distant pole to the forward-path transfer function.
Less sensitive to high frequency noise.
Distant pole
The transient response of a system can be selected by choosing an appropriate closed-loop pole location on the s-plane.
If this point is on the root locus, then a simple gain adjustment is all that is required in order to meet the transient response
specification.
If the closed loop root locus doesn’t go through the desired point, it needs to be reshaped (add poles and zeros in the forward path).
One way to speed up the original system is to add a single zero to the forward path.
𝐺 𝑐 𝑠 =𝑠+ 𝑧 𝑐
CEN455: Dr. Nassim Ammour

25. Ideal Derivative Compensation (PD) Improving Transient Response
see how it affects by an example of a system operating with a damping ratio of 0.4:
(a) uncompensated (b) compensated, zero at -2 (c) zero at -3 (d) zero at -4
Observations and facts:
In each case gain K is chosen such that percent overshoot is same.
Compensated poles have more negative real part (smaller settling time) and larger imaginary part (smaller peak time).
Zero placed farther from the dominant poles, compensated dominant poles move closer to the origin.
CEN455: Dr. Nassim Ammour

26. Example31
Given the system of Figure (a), design an ideal derivative compensator to yield a 16% overshoot, with a threefold reduction in settling time.
SOLUTION
Fig (a) Feedback control system
the performance of the uncompensated system operating with 16% overshoot.
Root-Locus for the uncompensated system fig (b):
along damping ratio line
16% Overshoot
ζ = 0.504,
-1,205 ±j2.064.
Dominant second-order
pair of poles
Settling time
𝑇 𝑠 = 4 ζ 𝜔 𝑛 = =3.320
The gain of the dominant poles
𝑘=43.35
Third pole at
Fig (b) Root locus for uncompensated system
CEN455: Dr. Nassim Ammour

27. Example32 We proceed to compensate the system
1. location of the compensated system's dominant poles.
Fig (a) Compensated dominant pole superimposed
over the uncompensated root locus
𝑇 𝑠 𝑛𝑒𝑤 = 𝑇 𝑠 3 =1.107
threefold reduction in the settling time
𝜎= 4 𝑇 𝑠 = =3.613
real part of the compensated system's
dominant, second-order pole
Imaginary part of the compensated system's dominant, second-order pole
𝜔 𝑑 =3.613 𝑡𝑎𝑛 180 𝑜 − =6.193
2. design the location of the compensator zero
- The angle contribution of poles for the desired pole location: −
In order to achieve -180 the angle contribution of the placed zero should be 95.6.
From the figure:
−𝜎 =tan⁡(180−95.6)
𝜎=3.006
Zero contribution angle > → zero position less than desired pole real part.
CEN455: Dr. Nassim Ammour
Fig (b) Evaluating the location of the compensating zero

28. Example33 Improvement in the transient response
Root-Locus After Compensation
Fig (b) Uncompensated and compensated system step responses
Fig (a) Root locus for the compensated system
CEN455: Dr. Nassim Ammour

29. Example34 Implementation
Once the compensating zero located, the ideal derivative compensator used to improve the transient response is implemented with a proportional-plus-derivative (PD) controller.
The transfer function of the controller is:
loop-gain value.
equal the negative of the compensator zero,
Two drawbacks:
requires an active circuit to perform the differentiation.
differentiation is a noisy process.
Implementation of ideal differentiator is expensive. So we may use the next technique: Lead Compensation
CEN455: Dr. Nassim Ammour

30. Lead Compensation
Basic Idea: The difference between 180° and the sum of the angles must be the angular contribution required of the compensator.
Example: looking at the Figure, we see that:
𝜃 2 − 𝜃 1 − 𝜃 3 − 𝜃 4 + 𝜃 5 = 2𝑘
𝑤ℎ𝑒𝑟𝑒 𝜃 1 − 𝜃 2 = 𝜃 𝑐 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑐𝑜𝑛𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑚𝑝𝑒𝑛𝑠𝑎𝑡𝑜𝑟
There are infinitely many choices of zc and pc providing same 𝜃 𝑐
The choice from infinite possibilities affects:
Static Error Constants.
Required gain to reach the design point on the root locus.
Difficulty in justification a second order approximation.
CEN455: Dr. Nassim Ammour

31. Example41
Design three lead compensators for the system in Figure to reduce the settling time by a factor of 2 while maintaining 30% overshoot.
SOLUTION
Characteristics of the uncompensated system operating at 30% overshoot
Dominant second-order
pair of poles
damping ratio
settling time
30%
Overshoot
𝑇 𝑠 = =3.972 s
ζ = 0.358,
-1,007 ±j2.627.
along damping ratio line
From pole's real part
Design point
real part of the desired pole location
twofold reduction
in settling time
𝑇 𝑠 = =1.986 s
−ζ𝜔 𝑛 = −4 𝑇 𝑠 =−2.014
Imaginary part of the desired pole location
𝜔 𝑑 =−2.014 tan =5.252
CEN455: Dr. Nassim Ammour

32. Example42 Lead compensator Design.
Place the zero on real axis at -5 (arbitrarily as possible solution).
sum the angles (this zero and uncompensated system's poles and zeros),
the angular contribution required
from the compensator pole
resulting angle
𝜃 𝑐 = −
𝜃 𝑐 = − = −7.31 0
location of the
compensator pole
compensator pole
𝑝 𝑐 −2.014 =tan⁡( )
𝑝 𝑐 =42.96
From geometry
in fig(a)
Fig (a) 5-plane picture used to calculate the location
of the compensator pole
approximation is not
valid for case C
CEN455: Dr. Nassim Ammour
Fig (c) Uncompensated system and lead compensation responses (zeros at a:-5, b:-4 c: -2)
Fig (b) Compensated system root locus

33. Improving Steady-State Error and Transient Response
Combine the design techniques to obtain improvement in steady-state error and transient response independently.
First improve the transient response.(PD or lead compensation).
Then improve the steady-state response. (PI or lag compensation).
Two Alternatives
PID (Proportional-plus-Integral-plus-Derivative) (with Active Elements).
Lag-Lead Compensator. (with Passive Elements).
CEN455: Dr. Nassim Ammour

34. PID Controller Design
Transfer Function of the compensator (two zeros and one pole):
𝐺 𝑐 𝑠 = 𝑘 𝑘 2 𝑠 + 𝑘 3 𝑠= 𝑘 1 𝑠+ 𝑘 2 + 𝑘 3 𝑠 2 𝑠 = 𝑘 3 ( 𝑠 𝑘 1 𝑘 3 𝑠+ 𝑘 2 𝑘 3 ) 𝑠
Design Procedure (Fig (a) )
Fig (a) PID controller implementation
From the requirements figure out the desired pole location to meet transient response specifications.
Design the PD controller to meet transient response specifications.
Check validity (all requirements have been met) of the design by simulation.
Design the PI controller to yield the required steady-state error.
Determine the gains, 𝑘 1 , 𝑘 2 𝑎𝑛𝑑 𝑘 3 (Combine PD and PI).
Simulate the system to be sure all requirements have been met.
Redesign if simulation shows that requirements have not been met.
CEN455: Dr. Nassim Ammour

35. Example51
Given the system of Figure (a), design a PID controller so that the system can operate with a peak time that is two-thirds that of the uncompensated system at 20% overshoot and with zero steady-state error for a step input.
Fig (a) Uncompensated feedback control system
SOLUTION
Evaluation of the uncompensated system
20%
Overshoot
ζ = 0.456,
along damping ratio line
Dominant second-order
pair of poles
-5,415 ±j10.57 with gain of
damping ratio
Peak time
𝑇 𝑝 =0.297 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
between - 8 and -10 for a gain
equivalent to that at the dominant poles
A third pole at
To reduce the peak time to two-thirds. (find the compensated system's dominant pole location)
The imaginary part
𝜔 𝑑 = 𝜋 𝑇 𝑝 = 𝜋 ( 2 3 )(0.297) =15.87
The real part
𝜎= 𝜔 𝑑 tan⁡( ) =−8.13
tan = - tan⁡(180− )
CEN455: Dr. Nassim Ammour

36. Example52 Design of the compensator
(sum of angles from the uncompensated system's poles and zeros to the desired compensated dominant pole is − )
the required contribution
from the compensator zero 𝑧 𝑐
−180 0 =
Fig (a) Calculating the PD compensator zero
From geometry
in Fig(a)
compensating
zero's location.
𝑧 𝑐 −8.13 =𝑡𝑎𝑛
𝑧 𝑐 =55.92
the PD controller.
𝐺 𝑃𝐷 (𝑠)=(𝑠+55.92)
gain at the design point
𝑘=5.34
The PD-compensated system is simulated. Fig (b) (next slide) shows the reduction in peak time and the improvement in steady-state error over the uncompensated system. (step 3 and 4)
Fig (b) Root locus for PD-compensated system
CEN455: Dr. Nassim Ammour

37. Example53
A PI controller is used to reduce the steady-state error to zero
(for PI controller the zero is placed close to the origin)
PI controller is used as
𝐺 𝑃𝐼 (𝑠)= 𝑠+0.5 𝑠
Searching the 0.456
damping ratio line
−7.516±𝑗 𝑤𝑖𝑡ℎ 𝑎𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑒𝑑 𝑔𝑎𝑖𝑛 𝑘=4.6
we find the dominant,
second-order poles
Fig (a) Root locus for PID-compensated system
Now we determine the gains (the PID parameters),
𝐺 𝑃𝐼𝐷 𝑠 = 𝑘 𝑠 𝑠+0.5 𝑠 = 4.6 𝑠 𝑠+0.5 𝑠
= 𝑠 +4.6 𝑠= 𝑘 1 + 𝑘 2 1 𝑠 + 𝑘 3 𝑠
Matching:
𝑘 1 =256.5, 𝑘 2 =128.6, 𝑘 3 =4.6
CEN455: Dr. Nassim Ammour
Fig (b) Step responses for uncompensated, PD compensated, and PID compensated systems

38. Lag-Lead Compensator Design (Cheaper solution then PID)
First design the lead compensator to improve the transient response. Next we design the lag compensator to meet the steady-state error requirement.
Design procedure:
Determine the desired pole location based on specifications. (Evaluate the performance of the uncompensated system).
Design the lead compensator to meet the transient response specifications.(zero location, pole location, and the loop gain).
Evaluate the steady state performance of the lead compensated system to figure out required improvement.(simulation).
Design the lag compensator to satisfy the improvement in steady state performance.
Simulate the system to be sure all requirements have been met. (If not met redesign)
CEN455: Dr. Nassim Ammour

39. Example61
Design a lag-lead compensator for the system of Figure so that the system will operate with 20% overshoot and a twofold reduction in settling time. Further, the compensated system will exhibit a tenfold improvement in steady-state error for a ramp input.
SOLUTION
Fig (a) Uncompensated system
Step 1: Evaluation of the uncompensated system
20%
Overshoot
ζ = 0.456,
along damping ratio line
Dominant second-order
pair of poles
-1,794 ±j3.501 with gain of
damping ratio
Step 2 : Lead compensator design (selection of the location of the compensated system's dominant poles).
Twofold reduction
of settling time
the real part of
the dominant pole
−ζ 𝜔 𝑛 =− =−3.588
the imaginary part of
𝜔 𝑑 = ζ 𝜔 𝑛 tan⁡( ) =7.003
lead compensator design.
𝑧 𝑐 =−6
Arbitrarily select a location
for the lead compensator zero.
- compensator zero coincident with the open-loop pole to eliminate a zero and leave the lead-compensated system with three poles. (same number that the uncompensated system has)
CEN455: Dr. Nassim Ammour

40. Example62 Finding the location of the compensator pole.
Sum the angles to the design point from the uncompensated system's poles and zeros and the compensator zero and get °.
The difference between 180° and this quantity is the angular contribution required from the compensator pole (—15.35°).
Using the geometry shown in Figure (b)
Compensator pole.
𝑝 𝑐− = tan ( ° )
𝑝 𝑐 =−29.1
Fig (a) Root locus for uncompensated system
- The complete root locus for the lead-compensated system is sketched in Figure (c)
Fig (b) Evaluating the compensator pole
CEN455: Dr. Nassim Ammour
Fig (c) Root locus for lead-compensated system

41. Example63
Steps 3 and 4: Check the design with a simulation. (The result for the lead compensated system is shown in Figure(a) and is satisfactory.)
inversely proportional to the steady-state error
Step 5: design the lag compensator to improve the steady-state error.
uncompensated system's open-loop transfer function
static error
constant
𝐺 𝑠 = 𝑠 𝑠+6 (𝑠+10)
𝑘 𝑣 =3.201
Need of tenfold
improvement
the addition of lead compensation has improved
the steady-state error by a factor of 2.122
=4.713
static error
constant
lag compensator factor
for steady-state error
improvement
𝐺 𝐿𝐶 𝑠 = 𝑠 𝑠+10 (𝑠+29.1)
𝑘 𝑣 =6.794
Step 6: We arbitrarily choose the lag compensator pole at 0.01,
lag-lead-compensated
Open loop TF
lag
compensator
lag compensator
zero
𝐺 𝐿𝑎𝑔 𝑠 = (𝑠 ) 𝑠+0.01
𝐺 𝐿𝐿𝐶 𝑠 = 𝐾 (𝑠 ) 𝑠 𝑠+10 (𝑠+29.1) 𝑠+0.01
𝑧 𝑐 = 𝑝 𝑐 𝑘 𝑣𝑛 𝑘 𝑣𝑜 =
The uncompensated system pole at - 6 canceled the lead compensator zero at -6.
Drawing the complete root locus for the lag-lead-compensated system and by searching along the damping ratio line
closed-loop
dominant poles
𝑝 𝑐 =−3.574 ±𝑗 with a gain of
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42. Example64
- Fig (b) shows the complete draw of the lag-lead-compensated root locus.
- The lag-lead compensation has indeed increased the speed of the system (settling time or the peak time).
Step 7: The final proof of our designs is shown by the simulations of Figure (b)
Fig (a) Root locus for lag-lead-compensated system
CEN455: Dr. Nassim Ammour
Fig (b) Improvement in step response for lag-lead-compensated system

43. Feedback Compensation
Compensator 𝐻 𝑐 𝑠 is used at the minor feedback to improve transient response and steady-state response independently (as opposed to the cascade compensators we have seen up to this point).
More complicated than cascade.
Generally provide faster response.
Can be used in cases where noise is a concern if we use cascade compensators.
May not require additional gain.
Fig (a) Generic control system with feedback compensation.
CEN455: Dr. Nassim Ammour

44. Two Approaches for Feedback Compensation
The design of feedback compensation consists of finding the gains, such as 𝐾, 𝐾 1 𝑎𝑛𝑑 𝐾 𝑓 . Two approaches:
Similar to cascade compensation. Consider compensation as adding poles and zeros to feedback section for the equivalent system.
Blocks in cascade (product of blocks)
Moving a summing point ahead of a block
𝐾 1
1 𝐾
𝐺 1 (𝑠)
𝐾 𝑓 𝐻 𝑐 (𝑠)
1 𝐺 2 (𝑠)
𝐺 2 (𝑠) 𝐾
Moving a pickoff point behind a block 1
𝐾 𝐾 1 𝐺 1 (𝑠)𝐺 2 (𝑠)
𝐾 𝑓 𝐻 𝑐 (𝑠) 𝐾𝐺 2 (𝑠)
𝐾 𝐾 1 𝐺 1 (𝑠)𝐺 2 (𝑠)
𝐾 𝑓 𝐻 𝑐 (𝑠) 𝐾𝐺 2 (𝑠) +1
Two blocks in parallel (sum of blocks)
2. First design the minor loop then design the major loop.
The minor loop is designed to change the open loop poles and open loop transient-response.
Loop gain is used to adjust the closed loop performance.
CEN455: Dr. Nassim Ammour

45. Feedback Compensation
Approach 1
The first approach consists of reducing Figure (a) to Figure (b).
The loop gain, G(s)H(s), is
𝐺 𝑠 𝐻 𝑠 = 𝐾 1 𝐺 1 𝑠 [ 𝐾 𝑓 𝐻 𝑐 𝑠 +𝐾 𝐺 2 𝑠 ]
Without feedback, 𝐾 𝑓 𝐻 𝑐 𝑠 , the loop gain is
Fig (a) Generic control system with feedback compensation.
𝐺 𝑠 𝐻 𝑠 = 𝐾𝐾 1 𝐺 1 𝑠 𝐺 2 𝑠
The effect of adding feedback is to replace the poles and zeros of 𝐺 2 𝑠
with the poles and zeros of 𝐾 𝑓 𝐻 𝑐 𝑠 +𝐾 𝐺 2 𝑠
Similar to cascade compensation in that we add new poles and zeros via H(s) to reshape the root locus to go through the design point.
Fig (b) Equivalent block diagram
Approach 2
The second approach allows us to use feedback compensation to design a minor loop's transient response separately from the closed-loop system response.
CEN455: Dr. Nassim Ammour

46. Example71
Given the system of Figure (a), design rate feedback compensation, as shown in Figure (b), to reduce the settling time by a factor of 4 while continuing to operate the system with 20% overshoot.
SOLUTION
System
First design a PD compensator.
- For the uncompensated system, Search along the 20% overshoot line (𝜁 = 0.456)
The angle of the 20%
overshoot line
180 ° −arccos⁡(𝜁)= °
the dominant
poles
−1.809±𝑗 (𝑠𝑒𝑒 𝑓𝑖𝑔 𝑒 )
system with rate feedback compensation
- The settling time is 2.21 seconds and must be reduced by a factor of 4 to 0.55 second.
Next determine the location of the dominant poles for the compensated system.
Zero at 1/Kf
equivalent compensated system;
- To achieve a fourfold decrease in the settling time, the real part of the desired pole must be increased by a factor of 4.
Real part of
Compensated pole
4 −1.809 =−7.236
Imaginary part of
Compensated pole
𝑤 𝑑 =−7.236 tan ° =14.12
CEN455: Dr. Nassim Ammour
(e) Root locus for uncompensated system
equivalent compensated system showing unity feedback

47. Example72
Compensated dominant
pole position
𝑝 𝑐 =−7.236±𝑗 14.12
Sum of the angles from the uncompensated system's poles (add zero to yields 180 ° )
compensator
zero contribution
𝜃= − °
𝜃 𝑧 = −180 ° ° = °
Using the geometry shown in Figure (a)
Compensator's
zero location
− 𝑧 𝑐 =tan⁡( 180 ° − ° )
𝑧 𝑐 =5.42
(a) Finding the compensator zero
The root locus for the equivalent compensated system (fig (c) previous slide) is shown in Figure (b)
The gain at
the design point,
𝐾 1 𝐾 𝑓 = 256.7
Since 𝐾 𝑓 is the reciprocal
of the compensator zero,
𝐾 𝑓 = 1 𝑧 𝑐 = 0.185
𝐾 1 = 1388
steady-state error characteristic (fig (d) slide 42 )
(b) Root locus for the
compensated system
𝐾 𝑣 = 𝐾 𝐾 1 𝐾 𝑓 =4.18
CEN455: Dr. Nassim Ammour

48. with a settling time of 0.75 second
Example73
The closed-loop transfer function is (fig (d) slide 42 )
𝑇 𝑠 = 𝐺(𝑠) 1+𝐺 𝑠 𝐻(𝑠) = 𝐾 1 𝑠 𝑠 𝐾 1 𝐾 𝑓 𝑠+ 𝐾 1
The results of the simulation are shown in Figure (a) and (b)
over-damped response
with a settling time of 0.75 second
The settling time is 2.21 seconds
(a) Step response for uncompensated system
(b) Step response for the compensated system
CEN455: Dr. Nassim Ammour

49. Physical Realization of Compensation
Active-Circuit Realization
𝑍 1 (𝑠)and 𝑍 2 (𝑠)are used as a building block to implement the compensators and controllers, such as PID controllers.
The transfer function of an inverting operational amplifier
𝑉 0 (𝑠) 𝑉 𝑖 (𝑠) =− 𝑍 2 (𝑠) 𝑍 1 (𝑠)
Fig (a) Operational amplifier for transfer function realization
CEN455: Dr. Nassim Ammour

50. Table1 summarizes the realization of PI, PD, and PID controllers as well as lag, lead, and lag-lead compensators using Operational amplifiers.
Fig (a) : lag-lead compensator can be formed by cascading the lag compensator with the lead compensator.
Fig (a) Lag-lead compensator implemented with operational amplifiers
CEN455: Dr. Nassim Ammour
Table 1 Active realization of controllers and compensators, using an operational amplifier

51. Example8 Implement the PID controller of Example 5 SOLUTION
𝐺 𝑐 𝑠 = 4.6 𝑠 𝑠+0.5 𝑠
The transfer function of the PID controller is
𝐺 𝑐 𝑠 =𝑠 𝑠
which can be put in the form
Comparing the PID controller in Table 1 with this equation we obtain the following three relationships:
𝑅 2 𝑅 𝐶 1 𝐶 2 =56.42
𝑅 2 𝐶 1 = 1
1 𝑅 1 𝐶 2 =27.96
Since there are four unknowns and three equations
we arbitrarily select a practical value:
Fig (a) PID controller
𝐶 2 =0.1 𝜇𝐹
𝑅 1 = 𝑘Ω, 𝑅 2 = 𝑘Ω 𝑎𝑛𝑑 𝐶 1 =5.59𝜇𝐹
The complete circuit is shown in Figure (a) where the circuit element values have been rounded off.
CEN455: Dr. Nassim Ammour

52. Passive-Circuit Realization
Lag, lead, and lag-lead compensators can also be implemented with passive networks (Table 2) .
CEN455: Dr. Nassim Ammour
TABLE 2 Passive realization of compensators

53. Example9
Realize the lead compensator designed in Example 4 (Compensator b zero at -4).
SOLUTION
𝐺 𝑐 𝑠 = 𝑠+4 𝑠+20.09
The transfer function of the lead compensator is
Comparing the transfer function of a lead network shown in Table 2 with The equation, we obtain the following two relationships:
1 𝑅 1 𝐶 = 𝑎𝑛𝑑 𝑅 1 𝐶 + 1 𝑅 2 𝐶 =20.09
Since there are three network elements and two equations, we may select one of the element values arbitrarily
𝑅 1 =250 𝑘Ω 𝑎𝑛𝑑 𝑅 2 =62.2 𝑘Ω
𝐶=1 𝜇𝐹
CEN455: Dr. Nassim Ammour