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Digital control systems (dcs) lecture 18-19-20
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Transcript of Digital control systems (dcs) lecture 18-19-20
1
Digital Control Systems (DCS)Lecture-18-19-20
Introduction to Digital Control Systems
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Lecture Outline• Introduction
• Difference Equations
• Review of Z-Transform
• Inverse Z-transform
• Relations between s-plane and z-plane
• Solution of difference Equations
3
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.
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Structure of a Digital Control System
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Examples of Digital control SystemsClosed-Loop Drug Delivery System
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Examples of Digital control Systems
Aircraft Turbojet Engine
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• 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.
Difference Equation vs Differential Equation
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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 �̈� (𝑡 )+𝐷 �̇� (𝑡 )+𝐾𝑦 (𝑡)
�̈� (𝑡 )= 1𝑚 𝐹 (𝑡 )− 𝐷
𝑚 �̇� (𝑡 ) 𝐾𝑚 𝑦 (𝑡 )
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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 , ,… and , ,… are constant. • is forcing function
𝑦 (𝑘+𝑛 )+𝑎𝑛−1 𝑦 (𝑘+𝑛−1 )+…+𝑎1 𝑦 (𝑘+1 )+𝑎0 𝑦 (𝑘 )=𝑏𝑛𝑢 (𝑘+𝑛 )+𝑏𝑛− 1𝑢 (𝑘+𝑛−1 )+…+𝑏1 𝑢 (𝑘+1 )+𝑏0 𝑢 (𝑘 )
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Difference Equations
• Example-1: For each of the following difference equations, determine the (a) order of the equation. Is the equation (b) linear, (c) time invariant, or (d) homogeneous?
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Difference Equations
• Example-1: For each of the following difference equations, determine the (a) order of the equation. Is the equation (b) linear, (c) time invariant, or (d) homogeneous?
Solution: a) The equation is second order. b) All terms enter the equation linearlyc) All the terms if the equation have constant coefficients.
Therefore the equation is therefore LTI. d) A forcing function appears in the equation, so it is
nonhomogeneous.
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Difference Equations
• Example-1: For each of the following difference equations, determine the (a) order of the equation. Is the equation (b) linear, (c) time invariant, or (d) homogeneous?
Solution: a) The equation is 4th order. b) All terms are linearc) The second coefficient is time dependent d) There is no forcing function therefore the equation is
homogeneous.
13
Difference Equations
• Example-1: For each of the following difference equations, determine the (a) order of the equation. Is the equation (b) linear, (c) time invariant, or (d) homogeneous?
Solution: a) The equation is 1st order. b) Nonlinearc) Time invariantd) Homogeneous
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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.
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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 )=𝑢𝑜+𝑢1 𝑧−1+𝑢2 𝑧− 2+𝑢𝑘 𝑧−𝑘
𝑈 (𝑧 )=∑𝑘=0
∞
𝑢𝑘 𝑧−𝑘
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Z-Transform
• Example-2: Obtain the z-transform of the sequence
,...0 ,0 ,0 ,4 ,0 ,2 ,3 ,1 ,10 kku
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Relation between Laplace Transform and Z-Transform
• The Laplace Transform of above equation is
• And the Z-transform of is given as
• Comparing (A) and (B) yields
𝑢∗ (𝑡 )=𝑢𝑜 𝛿 (𝑡 )+𝑢1 𝛿 (𝑡−𝑇 )+𝑢2 𝛿 (𝑡−2𝑇 )+…+𝑢𝑘 𝛿 (𝑡−𝑘𝑇 )
𝑈∗ ( 𝑠)=𝑢𝑜+𝑢1𝑒−𝑠𝑇+𝑢2 𝑒−2𝑠𝑇+…+𝑢𝑘 𝑒−𝑘𝑠𝑇
𝑈∗ ( 𝑠)=∑𝑘=0
∞
𝑢𝑘𝑒−𝑘𝑠𝑇 (𝐴)
𝑧=𝑒𝑠𝑇
𝑈 (𝑧 )=∑𝑘=0
∞
𝑢𝑘 𝑧−𝑘(𝐵)
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Conformal Mapping between s-plane to z-plane
• Where .
• Then in polar coordinates is given by
𝑧=𝑒(𝜎+ 𝑗𝜔 )𝑇
𝑧=𝑒𝜎 𝑇 𝑒 𝑗𝜔 𝑇
∠ 𝑧=𝜔𝑇|𝑧|=𝑒𝜎𝑇
𝑧=𝑒𝑠𝑇
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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
– Case-2: Imaginary Pole in s-plane
– Case-3:Complex Poles
𝑠−𝑝𝑙𝑎𝑛𝑒 𝑧−𝑝𝑙𝑎𝑛𝑒
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Conformal Mapping between s-plane to z-plane
• Case-1: Real pole in s-plane
• We know
• Therefore
∠ 𝑧=𝜔𝑇|𝑧|=𝑒𝜎𝑇
|𝑧|=𝑒𝜎𝑇 ∠ 𝑧=0
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Conformal Mapping between s-plane to z-plane
When
∠ 𝑧=𝜔𝑇|𝑧|=𝑒𝜎𝑇
|𝑧|=𝑒0𝑇=1∠ 𝑧=0𝑇=0
𝑠=0
𝑠−𝑝𝑙𝑎𝑛𝑒 𝑧−𝑝𝑙𝑎𝑛𝑒
1
Case-1: Real pole in s-plane
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Conformal Mapping between s-plane to z-plane
When
∠ 𝑧=𝜔𝑇|𝑧|=𝑒𝜎𝑇
|𝑧|=𝑒−∞𝑇=0∠ 𝑧=0
−∞
𝑠−𝑝𝑙𝑎𝑛𝑒 𝑧−𝑝𝑙𝑎𝑛𝑒
0
Case-1: Real pole in s-plane
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Conformal Mapping between s-plane to z-plane
Consider
∠ 𝑧=𝜔𝑇|𝑧|=𝑒𝜎𝑇
|𝑧|=𝑒−𝑎𝑇
∠ 𝑧=0
−𝑎
𝑠−𝑝𝑙𝑎𝑛𝑒 𝑧−𝑝𝑙𝑎𝑛𝑒
1
Case-1: Real pole in s-plane
0
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Conformal Mapping between s-plane to z-plane
• Case-2: Imaginary pole in s-plane
• We know
• Therefore
∠ 𝑧=𝜔𝑇|𝑧|=𝑒𝜎𝑇
|𝑧|=1 ∠ 𝑧=±𝜔 𝑇
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Conformal Mapping between s-plane to z-plane
Consider
∠ 𝑧=𝜔𝑇|𝑧|=𝑒𝜎𝑇
|𝑧|=𝑒0𝑇=1∠ 𝑧=𝜔𝑇
𝑠= 𝑗𝜔
𝑠−𝑝𝑙𝑎𝑛𝑒 𝑧−𝑝𝑙𝑎𝑛𝑒
1
−1
−1
1
𝜔 𝑇
Case-2: Imaginary pole in s-plane
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Conformal Mapping between s-plane to z-plane
When
∠ 𝑧=𝜔𝑇|𝑧|=𝑒𝜎𝑇
|𝑧|=𝑒0𝑇=1∠ 𝑧=−𝜔𝑇
𝑠=− 𝑗 𝜔
𝑠−𝑝𝑙𝑎𝑛𝑒 𝑧−𝑝𝑙𝑎𝑛𝑒
1
−1
−1
1
−𝜔𝑇
Case-2: Imaginary pole in s-plane
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Conformal Mapping between s-plane to z-plane
When
∠ 𝑧=𝜔𝑇|𝑧|=𝑒𝜎𝑇
|𝑧|=𝑒0𝑇=1∠ 𝑧=± 𝜋
− 𝑗 𝜔𝑇
𝑠−𝑝𝑙𝑎𝑛𝑒 𝑧−𝑝𝑙𝑎𝑛𝑒
1
−1
−1
1
𝜋
𝝎𝑻=𝝅
𝑗 𝜔𝑇
Case-2: Imaginary pole in s-plane
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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 .
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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).
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Conformal Mapping between s-plane to z-plane
∠ 𝑧=𝜔𝑇|𝑧|=𝑒𝜎𝑇
|𝑧|=𝑒𝜎𝑇
∠ 𝑧=±𝜔 𝑇
𝑠−𝑝𝑙𝑎𝑛𝑒 𝑧−𝑝𝑙𝑎𝑛𝑒
1
−1
−1
1
Case-3: Complex pole in s-plane
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Mapping regions of the s-plane onto the z-plane
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Mapping regions of the s-plane onto the z-plane
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Mapping regions of the s-plane onto the z-plane
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Example-3
• Map following s-plane poles onto z-plane assume (T=1). Also comment on the nature of step response in each case.
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z-Transforms of Standard Discrete-Time Signals
• The following identities are used repeatedly to derive several important results.
∑𝑘=0
𝑛
𝑎𝑘=1−𝑎𝑛+1
1−𝑎 ,𝑎≠1
∑𝑘=0
∞
𝑎𝑘= 11−𝑎 ,|𝑎|≠1
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z-Transforms of Standard Discrete-Time Signals
• Unit Impulse
• Z-transform of the signal
𝛿 (𝑘 )={1 ,0 ,𝑘=0𝑘≠0
𝛿 ( 𝑧 )=1
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z-Transforms of Standard Discrete-Time Signals
• Sampled Step
• or
• Z-transform of the signal
𝑢 (𝑘 )={1 ,0 ,𝑘≥0𝑘<0
𝑈 (𝑧 )=1+ 𝑧−1+𝑧−2+𝑧−3+…+𝑧−𝑘=∑𝑘=0
𝑛
𝑧−𝑘
𝑢 (𝑘 )= {1,1 ,1,1,… }
𝑈 (𝑧 )= 11−𝑧− 1=
𝑧𝑧−1
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z-Transforms of Standard Discrete-Time Signals
• Sampled Ramp
• Z-transform of the signal
𝑟 (𝑘 )={𝑘 ,0 ,𝑘≥0𝑘<0
𝑈 (𝑧 )= 𝑧( 𝑧−1 )2
𝑟 (𝑘 )
𝑘0 1 2 3
……
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z-Transforms of Standard Discrete-Time Signals
• Sampled Parabolic Signal
• Then
𝑢 (𝑘 )={𝑎𝑘 ,0 ,
𝑘≥0𝑘<0
𝑈 (𝑧 )=1+𝑎𝑧−1+𝑎2 𝑧− 2+𝑎3 𝑧−3+…+𝑎𝑘 𝑧−𝑘=∑𝑘=0
𝑛
(𝑎𝑧 )−𝑘
𝑈 (𝑧 )= 11−𝑎𝑧−1=
𝑧𝑧−𝑎
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Exercise
• 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 , h𝑜𝑡 𝑒𝑟𝑤𝑖𝑠𝑒
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Exercise
• 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× 𝑧𝑧−1 +4
𝐹 ( 𝑧 )=6 𝑧−4𝑧−1
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Exercise• Find the z-transform of following causal sequences.
𝑓 (𝑘 )={ 4 ,𝑘=2,3 ,…0 , h𝑜𝑡 𝑒𝑟𝑤𝑖𝑠𝑒
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−2𝒵 {1 (𝑘−2 ) }
𝐹 ( 𝑧 )=4 𝑧−2 𝑧𝑧−1=
4𝑧 (𝑧−1)
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Inverse Z-transform
1. Long Division: We first use long division to obtain as many terms as desired of the z-transform expansion.
2. 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).
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Inverse Z-transform
• Example-4: Obtain the inverse z-transform of the function
• Solution• 1. Long Division
𝐹 ( 𝑧 )= 𝑧+1𝑧2+0.2𝑧+0.1
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Inverse Z-transform• 1. Long Division
• Thus
• Inverse z-transform
𝐹 ( 𝑧 )= 𝑧+1𝑧2+0.2𝑧+0.1
𝐹 ( 𝑧 )=0+𝑧− 1+0.8 𝑧− 2−0.26 𝑧− 3+…
𝑓 (𝑘 )= {0 ,1 ,0.8 ,−0.26 ,… }
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Inverse Z-transform
• Example-5: Obtain the inverse z-transform of the function
• Solution• 2. Partial Fractions
𝐹 ( 𝑧 )= 𝑧+1𝑧2+0.3𝑧+0.02
𝐹 ( 𝑧 )𝑧 = 𝑧+1
𝑧( 𝑧2+0.3 𝑧+0.02)
𝐹 ( 𝑧 )𝑧 = 𝑧+1
𝑧( 𝑧2+0.1 𝑧+0.2𝑧+0.02)
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Inverse Z-transform𝐹 ( 𝑧 )
𝑧 = 𝑧+1𝑧 ( 𝑧+0.1)(𝑧+0.2)
𝐹 ( 𝑧 )𝑧 = 𝐴
𝑧 + 𝐵𝑧+0.1
+ 𝐶𝑧+0.2
5002.01
2.01.01)0()(
0
FzzFzA
z
90)2.01.0)(1.0(
11.0)2.0)(1.0(
11)1.0()()1.0(1.01.0
zz zz
zz
zzzFzB
40)1.02.0)(2.0(
12.0)2.0)(1.0(
11)2.0()()2.0(2.02.0
zz zz
zz
zzzFzC
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Inverse Z-transform𝐹 (𝑧 )
𝑧 = 50𝑧 −
90𝑧+0.1
+ 40𝑧+0.2
𝐹 ( 𝑧 )=50− 90 𝑧𝑧+0.1+
40 𝑧𝑧+0.2
• Taking inverse z-transform (using z-transform table)
𝑓 (𝑘 )=50 𝛿 (𝑘 )−90 (−0.1 )𝑘+40 (−0. 2 )𝑘
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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)
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Inverse Z-transform• Example-6: Obtain the inverse z-transform of the
function
• Solution• The most convenient method to obtain the partial fraction expansion
of a function with simple real roots is the method of residues.• The residue of a complex function F(z) at a simple pole zi is given by
• This is the partial fraction coefficient of the ith term of the expansion
𝐹 ( 𝑧 )= 𝑧(𝑧+0.1)(𝑧+0.2)(𝑧+0.3)
𝐴𝑖= (𝑧−𝑧𝑖 ) 𝐹 (𝑧 )|𝑧→𝑧 𝑖
F ( z )=∑𝑖=1
𝑛 𝐴𝑖
𝑧−𝑧 𝑖
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Z-transform solution of Difference Equations
• By a process analogous to Laplace transform solution of differential equations, one can easily solve linear difference equations.
1. The equations are first transformed to the z-domain (i.e., both the right- and left-hand side of the equation are z-transformed) .
2. Then the variable of interest is solved.
3. Finally inverse z-transformed is computed.
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Z-transform solution of Difference Equations
• Example-7: Solve the linear difference equation
• With initial conditions
• 1. Taking z-transform of given equation
• 2. Solve for X(z)
𝑥 (𝑘+2 )− 32 𝑥 (𝑘+1 )+ 1
2 𝑥 (𝑘 )=1(𝑘)
Solution
[𝑧 2 𝑋 (𝑧 )−𝑧2 𝑥 (0 )−𝑧𝑥 (1 ) ]− 32 𝑧𝑋 ( 𝑧 )+ 1
2 𝑋 ( 𝑧 )= 𝑧𝑧−1
[ 𝑧¿¿ 2− 32 𝑧+ 1
2 ]𝑋 (𝑧 )= 𝑧𝑧 −1+𝑧 2+(
52 −
32 ) 𝑧 ¿
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Z-transform solution of Difference Equations
• Then
• 3. Inverse z-transform
[ 𝑧¿¿ 2− 32 𝑧+ 1
2 ]𝑋 ( 𝑧 )= 𝑧𝑧 −1+𝑧 2+(
52 −
32 ) 𝑧 ¿
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−1 )2+
𝐵𝑧−1
+𝐶
𝑧−0.5
𝑋 (𝑧 )= 2𝑧( 𝑧−1 )2
+𝑧
𝑧−0.5
𝑥 (𝑘)=2𝑘+(0.5)𝑘