Digital Control of Electric...

Czech Technical University in Prague – Faculty of Electrical Engineering

Digital Control of Electric Drives

Ver.1.01

Digital Controller

J. Zdenek 2017

BE1M14DEP Digital Control of Electric Drives - 10 2

Digital vs. Analog Control

DIGITAL CONTROL – ADVANTAGES:

• Possible implementation of complex control algorithms (adaptive regulation), program solution

• Long-term stability of parameters and high accuracy - it does not depend on temperature and aging of components (this only applies to digital parts, not analogue control units).

• Modification (upgrade) of the control algorithm only by changing software (SW), hardware (HW) does not change (updating, enhancement of properties, removal of defects from development without change of HW).

• Easy communication interconnection with other blocks of distributed control system from multiple computers

• Easier set-up of serial product parameters (the debugged program works in the same piece of the controller as always - if HW is OK).

• Easy connection to human communication devices (control and display panels, visualization).


Digital vs. Analog Control (Cont.)

DIGITAL CONTROL – ADVANTAGES:

• Easily connect to global computer networks (Internet).• Remote diagnostics, parameter modification and data transfer (over the

Internet or radio digital networks - mobile phone systems).• The ability to customize properties by changing the software even after

delivery to the customer.

• The fast-expanding option (price and function) digital control can be implemented in HW gate arrays of the FPGA using Hardware Description Language (HDL) programming languages - a high FPGA speed and the ability to customize the necessary digital blocks.

• Lower price of the final product.



DIGITAL CONTROL – DISADVANTAGES:• For the development of digital control systems, additional knowledge of the

staff (beyond knowledge of control technology), ie design of HW control computers, digital signal processing, real-time programming, and program debugging methodology are required.

• For the development of digital control systems, appropriate instrumentation and software equipment (compilers and debuggers, code and HW simulatorsand emulators, logic analyzers, run-time monitors, etc.) are required.

• The view into the control computer operation is very poor (compared to the analog signal oscilloscope).

• The dynamic properties of the control algorithm depend on the sampling period T (see below). With its shrinking, the dynamic properties improve (closer to the analog solution), but there is a rapid increase in computer computational demands and the size of its data memory. This led to the establishment of a special class of computers - the so-called Signal Processors (DSP). These computers include features typically used in digital control technology.



ANALOG CONTROL – ADVANTAGES:

• Good dynamic properties.• For the development of analogue control, knowledge of the construction of

computer technology is not required.• View into an analog controller operation more prominent (oscilloscope).

• Simpler development equipment required (oscilloscope, multimeter, ...).



ANALOG CONTROL – DISADVANTAGES:

• Complex and may be impossible implementation of complex control algorithms (adaptive control).

• Poor long-term stability of the parameters and lower accuracy - depends on the temperature and aging of the components.

• The control algorithm is defined by interconnection of components, it can not be easily changed.

• Difficult connection to computer systems.

• For more complicated solutions, it is difficult to set the parameters of individual pieces of product (each piece must be set individually).

• Remote diagnostics and customization are impossible.

• For more sophisticated systems, a higher price


Transition of analog LTI to digital LTI


Digital controller (regulator)

...)3()2()1(...)2()1()()( 321210 −−−−−−−+−+−+= kubkubkubkeakeakeaku

Difference equation

...))((...)1()( 22

110

33

22

11 +++=+++ −−−−− zazaazEzbzbzbzU

After Z-transformation

After modification – Transfer function:

...1

...

)(

)()(

33

22

11

22

110

−−−

−−

++++++==zbzbzb

zazaa

zE

zUzC Discrete transfer function C(z)

makes possible to analyze propertiesof discrete systems in

z – complex plane



• k – sample serial number, actual time is kT if sampling period is T,• t – time of continuous signals,• w(k), w(t) – required value,• e(k(, e(t) – error (control deviation),• u(k), u(t) – action value,• y(k), y(t) – regulated (controlled) value,

Simplification:Transfer function

of sensor is 1



M005-PO1-dig-analPS-one-step-ovsh-euk

Timer

One Step Scope-01

F(s) - Plant

den(s)num(s)

Sum

+

-

Num: 1Num: 1Den: 1 + 0.1*sDen: 1 + 0.1*s

FOH

S/H

ADC

S/H

C(z) - Controller

den(z)num(z)

Mux

(2)(2)(1)(1) (3)(3) (4)(4) (5)(5) (6)(6)(7)(7) (8)(8)

(9)(9)

Digital PS Controller, Analog Plant (LR)Digital PS Controller, Analog Plant (LR)

Model: Scilab/Scicos

Example -> runFunction blocks of regulator and controlled process (plant)

are defined by transfer functions invariable z (controller) and s (controlled process).



Model with disturbance variable

Disturbance variable is used for modeling of process changes(e.g. loading torque changes) or for modeling

of random noise is system (e.g. noise of sensor)


Discrete controller design

Two methods:

Controller structure and its parameters are unknown beforehand and they are defined duringsynthesis by using of mathematical methods. This approach requires sufficiently accuratemathematical model of controlled process.

Controller structure is known beforehandand design task is to define its parameters. Thisapproach does not necessarily require mathematical model of controlled process. A processparameters are identified by measurement under specified conditions. Such a method is calledcontroller “tuning“.Such a methods for PID regulator structure are:Method of stable oscillation – Ziegler, Nichols (1952)Method of transition response – originally also Ziegler,Nichols


Discrete controller design

To design controller in continuous area(s-complex plane, Laplace transformation).The designed controller to digitize and using suitable sampling period T to reach similarbehavior as in designed analog controller.

Two methods:

To design controller in discrete area(z – complex plane, Z-transformation). Using this methodit is possible in some cases to implement discrete controller for large sampling periods too(with respect to time constants of controlled process (plant))


Discrete controller monitored properties

• controlled variable y(t) ability to trace required variable w(k) by suitable manner,

• ability to suppress influence of disturbance variable d(t) on the controlled variable y(t)

• regulation loop stability, i.e. its immunity against y(t) variable oscillations in the case theprocess or controller parameters are changed.

Similarly as in continuous controllers these are mainly:

Note to the regulator stability problem:

Discrete controller parameters inside computer (variables) are totally stable, there is no driftand no offset. But implemented algorithm could be for some parameter values unstable(influence of used algorithm and finite accuracy of used variable types). In the discrete control system is moreover range of analog parts (parts of ADC and DAC controllers, analog filter, sensors etc.) which parameters are changed with temperatureand ageing.


Continuous PID controller

Continuous PID controller is often use beforehand known structure of regulation block:

Complete PID controller consists of proportional (P), integral (I) a derivative (D) terms. Relation between u(t) (action variable) and e(t) (error variable) has basic form in time(parallel form):

++= ∫

t

DI dt

tdeTdrre

TteKtu

0

0

)()(

1)()(

PID (L-transformation):

)(1

1)( 0 sEsTsT

KsU DI

++=

PID transfer function:

++== sT

sTK

sE

sUsC D

I

11

)(

)()( 0

Ko – regulator gain,Ti – integral constant,Td – derivative constant.


PSD controller by discretization of PID differential equation

Derivative approximation by backward difference:

T

kxkx

dt

txd )1()()( −−≈

Integral approximation by numerical integration (sum of trapezoids)

[ ]∫ ∑=

−+≈kT k

i

ixixT

dttx0 0

)1()(2

)(

By substitution to the PID equation:

perioda vzorkovací−T

[ ] [ ])1()(()(1

0 −−+++= ∑ kekeT

TKieie

kD1)-

T2

TKe(k)Ku(k)

I00

T – sampling period



PSD incremental form (calculates ) (trapezoidal integration)

)1()()( −−=∆ kukuku

[ ] [ ])2()1(()1(1

10 −−−++−+= ∑

−

kekeT

TKieie

kD2-

T2

TK1)-e(kK1)-u(k

I00

After grouping of coefficients with the same time samples:

)2()1()(

)2((2

1

210 −+−+=

=−+

−+−+

++==∆

keqkeqkeq

keT

Tie

T

T

T

T DDD0

I0

I0 K1)-

T2

TKe(k)

T2

T1K1)-u(k-u(k) u(k)

T

Tq

T

TKq

T

TKq

D

D

D

0

I

I

K

T2

T

T2

T1

=

+−−=

++=

2

01

00

21

)(ku∆



Final PSD difference equation (suitable for recurrent calculation in computer):

)2()1()()1()( 210 −+−++−== keqkeqkeqkuku

After Z-transformation:

22

110

1 )()()()()( −−− +++= zzEqzzEqzEqzzUzU

PSD controller transfer function:

1

22

110

1)(

)()( −

−−

−++==

z

zqzqq

zE

zUzC


PSD controller by Bilinear (Tustin) transformation

Bilinear transformation uses substitution:

)1(

)1(

2 −+=

z

zTs

Substitution is used in PID controller Laplace s-domain:

perioda vzorkovací−T

)(1

1)( 0 sEsTsT

KsU DI

++=

Result of Bilinear transformation is identical to approximation of PID controllerdifferential equation using trapezoidal integration method and difference.




PSD positional form (calculates u(k) ) (trapezoidal integration) :

[ ] [ ]

[ ] [ ] [ ])1()(()(()(

)1()(()(

1

10

10

−−+++++

=−−+++=

∑

∑−

kekeT

TKieieieie

kekeT

TKieie

kD

kD

1)-T2

TK1)-

T2

TKe(k)K

1)-T2

TKe(k)Ku(k)

I0

I00

I00

After grouping of coefficients with the same time samples:

[ ]∑−

−+++

++=

1

1

(()(k

DD keT

TKieie

T

T1)-

T2

T1)-

T2

TKe(k)

T2

T1Ku(k)

II0

I0

To define:

[ ]∑−

+=1

1

()(k

ieie 1)-1)-sum(k



PSD positional form (calculates u(k) ) (trapezoidal integration) cont:

And we get the expression suitable for repeated calculation by computer:

1)-q1)-sum(ke(k)u(k) 2 keqq (10 ++=

Where:

−=

=

++=

T

T

T

TKq

T

TKq

T

T

T

TKq

D

I

I

D

I

2

2

2

02

01

00 1

And:

[ ]∑−

+=1

1

()(k

ieie 1)-1)-sum(k



PSD incremental form (calculates ) (backward rectangular integration))(ku∆

)2()1()()1()( 210 −+−++−= keqkeqkeqkuku

T

TKq

T

TKq

T

T

T

TKq DDD

I020100 ,

21, =

+−=

++= 1

Where:

PSD positional form (calculates u(k) ) (backward rectangular integration)

1)-keq)sum(k-qe(k)qu(k) (1 210 ++=

∑−

=1

1

)(1k

ie)sum(k-

−−==

++=

T

T

T

TKq

T

TKq

T

T

T

TKq DD

I

D

I 2,

2,

2 020100 1

Where:


PSD controller setting

Unit step function response method for controlled process in open loop:

• Disconnect feedback.• Bypass regulator (switch off regulator).• Generate unit step to the controlled process input.• Record unit step controlled process response.• In the inflection point of curve draw tangent line.• From graph read following values:

• K – static gain,• Tu – delay time,• Tn – rise time.

• Based on the formula (according relevant author) calculate initial PID parameters:• K0 – gain, • Ti – integral constant,• Td – derivative constant.

• Select sampling period T of discrete (digital) controller.

Approximate T value may by selected as , where is maximumtime constant of regulated system. 10

maxτ≈T maxτ


PSD controller setup (cont.)

• Based on selected PSD controller formula calculate q0, q1 and q2 coefficients itsdifference equation from K0, TI, TD a T values.

• Connect feedback loop and setup calculated q0, q1 a q2 coefficient to the PSD algorithm

• Tune PSD coefficients based on regulated system response on the unit step functionto be quality of PSD behavior near the required one.(aperiodic, faster with overshoot, …).

Unit step function response method for controlled process in open loop:


PSD setup (Unit step function response )

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

0.0

0.2

0.4

0.6

0.8

1.0

One Step Response Ziegler-Nichols Method

t [s]

y (t

) [1

]

Unit step function response

Inflection point

Tangent line Unit step

Static gain

Delay time Rise time


PSD setup (Unit step function response )

Ziegler-Nichols: [1],[2]PI:

,3,1

9,00 uIu

n TTT

T

KK ==

Ziegler-Nichols-Seborg: [1],[2]PID (overshoot):

uDuIu

n TTTTT

T

KK 32,1,2,

2

133,00 === π

[1] O’Dwyer A.: A Summary of PI and PID Controller Tuning Rules for Processeswith Time Delay. Part 2: PID Controller Tuning Rules. IFAC Workshop’00,Terrassa, Spain April 2000, str.242-247

And range of more formulae for PID controller parameters calculation


Discrete system stability

Continuous linear system (LTI) with feedback is stable if transfer function poleswith closed feedback are in complex left s-half plane

Relation between complex plane sand complex plane zis defined as:

vzorkováníperiodaTez sT −= ,

Position of poles in complex z-plane depends on sampling period T. A change of samplingperiod will influence position of poles in z-plane and therefore it changes discrete systemresponse and stability



Transformation of s-plane to z-plane by formula sTez =

Komplexní rovina s Komplexní rovina z

Re{z}

Im{z}

1

σ

jω

σ < 0 0 0

-1

j

- j

z = esT

|z| = eσT

< 1

Imaginary axis of s-plane (i.e. σ = 0) is transformed to the boundary of unit circle |z| = 1.

Complex s-plane Complex z-plane


Discrete system stability analysis (Hint of method)


Discrete system stability analysis (Hint of method) cont.

Transfer function of closed loop linear discrete system is defined as (see previous page):

)(1

)(

)(

)()(

zH

zH

zW

zYzG

+==

System stability is assessed from pole positions of transfer function G(z) i.e. from rootsof characteristic equation of G(z)denominator.

0)(1 =+ zH

Linear discrete system is:• stable– if all characteristic equation roots (closed loop transfer function poles) are inside

of unit circle in z-plane. • critically stable– if one characteristic equation root is equal to 1 or complex conjugate rootspair is on the unit circle.


Damped oscillating process (example)

Damped oscillating process, poles are inside unit circle (Oscillating is discrete PC controller)

0.00 0.05 0.10 0.15 0.20 0.25

0.0

0.2

0.4

0.6

0.8

1.0

1.2

Root Locus - Stability, F(z) K = 4.2

t [t]

y (t

) [1

]

)()()( sFsFzC FOH ++

3.4=K


Stability limit (example)

0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

0.0

0.5

1.0

1.5

2.0

2.5

Root Locus - Stability, H(z) = C(z)*Z[F(s)], K = 4.7

t [s]

y (t

) [1

]

[ ])()()( sFsFZzC FOH+7.4=K


Unstable system (example)

0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

0.0

0.5

1.0

1.5

2.0

2.5

Root Locus - Stability, H(z) = C(z)*Z[F(s)], K = 4.7

t [s]

y (t

) [1

]

85.4=K

[ ])()()( sFsFZzC FOH+


Influence of action element limitation

Action element limitation


Influence of action element limitation (No limitation)

No limitation


Influence of action element limitation – Wind-up effect

With limitation


Anti-Wind-up schema


Influence of action element limitation – Wind-up effect

With limitation, no Anti-Wind-up


Action element limitation and Anti Wind-up schema

With limitation, with Anti-Wind-up


Anti-Wind-up schema (possible solution)

Digital Control of Electric Drives

Digital controller

END

Czech Technical University in Prague – Faculty of Electrical Engineering

Digital Control of Electric...

Documents

Transcript of Digital Control of Electric...