ELG4157: Digital Control Systems

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ELG4157: Digital Control Systems Discrete Equivalents Z-Transform Stability Criteria Steady State Error Design of Digital Control Systems

1

Advantages and Disadvantages • Improved sensitivity. • Use digital components. • Control algorithms easily modified. • Many systems inherently are digital.

• Develop complex math algorithms. • Lose information during conversions due to technical problems. • Most signals continuous in nature.

Digitization • The difference between the continuous and digital systems is that the digital system operates on samples of the sensed plant rather than the continuous signal and that the control provided by the digital controller D(s) must be generated by algebraic equations. • In this regard, we will consider the action of the analog-to-digital (A/D) converter on the signal. This device samples a physical signal, mostly voltage, and convert it to binary number that usually consists of 10 to 16 bits. • Conversion from the analog signal y(t) to the samples y(kt), occurs repeatedly at instants of time T seconds apart. • A system having both discrete and continuous signals is called sampled data system. • The sample rate required depends on the closed-loop bandwidth of the system. Generally, sample rates should be about 20 times the bandwidth or faster in order to assure that the digital controller will match the performance of the continuous controller. 3

Digital Control System D

A ADC

+

A

D Micro Processor

DAC

Correction Element

Process

-

Clock

Measurement

A: Analog D: Digital 4

Continuous Controller and Digital Control y(t)

R(t) +

Gc(s)

Plant

-

Continuous Controller r(t)

A/D

Digital Controller

+

Digital Controller

r(kT) D/A and Hold

p(t)

y(t) Plant

-

m(kT)

D/A

m(t) 5

Applications of Automatic Computer Controlled Systems • Most control systems today use digital computers (usually microprocessors) to implement the controllers). Some applications are: • Machine Tools • Metal Working Processes • Chemical Processes • Aircraft Control • Automobile Traffic Control • Automobile Air-Fuel Ratio • Digital Control Improves Sensitivity to Signal Noise. 6

Digital Control System • Analog electronics can integrate and differentiate signals. In order for a digital computer to accomplish these tasks, the differential equations describing compensation must be approximated by reducing them to algebraic equations involving addition, division, and multiplication. • A digital computer may serve as a compensator or controller in a feedback control system. Since the computer receives data only at specific intervals, it is necessary to develop a method for describing and analyzing the performance of computer control systems. • The computer system uses data sampled at prescribed intervals, resulting in a series of signals. These time series, called sampled data, can be transformed to the s-domain, and then to the z-domain by the relation z = ezt. • Assume that all numbers that enter or leave the computer has the same fixed period T, called the sampling period. • A sampler is basically a switch that closes every T seconds for one instant of time. 7

Sampler r(t) Continuous

0

T

r*(t) Sampled

2T

3T

4T

r(3T) r(4T)

r(2T) r(kT)

r(T) T

2T

3T

Zero-order Hold Go(s)

P(t)



1 1  sT 1  e  sT G0 ( s)   e  s s s



4T 8

Modeling of Digital Computer r(t)

e(t) -

A/D

u*(t)

e*(t) Computer

u(t) D/A

Process

Measure

Sampling analysis Expression of the sampling signal

x * (t )  x(t )   T (t )  x(t ) 





k 0

k 0

 (t  kT)   x(kT) (t  kT)

c(t)

Analog to Digital Conversion: Sampling An input signal is converted from continuous-varying physical value (e.g. pressure in air, or frequency or wavelength of light), by some electro-mechanical device into a continuously varying electrical signal. This signal has a range of amplitude, and a range of frequencies that can present. This continuously varying electrical signal may then be converted to a sequence of digital values, called samples, by some analog to digital conversion circuit. • There are two factors which determine the accuracy with which the digital sequence of values captures the original continuous signal: the maximum rate at which we sample, and the number of bits used in each sample. This latter value is known as the quantization level 10

Zero-Order Hold • The Zero-Order Hold block samples and holds its input for the specified sample period. • The block accepts one input and generates one output, both of which can be scalar or vector. If the input is a vector, all elements of the vector are held for the same sample period. • This device provides a mechanism for discretizing one or more signals in time, or resampling the signal at a different rate. • The sample rate of the Zero-Order Hold must be set to that of the slower block. For slow-to-fast transitions, use the unit delay block.

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The z-Transform The z-Transform is used to take discrete time domain signals into a complexvariable frequency domain. It plays a similar role to the one the Laplace transform does in the continuous time domain. The z-transform opens up new ways of solving problems and designing discrete domain applications. The ztransform converts a discrete time domain signal, which is a sequence of real numbers, into a complex frequency domain representation. 

r * (t )   r (kT )  (t  kT ) k 0

For a signal t  0, Using the Laplace transforms, we have 

{r * (t )}   r (kT )e  ksT k 0

z  e sT 

Z {r (t )}  Z {r * (t )}   r (kT ) z k k 0

U ( z) 

z z 1 

Z { f (t )}  F ( z )   f (kT ) z  k k 0

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Transfer Function of Open-Loop System r(t)

T=1

r*(t)

Zero-order Hold Go(s)

Process

(1  e  st ) 1 Go ( s )  ; G p ( s)  s s ( s  1) Y (s) 1  e  st  Go ( s )G p ( s )  G ( s )  2 R * ( s) s ( s  1) 1 1 1 Expanding into partial fraction : G ( s )  1  e  st ( 2   ) s s 1 s 0.3678 z  0.2644 G( z)  2 z  1.3678 z  0.3678





13

14

Z-Transform Z-transform method: Partial-fraction expansion approaches If :

Kn K1 K2 A(s)      ( s  a1 )( s  a2 )    ( s  an ) s  a1 s  a2 s  an

X(s) 

n

Then : X ( z ) 

 z e

Ki z

i 1

Example:

 ai T

 5( s  4)  5  10 z 15 z 5z 10 15 Z  Z        s s 1 s  2  z 1 z  e T z  e  2T    s( s  1)( s  2) 

Inverse Z-transform method: Partial-fraction expansion approaches

If :

X(z) 

A(z) (z  e n

then : X (kT ) 

 a1T



)( z  e

 a2T

)    (s  e

 a nT

 )

K1 z z e

 a1T



K2 z se

 a2T

 

K i e  ai kT

i 1

Example:

2T  z ( 1  e )  1  z z  1  2kT x(kT)  Z   Z   1  e   z 1   2T z  e 2T    ( z  1)( z  e ) 

Closed-Loop Feedback Sampled-Data Systems r(t)

R(z)

E(z)

G(z)

Y(z)

Y(z)

R(z)

D(z)

E(z)

G(z)

Y(z)

Y(z)

Y ( z) G( z ) G ( z ) D( z )  T ( z)   R( z ) 1  G ( z ) 1  G ( z ) D( z )

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Now Let us Continue with the Closed-Loop System for the Same Problem

Y ( z) G( z) 0.3678 z  0.2644   2 R( z ) 1  G ( z ) z  z  0.6322 z Assume an a unit step input : R( z )  z 1 z (0.3678 z  0.2644) 0.3678 z 2  0.2644 z Y ( z)   3 2 ( z  1)( z  z  0.6322) z  2 z 2  1.6322 z  0.6322 Y ( z )  0.3678 z 1  z 2  1.4 z 3  1.4 z 4  1.147 z 5

17

Stability • The difference between the stability of the continuous system and digital system is the effect of sampling rate on the transient response. • Changes in sampling rate not only change the nature of the response from overdamped to underdamped, but also can turn the system to an unstable. • Stability of a digital system can be discussed from two perspectives: • z-plane • s-plane

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Stability Analysis in the z-Plane A linear continuous feedback control system is stable if all poles of the closed-loop transfer function T(s) lie in the left half of the s-plane. In the left-hand s-plane,   0; therefore, the related magnitude of z varies between 0 and 1. Accordingly the imaginary axis of the s-plane corresponds to the unit circle in the z-plane, and the inside of the unit circle corresponds to the left half of the s-plane. A sampled system is stable if all the poles of the closed-loop transfer function T(z) lie within the unit circle of the z-plane.

ze

sT

e

(  j )T

z  eT z  T 19

The Stability Analysis

Critical stability Im z-plane

The graphic expression of the stability condition for the sampling control systems

The stability criterion

Unstable zone

1

Re

Stable zone

In the characteristic equation 1+GH(z)=0, substitute z with

z

s 1 s 1

—— Bilinear transformation

We can analyze the stability of the sampling control systems the same as we did in chapter 3 (Routh criterion in the s-plane) .  Proof : suppose w    j , z  x  jy , then :   z  1 x  jy  1  x  1  jy  x 2  y 2  1 2y   s    j       j  z  1 x  jy  1  x  1  jy  ( x  1) 2  y 2 ( x  1) 2  y 2   2 2 2 2   0  x  y  1  0  x  y 1   ( for the left half of the s-plane)  (inside the unit circle of the z-plane) 

        

The Stability Analysis 1  G( z )  1 

0.632 Kz z  1.368 z  0.368 2

0

Determine K for the stable system Solution:

1

0.632 Kz z  1.368 z  0.368 2

Make

z

s 1 s 1

 0  0.632 Ks  1.264 s  (2.736  0.632 K )  0

In terms o f the Routh criterion :

0.632 K 1.264 2.736  0.632 K

We have: 0 < K < 4.33

2.736  0.632 K

Example: Stability of a closed-loop system r(t)

Go(s)

Gp(s)

Y(t)

K K (0.3678 z  0.2644) K (az  b) ; G( z)   s ( s  1) z 2  1.3678 z  0.3678 z 2  (1  a ) z  a The poles of the losed - loop transfer function t(z) are the roots of the equation [1  G(z)]  0 : z 2  (1  a ) z  a  Kaz  Kb  0 G p (s) 

K  1; z 2  z  0.6322  ( z  0.5  j 0.6182)( z  0.5  j 0.6182)  0 The systemis stable because the roots lie within the unit circle, When K  10 z 2  2.310 z  3.012  ( z  1.115  j1.295) ( z  1.115  j1.295) (unstable) This systemis stable for : 0  K  2.39 Second - order sampled systemis unstable for increased gain where the continuous22is stable for all values of gain.

Example

23

The Steady State Error Analysis ess  lim ( z  1) E ( z ) z 1

E ( z )  R( z )  c( z )  R( z ) 

R( z )G( z ) R( z )  1  G( z ) 1  G( z )

ess  lim ( z  1) E ( z )  lim ( z  1) z 1

 1  * 1  K p  T  *  Kv  T2  *  K a

r

z 1

r (t )  t

 R( z )  2



G(s)

R( z ) 1  G( z)

r (t )  1(t )  R( z )  r (t )  t

e

 R( z ) 

z ; z 1 Tz ( z  1)

K *p  lim G ( z ) z 1

; 2

T 2 z ( z  1) ( z  1) 3

K v*  lim ( z  1)G ( z ) z 1

; K a*  lim ( z  1) 2 G ( z ) z 1

c

Example e

r -

Z.O.H

T  1s 1) Determine K for the stable system.

G (s)

c

K G( s)  s( s  5)

2) If r(t) = 1+t, determine ess=? Solution

1) 1  e Ts K  G( z)  Z    s s ( s  5 )    K   (1  e ) Z  2   s ( s  5)  K  K K Ts 5 5  (1  e ) Z  2   25  s s s5   Ts

Kz Kz  KTz    5 5 25  (1  z )    ( z  1) 2 z  1 z  e 5T    T 1 1

K z 2  2.2067 z  0.2135   5 ( z  1)( z  0.0067 )

The charecteristic equation of the system: K z 2  2.2067 z  0.2135 1  G( z)  1   0 5 ( z  1)( z  0.0067 ) (5-K ) z 2  (2.2067 K  5.0335 ) z  (0.0335  0.2135 K )  0 z

s 1  0.9932 w  (9.993  1.573 K ) w  (10 .067  2.4202 K )  0 s 1

0  K  4.16 2)

2 K z  2.2067 z  0.2135 K *p  lim G ( z )  lim   z 1 z 1 5 ( z  1)( z  0.0067 )

K v*

K z 2  2.2067 z  0.2135  lim ( z  1)G ( z )  lim    0.2 K z 1 z 1 5 ( z  0.0067 )

ess 

1 1  K *p



T K v*

 0

T 0.2 K

 T 1

5 K

Steady State Error and System Type

1) For unity feedback in figure below,

2)

Design of Digital Control Systems The Procedure:

• Start with continuous system. • Add sampled-data system elements. • Chose sample period, usually small but not too small. Use sampling period T = 1 / 10 fB, where fB = B / 2 and B is the bandwidth of the closed-loop system. – Practical limit for sampling frequency: 20 ˂ s / B ˃40 • Digitize control law. • Check performance using discrete model or SIMULINK.

30

31

Start with a Continuous Design D(s) may be given as an existing design or by using root locus or bode design.

E(z) r(t)

R(z)

D(z)

G(z)

Y(z)

Y(z)

32

Add Samples Necessary for Digital Control • Transform D(s) to D(z): We will obtain a discrete system with a similar behavior to the continuous one. • Include D/A converter, usually a zero-order-device. • Include A/D converter modeled as an ideal sampler.

• And an antialiasing filter, a low pass filter, unity gain filter with a sharp cutoff frequency. • Chose a sample frequency based on the closed-loop bandwidth B of the continuous system. 33

Closed-Loop System with Digital Computer Compensation

Y ( z) G ( z ) D( z )  T ( z)  R( z ) 1  G ( z ) D( z ) U ( z) The tranfer function of the computer is  D( z ) E( z) Consider the second order systemwith a zero - order hold and a plant 1 0.3678z  0.7189 k ( z  0.3678) Gp( s )  when T  1; G ( z )  ; If we select D( z )  z  1z  0.3678 s ( s  1) (z  r) We cancer the pole of G ( z )at z  0.3678 and have the two parameters r and K . 1.359 z  0.7189 0.5 z  0.7189 D( z )  ; G ( z ) D( z )  z  0.240 z  1z  0.240 1  A  K a sa zA Gc ( s )  K ; D( z )  C ; Z {Gc ( s )}  D( z ); A  e  aT ; B  e bT ; C 1  B  sb zB b

34

Compensation Networks (10.3; page 747) The compensation network, Gc(s) is cascaded with the unalterable process G(s) in order to provide a suitable loop transfer function Gc(s)G(s)H(s). Compensation R(s)

Gc(s)

+

G(s)

Y(s)

j H(s) M

Gc ( s) 

K  ( s  zi ) i 1 N

 ( s  pi )

 -p

-z

j 1

K (s  z) First  order compensator  Gc ( s )  ( s  p) When z p, the network is called a phase - lead network

35

Closed-Loop System with Digital Computer Compensation There are two methods of compensator design: (1) Gc(s)-to-D(z) conversion method, and (2) Root locus z-plane method.

The Gc(s)-to-D(z) Conversion Method

sa Gc ( s )  K (First - Order Compensator) sb zA D( z )  C (Digital Controller) zB Z {Gc ( s )}  D( z ) (z - transform) Ae

 aT

;B  e

bT

 1  A a ;C  K when s  0 1  B  b 36

The Frequency Response The frequency response of a system is defined as the steady-state response of the system to a sinusoidal input signal. The sinusoid is a unique input signal, and the resulting output signal for a linear system, as well as signals throughout the system, is sinusoidal in the steady-state; it differs form the input waveform only in amplitude and phase.

37

Phase-Lead Compensator Using Frequency Response A first-order phase-lead compensator can be designed using the frequency response. A lead compensator in frequency response form is given by

Gc( s )

 1     s     1   s 

p

1 

z

1 

m

z p

 

sin m

1 1

In frequency response design, the phase-lead compensator adds positive phase to the system over the frequency range. A bode plot of a phase-lead compensator looks like the following

Phase-Lead Compensator Using Frequency Response Additional positive phase increases the phase margin and thus increases the stability of the system. This type of compensator is designed by determining alfa from the amount of phase needed to satisfy the phase margin requirements. Another effect of the lead compensator can be seen in the magnitude plot. The lead compensator increases the gain of the system at high frequencies (the amount of this gain is equal to alfa. This can increase the crossover frequency, which will help to decrease the rise time and settling time of the system.

Phase-Lag Compensator Using Root Locus A first-order lag compensator can be designed using the root locus. A lag compensator in root locus form is given by

G c( s )

( s  z) ( s  p)

where the magnitude of z is greater than the magnitude of p. A phase-lag compensator tends to shift the root locus to the right, which is undesirable. For this reason, the pole and zero of a lag compensator must be placed close together (usually near the origin) so they do not appreciably change the transient response or stability characteristics of the system. When a lag compensator is added to a system, the value of this intersection will be a smaller negative number than it was before. The net number of zeros and poles will be the same (one zero and one pole are added), but the added pole is a smaller negative number than the added zero. Thus, the result of a lag compensator is that the asymptotes' intersection is moved closer to the right half plane, and the entire root locus will be shifted to the right.

Lag or Phase-Lag Compensator using Frequency Response A first-order phase-lag compensator can be designed using the frequency response. A lag compensator in frequency response form is given by

G c( s )

 1     s     1   s 

The phase-lag compensator looks similar to a phase-lead compensator, except that a is now less than 1. The main difference is that the lag compensator adds negative phase to the system over the specified frequency range, while a lead compensator adds positive phase over the specified frequency. A bode plot of a phase-lag compensator looks like the following

Example: Design to meet a Phase Margin Specification Based on Chapter 10 (Dorf): Example 13.7

G p (s) 

1740 . We will attempt to design Gc ( s ) so that we achieve s (0.25s  1)

a phase margin of 45o with a crossover frequeny c  125 rad/s (Fig 10.10). Using the Bode diagram of G p ( s ), we find that the phase margin is 2 o (Eq 10.24). Based on 10.4, we find that the required pole - zero ratio is   6.25 (Eq 10.18). 1 2

c  ab  ; a  50; and b  312; Gc ( s ) 

K ( s  50) ( s  312)

We select K in order to yield GGc ( jω)  1 When   c  125 rad/s. Then K  5.6. Now the compensator Gc ( s ) is to be realized by D( z ). Set T  0.001 second. We have 4.85( z  0.95) ( z  0.73) If we select another value for the sampling period, the the coefficient of D( z ) would differ! A  e 0.05  0.95, B  e -0.312  0.73, and C  4.85; D( z ) 

42

The Root Locus of Digital Control Systems

R(s) D(z)

+ -

Zero Order hold

Y(s) KGp(s)

Y ( z) KG ( z ) D( z )  ; 1  KG ( z ) D( z )  0 (Characteristic equation) R( z ) 1  KG ( z ) D( z ) Plot the root locus for the characteri stic equation of the sampled systemas K varies. 1. The root locus starts at the poles and progresses to the zeros. 2. The root locus lies on a section of the real axis to the left of an odd number of poles and zeros. 3. The root locus is symmetrical with respect to the horizontal real axis. 4. 1  KG ( z ) D( z )  0 or KG ( z ) D( z )  1 and KG ( z ) D( z )  180o  k 360o 43

Root Locus of a Second Order System K increasing Im {z}

Unstable

Root locus Unit circle

One zero At z = -1 -3

-2

-1

Re {z} 0

2 poles at z=1

K ( z  1) 0 2 ( z  1) Let z   and solve for K

1  KG ( z )  1 

(  1) 2 K   F ( ) (  1) dF ( )  0;  1  3;  2  1 d

44

Design of a Digital Controller In order to achieve a specified response utilizing a root locus method, ( z  a) we will select a controller D( z )  ( z  b) Use (z - a) to cancel one pole at G(z) that lies on the positive real axis of the z - plane. Select (z - b) so that the locus of the compensated systemwill give a set of complex roots at a desired point within the unit circle on the z - plane.

45

Example: Design of a digital compensator Let us design a compensator D(z) that will result in a stable system when G p ( s ) is as described in Example 13.8. With D( z )  1, we have unstable system.Select D( z ) 

za z b

K ( z  1)( z  a ) KG ( z ) D( z )  ( z  1) 2 ( z  b) If we select a  1 and b  0.2, k ( z  1) we have KG ( z ) D( z )  ( z  1)( z  0.2) Using the equation for F ( ), we obtain the entry point as z  -2.56. The root locus is on the unit circle at K  0.8. Thus the systemis stable for K  0.8. 46

If the systemperformance were inadequate , we would improve the root locus by selecting a  1 and b  - 0.98 so that K ( z  1) K KG ( z ) D( z )   ( z  1)( z  0.98) ( z  1) Then the root locus would lie on the real axis of the z - plane. When K  1, the root of the characteri stic equation is at the origin.

47

Im{z} K=0.8 K increasing

Unit circle

Re{z} -1

0.2

+1

Entry point at z = -2.56

Root locus 48

P13.10 Dorf 1 ; T  0.1; D( z )  K s ( s  10) 0.0037 z  0.0026 (a) The transfer function G ( z ) D( z )  K 2 z  1.368 z  0.3679 0.0037 z  0.0026 (b) The closed - loop systemcharacteri stic equation is 1  K 2 0 z  1.368 z  0.3679 (c) Using root locus method, maximum value of K is 239. (d) Using Figure 13.19 for T/τ  1 and maximum overshoot of 0.3, we find K  75. 0.2759 z  0.1982 (e) When K  75; T ( z )  2 z  1.092 z  0.5661 (f) When K  119.5, the poles are z  0.4641  j 0.6843. The overshoot is 0.55. G p (s) 

49

P13.11 Dorf

sa sb By using Bode Plot, we may select a  0.7, b  0.1, and K  150. The compensated systemovershoot and safety - state tracking error (for a ramp input) are PO  30% and ess  0.01. (a) Gc ( s )  K

(b) Use Gc ( s ) to D( z ) method (T  0.1) : D( z )  C

zA z  0.9324  155.3 zB z  0.99

1 A a A  e ; B  e ;C K 1 B b A  e 0.007  0.9324; B  e 0.01  0.99; C  155.3  aT

bT

(d) Use Gc ( s ) to D( z ) method (T  0.01) : D( z )  C

zA z  0.993  150 zB z  0.999

1 A a A  e ; B  e ;C K 1 B b A  e 0.07  0.993; B  e 0.01  0.999; C  150  aT

bT

50

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ELG4157: Digital Control Systems

ELG4157: Digital Control Systems Discrete Equivalents Z-Transform Stability Criteria Steady State Error Design of Digital Control Systems 1 Advanta...

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