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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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Basic Elements of Electrical Systems
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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Basic Elements of Electrical Systems
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)
"!