1. Control Systems With Embedded Implementation (CSEI)
Lecture-1
Introduction to the Digital Control Systems
Dr. Imtiaz Hussain
Assistant Professor
URL :

2. Lecture Outline Introduction Difference Equations
Review of Z-Transform
Inverse Z-transform
Relations between s-plane and z-plane
Solution of difference Equations

3. Professor of Electrical Engineering
Recommended Book
M.S. Fadali, “Digital Control Engineering – Analysis and Design”, Elsevier, ISBN: 13:
Professor of Electrical Engineering
Area of Specialization: Control Systems

4. Introduction
Digital control offers distinct advantages over analog control that explain its popularity.
Accuracy: Digital signals are more accurate than their analogue counterparts.
Implementation Errors: Implementation errors are negligible.
Flexibility: Modification of a digital controller is possible without complete replacement.
Speed: Digital computers may yield superior performance at very fast speeds
Cost: Digital controllers are more economical than analogue controllers.

5. Structure of a Digital Control System

6. Examples of Digital control Systems
Closed-Loop Drug Delivery System

7. Examples of Digital control Systems
Aircraft Turbojet Engine

8. Difference Equation vs Differential Equation
A difference equation expresses the change in some variable as a result of a finite change in another variable.
A differential equation expresses the change in some variable as a result of an infinitesimal change in another variable.

9. Differential Equation
Following figure shows a mass-spring-damper-system. Where y is position, F is applied force D is damping constant and K is spring constant.
Rearranging above equation in following form
𝐹 𝑡 =𝑚 𝑦 𝑡 +𝐷 𝑦 𝑡 +𝐾𝑦(𝑡)
𝑦 𝑡 = 1 𝑚 𝐹 𝑡 − 𝐷 𝑚 𝑦 𝑡 𝐾 𝑚 𝑦(𝑡)

10. Differential Equation
Rearranging above equation in following form
𝑦 𝑡 = 1 𝑚 𝐹 𝑡 − 𝐷 𝑚 𝑦 𝑡 𝐾 𝑚 𝑦(𝑡) 𝑑𝑡
1 𝑚
− 𝐷 𝑚
− 𝐾 𝑚 𝑦
𝐹(𝑡)

11. Difference Equation 𝑦(𝑘+2)= 1 𝑚 𝐹(𝑘)− 𝐷 𝑚 𝑦(𝑘+1) 𝐾 𝑚 𝑦(𝑘) 𝑦(𝑘) 𝐹(𝑘)
𝑦(𝑘+2)= 1 𝑚 𝐹(𝑘)− 𝐷 𝑚 𝑦(𝑘+1) 𝐾 𝑚 𝑦(𝑘)
1 𝑧
1 𝑚
− 𝐷 𝑚
− 𝐾 𝑚
𝑦(𝑘+2)
𝑦(𝑘)
𝐹(𝑘)
𝑦(𝑘+1)

12. Difference Equations
Difference equations arise in problems where the independent variable, usually time, is assumed to have a discrete set of possible values.
Where coefficients 𝑎 𝑛−1 , 𝑎 𝑛−2 ,… and 𝑏 𝑛 , 𝑏 𝑛−1 ,… are constant.
𝑢(𝑘) is forcing function
𝑦 𝑘+𝑛 + 𝑎 𝑛−1 𝑦 𝑘+𝑛−1 +…+ 𝑎 1 𝑦 𝑘+1 + 𝑎 0 𝑦 𝑘 = 𝑏 𝑛 𝑢 𝑘+𝑛 + 𝑏 𝑛−1 𝑢 𝑘+𝑛−1 +…+ 𝑏 1 𝑢 𝑘+1 + 𝑏 0 𝑢 𝑘

13. Difference Equations
Example-1: For each of the following difference equations, determine the order of the equation. Is the equation (a) linear, (b) time invariant, or (c) homogeneous?
𝑦 𝑘 𝑦 𝑘 𝑦 𝑘 =𝑢 𝑘
𝑦 𝑘+4 +sin⁡(0.4𝑘)𝑦 𝑘 𝑦 𝑘 =0
𝑦 𝑘+1 =−0.1 𝑦 2 𝑘

14. Z-Transform
Difference equations can be solved using classical methods analogous to those available for differential equations.
Alternatively, z-transforms provide a convenient approach for solving LTI equations.
The z-transform is an important tool in the analysis and design of discrete-time systems.
It simplifies the solution of discrete-time problems by converting LTI difference equations to algebraic equations and convolution to multiplication.
Thus, it plays a role similar to that served by Laplace transforms in continuous-time problems.

15. Z-Transform
Given the causal sequence {u0, u1, u2, …, uk}, its z-transform is defined as
The variable z −1 in the above equation can be regarded as a time delay operator.
𝑈 𝑧 = 𝑢 𝑜 + 𝑢 1 𝑧 −1 + 𝑢 2 𝑧 −2 + 𝑢 𝑘 𝑧 −𝑘
𝑈 𝑧 = 𝑘=0 ∞ 𝑢 𝑘 𝑧 −𝑘

16. Z-Transform
Example-2: Obtain the z-transform of the sequence

17. Relation between Laplace Transform and Z-Transform
Given the impulse train representation of a discrete-time signal
The Laplace Transform of above equation is
Let z be defined by
𝑢 ∗ 𝑡 = 𝑢 𝑜 𝛿 𝑡 + 𝑢 1 𝛿 𝑡−𝑇 + 𝑢 2 𝛿 𝑡−2𝑇 +…+ 𝑢 𝑘 𝛿 𝑡−𝑘𝑇
𝑢(𝑡)
𝑈(𝑠)
𝑈 ∗ (𝑠)
𝑢 ∗ (𝑡)
𝑢 ∗ 𝑡 = 𝑘=0 ∞ 𝑢 𝑘 𝛿 𝑡−𝑘𝑇
𝑈 ∗ 𝑠 = 𝑢 𝑜 + 𝑢 1 𝑒 −𝑠𝑇 + 𝑢 2 𝑒 −2𝑠𝑇 +…+ 𝑢 𝑘 𝑒 −𝑘𝑠𝑇
𝑈 ∗ 𝑠 = 𝑘=0 ∞ 𝑢 𝑘 𝑒 −𝑘𝑠𝑇
𝑧= 𝑒 𝑠𝑇

18. Conformal Mapping between s-plane to z-plane
Where 𝑠=𝜎+𝑗𝜔. Then 𝑧 in polar coordinates is given by
Therefore,
𝑧= 𝑒 𝑠𝑇
𝑧= 𝑒 (𝜎+𝑗𝜔)𝑇
𝑧= 𝑒 𝜎𝑇 𝑒 𝑗𝜔𝑇
∠𝑧=𝜔𝑇
𝑧 = 𝑒 𝜎𝑇

19. Conformal Mapping between s-plane to z-plane
We will discuss following cases to map given points on s-plane to z-plane.
Case-1: Real pole in s-plane (𝑠=𝜎)
Case-2: Imaginary Pole in s-plane (𝑠=𝑗𝜔)
Case-3: Complex Poles (𝑠=𝜎+𝑗𝜔)
𝑠−𝑝𝑙𝑎𝑛𝑒
𝑧−𝑝𝑙𝑎𝑛𝑒

20. Conformal Mapping between s-plane to z-plane
Case-1: Real pole in s-plane (𝑠=𝜎)
We know
Therefore
∠𝑧=𝜔𝑇
𝑧 = 𝑒 𝜎𝑇
𝑧 = 𝑒 𝜎𝑇
∠𝑧=0

21. Conformal Mapping between s-plane to z-plane
∠𝑧=𝜔𝑇
𝑧 = 𝑒 𝜎𝑇
Case-1: Real pole in s-plane (𝑠=𝜎)
When 𝑠=0
𝑧 = 𝑒 0𝑇 =1
∠𝑧=0𝑇=0
𝑠=0 1
𝑠−𝑝𝑙𝑎𝑛𝑒
𝑧−𝑝𝑙𝑎𝑛𝑒

22. Conformal Mapping between s-plane to z-plane
∠𝑧=𝜔𝑇
𝑧 = 𝑒 𝜎𝑇
Case-1: Real pole in s-plane (𝑠=𝜎)
When 𝑠=−∞
𝑧 = 𝑒 −∞𝑇 =0
∠𝑧=0 −∞
𝑠−𝑝𝑙𝑎𝑛𝑒
𝑧−𝑝𝑙𝑎𝑛𝑒

23. Conformal Mapping between s-plane to z-plane
∠𝑧=𝜔𝑇
𝑧 = 𝑒 𝜎𝑇
Case-1: Real pole in s-plane (𝑠=𝜎)
Consider 𝑠=−𝑎
𝑧 = 𝑒 −𝑎𝑇
∠𝑧=0 1 −𝑎
𝑠−𝑝𝑙𝑎𝑛𝑒
𝑧−𝑝𝑙𝑎𝑛𝑒

24. Conformal Mapping between s-plane to z-plane
Case-2: Imaginary pole in s-plane (𝑠=±𝑗𝜔)
We know
Therefore
∠𝑧=𝜔𝑇
𝑧 = 𝑒 𝜎𝑇
𝑧 =1
∠𝑧=±𝜔𝑇

25. Conformal Mapping between s-plane to z-plane
∠𝑧=𝜔𝑇
𝑧 = 𝑒 𝜎𝑇
Case-2: Imaginary pole in s-plane (𝑠=±𝑗𝜔)
Consider 𝑠=𝑗𝜔
𝑧 = 𝑒 0𝑇 =1
∠𝑧=𝜔𝑇 1
𝑠=𝑗𝜔 𝜔𝑇 −1 1 −1
𝑠−𝑝𝑙𝑎𝑛𝑒
𝑧−𝑝𝑙𝑎𝑛𝑒

26. Conformal Mapping between s-plane to z-plane
∠𝑧=𝜔𝑇
𝑧 = 𝑒 𝜎𝑇
Case-2: Imaginary pole in s-plane (𝑠=±𝑗𝜔)
When 𝑠=−𝑗𝜔
𝑧 = 𝑒 0𝑇 =1
∠𝑧=−𝜔𝑇 1 −1
−𝜔𝑇 1
𝑠=−𝑗𝜔 −1
𝑠−𝑝𝑙𝑎𝑛𝑒
𝑧−𝑝𝑙𝑎𝑛𝑒

27. Conformal Mapping between s-plane to z-plane
∠𝑧=𝜔𝑇
𝑧 = 𝑒 𝜎𝑇
Case-2: Imaginary pole in s-plane (𝑠=±𝑗𝜔)
When 𝑠=±𝑗 𝜔 𝑇
𝑧 = 𝑒 0𝑇 =1
∠𝑧=±𝜋
𝝎𝑻=𝝅
𝑗 𝜔 𝑇 1 𝜋 −1 1
−𝑗 𝜔 𝑇 −1
𝑠−𝑝𝑙𝑎𝑛𝑒
𝑧−𝑝𝑙𝑎𝑛𝑒

28. Conformal Mapping between s-plane to z-plane
Anything in the Alias/Overlay region in the S-Plane will be overlaid on the Z-Plane along with the contents of the strip between ±𝑗 𝜋 𝑇 .

29. Conformal Mapping between s-plane to z-plane
In order to avoid aliasing, there must be nothing in this region, i.e. there must be no signals present with radian frequencies higher than w = p/T, or cyclic frequencies higher than f = 1/2T.
Stated another way, the sampling frequency must be at least twice the highest frequency present (Nyquist rate).

30. Conformal Mapping between s-plane to z-plane
∠𝑧=𝜔𝑇
𝑧 = 𝑒 𝜎𝑇
Case-3: Complex pole in s-plane (𝑠=𝜎±𝑗𝜔)
𝑧 = 𝑒 𝜎𝑇
∠𝑧=±𝜔𝑇 1 −1 1 −1
𝑠−𝑝𝑙𝑎𝑛𝑒
𝑧−𝑝𝑙𝑎𝑛𝑒

31. Mapping regions of the s-plane onto the z-plane

32. Mapping regions of the s-plane onto the z-plane

33. Mapping regions of the s-plane onto the z-plane

36. Example-3
Map following s-plane poles onto z-plane assume (T=1). Also comment on the nature of step response each case.
𝑠=−3
𝑠=±4𝑗
𝑠=±𝜋𝑗
𝑠=±2𝜋𝑗
𝑠=−10±5𝑗

37. z-Transforms of Standard Discrete-Time Signals
The following identities are used repeatedly to derive several important results.
𝑘=0 𝑛 𝑎 𝑘 = 1− 𝑎 𝑛+1 1−𝑎 , 𝑎≠1
𝑘=0 ∞ 𝑎 𝑘 = 1 1−𝑎 , 𝑎 ≠1

38. z-Transforms of Standard Discrete-Time Signals
Unit Impulse
Z-transform of the signal
𝛿 𝑘 = 1, 0, 𝑘=0 𝑘≠0
𝛿 𝑧 =1

39. z-Transforms of Standard Discrete-Time Signals
Sampled Step or
Z-transform of the signal
𝑢 𝑘 = 1, 0, 𝑘≥0 𝑘<0
𝑢 𝑘 = 1, 1, 1,1, … 𝑘≥0
𝑈 𝑧 =1+ 𝑧 −1 + 𝑧 −2 + 𝑧 −3 +…+ 𝑧 −𝑘 = 𝑘=0 𝑛 𝑧 −𝑘
𝑈 𝑧 = 1 1− 𝑧 −1 = 𝑧 𝑧− 𝑧 <1

40. z-Transforms of Standard Discrete-Time Signals
Sampled Ramp
Z-transform of the signal
𝑟 𝑘 = 𝑘, 0, 𝑘≥0 𝑘<0
𝑟 𝑘 𝑘 1 2 3 ……
𝑈 𝑧 = 𝑧 𝑧−1 2

41. z-Transforms of Standard Discrete-Time Signals
Sampled Parabolic Signal
Then
𝑢 𝑘 = 𝑎 𝑘 , 0, 𝑘≥0 𝑘<0
𝑈 𝑧 =1+ 𝑎𝑧 −1 + 𝑎 2 𝑧 −2 + 𝑎 3 𝑧 −3 +…+ 𝑎 𝑘 𝑧 −𝑘 = 𝑘=0 𝑛 (𝑎𝑧) −𝑘
𝑈 𝑧 = 1 1− 𝑎𝑧 −1 = 𝑧 𝑧−𝑎 𝑧 <1

42. Inverse Z-transform
Long Division: We first use long division to obtain as many terms as desired of the z-transform expansion.
Partial Fraction: This method is almost identical to that used in inverting Laplace transforms. However, because most z-functions have the term z in their numerator, it is often convenient to expand F(z)/z rather than F(z).

43. Inverse Z-transform
Example-4: Obtain the inverse z-transform of the function
Solution
1. Long Division
𝐹 𝑧 = 𝑧+1 𝑧 𝑧+0.1

44. Inverse Z-transform 1. Long Division Thus Inverse z-transform
𝐹 𝑧 = 𝑧+1 𝑧 𝑧+0.1
1. Long Division
Thus
Inverse z-transform
𝐹 𝑧 =0+ 𝑧 − 𝑧 −2 −0.26 𝑧 −3 +…
𝑓 𝑘 = 0, 1, 0.8, −0.26, …

45. Inverse Z-transform
Example-5: Obtain the inverse z-transform of the function
Solution
2. Partial Fractions
𝐹 𝑧 = 𝑧+1 𝑧 𝑧+0.02
𝐹 𝑧 𝑧 = 𝑧+1 𝑧( 𝑧 𝑧+0.02)
𝐹 𝑧 𝑧 = 𝑧+1 𝑧( 𝑧 𝑧+0.2𝑧+0.02)

46. Inverse Z-transform 𝐹 𝑧 𝑧 = 𝑧+1 𝑧(𝑧+0.1)(𝑧+0.2)
𝐹 𝑧 𝑧 = 𝑧+1 𝑧(𝑧+0.1)(𝑧+0.2)
𝐹 𝑧 𝑧 = 𝐴 𝑧 + 𝐵 𝑧 𝐶 𝑧+0.2

47. Inverse Z-transform 𝐹 𝑧 𝑧 = 50 𝑧 − 90 𝑧+0.1 + 40 𝑧+0.2
𝐹 𝑧 𝑧 = 50 𝑧 − 90 𝑧 𝑧+0.2
𝐹 𝑧 =50− 90𝑧 𝑧 𝑧 𝑧+0.2
Taking inverse z-transform (using z-transform table)
𝑓 𝑘 =50𝛿 𝑘 −90 −0.1 𝑘 +40 −0.2 𝑘

48. Inverse Z-transform
Example-6: Obtain the inverse z-transform of the function
Solution
2. Partial Fractions
𝐹 𝑧 = 𝑧 (𝑧+0.1)(𝑧+0.2)(𝑧+0.3)

49. Z-transform solution of Difference Equations
By a process analogous to Laplace transform solution of differential equations, one can easily solve linear difference equations.
The equations are first transformed to the z-domain (i.e., both the right- and left-hand side of the equation are z-transformed) .
Then the variable of interest is solved.
Finally inverse z-transformed is computed.

50. Z-transform solution of Difference Equations
Example-7: Solve the linear difference equation
With initial conditions 𝑥 0 =1, 𝑥 1 = 5 2
1. Taking z-transform of given equation
2. Solve for X(z)
𝑥 𝑘+2 − 3 2 𝑥 𝑘 𝑥 𝑘 =1(𝑘)
Solution
𝑧 2 𝑋 𝑧 − 𝑧 2 𝑥 0 −𝑧𝑥 1 − 3 2 𝑧𝑋 𝑧 𝑋 𝑧 = 𝑧 𝑧−1
[𝑧 2 − 3 2 𝑧+ 1 2 ]𝑋 𝑧 = 𝑧 𝑧−1 + 𝑧 2 +( 5 2 − 3 2 )𝑧

51. Z-transform solution of Difference Equations
[𝑧 2 − 3 2 𝑧+ 1 2 ]𝑋 𝑧 = 𝑧 𝑧−1 + 𝑧 2 +( 5 2 − 3 2 )𝑧
Then
3. Inverse z-transform
𝑋 𝑧 = 𝑧[1+(𝑧+1)(𝑧−1)] (𝑧−1)(𝑧−1)(𝑧−0.5) = 𝑧 3 𝑧−1 2 (𝑧−0.5)
𝑋 𝑧 𝑧 = 𝑧 2 𝑧−1 2 (𝑧−0.5) = 𝐴 𝑧− 𝐵 𝑧−1 + 𝐶 𝑧−0.5
𝑋(𝑧)= 2𝑧 𝑧− 𝑧 𝑧−0.5
𝑥(𝑘)=2𝑘+ (0.5) 𝑘

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