1. IntroductionLecture 1: Basic Ideas & Terminology
DIGITAL CONTROL SYSTEM
자동제어
PROF. KALYANA C. VELUVOLU (별루볼루 교수)
School of Electronics Engineering (전자공학부 IT 대학)

2. WEEK 7
ANALYSIS TECHNIQUES

3. Steady State Analysis Root-Locus Technique Frequency Response
TOPICS
Steady State Analysis
Root-Locus Technique
Frequency Response

4. STEADY- STATE ERROR OF STABLE SYSTEM
For the unity-feedback sampled-data system,

5. If the system is stable, the final-value theorem can be applied to obtain the steady-state or final value of the output and the error at the sampling instants: 𝑐 ∗ ∞ = lim 𝑡→∞ 𝑐 ∗ 𝑡 = lim 𝑧→1 1− 𝑧 −1 𝐺 𝑧 1+𝐺 𝑧 𝑅(𝑧) 𝑒 ∗ ∞ = lim 𝑡→∞ 𝑒 ∗ 𝑡 = lim 𝑧→1 1− 𝑧 −1 𝑅 𝑧 1+𝐺 𝑧

6. ROOT-LOCUS TECHNIQUE
Recall that the stability of a given sampled-data system can be determined by examining the roots obtained from the characteristic equation 𝐼+𝐺 𝑧 𝐻 𝑧 =0.
The root-locus method is a plot of the roots of the characteristic equation of the closed-loop system as a function of a gain constant. It is based on that the poles of 𝐶(𝑧) 𝑅 𝑧 𝑜𝑟 𝐶(𝑧) are related to the zeros and poles of the open-loop transfer function G(z)H(z) and the gain.
The roots of the characteristic equation of the system can be obtained directly, resulting in a complete and accurate solution of the transient and steady-state response. Further, the root locus diagram can be used for designing controllers.

7. Procedure outline
Derive the open-loop transfer function G(z)H(z) of the system.
Factorize the numerator and denominator of the transfer function into linear factors of the form z+a.
Plot the zeros and poles of the open-loop transfer function in the z plane, where z= 𝜎 𝑧 + 𝑗𝜔 𝑧 .
The plotted zeros and poles of the open-loop function determine the root of the characteristic equation of the closed-loop system 1+G(z)H(z)=0.
The root-locus diagram can be used in two ways. If the gain of the open-loop system is predetermined, the location of the exact roots of 1+G(z)H(z)=0 is immediately known.
If the location of the roots (or ξ) is specified, the required value of k can be determined.

8. Once the roots have been found, the system’s time response can be calculated by taking the inverse Z transform.
If the response does not meet the desired specifications, determine the shape that the root-locus must have to meet these specifications.
Synthesize a compensator that must be inserted into the system, if other than gain adjustment is required, to make the required modification on the original locus.

9. Rules for Drawing Root Locus
𝐺 𝑧 𝐻 𝑧 = 𝐾 𝑧− 𝑧 1 … 𝑧− 𝑧 𝑖 …(𝑧− 𝑧 𝑤 ) 𝑧− 𝑝 1 … 𝑧− 𝑝 𝑐 …(𝑧− 𝑝 𝑛 )
Two conditions to be satisfied on the root locus:
Magnitude condition:
𝐺 𝑧 𝐻(𝑧) =1
Angle condition:
<𝐺 𝑧 𝐻 𝑧 = 1+2ℎ 𝑓𝑜𝑟 𝑘>0 ℎ 𝑓𝑜𝑟𝑘<0
Where h is any integer.

10. Rules:
The number of branches is equal to the number of poles of the open-loop transfer functions.
For positive K, the root locus exists on portion of the real axis, the sum of poles and zeros to the right is an odd integer. For negative K, the root locus exists on those portions, the sum of the poles and zeros to the right is an even integer (including zero).
The root locus starts at the open-loop poles and terminates at the open-loop zeros or at infinity.
The angles of the asymptotes of the root locus that end at infinity are determined by, for k>0,
𝛾= (1+2ℎ) 𝑛𝑜. 𝑜𝑓 𝑝𝑜𝑙𝑒𝑠 𝑜𝑓 𝐺 𝑧 𝐻 𝑧 −[𝑛𝑜. 𝑜𝑓 𝑧𝑒𝑟𝑜𝑠 𝑜𝑓 𝐺 𝑧 𝐻 𝑧 ]

11. and, for k<0, 𝛾= ℎ 360 0 𝑛𝑜. 𝑜𝑓 𝑝𝑜𝑙𝑒𝑠 𝑜𝑓 𝐺 𝑧 𝐻 𝑧 −[𝑛𝑜
and, for k<0, 𝛾= ℎ 𝑛𝑜. 𝑜𝑓 𝑝𝑜𝑙𝑒𝑠 𝑜𝑓 𝐺 𝑧 𝐻 𝑧 −[𝑛𝑜. 𝑜𝑓 𝑧𝑒𝑟𝑜𝑠 𝑜𝑓 𝐺 𝑧 𝐻(𝑧) 5. The real-axis intercept of the asymptotes is 𝑧 0 = 𝑐=1 𝑛 𝑅𝑒 𝑃 𝑐 − 𝑖=1 𝑤 𝑅𝑒( 𝑧 𝑖 ) 𝑛−𝑤 6. The breakaway point between two poles (or the breakaway-in point between two zeros) on the real axis can be determined by taking the derivative of the gain k and equate this derivative to zero and find the roots of the resulting equation.

12. For k>0 the angle of departure from a complex pole is equal to minus the sum of the angles from the other poles plus the sum of the angles from the zeros. For k<0 the departure angle is from that obtained for k>0. For k>0 the angle of approach to a complex zero is equal to the sum of the angles from the poles minus the sum of the angles from the other zeros minus For k<0 the approach angle is from that obtained for k>0. 8. The unit circle crossing of the root locus can be determined by setting up the Jury’s array from closed-loop characteristic equation. Determine the range of values that K must have to satisfy the necessary and sufficient conditions for a stable system. 9. The root loci are symmetrical about the real axis.

13. 10. The value of k on any given point of the root locus can be made by applying the magnitude condition as follows: 𝑘= 𝑧− 𝑝 1 … 𝑧− 𝑝 𝑐 … 𝑧− 𝑝 𝑛 𝑧− 𝑧 1 … 𝑧− 𝑧 𝑖 … 𝑧− 𝑧 𝑤 11. The selection of the dominant roots is based on the specifications that give the required system performance. These value in turn are mapped into the domain to determine the location of the desired dominant roots in the z plane. The gain k for these roots is determined by means of the magnitude condition. The remaining roots are then determined to satisfy the same magnitude condition.

14. Example: The second-order characteristic equation 𝑧 2 −0.2𝐴𝑧+0.1𝐴=0
Is partitioned to put it into the format as follows:
𝑧 2 −0.2𝐴𝑧+0.1𝐴=0.2A(z-0.5)=-k(z-0.5)
Where k=-0.2A. It can be rearranged to yield
𝑃 𝑧 = 𝑘(𝑧−0.5) 𝑧 2 =−1

15. The construction rules are as follows.
Number of branches of the root locus is n-2
For k>0 (A<0) the real-axis locus exists between z=0.5 and z=∞, and for k<0 (A>0) the real-axis locus exists between z=0.5, and one ends at k=±∞
The branches starts, with k=0, at the poles of P(z). One end at z=0.5, and one ends at k=±∞.
Asymptotes for k>0:
𝛾= (1+2ℎ) −1 = 1+2ℎ →𝛾= 180 0
and for k<0
𝛾= ℎ −1 =ℎ →𝛾= ( 0 0 )
5. Not applicable.

16. 6. There is no breakaway point for k>0. For k<0, 𝑊 𝑧 = 𝑧 2 −𝑧+0
6. There is no breakaway point for k>0. For k<0, 𝑊 𝑧 = 𝑧 2 −𝑧+0.5 =𝑘 Take the derivative and set it equal to zero yields 𝑑𝑊(𝑧) 𝑑𝑧 = 𝑧(𝑧−1) (−𝑧+0.5 ) 2 =0 which yields 𝑧 1,2 =0, 1. 𝑧 1 =0 is the breakaway point and 𝑧 2 =1 is the break-in point. 7. Not applicable. 8. The UC intersections is determined by the Jury’s stability test for the values of k= 2 3 & 𝑘=−2. 9. The root locus is symmetrical about the real axis. 10. The value of k at each point of a locus can be determined by the magnitude condition. 11. Not applicable.

18. s-, z-, w-PLANE TIME-RESPONSE
Using the continuous-time plant for the unity-feedback control system, the effect of the choice of T on the degree of correlation between the s, z, and w planes can be illustrated. Let
𝐺 𝑥 𝑠 = 4.2 𝑠(𝑠+1)(𝑠+5)
Assume the plant is proceeded with an ideal sampler and Z.O.H. Then, the z-transform of the plant is given by
𝐺 𝑧 = 𝐾 𝑧 𝐺 ′ (𝑧)
=4.2 1− 𝑧 −1 𝑍 1 𝑠 2 (𝑠+1)(𝑠+5)
=4.2 5𝑇−6𝑧+6 25(𝑧−1) + (24𝑧−25 𝑒 −5𝑇 + 𝑒 −𝑇 (𝑧−1) 100(𝑧− 𝑒 −𝑇 )(𝑧− 𝑒 −5𝑇 )

19. The value of K=4. 2 is for dominant closed-up roots having ξ=0. 45
The value of K=4.2 is for dominant closed-up roots having ξ=0.45. The corresponding w transform, G(w)= 𝐾 𝑤 𝐺 ′ (𝑤), are obtained for various values of T. The gain ( 𝐾 𝑧 𝑜𝑟 𝐾 𝑤 ) and the zeros and poles of G(.) for each value of T are given in the table below.

20. The gain and poles of G(.) for various values of T

21. Comments:
To accurately duplicate the performance of a continuous-time system, having 𝐺 𝑥 (𝑠) as its plant, by a sampled-data unity feedback system and a ZOH device, the dominant poles and zeros of 𝐶(𝑧) 𝑅(𝑠) must satisfy the conditions of 𝜔 𝑠𝑝 𝑇 2< and 𝛼 2 ≪2. therefore, the smallest value of T should be chosen as constrained by the processing time of the digital controller.
A bilinear transformation results in G(w) having the same number of zeros and poles. Thus, the third-order plant 𝐺 𝑥 (𝑠) of this example results in G(w) having three zeros and three poles. As T approaches 0, the poles will converge to the poles of the continuous-time systems and the zeros will converge to infinity.

22. FREQUENCY RESPONSE
Frequency response for sampled-data systems can be determined by two methods:
In the z domain, substituting 𝑧= 𝑒 𝑠𝑇 = 𝑒 𝑗𝜔𝑇 into 𝐶(𝑧) 𝑅 𝑧 :
𝐶( 𝑒 𝑗𝜔𝑇 ) 𝑅( 𝑒 𝑗𝜔𝑇 ) = 𝑀 𝑧 < 𝛼 𝑧
Where 𝑀 𝑧 is the magnitude and 𝛼 𝑧 is the phase;
(2) In the w domain, substituting 𝑤=𝑗 𝜔 𝑤 𝑖𝑛𝑡𝑜 𝐶(𝑤) 𝑅 𝑤 :
𝐶(𝑗 𝜔 𝑤 ) 𝑅(𝑗 𝜔 𝑤 ) = 𝑀 𝑤 < 𝛼 𝑤

23. Example: Suppose
𝐺 𝑥 𝑠 = 4.2 𝑠(𝑠+1)(𝑠+5)
Plot 𝑀 𝑠 , 𝑣𝑠. 𝜔 𝑎𝑛𝑑 𝛼 𝑠 , 𝑣𝑠. 𝜔 for the continuous-time system configuration, where [ 𝐶(𝑗𝜔) 𝑅(𝑗𝜔)]= 𝑀 𝑠 < 𝛼 𝑠
Plot 𝑀 𝑧 , 𝑣𝑠. 𝜔, 𝛼 𝑧 𝑣𝑠. 𝜔, 𝑀 𝑤 𝑣𝑠. 𝜔 𝑤 𝑎𝑛𝑑 𝛼 𝑤 𝑣𝑠. 𝜔 𝑤 for sampled-data configuration where T=0.01sec. Also, determine the bandwidth 𝜔 𝑏 for all three domains.
SOLUTION: (a) The dominant poles for the continuous-time system are 𝑃 1,2 =−0.404 ±𝑗 Thus, the value of 𝜔 𝑚 is in the neighborhood of 𝜔 𝑑 =0.8. the frequency-response plots of 𝑀 𝑠 , 𝑣𝑠. 𝜔 𝑎𝑛𝑑 𝛼 𝑠 , 𝑣𝑠. 𝜔 are shown on next slide.
The plots for the z- and w- domains are very similar so they are not shown.

25. SUMMARY Transfer Functions of sampled-data system:
Need to be able to derive z-domain representation for different configurations
Zero-order Hold Function:
Essential device for sampled-data system.
Stability Test: Jury test and w-plane method
Steady State analysis
Root-locus Technique:
Need to be able to draw simple root-locus diagrams and understand how to use them for stability analysis and control design (more on control design in next lecture)
Frequency Response:
Need to be able to draw frequency response plots on s-, z- and w-planes.

26. DESIGN TECHNIQUES Topics: Digital Feedback Design
S-plane Method
z-plane Method
W-plane Method
Lead-lag Compensation
PID Control

27. Three Design Approaches
S-plane Design: Design a continuous-time controller and then approximate it using a digital controller.
Advantage: Simple, tools readily available.
Disadvantages: Approximation errors, no guarantee on stability & performance after digital implementation.
z-plane Design: Design a digital controller directly for a digital system.
Advantages: No approx. errors, direct implementation.
Disadvantages: Design specifications need to be in the z-domain; Design tools need to be modified.
W-plane Design: Transform the digital system to the w-plane and apply continuous-time design methods
Advantages: No implementation errors, tools available
Disadvantages: Extra transformations.
Dr. Kalyana Veluvolu

28. DESIGN IN s-PLANE
Consider the unity feedback sampled-data system below
Dr. Kalyana Veluvolu

29. Use the first-order 𝑃𝑎𝑑𝑒 ′ approximation 𝑒 −𝑠𝑇 ≈ 1+(−𝑠𝑇/2) 1−(−𝑠𝑇/2) For the ZOH device, when T is small. 𝐺 𝑧𝑜 𝑠 = 1− 𝑒 −𝑠𝑇 𝑠 ≈ 2𝑇 𝑇𝑠+2 = 𝐺 𝑝𝑎 𝑠 Approximate model for sampler and ZOH 𝐺 𝐴 𝑠 = 1 𝑇 𝐺 𝑝𝑎 𝑠 = 2 𝑇𝑠+2
Dr. Kalyana Veluvolu

30. Dr. Kalyana Veluvolu

31. The design procedure is to first obtain a closed-loop model [ 𝐶 𝑠 𝑅 𝑠 ] 𝑇 for the approximate system: 𝐺_𝑃𝐶(𝑠)= 𝐺_𝐴(𝑠)𝐺_𝑥(𝑠)= 2 𝐾 𝑥 /𝑇 𝑠(𝑠+1)(𝑠+ 2 𝑇 ) So for T = 0.1 s, we get 𝐶(𝑠) 𝑅(𝑠) 𝑇 = 𝐺 𝑃𝐶 (𝑠) 1+ 𝐺 𝑃𝐶 (𝑠) = 20 𝐾 𝑥 𝑠 𝑠 2 +20𝑠+20 𝐾 𝑥 For this it is assumed that the desired value of 𝜁 for the dominant roots is This gives 𝐾 𝑥 =
Dr. Kalyana Veluvolu

32. Dr. Kalyana Veluvolu

33. Thus for a unit-step input, 𝐶(𝑠) 𝑇 = 9.543 𝑠 𝑠 3 +21 𝑠 2 +20𝑠+9.543
𝐶(𝑠) 𝑇 = 𝑠 𝑠 𝑠 2 +20𝑠+9.543
= 𝑠(𝑠 ±𝑗0.4883)(𝑠+20.03)
The step response is shown on the next slide.
Notice that the approximation error is virtually negligible. This is because T is very small. The cost is more computation. If T is larger, the ZOH unit may degrade the degree of stability. This can be compensated in two ways:
Modifying the gain
Inserting a compensator
Dr. Kalyana Veluvolu

34. Dr. Kalyana Veluvolu

35. DESIGN IN z-PLANE:
This is an exact approach, requiring the z-transfer function of the forward loop. For the same example, we have 𝐺 𝑧 =𝑍 𝐾 𝑥 (1− 𝑒 −𝑠𝑇 ) 𝑠 2 (𝑠+1)
= 1− 𝑧 −1 𝑍 𝐾 𝑥 𝑠 2 (𝑠+1)
= 𝐾 𝑧 𝑇−1+ 𝑒 −𝑇 𝑧+(1−𝑇 𝑒 −𝑇 − 𝑒 −𝑇 ) 𝑧 2 − 1− 𝑒 −𝑇 + 𝐾 𝑥 −𝑇 𝐾 𝑥 − 𝐾 𝑥 𝑒 −𝑇 𝑧+ 𝑒 −𝑇 + 𝐾 𝑥 − 𝐾 𝑥 (𝑇+1) 𝑒 −𝑇
Thus for T = 0.1 s,
𝐺 𝑧 𝑧 = 𝐾 𝑥 (𝑧 ) (𝑧−1)(𝑧−0.9048)
Dr. Kalyana Veluvolu

36. However, this technique requires the s-domain design specifications to be transformed into the z-domain. For our example, we may assume that we want the closed-loop system to have two dominant poles at (as in the previous design) 𝑠 1,2 =−0.4875±𝑗 Which correspond to 𝜁 = Translating these poles to the z-domain using 𝑧= 𝑒 𝑠𝑇 : 𝑧 1,2 =0.9513±𝑗 The control design can then be done using the root-locus technique (or other method like pole-placement & frequency domain design).
Dr. Kalyana Veluvolu

37. Root-locus diagram (Global):
Dr. Kalyana Veluvolu

38. Root-locus diagram (for the region of interest):
The loop gain is determined to be 𝐾 𝑥 =0.478(similar as above)
Dr. Kalyana Veluvolu