1. Digital and Non-Linear Control
Digital Control and Z Transform

2. Introduction
Digital control offers distinct advantages over analog control that explain its popularity.
Accuracy: Digital signals are more accurate than their analogue counterparts.
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.

3. Structure of a Digital Control System

4. Examples of Digital control Systems
Aircraft Turbojet Engine

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

6. 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)

7. 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)

8. 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)

9. Difference Equations
Difference equations arise in problems where the 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

10. Difference Equations
Example 1: For the given difference equation, determine the (a) order of the equation. Is the equation (b) linear, (c) time invariant, or (d) homogeneous?
𝑦 𝑘 𝑦 𝑘 𝑦 𝑘 =𝑢 𝑘
Solution:
The equation is second order.
All terms enter the equation linearly
All the terms if the equation have constant coefficients. Therefore the equation is therefore LTI.
A forcing function appears in the equation, so it is nonhomogeneous.

11. Z-Transform
Difference equations can be solved using z-transforms which 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.

12. 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

13. Z-Transform 𝑈 𝑧 =1+1 𝑧 −1 +3 𝑧 −2 +2 𝑧 −3 +…
Example 2: Obtain the z-transform of the sequence
𝑈 𝑧 =1+1 𝑧 −1 +3 𝑧 −2 +2 𝑧 −3 +…

14. Laplace Transform and Z-Transform
Given the sampled impulse train of a signal
𝑢(𝑡)
𝑈(𝑠)
𝑈 ∗ (𝑠)
𝑢 ∗ (𝑡)
Sampled impulse train:
𝑢 ∗ 𝑡 = 𝑢 𝑜 𝛿 𝑡 + 𝑢 1 𝛿 𝑡−𝑇 + 𝑢 2 𝛿 𝑡−2𝑇 +…+ 𝑢 𝑘 𝛿 𝑡−𝑘𝑇
𝑢 ∗ 𝑡 = 𝑘=0 ∞ 𝑢 𝑘 𝛿 𝑡−𝑘𝑇

15. Relationship Between Laplace Transform and Z-Transform
The Laplace Transform of above equation is
The Z-transform of 𝑢 ∗ 𝑡 is given as
Comparing (A) and (B) yields
𝑢 ∗ 𝑡 = 𝑢 𝑜 𝛿 𝑡 + 𝑢 1 𝛿 𝑡−𝑇 + 𝑢 2 𝛿 𝑡−2𝑇 +…+ 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 (𝐴)
𝑈 𝑧 = 𝑢 𝑜 + 𝑢 1 𝑧 −1 + 𝑢 2 𝑧 −2 +…+ 𝑢 𝑘 𝑧 −𝑘
𝑈 𝑧 = 𝑘=0 ∞ 𝑢 𝑘 𝑧 −𝑘 (𝐵)
𝑧= 𝑒 𝑠𝑇 where 𝑇 is the sample period

16. A Note
In general, given a transfer function in s-domain you cannot just replace 𝑠 by s= 𝑙𝑛𝑧 𝑇 (from 𝑧= 𝑒 𝑠𝑇 ) to get its z-domain transfer function.
The reason is that 𝑧= 𝑒 𝑠𝑇 is true with the sampled signal.

17. Conformal Mapping between s-plane to z-plane
Where 𝑠=𝜎+𝑗𝜔 for real number 𝜎 and real number 𝜔.
Then 𝑧 in polar coordinates is given by
𝑧= 𝑒 𝑠𝑇
𝑧= 𝑒 (𝜎+𝑗𝜔)𝑇
𝑧= 𝑒 𝜎𝑇 𝑒 𝑗𝜔𝑇
∠𝑧=𝜔𝑇
𝑧 = 𝑒 𝜎𝑇

18. 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

19. 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

20. 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

21. 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

22. 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

23. 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

24. 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

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

26. 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
𝑘=0 ∞ 𝑎 𝑘 = 1 1−𝑎 𝑎 <1

27. 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

28. 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 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 <1

29. 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

30. z-Transforms of Standard Discrete-Time Signals
Sampled Exponential 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 𝑎 𝑧 −1 <1

31. Example 3 Find the z-transform of following causal sequences.
𝑓 𝑘 =2×1 𝑘 +4×𝛿 𝑘 , 𝑘=0,1,2,…
𝑓 𝑘 = 4, 𝑘=2,3,… 0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

32. Example 3 Find the z-transform of following causal sequences.
𝑓 𝑘 =2×1 𝑘 +4×𝛿 𝑘 , 𝑘=0,1,2,…
Solution: Using Linearity Property
𝐹 𝑧 =𝒵{2×1 𝑘 +4×𝛿 𝑘 }
𝐹 𝑧 =2×𝒵 1 𝑘 +4×𝒵{𝛿 𝑘 }
𝐹 𝑧 =2× 𝑧 𝑧−1 +4
𝐹 𝑧 = 6𝑧−4 𝑧−1

33. Example 3 Find the z-transform of following causal sequences.
𝑓 𝑘 = 4, 𝑘=2,3,… 0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
Solution: The given sequence is a sampled step starting at k-2 rather than k=0 (i.e. it is delayed by two sampling periods). Using the delay property, we have
𝐹 𝑧 =𝒵{4×1 𝑘−2 }
𝐹 𝑧 =4 𝑧 −2 𝒵{1 𝑘 }
𝐹 𝑧 =4 𝑧 −2 𝑧 𝑧−1 = 4 𝑧(𝑧−1)

34. Inverse Z-transform
Partial Fraction Expansion: This method is very similar 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).

35. Inverse Z-transform
Example 4: Obtain the inverse z-transform of the function
Solution
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)

36. 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

37. 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