modeling of electrical systems

7/24/2019 Modeling of Electrical Systems

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Mathematical Modeling of control systems

1- Electrical and electronic Systems

1


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Basic Elements of Electrical Systems

The time domain expression relating voltage and currentfor the resistor is given by Ohms law i-e

Rtitv RR )()( =

The Laplace transform of the above euation is

RsIsVRR

)()( =

Resistor

!


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The time domain expression relating voltage and currentfor the Capacitor is given as:

dttiC

tv cc = )()( 1

The Laplace transform of the above equation (assumingthere is no charge stored in the capacitor) is

)()( sICs

sV cc1

=

"


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The time domain expression relating voltage and currentfor the inductor is given as#

dt

tdiLtv

L

L

)()( =

The Laplace transform of the above euation (assumingthere is no energy stored in inductor) is

)()( sLsIsVLL

=

$


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V- and -V relations of main elements

Compone

nt !"mbol

V-

Relation -V Relation

%esistor

&apacitor

'nductordt

tdiLtv

L

L

)()( =

dttiC

tv cc = )()( 1

RtitvRR

)()( =R

tvti

R

R

)()( =

dt

tdvCti

c

c

)()( =

dttvL

tiLL = )()(

1


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#irchho$%s La&:Kirchhoffs current law (node law): states that the algebraic sum of all

currents entering and leaing a node is !ero"

Kirchhoffs oltage law (loo# or mesh law): states that at any gien

instant the algebraic sum of the oltages around any loo# in an electrical circuit

is !ero"

$ mathematical model of an electrical circuit can be obtained by a##lying one

or both of Kirchhoffs laws to it"

!teps to get Transfer 'unction:

1" $##ly Kirchhoffs law (%ode or &oo# &aw) and write the differentiale'uations for the circuit"

" hen ta*e the &a#lace transforms of the differential e'uations"

+" ,inally sole for the transfer function"

" .raw the bloc* diagram


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xample*:

The two-port networ* shown in the following +gure

has vi(t)as the input voltage and vo(t)as the outputvoltage, ind the transfer function .o(s)/.i(s) of the

networ*,

0

i(t)vi( t) v!(t)

+= dttiC

Rtitvi )()()( 1

= dttiC

tvo )()( 1

1-a##ly Kirchhoff law


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-a*ing &a#lace transform of both e'uations/ considering

initial conditions to !ero"

)()()( sICsRsIsVi1

+= )()( sICs

sVo

1

=

)()( sIsCsVo =

))(()(Cs

RsIsVi1

+=


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2ubstitute I(s) in euation on left

3

))(()(

Cs

RsCsVsV oi1

+=

)1

(

1

)(

)(.

CsRCs

sV

sVFT

i

o

+

==

RCsRCs

101 ==+

The system has one pole at

RCssV

sV

i

o

+

=

1

1

)(

)( .o(s)

.i(s)


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xample*:another solution

The euations of this %& circuitsare4

The Laplace transform of theseeuation are4

The above 5uations give amathematical model of the %& circuit,

16


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&ombining the above two bloc*s we get

7loc* diagrams of these euations are4

11


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he transfer function T.F=0(120"3) of this unit feedbac* system or

45 circuit is6

1!


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xample+: ,btain the transfer functionof the given RLC Circuit

9 transfer-function model of the circuit can be obtained by

ta*ing the Laplace transforms of 5uations (a) and (b) withthe assumption of :ero initial condition; we obtain

9pplying


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The transfer function; T = Eo(s)/Ei(s),of this %L& circuit

can be obtain as4

Ta*ing the I(s) common in euation (c); will geteuation (e);

>ivide euation (d) by (e); inally; ?ultiply and divided by CS.

@ence; the transfer function; T = Eo(s)/Ei(s),ofthe %L& circuit after simpli+cation is

1$


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Transfer 'unction:

n deriving transfer functions for electrical circuits. it is

convenient to &rite the Laplace-transformed equations

directl". &ithout &riting the di$erential equations

Remember that the impedance approach is valid onl" if the

initial conditions involved are all /eros

!ince the transfer function requires /ero initial conditions. the

impedance approach can be applied to obtain the transfer

function of the electrical circuit

This approach greatl" simpli0es the derivation of transfer

functions of electrical circuits

1


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lements transformation using impedance approach

i%(t)

v%(t)

A

-

'%(2)

.%(2)

A

-

B%C%

Transformation

iL(t)

vL(t)

A

-

'L(2)

.L(2)

A

-

BLCL2

ic(t)

vc(t)

A

-

'c(2)

.c(2)

A

-

B&(2)C1/&2

1


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xample1: repeat example-+ and obtainthe T' using the mpedance pproach

The transfer function; Eo(s)/Ei(s); can be obtain by applying the voltage-divider rule; hence

he +rst step is to transform this %L& circuit into the euivalent impedance for

10


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Cascaded lements

&onsider the system shown below, 9ssume that eiis the input and eo

is the output,

The capacitances C1 and C! are not charged initially,

't will be shown that the second stage of the circuit (R!C! portion)

produces a loading e=ect on the +rst stage (R1C1 portion),

1


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The euations for this system are4

(a)

(b)

(c)

Ta*ing the Laplace transforms of 5uations (a); (b) and (c); using :ero

initial conditions; we obtain(d)

(e)

(e) 13


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5liminating I1(s) from 5uations (d) and (e) and writing Ei(s)

in terms of I2(s), we +nd the transfer function between Eo(s)

and Ei(s) to be

!6


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xample3: repeat example2 using thempedance pproach

Obtain the transfer function Eo(s)/Ei(s) by use of the compleximpedance approach, (&apacitors C1 and C! are not charged initially,)

The circuit shown in igure (a) can beredrawn as that shown in igure (b);which can be further modi+ed toigure (c),

(a) (b)

(c)

!1


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'n the system shown the current ' isdivided into two currents I1 and I! , Dotingthat

!!


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Substituting 71 8 41/ 7 8 1(51S)/ 7+ 8 4 / and 7 8 1(5S) into this

last e'uation/ we get

!"


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lectronic !"stems4odeling

!$


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,perational mpli0ers modeling

!

1

2

Z

Z

V

V

in

out =1

21Z

Z

V

V

in

out+=

nverting ampli0ers 5on inverting ampli0ers


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nverting mpli0er#

2ince only a negligible current Eows

into the ampli+er; the current i1must be eual to current i! , Thus4

2ince ; hence we have4

Thus the circuit shown is an inverting ampli+er, 'f R1 C R! ; then the op-

amp circuit shown acts as a sign inverter, !


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5on-inverting mpli0ers:

where < is the di=erential gain of the

ampli+er, rom this euation; we get

This euation gives the output voltage eo, 2ince e

oand e

ihave the

same signs; this op-amp circuit is non-inverting, !0


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xample*: ,btain the Transfer 'unction of thegiven inverting ampli0er

Doting that the current Eowing intothe ampli+er is negligible; we have4

@ence

Ta*ing the Laplace transform; we get

@ence the transfer function

of this inverting ampli+er is4

!


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,btaining the T' of example* using mpedance pproach:

!3


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xample+: repeat example* and obtainthe T' b" mpedance pproach

The complex impedances B1(s) andB!(s) for this circuit are4

The transfer function Eo(s)/Ei(s) is;therefore; obtained as

"6


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xample+: 'ind the transfer function. Vo(s)6Vi(s)

The transfer function of the invertingampli+er circuit is given by

2ince the admittances of parallelcomponents add; B1(s) is the reciprocal of

the sum of the admittances; or

or B!(s) the impedances add; or

The T can be obtain by putting thevalues of B1(s) and B!(s) from e (b) and(c) in e (a),

@ence the transfer function after

(a)

(c)

(b)

"1


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xample1: 'ind the transfer function. Vo(s)6Vi(s)

The transfer function of the non-invertingampli+er circuit is given by

Fe +nd each of the impedance functions;B1(s) and B!(s); and then substitute them intoT euation (a) of non-inverting ampli+er,Thus

(a)

(b)

(c)

"!

modeling of electrical systems

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