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CCHHAAPPTTEERR 33
MMOODDEELLLLIINNGG OOFF PPOOWWEERR SSYYSSTTEEMMSS FFOORR AAGGCC WWIITTHH
IINNTTEEGGRRAALL CCOONNTTRROOLL AANNDD OOPPTTIIMMAALL CCOONNTTRROOLL
33..11 IINNTTRROODDUUCCTTIIOONN
For AGC studies it is necessary to obtain appropriate models of the
interconnected power systems. In present research work, models of the following
types of power systems have been considered for AGC studies [3-4, 6-8, 11, 35-42].
1) Two area thermal-thermal (non reheat)
2) Two area thermal-thermal (reheat)
3) Two area thermal-hydro
4) Three area thermal-thermal-hydro
The models of above mentioned interconnected power systems with integral
control scheme, state space modeling of these power systems to design optimal
controllers and stability studies of these power system models have been dealt with
in this chapter. The discrete versions of these power system models have also been
obtained.
The models of above mentioned interconnected power systems have been
used subsequently in Chapter 5 for illustrating the application of artificial neural
networks as controllers for AGC.
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33..22..11 MMOODDEELL OOFF AA TTWWOO AARREEAA TTHHEERRMMAALL--TTHHEERRMMAALL ((NNOONN RREEHHEEAATT))
PPOOWWEERR SSYYSSTTEEMM WWIITTHH IINNTTEEGGRRAALL CCOONNTTRROOLLLLEERR
Perturbed model of a two area thermal-thermal (non reheat) power system
with conventional integral controller scheme is shown in Fig. 3.1.
1------------1 + sTg1
1------------1 + sTt1
Kp1------------1 + sTp1
1-----
S
1------------1 + sTg2
1------------1 + sTt2
Kp2------------1 + sTp2
-1
-----
S
-1
-KT
∆Pg1 ∆Pt1
∆Pg2 ∆Pt2
∆PD1
∆PD2
∆Ptie(1,2)
−
−
+
−
+
−
+
−
u1ACE1
1-----R1
−∆ f 1
∆ f 2
B1
1-----
S +
1-----R2
2πT0
B2
+
+
+
+
GovernorSteam Turbine
Non Reheat Power System
GovernorSteam Turbine
Non Reheat Power System
AREA 1(THERMAL NON REHEAT)
TIE LINE
12
3
AREA 2(THERMAL NON REHEAT)
45
6
7
8
9
u2
x1x2x3
x4x5x6
x7
LoadDisturbance
(d1)
LoadDisturbance
(d2)
Integral Controller
Integral Controller
-KT
ACE2
Fig. 3.1: Two area thermal-thermal (non reheat) system with integral controller
The system state equations with reference to transfer function blocks from 1
to 7 (equations for 71xto x ) are same as in state space model of the same power
system (Section 3.3.1, Fig. 3.5). With integral control, the equations for control inputs
21&uu are as given below:
Area 1 (Block 8):
)()( 71111 x x BK ACE K u T T +−=−=
Area 2 (Block 9):
)()( 74222 x x BK ACE K u T T −−=−=
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33..22..22 MMOODDEELL OOFF AA TTWWOO AARREEAA TTHHEERRMMAALL––TTHHEERRMMAALL ((RREEHHEEAATT)) PPOOWWEERR
SSYYSSTTEEMM WWIITTHH IINNTTEEGGRRAALL CCOONNTTRROOLLLLEERR
Perturbed model of a two area thermal-thermal (reheat) power system with
conventional integral controller scheme is shown in Fig. 3.2.
Kp1----------1+sTp1
1----S
Kp2----------1+sTp2
-1-1
∆PD1
∆PD2
∆Ptie(1,2)
−
−
+
−
+
−
+
−
ACE1
1-----R1
−
∆ f 1
∆ f 2
B1
+
ACE2
1-----R2
B2
+
+
+
+
+
−
AREA 1(THERMAL)
TIE LINE
∆Pt21----S
Σ Σ Σ
Σ
ΣΣΣ
1-----------1+sTg1
1----------1+sTt1
∆Pg1 ∆Pt11+sKr1Tr1---------------
1 + sTr1
∆Pr1
AREA 2(THERMAL)
1-----------1+sTg2
1----------1+sTt2
∆Pg2 1+sKr2Tr2---------------
1 + sTr2
∆Pr2
2πΤ0−−−−
S
Power SystemSteam
TurbineReheater StageGovernorIntegral Controller
Power SystemSteam
TurbineReheater StageGovernorIntegral Controller
-KT
x1x2x3x4
x5x6x7x8
x9
12
34
5678
9
10
11
u1
u2-KT
Fig. 3.2: Two-area thermal-thermal (reheat) power system with integral controller
The system state equations with reference to transfer function blocks from 1
to 9 (equations for 91 xto x ) are same as in the state space model of the same power
system (Section 3.3.2, Fig. 3.6). With integral control, the equations for control inputs
21&uu are as given below:
Area 1 (Block 10):
)()(91111x x BK ACE K u T T +−=−=
Area 2 (Block 11):
)()(95222 x x BK ACE K u T T −−=−=
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33..22..33 MMOODDEELL OOFF AA TTWWOO AARREEAA TTHHEERRMMAALL––HHYYDDRROO PPOOWWEERR SSYYSSTTEEMM WWIITTHH
IINNTTEEGGRRAALL CCOONNTTRROOLLLLEERR
Perturbed model of a two area thermal-hydro power system with
conventional integral controller scheme is shown in Fig. 3.3.
1------------1 + sTg1
1------------1 + sTt1
Kp1------------1 + sTp1
1-----
S
1----------1 + sT1
1 - sTw-------------1+0.5sTw
Kp2------------1 + sTp2
-1-1
-KT
∆Pg1 ∆Pt1
∆Ptw
∆PD1
∆PD2
∆Ptie(1,2)
−
−
+
−
+
−
+
−
ACE1
1-----R1
−∆ f 1
∆ f 2
B1
1-----
S +
ACE2
1-----R2
2πTo
------
s
B2
+
+
+
+
+
−
GovernorSteam Turbine
Non Reheat Power System
Governorstage 2 Water Turbine Power System
AREA 2(HYDRO)
AREA 1(THERMAL NONREHEAT)
TIE LINE
Σ Σ Σ
Σ
Σ
ΣΣ1+sT2
-----------1 + sT3
Governorstage 1
∆PG2∆PG1
Integral Controller
Integral Controller
x1x2x3
x4x5x6x7
x8
9
u1
u2
10
12
3
4567
-KH
8
Fig. 3.3: Two area thermal-hydro power system with integral controller
The system state equations with reference to transfer function blocks from 1
to 8 (equations for 81 xto x ) are same as in the state space model of the same power
system (Section 3.3.3, Fig. 3.7). With integral control, the equations for control inputs
21&uu are as given below:
Area 1 (Block 9):
)()(81111 x x BK ACE K u T T +−=−=
Area 2 (Block 10):
)()( 84222 x x BK ACE K u H H −−=−=
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33..22..44 MMOODDEELL OOFF AA TTHHRREEEE AARREEAA TTHHEERRMMAALL––TTHHEERRMMAALL--HHYYDDRROO PPOOWWEERR
SSYYSSTTEEMM WWIITTHH IINNTTEEGGRRAALL CCOONNTTRROOLLLLEERR
Perturbed model of a three area thermal-thermal-hydro power system with
conventional integral controller scheme is shown in Fig. 3.4.
1------------1 + sTg1
1------------1 + sTt1
Kp1------------1 + sTp1
1-----S
1------------1 + sTg2
1------------1 + sTt2
Kp2------------1 + sTp2
-KT
∆Pg1 ∆Pt1
∆Pg2 ∆Pt2
∆PD1
∆PD2
∆Ptie(1)
−
−
+
−
+
−
+
ACE1
1-----R1
−
∆ f 1
∆ f 2
B1
1-----S
-KT+
ACE2
+
+
+
+
GovernorSteam Turbine
Non Reheat Power System
GovernorSteam Turbine
Non Reheat Power System
AREA 1(THERMAL NON REHEAT)
1 + sT2------------1 + sT3
1 - sTw---------------1 + 0.5 sTw
Kp3------------1 + sTp3
∆PG2 ∆Ptw−
+
−
−
∆ f31-----S
-KH+
u3ACE3+
+
Water Turbine Power System
------------s
2πT12 ------------s
2πT13
------------s
2πT23
1-----R2
B2
∆PD31
-----R3
B3
+
−
+ −
+
+
+ −
+
+
a12 = -1
a13 = -1
a23 = -1
+
+
−
∆Ptie(2)
∆Ptie(3)
AREA 3(HYDRO)
1------------1 + sT1
Governorstage 1
∆PG1
AREA 2(THERMAL NON REHEAT)
4
5
6
78910
12
3
13
1112
14
15
16
u2
u1
x1x2x3
x4x5x6
x7x8x9x10
x11x12
x13
TIE LINE
TIE LINE
TIE LINE
Governorstage 2
Fig. 3.4: Three area thermal-thermal-hydro power system with integral controller
The system state equations with reference to transfer function blocks from 1
to 13 (equations for 131 xto x ) are same as in the state space model of the same power
system (Section 3.3.4, Fig. 3.8). With integral control, the control inputs 321&, uuu are:
Area 1 (Block 14):
)()(12111111x x x BK ACE K u T T ++−=−=
Area 2 (Block 15):
)()( 13114222 x x x BK ACE K u T T +−−=−=
Area 3 (Block 16):
)()(13127333x x x BK ACE K u H H −−−=−=
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33..33 SSTTAATTEE SSPPAACCEE RREEPPRREESSEENNTTAATTIIOONN OOFF PPOOWWEERR SSYYSSTTEEMM MMOODDEELLSS FFOORR
OOPPTTIIMMAALL CCOONNTTRROOLL
33..33..11 SSTTAATTEE SSPPAACCEE MMOODDEELL OOFF TTWWOO AARREEAA TTHHEERRMMAALL--TTHHEERRMMAALL ((NNOONN RREEHHEEAATT)) PPOOWWEERR SSYYSSTTEEMM
For the two area thermal–thermal (non reheat) power system, the state space model
with full state feedback (9 state feedback) has been developed as shown in Fig. 3.5.
1------------1 + sTg1
1------------1 + sTt1
Kp1------------1 + sTp1
1----S
Kp2------------1 + sTp2
-1
1-----S
-1
∆Pg1 ∆Pt1
∆PD1
∆PD2
∆Ptie(1,2)
−
−
+
−
+
−
+
−
ACE1
1-----R1
−
∆ f 1
∆ f 2
B1
+
ACE2
1-----R2
2πT0
B2
+
+
+
+
+
−
AREA 2(THERMAL)
AREA 1(THERMAL)
TIE LINE
1------------1 + sTg2
1------------1 + sTt2
∆Pg2 ∆Pt2
x1 x2 x3
x7
x5 x6
x8
x9 x4
u1
u2
d1
d2
1----S
123
6 5 4
7
8
9
Σ Σ Σ
Σ
ΣΣΣ
Fig. 3.5: State space model of two area thermal-thermal (non reheat) power system
Different variables have been defined as:
State Variables:
11f x ∆= 12
Pt x ∆= 13 Pg x ∆= 24f x ∆= 25 Pt x ∆= 26 Pg x ∆=
)2,1(7 tieP x ∆= = dt ACE x 18 = dt ACE x 29
Control inputs: 1u and 2
u
Disturbance inputs: 11 DPd ∆= and 22 DPd ∆=
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State equations:
From the transfer function blocks labeled from 1 to 9 (Fig. 3.5);
For block 1:
)(1721111
d x xK xT xPP
−−=+
i.e., 1
1
1
7
1
1
2
1
1
1
1
1
1d
T
K x
T
K x
T
K x
T x
P
P
P
P
P
P
P
−−+−=
For block 2:
3212 x xTt x =+
i.e., 3
1
2
1
2
11 x
Tt x
Tt x +−=
For block 3:
11
1
313
1u x
R xTg x +
−=+
i.e., 1
1
3
1
1
11
3
111u
Tg x
Tg x
Tg R x +−
−=
For block 4:
)(2752424 d x xK xT x PP −+=+
i.e., 2
2
2
7
2
2
5
2
2
4
2
41 d
T K x
T K x
T K x
T x
P
P
P
P
P
P
P
−++−=
For block 5:
6525 x xTt x =+
i.e., 6
2
5
2
5
11 x
Tt x
Tt x +−=
For block 6:
24
2
6261 u x
R xTg x +−=+
i.e., 2
2
6
2
4
22
6
111u
Tg x
Tg x
Tg R x +−−=
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For block 7:
4
0
1
0
722 xT xT x π π −=
For block 8:
7118 x x B x +=
For block 9:
7429 x x B x −=
Above equations are arranged in vector matrix form called as the ‘State Equation’:
d Bu Ax x Γ ++= ,
where, A(9×9) is State Matrix, B(9×2) is Control Matrix & (9×2) is Disturbance
Matrix
The matrices A, B and are:
−
−
=Γ
=
−
−
−−
−
−
−−
−
−−
=
00
00
00
00
00
0
00
00
0
;
00
00
00
10
00
00
01
00
00
;
00100000
00100000
000002002
0001
01
000
00011
0000
0001
000
0000001
01
00000011
0
0000001
2
2
1
1
2
1
2
1
00222
22
2
2
2
2
2
111
11
1
1
1
1
1
P
P
P
P
P
P
P
P
P
P
P
P
P
P
T K
T
K
Tg
Tg
B
B
B
T T
TgTg R
Tt Tt
T
K
T
K
T
TgTg R
Tt Tt
T
K
T
K
T
A
π π
The State Vector ‘x’ (9×1), Control Vector ‘u’ (2×1) and the Disturbance Vector ‘d’ (2×1) are:
[ ]T x x x x x x x x x x987654321= ; u=
2
1
u
u; d =
2
1
d
d
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DESIGN OF OPTIMAL CONTROLLER
In optimal control, the control inputs are chosen as a linear combination of
feedback from all the nine system states (921
,........,, x x x ) as given below:
9192121111.......... xk xk xk u +++=
9292221212.......... xk xk xk u +++=
where, ‘K’ (2×9) is the feedback gain matrix given by;
=
292827262524232221
191817161514131211
k k k k k k k k k
k k k k k k k k k K
The system State Equation is:
Bu Ax x += ……. (For a step load change of a constant magnitude, ‘ d Γ . ’ = 0)
The output equation is:
DuCx y +=
However, for a feedback control system, the matrix D is assumed zero.
Hence; Cx y = where C (2 × 9) is the Output Matrix.
Finally, the state space model of the system under consideration takes a form as;
Bu Ax x += and Cx y =
The control inputs are linear combinations of system states given by; Kxu −=
Determination of the Feedback Gain Matrix (K):
The design of an optimal controller is to determine the feedback matrix ‘K’ in
such a way that a certain Performance Index (PI) is minimized while transferring the
system from an initial arbitrary state0)0( ≠ x
to origin in infinite time i.e.,0)( =∞ x
.Generally the PI is chosen in quadratic form as:
( )∞
+=
02
1dt u Ru xQ xPI
T T
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where, ‘Q’ is a real, symmetric and positive semi-definite matrix called as ‘state
weighting matrix’ and ‘R’ is a real, symmetric and positive definite matrix called as
‘control weighting matrix’.
The matrices Q and R are determined on the basis of following system requirements.
1) The excursions (deviations) of ACEs about steady values are minimized. In this
model, these excursions are;
711)2,1(111 x x BP f B ACE tie +=+∆= and742)2,1(222 x x BP f B ACE tie −=−∆=
2) The excursions of dt ACE about steady values are minimized. In this model,
these excursions are 8 x & 9
x .
3) The excursions of control inputs 21uand u about steady values are minimized.
Under these considerations, the PI takes a form;
( ) ( ) ( ) ( ) ( ) ( )[ ]∞
++++−++=
0
2
2
2
1
2
9
2
8
2
742
2
7112
1dt uu x x x x B x x BPI
i.e., [ ]∞
++++−+++=
0
2
2
2
1
2
9
2
8742
2
4
2
2
2
7711
2
1
2
1 2222
1dt uu x x x x B x B x x x B x BPI
This gives the matrices Q (9×9) and R (2×2) as:
−
−
=
100000000
010000000
0020000
000000000
000000000
0000000
000000000
000000000
0000000
21
2
2
2
1
2
1
B B
B B
B B
Q
=
10
01 R
The matrices A, B, Q & R are known.
The optimal control is given by Kxu −=
‘K’ is the feedback gain matrix given by;
S B RK T 1−
=
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where, ‘S’ is a real, symmetric and positive definite matrix which is the unique
solution of matrix Riccati Equation:
01
=+−+− QS BSBRSAS A
T T
The closed loop system equation is; x A x BK AKx B Ax x C =−=−+= )()(
The matrix )( BK A AC
−= is the closed loop system matrix. The stability of closed
loop system can be tested by finding eigenvalues of C A .
SYSTEM ANALYSIS USING MATLAB
After substituting values of parameters as given in Appendix - I, the state equationsand the matrices A, B, Q & R are:
[ ]17211605.0 d x x x x −−+−=
3225.25.2 x x x +−=
13135.125.122083.5 u x x x +−−=
[ ]27544605.0 d x x x x −++−=
655 5.25.2 x x x +−=
26465.125.122083.5 u x x x +−−=
41744422.044422.0 x x x −=
718 425.0 x x x +=
749425.0 x x x −=
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=
−
−
−−
−
−
−−
−
−−
=
00
00
00
5.12000
00
05.12
00
00
;
00100425.0000
00100000425.0
000004442.0004442.0
0005.1202083.50000005.25.20000
0060605.0000
0000005.1202083.5
0000005.25.20
0060000605.0
B A
=
−
−
=1001;
100000000
010000000
00200425.000425.0
000000000
000000000
00425.000180625.0000
000000000
000000000
00425.000000180625.0
RQ
MATLAB program to obtain S, K & Ac:
(‘MATLAB 6’ software has been used to design the optimal controllers).
A = [-0.05 6 0 0 0 0 -6 0 0; 0 -2.5 2.5 0 0 0 0 0 0; -5.20833 0 -12.5 0 0 0 0 0 0; 0 0 0 -0.05 6 0
6 0 0; 0 0 0 0 -2.5 2.5 0 0 0; 0 0 0 -5.20833 0 -12.5 0 0 0; 0.44422 0 0 -0.44422 0 0 0 0 0;
0.425 0 0 0 0 0 1 0 0; 0 0 0 0.425 0 0 -1 0 0]
B = [0 0; 0 0; 12.5 0; 0 0; 0 0; 0 12.5; 0 0; 0 0; 0 0]
Q = [0.180625 0 0 0 0 0 0.425 0 0; 0 0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 0 0; 0 0 0 0.180625 0 0 -
0.425 0 0; 0 0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 0 0; 0.425 0 0 -0.425 0 0 2 0 0; 0 0 0 0 0 0 0 1 0; 0 0
0 0 0 0 0 0 1]
R = [1 0; 0 1]
S = care(A,B,Q,R)
K = inv(R)*B'*S
Ac = A-B*K
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eig(S)
eig(A)
eig(Ac)
Output of the MATLAB program:
−−
−−
−−−−
−−−
−−−−
−−−−−
−−−−
−−−−−
−−−
=
854.10088.01527.008.04615.03305.00008.03305.0
0088.0854.11527.00008.00226.008.04615.01025.0
1527.01527.06086.00219.00673.01025.00219.00673.0425.0
08.000219.00123.00664.00338.00016.00092.0005.0
4615.0008.00673.00664.03689.02122.00092.00571.00398.0
3305.00226.01025.00338.02122.01797.0005.00398.00462.0
008.00219.00016.00092.0005.00123.00664.00338.0
008.04615.00673.00092.00571.00398.00664.03869.02122.0
0226.03305.01025.0005.00398.00.0462-0.03380.21220.1797
S
Matrix ‘S’ is found to be real, positive definite & symmetric. Its all eigenvalues are
real and positive: 0 ; 0 ; 0.0045 ; 0.0287 ; 0.2385 ; 0.3482 ; 0.656 ; 2.0516 ; 2.111
−−−
−−−−=
102737.01538.08294.04226.002.01156.0063.0
012737.002.01156.0063.01538.08294.04226.0K
Hence the control inputs:
872737.0
602.0
51156.0
4063.0
31538.0
28294.0
14226.0
1x x x x x x x xu −++++−−−=
972737.0
61538.0
58294.0
44226.0
302.0
21156.0
1063.0
2x x x x x x x xu −−−−−++=
The closed loop system matrix ‘AC’ is:
−
−
−−−−−
−
−
−−−−
−
−
=
00100425.0000
00100000425.0
000004442.0004442.0
5.1204208.3423.143673.104908.102504.04444.17871.0
0005.25.20000
0060605.0000
05.124208.32504.04444.17871.0423.143673.104908.10
0000005.25.20
006000060.05-
C A
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The eigenvalues of open loop system matrix (state matrix) ‘A’ are:
0; 0; -13.068; -13.052; -0.38+3.189i; -0.38-3.189i; -0.991+2.262i; -0.991-2.262i; -1.2376
Two eigenvalues are zero and remaining have negative real parts indicating that, the
system is marginally stable before applying the optimal control strategy.The eigenvalues of closed loop system matrix ‘AC’ are:
-13.0594; -13.0758; -1.034+3.4078i; -1.034-3.4078i; -1.4791+2.5810i; -1.4791-2.581i;
-1.3521; -0.7439; -0.6887
All eigenvalues of ‘AC’ have negative real parts indicating that the system is
asymptotically stable after applying optimal control strategy.
33..33..22 SSTTAATTEE SSPPAACCEE MMOODDEELL OOFF TTWWOO AARREEAA TTHHEERRMMAALL--TTHHEERRMMAALL
((RREEHHEEAATT)) PPOOWWEERR SSYYSSTTEEMM
The state space model of two area thermal–thermal (reheat) power system,
with full state feedback (11 state feedback) has been developed as shown in Fig. 3.6.
Kp1----------1+sTp1
1----S
Kp2----------
1+sTp2
-1-1
∆PD1
∆PD2
∆Ptie(1,2)
−
−
+
−
+
−
+
−
ACE1
1-----R1
−
∆ f 1
∆ f 2
B1
+
ACE2
1-----R2
2πTo
-------
s
B2
+
+
+
+
+
−
AREA 1(THERMAL REHEAT)
TIE LINE
∆Pt2
x1
x9
x10
x11 x5
u1
u2
d1
d2
1----
S
12
5
9
10
11
Σ Σ Σ
Σ
ΣΣΣ
1-----------1+sTg1
1----------1+sTt1
∆Pg1∆Pt1
x3 x4
4 31+sKr1Tr1---------------1 + sTr1 x2
∆Pr1
AREA 2(THERMAL REHEAT)
61
-----------
1+sTg2
1----------
1+sTt2
∆Pg2
x7 x8
8 71+sKr2Tr2---------------
1 + sTr2 x6
∆Pr2
Fig. 3.6: State space model of two area thermal-thermal (reheat) power system
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State Variables:
11f x ∆= 12
Pt x ∆= 13 Pr∆= x 14 Pg x ∆= 25 f x ∆= 26 Pt x ∆=
27 Pr∆= x 28 Pg x ∆= )2,1(9 tieP x ∆= = dt ACE x 110 = dt ACE x 211
Control inputs: 21uand u
Disturbance inputs:11 DPd ∆= and
22 DPd ∆=
State equations:
For block 1:
)( 1921111 d x xK xT x PP −−=+
i.e., 1
1
19
1
12
1
11
1
1
1d
T
K x
T
K x
T
K x
T x
P
P
P
P
P
P
P
−−+−=
For block 2:
3212 x xTt x =+
i.e., 3
1
2
1
2
11 x
Tt x
Tt x +−=
For block 3:
4114313 xTr Kr x xTr x +=+
i.e.,
+−
−++−= 1
1
4
1
1
11
14
1
3
1
3
11111u
Tg x
Tg x
Tg RKr x
Tr x
Tr x
i.e., 1
1
14
1
1
1
3
1
1
11
13
11u
Tg
Kr x
Tg
Kr
Tr x
Tr x
Tg R
Kr x +
−+−−=
For block 4:
11
1
414
1u x
R xTg x +
−=+
i.e., 1
1
4
1
1
11
4
111u
Tg x
Tg x
Tg R x +−
−=
For block 5:
)( 2962525 d x xK xT x PP −+=+
i.e., 22
2
92
2
62
2
52
5
1
d T
K
xT
K
xT
K
xT xP
P
P
P
P
P
P−++−=
For block 6:
7626 x xTt x =+
i.e., 7
2
6
2
6
11 x
Tt x
Tt x +−=
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For block 7:
8228727 xTr Kr x xTr x +=+
i.e.,
+−−++−= 2
2
8
2
5
22
28
2
7
2
7
11111u
Tg x
Tg x
Tg RKr x
Tr x
Tr x
i.e., 211
2
28
2
2
2
7
2
5
22
27 u
Tg
Kr x
Tg
Kr
Tr x
Tr x
Tg R
Kr x +
−+−−=
For block 8:
25
2
828
1u x
R xTg x +−=+
i.e., 2
2
8
2
5
22
8
111u
Tg x
Tg x
Tg R x +−−=
For block 9:
5
0
1
0
9
22 xT xT x π π −=
For block 10:
91110 x x B x +=
For block 11:
95211 x x B x −=
The matrices A (11×11) and B (11×2) are:
=
−
−
−−
−−−
−
−
−−
−
−−
−
−−
=
00
00
002
10
2
20
00
00
0
1
1
0
1
1
00
00
;
0010002
0000
00100000001
0000000
20000
2
000
2
100
22
10000
000
2
2
2
1
2
10
22
20000
0000
2
1
2
100000
00
2
200
2
2
2
10000
0000000
1
100
11
1
0000000
1
1
1
1
1
10
11
1
00000000
1
1
1
10
00
1
1000000
1
1
1
1
Tg
Tg
Kr
Tg
Tg
r K
B
B
B
T T
TgTg R
Tg
Kr
Tr Tr Tg R
Kr
Tt Tt
PT
PK
PT
PK
PT
TgTg R
Tg
Kr
Tr Tr Tg R
Kr
Tt Tt
P
T
PK
P
T
PK
P
T
A
π π
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State Vector (x) = [ ]T x x x x x x x x x x x 1110987654321
Control Vector (u) =
2
1
u
u
DESIGN OF OPTIMAL CONTROLLER
The control inputs are:
11111101012121111 .......... xk xk xk xk u−−
++++=
11112101022221212 .......... xk xk xk xk u−−
++++=
where,
=
−−
−−
112102292827262524232221
111101191817161514131211
k k k k k k k k k k k
k k k k k k k k k k k K
Hence, Kxu −=
The system state equation is: Bu Ax x +=
The output equation is: Cx y =
Determination of the Feedback Gain Matrix (K):
( )∞
+=
02
1dt RuuQx xPI
T T
( ) ( ) ( ) ( ) ( ) ( )[ ]∞
++++−++=
0
2
2
2
1
2
11
2
10
2
952
2
9112
1dt uu x x x x B x x BPI
[ ]∞
++++−+++=
0
2
2
2
1
2
11
2
10952
2
5
2
2
2
9911
2
1
2
1 2222
1dt uu x x x x B x B x x x B x BPI
This gives the symmetric matrices Q (11×11) and R (2×2) as:
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−
−
=
10000000000
01000000000
002000000
00000000000
00000000000
00000000000
000000000
00000000000
00000000000
00000000000
000000000
21
2
2
2
1
2
1
B B
B B
B B
Q
= 10
01 R
SYSTEM ANALYSIS USING MATLAB
After substituting values of parameters as given in Appendix - I, the state equations
and the matrices A, B, Q & R are:
[ ]19211 605.0 d x x x x −−+−=
3225.25.2 x x x +−=
143131625.40625.41.073437.1 u x x x x +−−−=
14145.125.122083.5 u x x x +−−=
[ ]29655
605.0 d x x x x −++−=
7665.25.2 x x x +−=
28757 1625.40625.41.073437.1 u x x x x +−−−=
2858 5.125.122083.5 u x x x +−−=
51944422.044422.0 x x x −=
9110 425.0 x x x +=
9511425.0 x x x −=
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=
−
−
−−
−−−
−
−
−−
−−−
−
−−
=
00
00
00
5.120
1625.40
00
00
05.12
01625.4
00
00
;
001000425.00000
0010000000425.0
00000044422.000044422.0
0005.12002083.50000
0000625.41.0073437.10000
00005.25.200000
00600605.00000
00000005.12002083.5
00000000625.41.0073437.1
000000005.25.20
006000000605.0
B A
=
−
−
=10
01
10000000000
01000000000
002000425.0000425.0
00000000000
00000000000
00000000000
00425.0000180625.0000000000000000
00000000000
00000000000
00425.00000000180625.0
RQ
MATLAB program to obtain S, K & Ac:
A = [-0.05 6 0 0 0 0 0 0 -6 0 0; 0 -2.5 2.5 0 0 0 0 0 0 0 0; -1.734375 0 -0.1 -4.0625 0 0 0 0 0 0
0; -5.20833 0 0 -12.5 0 0 0 0 0 0 0; 0 0 0 0 -0.05 6 0 0 6 0 0; 0 0 0 0 0 -2.5 2.5 0 0 0 0; 0 0 0 0
-1.734375 0 -0.1 -4.0625 0 0 0; 0 0 0 0 -5.20833 0 0 -12.5 0 0 0; 0.44422 0 0 0 -0.44422 0 0 0
0 0 0; 0.425 0 0 0 0 0 0 0 1 0 0; 0 0 0 0 0.425 0 0 0 -1 0 0]
B = [0 0; 0 0; 4.1625 0; 12.5 0; 0 0; 0 0; 0 4.1625; 0 12.5; 0 0; 0 0; 0 0]
Q = [0.180625 0 0 0 0 0 0 0 0.425 0 0; 0 0 0 0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 0 0 0 0; 0 0 0 0 0 0
0 0 0 0 0; 0 0 0 0 0.180625 0 0 0 -0.425 0 0; 0 0 0 0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 0 0 0 0; 0 0 0
0 0 0 0 0 0 0 0; 0.425 0 0 0 -0.425 0 0 0 2 0 0; 0 0 0 0 0 0 0 0 0 1 0; 0 0 0 0 0 0 0 0 0 0 1]
R = [1 0; 0 1]
S = care(A,B,Q,R)
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K = inv(R)*B'*S
Ac = A-B*K
eig(A)
eig(S)eig(Ac)
Output of the MATLAB program:
−−−
−−−−
−−−−−
−−−−−−−−
−−−−−−−
−−−−
−−−−−−−
−−−−
−−−−−
−−−−−
=
5523.22827.03311.08619.08284.20729.16283.01766.05304.00436.00186.0
2827.05523.23311.01766.05304.00436.00186.08619.08284.20729.16283.0
3311.03311.07592.22909.00713.18116.00283.02909.00713.18116.00283.0
8619.01766.02909.04697.15859.44893.01815.00457.015.00581.00366.0
8284.25304.00713.15859.4348.147307.16626.015.04894.01752.01163.00729.10436.08116.04893.07307.12523.16208.00581.01752.00238.00095.0
6283.00186.00283.01815.06626.06208.04122.00366.01163.00095.00621.0
1766.08619.02909.00457.015.00581.00366.04697.15859.44893.01815.0
5304.08284.20713.115.04894.01752.01163.05859.4348.147307.16626.0
0436.00729.18116.00581.01752.00238.00095.04893.07307.12533.16208.0
0186.06283.00283.00366.01163.00095.00621.01815.06626.06208.04122.0
S
The matrix ‘S’ has been real, positive definite & symmetric. Its eigenvalues are:
0; 0; 0; 0.0293; 0.5654; 0.6722; 2.1545; 2.5134; 3.117; 16.6845; 17.094
All the eigenvalues of matrix ‘S’ are real and positive.The feedback gain matrix ‘K’ has been obtained as:
−−
−−−=
10823.07172.03995.20873.14896.0053.01621.00026.0026.0
01823.0053.01621.00026.0026.07172.03995.20873.14896.0K
Hence the control inputs are:
109876543211 823.0053.01621.00026.0026.07172.03995.20873.14896.0 x x x x x x x x x xu−++−−−+−−−=
119876543212823.07172.03995.20873.14896.0053.01621.00026.0026.0 x x x x x x x x x xu −−+−−−+−−−=
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The closed loop control matrix ‘AC’ has been obtained as:
−
−
−−−−−−−−−
−−−−−−−−−
−
−
−−−−−−−−
−−−−−−−−
−
−−
=
001000425.00000
0010000000425.0
0000004442.00004442.0
5.1202877.105355.39935.295918.133278.11663.00269.20323.03253.0
1625.404258.30773.10878.105261.47721.32208.06749.00108.01083.0
00005.25.200000
00600605.00000
05.122877.1663.00269.20323.03253.05355.39935.295918.133278.11
01625.44258.32208.06749.00108.01083.00773.10878.105261.47721.3
000000005.25.20
006000000605.0
C A
The eigenvalues of open loop system matrix ‘A’ are:
0; 0; -12.6985; -12.6923; -0.1716+2.5868i; -0.1716-2.5868i; -2.0178; -1.0223+0.7052i;-1.0223-0.7052i; -0.0968; -0.4068
Two eigenvalues are zero and remaining have negative real parts indicating that, the
system is marginally stable before applying the optimal control strategy.
The eigenvalues of closed loop system matrix ‘AC’ are:
-12.6995; -12.6933; -0.4538+2.6334i; -0.4538-2.6334i; -2.0199; -1.1766+1.0781i;
-1.1766-1.0781i; -0.839; -0.2703+0.1338i; -0.2703-0.1338i; -0.2937
The real parts of all the eigenvalues of ‘AC’ are negative indicating that, the system is
asymptotically stable after applying the optimal control strategy.
33..33..33 SSTTAATTEE SSPPAACCEE MMOODDEELL OOFF TTWWOO AARREEAA TTHHEERRMMAALL--HHYYDDRROO PPOOWWEERR
SSYYSSTTEEMM
State space model of two area thermal–hydro power system with full state feedback
(10 state feedback) has been developed as shown in Fig. 3.7.
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1
------------1 + sTg1
1
------------1 + sTt1
Kp1
------------1 + sTp1
1
-----S
1----------1 + sT1
1 - sTw-------------1+0.5sTw
Kp2------------1 + sTp2
-1
1-----S
-1
∆Pg1 ∆Pt1
∆Ptw
∆PD1
∆PD2
∆Ptie(1,2)
−
−
+
−
+
−
+
−
ACE1
1-----R1
−∆ f 1
∆ f 2
B1
1-----S
+
ACE2
1-----R2
2πT0
B2
+
+
+
+
+
−
1
AREA 2(HYDRO)
AREA 1(THERMAL)
TIE LINE
Σ Σ Σ
Σ
Σ
ΣΣ1+sT2-----------1 + sT3
∆PG2∆PG1
x1
23
7 6 5 4
8
9
10
x2 x3 x9
x10 x7 x6 x5
x4
x8
u1
u2
d1
d2
Fig. 3.7: State space model of two area thermal-hydro power system
State Variables:
11f x ∆= 12
Pt x ∆= 13 Pg x ∆= 24 f x ∆= Ptw x ∆=5 26 GP x ∆=
17 GP x ∆= )2,1(8 tieP x ∆=
= dt ACE x 19
= dt ACE x 210
Control inputs:21
uand u
Disturbance inputs: 11 DPd ∆= and 22 DPd ∆=
State equations:
For block 1:
)( 1821111 d x xK xT x PP −−=+
i.e., 1
1
18
1
12
1
11
1
1
1d
T
K x
T
K x
T
K x
T
xP
P
P
P
P
P
P
−−+−=
For block 2:
3212 x xTt x =+
i.e., 3
1
2
1
2
11 x
Tt x
Tt x +−=
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For block 3:
11
1
313
1u x
R xTg x +
−=+
i.e., 11
31
111
3
111
uTg xTg xTg R x+−
−=
For block 4:
)( 2852424 d x xK xT x PP −+=+
i.e., 2
2
28
2
25
2
24
2
4
1d
T
K x
T
K x
T
K x
T x
P
P
P
P
P
P
P
−++−=
For block 5:
6655 5.0 xTw x xTw x −=+
∴ 6655 25.0
1
5.0
1 x x
Tw x
Tw x −+−=
∴
+
−+−
−−+
−= 2
31
27
31
2
3
6
3
4
312
2655
112
22u
T T
T x
T T
T
T x
T x
T T R
T x
Tw x
Tw x
i.e., 2
31
27
331
26
3
54
312
25
2222222u
T T
T x
T T T
T x
T Tw x
Tw x
T T R
T x −
−+
++−=
For block 6:
727636 xT x xT x +=+
∴ 7
3
27
3
6
3
6
11 x
T
T x
T x
T x ++−=
+−−++−= 2
1
7
1
4
123
27
3
6
3
6
11111u
T x
T x
T RT
T x
T x
T x
i.e.,2
31
27
31
2
3
16
3
14
312
26
uT T
T x
T T
T
T x
T x
T T R
T x +
−+−−=
For block 7:
24
2
717
1
u x R xT x +−=+
i.e., 2
1
7
1
4
12
7
111u
T x
T x
T R x +−−=
For block 8:
40
10
8 22 xT xT x π π −=
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For block 9:
8119 x x B x +=
For block 10:
84210 x x B x −=
The matrices A(10×10) and B(10×2) are:
−
=
−
−
−−
−
−−
−
+
−
−
−−
−
−−
=
00
00
001
10
31
20
31
22
0
00
0
1
100
00
0010002
000
0010000001
0000000
2000
2
000
1
100
12
1000
000
31
2
3
1
3
10
312
2000
000
3
2
31
22
3
222
312
22
000
00
2
200
2
2
2
1000
0000000
1
10
11
1
0000000
1
1
1
10
00
1
100000
1
1
1
1
T
T T
T
T T
T
Tg
B
B
B
T T
T T R
T T
T
T T T T R
T T T T
T
T TwTwT T R
T
PT
PK
PT
PK
PT
TgTg R
Tt Tt
PT
PK
PT
PK
PT
A
π π
State Vector (x) = [ ]T x x x x x x x x x x 10987654321
Control Vector (u) =
2
1
u
u
DESIGN OF OPTIMAL CONTROLLER
The control inputs are:
101012121111 .......... xk xk xk u−
+++=
101022221212 .......... xk xk xk u−
+++=
where,
=
−
−
102292827262524232221
101191817161514131211
k k k k k k k k k k k k k k k k k k k k K
hence, Kxu −=
The system state equation is: Bu Ax x +=
The output equation is: Cx y =
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To determine the Feedback Gain Matrix (K):
( )∞
+=
02
1dt RuuQx xPI T T
( ) ( ) ( ) ( ) ( ) ( )
[ ]
∞
++++−++=
0
2
2
2
1
2
10
2
9
2
842
2
8112
1dt uu x x x x B x x BPI
[ ]∞
++++−+++=
0
2
2
2
1210
29842
24
2
228811
2
1
2
1 2222
1dt uu x x x x B x B x x x B x BPI
This gives the symmetric matrices Q (10×10) and R (2×2) as:
−
−
=
1000000000
0100000000
00200000
0000000000
0000000000
0000000000
00000000
0000000000
0000000000
00000000
21
2
2
2
1
2
1
B B
B B
B B
Q
=
10
01 R
SYSTEM ANALYSIS USING MATLAB
After substituting values of parameters as given in Appendix - I, the state equations
and the matrices A, B, Q & R are:
[ ]18211605.0 d x x x x −−+−=
322 5.25.2 x x x +−=
1313 5.125.122083.5 u x x x +−−=
[ ]28544605.0 d x x x x −++−=
276545 002106.01979.02.220008778.0 u x x x x x −−+−=
2001053.0
709894.0
61.0
4000439.0
6u x x x x ++−−=
274702053.002053.0008555.0 u x x x +−−=
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41844422.044422.0 x x x −=
819425.0 x x x +=
8410425.0 x x x −=
−=
−
−
−−
−−
−−
−
−−
−
−−
=
00
00
00
02053.00
001053.00
002106.00
00
05.12
00
00
001000425.0000
001000000425.0
00000044422.00044422.0
00002053.000008555.0000
00009894.01.00000439.0000
0001979.02.220008778.0000
00600605.0000
00000005.1202083.5
00000005.25.20
00600000605.0
B A
=
−
−
=10
01
10000000000100000000
002000425.000425.0
0000000000
0000000000
0000000000
00425.0000180625.0000
0000000000
0000000000
00425.0000000180625.0
RQ
MATLAB program to obtain S, K & Ac:
A = [-0.05 6 0 0 0 0 0 -6 0 0; 0 -2.5 2.5 0 0 0 0 0 0 0; -5.2083 0 -12.5 0 0 0 0 0 0 0; 0 0 0 -0.05 6 0 0 6
0 0; 0 0 0 0.0008778 -2 2.2 -0.1979 0 0 0; 0 0 0 -0.000439 0 -0.1 0.09894 0 0 0; 0 0 0 -0.008555 0 0 -
0.02053 0 0 0; 0.44422 0 0 -0.44422 0 0 0 0 0 0; 0.425 0 0 0 0 0 0 1 0 0; 0 0 0 0.425 0 0 0 -1 0 0]
B = [0 0; 0 0; 12.5 0; 0 0; 0 -0.002106; 0 0.001053; 0 0.02053; 0 0; 0 0; 0 0]
Q = [0.180625 0 0 0 0 0 0 0.425 0 0; 0 0 0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 0 0 0; 0 0 0 0.180625 0 0 0 -
0.425 0 0; 0 0 0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 0 0 0; 0.425 0 0 -0.425 0 0 0 2 0 0; 0 0
0 0 0 0 0 0 1 0; 0 0 0 0 0 0 0 0 0 1]
R = [1 0; 0 1]
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S = care(A,B,Q,R)
K = inv(R)*B'*S
Ac = A-B*K
eig(A)
eig(S)
eig(Ac)
Output of the MATLAB program:
−
−−−−−−
−
−−−
−
−
−
−
−
=
8.94.67.77.34908.81.303.02.0
4.67.74.57.2676621.04.03.0
7.74.59.81.243.757.87.21.06.06.0
7.347.261.249.3963.5207.298.903.02.0
90763.753.5205.11524.847.272.09.06.0
8.867.87.294.845.93.31.05.04.0
1.327.28.97.273.32.101.01.0
01.01.002.01.0001.01.0
3.04.06.03.09.05.01.01.05.03.0
2.03.06.02.06.04.01.01.03.03.0
S
The matrix ‘S’ has been real, positive definite & symmetric. Its eigenvalues are:
0; 0; 0; 0.3; 0.5; 1.5; 2.5; 7.3; 135.9; 1439.2
All the eigenvalues of matrix ‘S’ are real and positive.
The feedback gain matrix ‘K’ has been obtained as:
−−−−
−=
7877.0616.05555.06338.87167.116794.0224.00009.00059.00052.0
616.07877.09674.05813.09204.19205.01842.02095.01572.16597.0K
Hence the control inputs are:
109876543211616.07877.09674.05813.09204.19205.01842.02095.01572.16597.0 x x x x x x x x x xu −−−+−−−−−−=
19876543212 7877.0616.05555.06338.87167.116794.0224.00009.00059.00052.0 x x x x x x x x x xu −+−−−−−++=
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The closed loop system matrix ‘AC’ has been obtained as:
=
−
−
−−−−−−
−−−−−
−−−
−
−−−−−−−−−
−
−−
101000425.0000
001000000425.0
0000004442.0004442.0
0162.00126.00114.01978.02405.00139.00132.000001.00001.0
0008.00006.00006.00898.01123.00007.00007.0000
0017.00013.00012.01797.02247.29986.10013.0000
00600065.0000
7003.78466.90926.122665.70053.245056.113025.21187.154648.144543.13
00000005.25.20
00600000605.0
C A
The eigenvalues of open loop system matrix ‘A’ are:
0; 0; -13.06; -0.4002+2.8873i; -0.4002-2.8873i; -0.6199+1.2494i; -0.6199-1.2494i;
-0.0601+0.0207i; -0.0601-0.0207i; -2.0001
Two eigenvalues are zero and the remaining have negative real parts indicating that,
the system is marginally stable before applying the optimal control strategy.
The eigenvalues of closed loop system matrix ‘AC’ are:
-13.0676; -1.0386+3.1050i; -1.0386-3.1050i; -0.9071+1.4148i; -0.9071-1.4148i;
-2.0001; -0.7583; -0.0778+0.1130i; -0.0778-0.1130i; -0.1545The real parts of all the eigenvalues of ‘AC’ are negative indicating that, after
applying the optimal control strategy, the system is asymptotically stable.
33..33..44 SSTTAATTEE SSPPAACCEE MMOODDEELL OOFF TTHHRREEEE AARREEAA TTHHEERRMMAALL--TTHHEERRMMAALL--
HHYYDDRROO PPOOWWEERR SSYYSSTTEEMM
The state space model of three area thermal–thermal-hydro power system with full
state feedback (16 state feedback) has been developed as shown in Fig. 3.8.
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1------------1 + sTg1
1------------1 + sTt1
Kp1------------1 + sTp1
1-----S
1------------1 + sTg2
1------------1 + sTt2
Kp2------------1 + sTp2
∆Pg1 ∆Pt1
∆Pg2 ∆Pt2
∆PD1
∆PD2
∆Ptie(1)
−
−
+
−
+
−
+
ACE1
1-----R1
−
∆ f 1
∆ f 2
B1
+
ACE2
+
+
+
+
GovernorSteam Turbine
Non Reheat Power System
GovernorSteam Turbine
Non Reheat Power System
AREA 1(THERMAL NON REHEAT)
1 + sT2------------1 + sT3
1 - sTw---------------1 + 0.5 sTw
Kp3------------1 + sTp3
∆PG2 ∆Ptw−
+
−
∆ f3
+
ACE3+
+
Governor Water Turbine Power System
------------s
2πT12 ------------s
2πT13
------------s
2πT23
1-----R2
B2
∆PD31
-----
R3
B3
+
−
+ −
++
+ −
+
+
a12 = -1
a13 = -1
a23 = -1
+
+
−
∆Ptie(2)
∆Ptie(3)
x1 x2 x3
x4 x5 x6
x7 x8 x9
1------------1 + sT1
Governor
∆PG1
x10
x11
x12
x13
u1
x14
1-----S x15
u2
1-----S x16
u3
d2
d3
d1
AREA 3(HYDRO)
−
1214 3
456
78910
1112
13
15
16
AREA 2(THERMAL NON REHEAT)
Fig. 3.8: State space model of three area thermal-thermal-hydro power system
State Variables:
11 f x∆=
12 Pt x∆=
13 Pg x∆=
24 f x∆=
25 Pt x∆=
26 Pg x∆=
37 f x ∆= Ptw x ∆=8 29 GP x ∆= 110 GP x ∆=
)2,1(11 tieP x ∆= )3,1(12 tieP x ∆=
)3,2(13 tieP x ∆= = dt ACE x 114 = dt ACE x 215 = dt ACE x 316
Control inputs:321
, uand uu
Disturbance inputs:11 DPd ∆= , 22 DPd ∆= and 33 DPd ∆=
State equations:
For block 1:
)( 1121121111 d x x xK xT x PP −−−=+
i.e., 1
1
112
1
111
1
12
1
11
1
1
1d
T
K x
T
K x
T
K x
T
K x
T x
P
P
P
P
P
P
P
P
P
−−−+−=
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For block 2:
3212 x xTt x =+
i.e., 3
1
2
1
2
11 x
Tt x
Tt x +−=
For block 3:
11
1
313
1u x
R xTg x +
−=+
i.e., 1
1
3
1
1
11
3
111u
Tg x
Tg x
Tg R x +−−=
For block 4:
[ ]2131152424d x x xKp xTp x −−+=+
i.e., 2
2
213
2
211
2
25
2
24
2
4
1d
Tp
Kp x
Tp
Kp x
Tp
Kp x
Tp
Kp x
Tp x −−++
−=
For block 5:
6525x xTt x =+
i.e.,6
2
5
2
5
11 x
Tt x
Tt x +−=
For block 6:
24
2
626
1u x
R xTg x +−=+
i.e., 2
2
6
2
4
22
6
111u
Tg x
Tg x
Tg R x +−−=
For block 7:
[ ]3131283737
d x x xKp xTp x −++=+
i.e., 3
3
313
3
312
3
38
3
37
3
7
1d
Tp
Kp x
Tp
Kp x
Tp
Kp x
Tp
Kp x
Tp x −+++−=
For block 8:
99885.0 xTw x xTw x −=+
i.e., 3
31
210
331
29
3
87
313
28
2222222u
T T
T x
T T T
T x
T Tw x
Tw x
T T R
T x −
−+
++−=
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For block 9:
10210939 xT x xT x +=+
i.e., 3
31
210
31
2
3
9
3
7
313
29
11u
T T
T x
T T
T
T
x
T
x
T T R
T x +
−+−−=
For block 10:
37
3
10110
1u x
R xT x +−=+
i.e., 3
1
10
1
7
13
10
111u
T x
T x
T R x +−−=
For block 11:
41211211 22 xT xT xπ π −=
For block 12:
7131131222 xT xT x π π −=
For block 13:
7234231322 xT xT x π π −=
For block 14:
12111114 x x x B x ++=
For block 15:
13114215 x x x B x +−=
For block 16:
13127316x x x B x −−=
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The matrices A (16×16) and B (16×3) are:
−−
−
−
−
−
−−
−−−
−
+−
−
−−
−
−−
−−
−
−−
=
0001100003
000000
0001010000002
000
0000110000000001
00000000023
20023
2000
00000000013
20000013
2
00000000000012
20012
2
000000
1
100
13
1000000
000000
31
2
3
1
3
10
313
2000000
000000
3
2
31
22
3
222
313
22
000000
000
3
3
3
3000
3
3
3
1000000
0000000000
2
10
22
1000
0000000000
2
1
2
10000
000
2
20
2
200000
2
2
2
1000
0000000000000
1
10
11
1
0000000000000
1
1
1
10
0000
1
1
1
100000000
1
1
1
1
B
B
B
T T
T T
T T
T T R
T T
T
T T T T R
T
T T T
T
T TwTwT T R
T
Tp
Kp
Tp
Kp
Tp
Kp
Tp
TgTg R
Tt Tt
Tp
Kp
Tp
Kp
Tp
Kp
Tp
TgTg R
Tt Tt
Tp
Kp
Tp
Kp
Tp
Kp
Tp
A
π π
π π
π π
−
=
000
000
000
000
000
0001
100
31
200
31
22
00
000
0
2
10
000
000
00
1
1000
000
T
T T
T
T T
T
Tg
Tg
B
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State Vector (x) =
[ ]T x x x x x x x x x x x x x x x x 16151413121110987654321
Control Vector (u) =
3
2
1
uu
u
DESIGN OF OPTIMAL CONTROLLER
The control inputs are:
16161151512121111.......... xk xk xk xk u
−−++++=
16162151522221212.......... xk xk xk xk u
−−++++=
16163151532321313 .......... xk xk xk xk u−−
++++=
where,
=
−−−−−−−
−−−−−−−
−−−−−−−
13153143133123113103393837363534333231
12152142132122112102292827262524232221
11151141131121111101191817161514131211
K K K K K K K K K K K K K K K K
K K K K K K K K K K K K K K K K
K K K K K K K K K K K K K K K K
K
Hence, Kxu −=
The system state equation:
Bu Ax x +=
The output equation:
Cx y =
Determination of the Feedback Gain Matrix (K):
( )
∞
+=
021 dt RuuQx xPI T T
( ) ( ) ( )[ ]∞
++++++−−++−+++=
0
2
3
2
2
2
1
2
16
2
15
2
14
2
131273
2
131142
2
1211112
1dt uuu x x x x x x B x x x B x x x BPI
[ ]∞
++++−+++=
0
2
2
2
1
2
11
2
10952
2
5
2
2
2
9911
2
1
2
1 2222
1dt uu x x x x B x B x x x B x BPI
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This gives the symmetric matrices Q (16×16) and R (3×3) as:
=
−−
−
−−
−−
−
=
100
010
001
;
1000000000000000
0100000000000000
001000000000000000021100000000
00012100000000
00011200000000
0000000000000000
0000000000000000
0000000000000000
0000000000000
0000000000000000
0000000000000000
0000000000000
0000000000000000
0000000000000000
0000000000000
32
31
21
33
2
3
222
2
11
2
1
R
B B
B B
B B
B B B
B B B
B B B
Q
SYSTEM ANALYSIS USING MATLAB
After substituting values of parameters as given in Appendix - I, the state equations
and the matrices A, B, Q & R are:
[ ]11211211
605.0 d x x x x x −−−+−=
322 5.25.2 x x x +−=
13135.125.122083.5 u x x x +−−=
[ ]21311544
605.0 d x x x x x −−++−=
6555.25.2 x x x +−=
26465.125.122083.5 u x x x +−−=
[ ]31312877 605.0 d x x x x x −+++−=
3109878 002106.019789.02.22000877.0 u x x x x x −−+−=
310979 001053.009894.01.0000438.0 u x x x x ++−−=
310710 02053.002053.0008555.0 u x x x +−−=
411144422.044422.0 x x x −=
7112 44422.044422.0 x x x −=
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7413 44422.044422.0 x x x −=
1211114425.0 x x x x ++=
1311415425.0 x x x x +−=
1312716425.0 x x x x −−=
=
−−
−
−
−
−
−−
−−
−−
−
−−
−
−−
−−
−
−−−
00110000425.0000000
00101000000425.0000
00011000000000425.0
000000004442.0004442.0000
000000004442.0000004442.0
000000000004442.0004442.0
0000002053.000008555.0000000
0000009894.01.00000438.0000000
0000019789.02.22000877.0000000
0066000605.0000000
0000000005.1202083.5000
0000000005.25.20000
0060600000605.0000
0000000000005.1202083.5
0000000000005.25.20
0006600000000605.0
A
−=
000
000
000
000
000
000
020534.000
001053.000
002106.000
000
05.120
000
000
005.12
000
000
B
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=
−−
−
−−
−−
−
=
100
010
001
;
10000000000000000100000000000000
0010000000000000
000211000425.000425.0000
000121000425.000000425.0
000112000000425.000425.0
0000000000000000
0000000000000000
0000000000000000
000425.0425.00000180625.0000000
0000000000000000
0000000000000000
000425.00425.0000000180625.0000
0000000000000000
0000000000000000
0000425.0425.0000000000180625.0
RQ
MATLAB program to obtain S, K & Ac:
A = [-0.05 6 0 0 0 0 0 0 0 0 -6 -6 0 0 0 0; 0 -2.5 2.5 0 0 0 0 0 0 0 0 0 0 0 0 0; -5.2083 0 -12.5 0 0 0 0 0
0 0 0 0 0 0 0 0; 0 0 0 -0.05 6 0 0 0 0 0 6 0 -6 0 0 0; 0 0 0 0 -2.5 2.5 0 0 0 0 0 0 0 0 0 0; 0 0 0 -5.2083 0
-12.5 0 0 0 0 0 0 0 0 0 0; 0 0 0 0 0 0 -0.05 6 0 0 0 6 6 0 0 0; 0 0 0 0 0 0 0.0008778 -2 2.2 -0.19789 0 0
0 0 0 0; 0 0 0 0 0 0 -0.00044 0 -0.1 0.0989 0 0 0 0 0 0; 0 0 0 0 0 0 -0.0085 0 0 -0.02053 0 0 0 0 0 0;
0.4442 0 0 -0.4442 0 0 0 0 0 0 0 0 0 0 0 0; 0.4442 0 0 0 0 0 -0.4442 0 0 0 0 0 0 0 0 0; 0 0 0 0.4442 0 0
-0.4442 0 0 0 0 0 0 0 0 0; 0.425 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0; 0 0 0 0.425 0 0 0 0 0 0 -1 0 1 0 0 0; 0 0 0
0 0 0 0.425 0 0 0 0 -1 -1 0 0 0]
B = [0 0 0; 0 0 0; 12.5 0 0; 0 0 0; 0 0 0; 0 12.5 0; 0 0 0; 0 0 -0.0021; 0 0 0.001; 0 0 0.0205; 0 0 0; 0 0 0;
0 0 0; 0 0 0; 0 0 0; 0 0 0]
Q = [0.1806 0 0 0 0 0 0 0 0 0 0.425 0.425 0 0 0 0; 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0; 0 0 0 0.1806 0 0 0 0 0 0 -0.425 0 0.425 0 0 0; 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0.1806 0 0 0 0 -0.425 -0.425 0 0 0; 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0; 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0; 0.425 0 0 -0.425 0 0 0 0 0 0 2 1 -1 0 0 0; 0.425 0
0 0 0 0 -0.425 0 0 0 1 2 1 0 0 0; 0 0 0 0.425 0 0 -0.425 0 0 0 -1 1 2 0 0 0; 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0
0; 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0; 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1]
R = [1 0 0; 0 1 0; 0 0 1]
S = care(A,B,Q,R)
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K = inv(R)*B'*S
Ac = A-B*K
eig(S)
eig(A)eig(Ac)
Output of the MATLAB program:
=
−−
−−−−−−−−−−−
−−−−−−−−−−
−−−
−−
−−−−−−
−−−−−−−−
−−
−−
−−−−−−−−
−−−
−−−
−−−−
−−−−
−−−−
−−−
2554.124593.44593.41382.51382.503068.381133.946409.80003.30281.01707.01404.00281.01707.01404.0
4593.4908.30688.27076.18434.11358.03565.159299.401732.30514.10747.04486.03591.00053.0027.00112.0
4593.40688.2908.38434.17076.11358.03565.159299.401732.30514.10053.0027.00112.00747.04486.03591.0
1382.57076.18434.15.13095.6903.29247.132933.407813.43623.10039.01326.03639.00345.02453.02702.0
1382.58434.17076.13095.65.1903.29247.132933.407813.43623.10345.02453.02702.00039.01326.03639.0
01358.01358.0903.2903.29065.100000306.01127.00938.00306.01127.00938.03068.383565.153565.159247.139247.1303328.3734112.4623456.267647.80238.01472.01259.00238.01472.01259.0
1133.949299.409299.402933.402933.4004112.4624577.9562068.683254.220697.04038.02807.00697.04038.02807.0
6409.81732.31732.37813.47813.403456.262068.682592.74631.20229.01682.02081.00229.01682.02081.0
0003.30514.10514.13623.13623.107647.83254.224631.29933.00045.00256.00065.00045.00256.00065.0
0281.00747.00053.00039.00345.00306.00238.00697.00229.00045.00161.00889.00479.0001.00062.00057.0
1707.04486.0027.01326.02453.01127.01472.04038.01682.00256.00889.05081.0315.00062.00335.00292.0
1404.03591.00112.03639.02702.00938.01259.02807.02081.00065.00479.0315.02896.00057.00292.00142.0
0281.00053.00747.00345.00039.00306.00238.00697.00229.00045.0001.00062.00057.00161.00889.00479.0
1707.0027.04486.02453.01326.01472.01472.04038.01682.00256.00062.00335.00292.00889.05081.0315.0
1404.00112.03591.02702.03639.00938.01259.02807.02081.00065.00057.00292.00142.00.0479 0.3150.2896
S
The matrix ‘S’ has been real, positive definite & symmetric. Its eigenvalues are:
0; 0; 0; 0; 0; 0; 0.4; 0.4; 0.6; 1.1; 1.7; 2.1; 2.7; 6.6; 122; 1230.8
All the eigenvalues of matrix ‘S’ are real and positive.
The feedback gain matrix ‘K’ has been obtained as:
=
−−−−−−−−
−−−
−−−−
8675.03517.03517.03183.03183.000975.83587.105975.01983.00005.0003.00027.00005.0003.00027.0
3517.09337.00663.00483.04309.03826.02976.08709.02863.00564.02017.01107.15992.00129.00777.00717.0
3517.00663.09337.04309.00483.03826.02976.08709.02863.00564.00129.00777.00717.02017.01107.15992.0
K
Hence the control inputs are:
161514131211109
876543211
3517.00663.09337.04309.00483.03826.02976.08709.0
2863.00564.00129.00777.00717.02017.01107.15992.0
x x x x x x x x
x x x x x x x xu
−+−−−++−
−+−−−−−−=
161514131211109
876543212
3517.09337.00663.00483.04309.03826.02976.08709.0
2863.00564.02017.01107.15992.00129.00777.00717.0
x x x x x x x x
x x x x x x x xu
−−+−−−+−
−+−−−−−−=
1615141312109
876543213
8675.03517.03517.03183.03183.00975.83587.10
5975.01983.00005.0003.00027.00005.0003.00027.0
x x x x x x x
x x x x x x x xu
−++−−−−
−−+++++=
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The closed loop system matrix ‘AC’ is:
=
−−
−
−
−
−
−−−−−−−
−−−−−−
−−−−
−
−−−−−−−−−−−−−
−
−−
−−−−−−−−−−−−
−
−−−
000110000425.0000000
000101000000425.0000
000011000000000425.0
0000000004442.0004442.0000
0000000004442.0000004442.0
0000000000004442.0004442.0
0178.00072.00072.00065.00065.001868.02127.00123.00126.000001.00001.000001.00001.0
0009.00004.00004.00003.00003.000904.01109.00006.00006.0000000
0018.00007.00007.00007.00007.001808.02218.29987.10013.0000000
00066000605.00000003968.46718.118282.06035.03861.57825.47194.38858.10579.37052.00214.158837.136981.121616.0971.08966.0
00000000005.25.20000
00060600000605.0000
3968.48282.06718.113861.56035.07825.47194.38858.10579.37052.01616.0971.08966.00214.158837.136981.12
00000000000005.25.20
00006600000000605.0
A
The eigenvalues of open loop system matrix ‘A’ are:
0; 0; 0; -13.0601; -13.0445; -0.2686+3.602i; -0.2686-3.602i; -0.2372+3.1718i;
-0.2372-3.1718i; -0.7827+1.6214i; -0.7827-1.6214i; -1.4683; -2.0; -0.0843; -0.0362; 0
Four eigenvalues are zero and the remaining have negative real parts indicating
that, the system is marginally stable before applying the optimal control strategy.
The eigenvalues of closed loop system matrix ‘AC’ are:
-13.0677; -13.0514; -0.9698+3.8136i; -0.9698-3.8136i; -0.8646+3.3335i;
-0.8646-3.3335i; -1.1245+1.8141i; -1.1245-1.8141i; -1.685; -2; -0.7362; -0.7338;
-0.0741+0.1067i; -0.0741-0.1067i; -0.1492; 0
It can be seen that after applying optimal control, the system is asymptotically
stable.
33..44 EEQQUUAATTIIOONNSS OOFF PPOOWWEERR SSYYSSTTEEMM MMOODDEELLSS IINN DDIISSCCRREETTEE FFOORRMM FFOORR
IINNTTEEGGRRAALL CCOONNTTRROOLL AANNDD OOPPTTIIMMAALL CCOONNTTRROOLL
State equations and control equations for power system models under consideration
have been obtained in sections 3.2.1 to 3.2.4 for integral control and in sections 3.3.1to 3.3.4 for optimal control. For development of ANN controllers with MATLAB
programming, the power system equations are required in discrete time form, which
have been obtained as given below. The sampling time is 0.01 second and ‘k’ denotes
the sample number (iteration number).
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33..44..11 EEQQUUAATTIIOONNSS FFOORR TTWWOO AARREEAA TTHHEERRMMAALL--TTHHEERRMMAALL ((NNOONN RREEHHEEAATT))
PPOOWWEERR SSYYSSTTEEMM
Equations for Integral Control:
[ ])()()(06.0)(9995.0)1( 17211 k d k xk xk xk x −−+=+
)(025.0)(975.0)1(322k xk xk x +=+
)(125.0)(875.0)(05208.0)1( 1313 k uk xk xk x ++−=+
[ ])()()(06.0)(9995.0)1( 27544 k d k xk xk xk x −++=+
)(025.0)(975.0)1(655 k xk xk x +=+
)(125.0)(875.0)(05208.0)1(2646 k uk xk xk x ++−=+
)()(0044422.0)(0044422.0)1( 7417 k xk xk xk x +−=+
)()(002.0)(00085.0)1( 1711 k uk xk xk u +−−=+
)()(002.0)(00085.0)1( 2742 k uk xk xk u ++−=+
Equations for Optimal Control:
[ ])()()(06.0)(9995.0)1(17211 k d k xk xk xk x −−+=+
)(025.0)(975.0)1(322k xk xk x +=+
)(125.0)(875.0)(05208.0)1(1313
k uk xk xk x ++−=+
[ ])()()(06.0)(9995.0)1( 27544 k d k xk xk xk x −++=+
)(025.0)(975.0)1(655 k xk xk x +=+
)(125.0)(875.0)(05208.0)1(2646 k uk xk xk x ++−=+
)()(0044422.0)(0044422.0)1( 7417 k xk xk xk x +−=+
)()(01.0)(00425.0)1( 8718 k xk xk xk x ++=+
)()(01.0)(00425.0)1( 9749 k xk xk xk x +−=+
)(8
)(7
2737.0)(6
02.0
)(5
1156.0)(4
063.0)(3
1538.0)(2
8294.0)(1
4226.0)1(1
k xk xk x
k xk xk xk xk xk u
−++
++−−−=+
)(9
)(7
2737.0)(6
1538.0
)(5
8294.0)(4
4226.0)(3
02.0)(2
1156.0)(1
063.0)1(2
k xk xk x
k xk xk xk xk xk u
−−−
−−++=+
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69
33..44..22 EEQQUUAATTIIOONNSS FFOORR TTWWOO AARREEAA TTHHEERRMMAALL--TTHHEERRMMAALL ((RREEHHEEAATT))
PPOOWWEERR SSYYSSTTEEMM
Equations for Integral Control:
[ ])()()(06.0)(9995.0)1( 19211 k d k xk xk xk x −−+=+
)(025.0)(975.0)1( 322 k xk xk x +=+
)(041625.0)(040625.0)(999.0)(0173437.0)1(14313 k uk xk xk xk x +−+−=+
)(125.0)(875.0)(052083.0)1(1414k uk xk xk x ++−=+
[ ])()()(06.0)(9995.0)1( 29655 k d k xk xk xk x −++=+
)(025.0)(975.0)1(766k xk xk x +=+
)(041625.0)(040625.0)(999.0)(0173437.0)1( 28757 k uk xk xk xk x +−+−=+
)(125.0)(875.0)(052083.0)1(2858k uk xk xk x ++−=+
)()(0044422.0)(0044422.0)1( 9519 k xk xk xk x +−=+
)()(002.0)(00085.0)1( 1911 k uk xk xk u +−−=+
)()(002.0)(00085.0)1( 2952 k uk xk xk u ++−=+
Equations for Optimal Control:
[ ])()()(06.0)(9995.0)1( 19211 k d k xk xk xk x −−+=+
)(025.0)(975.0)1(322k xk xk x +=+
)(041625.0)(040625.0)(999.0)(0173437.0)1(14313k uk xk xk xk x +−+−=+
)(125.0)(875.0)(052083.0)1(1414k uk xk xk x ++−=+
[ ])()()(06.0)(9995.0)1(29655 k d k xk xk xk x −++=+
)(025.0)(975.0)1(766 k xk xk x +=+
)(041625.0)(040625.0)(999.0)(0173437.0)1( 28757 k uk xk xk xk x +−+−=+
)(125.0)(875.0)(052083.0)1( 2858 k uk xk xk x ++−=+
)()(0044422.0)(0044422.0)1(9519 k xk xk xk x +−=+
)()(01.0)(00425.0)1( 109110 k xk xk xk x ++=+
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)()(01.0)(00425.0)1( 119511 k xk xk xk x +−=+
)()(823.0)(053.0)(1621.0)(0026.0
)(026.0)(7172.0)(3995.2)(0873.1)(4896.0)1(
109876
543211
k xk xk xk xk x
k xk xk xk xk xk u
−++−−
−+−−−=+
)()(823.0)(7172.0)(3995.2)(0873.1)(4896.0)(053.0)(1621.0)(0026.0)(026.0)1(
119876
543212
k xk xk xk xk xk xk xk xk xk xk u
−−+−−−+−−−=+
33..44..33 EEQQUUAATTIIOONNSS FFOORR TTWWOO AARREEAA TTHHEERRMMAALL--HHYYDDRROO PPOOWWEERR SSYYSSTTEEMM
Equations for Integral Control:
[ ])()()(06.0)(9995.0)1(18211k d k xk xk xk x −−+=+
)(025.0)(975.0)1( 322 k xk xk x +=+
)(125.0)(875.0)(052083.0)1(1313 k uk xk xk x ++−=+
[ ])()()(06.0)(9995.0)1(28544 k d k xk xk xk x −++=+
)(00002106.0)(001979.0)(022.0)(98.0)(000008778.0)1( 276545 k uk xk xk xk xk x −−++=+
)(00001053.0)(0009894.0)(999.0)(00000439.0)1(27646 k uk xk xk xk x +++−=+
)(0002053.0)(9997947.0)(00008555.0)1( 2747 k uk xk xk x ++−=+
)()(0044422.0)(0044422.0)1( 8418 k xk xk xk x +−=+
)()(002.0)(00085.0)1(1811 k uk xk xk u +−−=+
)()(0002.0)(000085.0)1( 2842 k uk xk xk u ++−=+
Equations for Optimal Control:
[ ])()()(06.0)(9995.0)1( 18211 k d k xk xk xk x −−+=+
)(025.0)(975.0)1(322 k xk xk x +=+
)(125.0)(875.0)(052083.0)1( 1313 k uk xk xk x ++−=+
[ ])()()(06.0)(9995.0)1( 28544 k d k xk xk xk x −++=+
)(00002106.0)(001979.0)(022.0)(98.0)(000008778.0)1( 276545 k uk xk xk xk xk x −−++=+
)(00001053.0)(0009894.0)(999.0)(00000439.0)1(27646k uk xk xk xk x +++−=+
)(0002053.0)(9997947.0)(00008555.0)1( 2747 k uk xk xk x ++−=+
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71
)()(0044422.0)(0044422.0)1( 8418 k xk xk xk x +−=+
)()(01.0)(00425.0)1(9819k xk xk xk x ++=+
)()(01.0)(00425.0)1( 108410 k xk xk xk x +−=+
)(616.0)(7877.0)(9674.0)(5813.0)(9204.1
)(9205.0)(1842.0)(2095.0)(1572.1)(6597.0)1(
109876
543211
k xk xk xk xk x
k xk xk xk xk xk u
−−−+−
−−−−−=+
)(7877.0)(616.0)(5555.0)(6338.8)(7167.11
)(6794.0)(224.0)(0009.0)(0059.0)(0052.0)1(
109876
543212
k xk xk xk xk x
k xk xk xk xk xk u
−+−−−
−−++=+
33..44..44 EEQQUUAATTIIOONNSS FFOORR TTHHRREEEE AARREEAA TTHHEERRMMAALL-- TTHHEERRMMAALL -- HHYYDDRROO
PPOOWWEERR SSYYSSTTEEMM
Equations for Integral Control:
[ ])()()()(06.0)(9995.0)1(11211211k d k xk xk xk xk x −−−+=+
)(025.0)(975.0)1(322k xk xk x +=+
)(125.0)(875.0)(052083.0)1(1313k uk xk xk x ++−=+
[ ])()()()(06.0)(9995.0)1(21311544k d k xk xk xk xk x −−++=+
)(025.0)(975.0)1( 655 k xk xk x +=+
)(125.0)(875.0)(052083.0)1(2646k uk xk xk x ++−=+
[ ])()()()(06.0)(9995.0)1(31312877k d k xk xk xk xk x −+++=+
)(00002106.0)(0019789.0)(022.0)(98.0)(000008778.0)1( 3109878 k uk xk xk xk xk x −−++=+
)(00001053.0)(0009894.0)(999.0)(00000438.0)1(310979 k uk xk xk xk x +++−=+
)(0002053.0)(999794.0)(00008555.0)1( 310710 k uk xk xk x ++−=+
)()(0044422.0)(0044422.0)1(114111
k xk xk xk x +−=+
)()(0044422.0)(0044422.0)1( 127112 k xk xk xk x +−=+
)()(0044422.0)(0044422.0)1(137413
k xk xk xk x +−=+
[ ] )()()()(425.0002.0)1(1121111k uk xk xk xk u +++−=+
[ ] )()()()(425.0002.0)1( 2131142 k uk xk xk xk u ++−−=+
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72
[ ] )()()()(425.00002.0)1( 3131273 k uk xk xk xk u +−−−=+
Equations for Optimal Control:
[ ])()()()(06.0)(9995.0)1(11211211k d k xk xk xk xk x −−−+=+
)(025.0)(975.0)1( 322 k xk xk x +=+
)(125.0)(875.0)(052083.0)1( 1313 k uk xk xk x ++−=+
[ ])()()()(06.0)(9995.0)1(21311544 k d k xk xk xk xk x −−++=+
)(025.0)(975.0)1( 655 k xk xk x +=+
)(125.0)(875.0)(052083.0)1(2646k uk xk xk x ++−=+
[ ])()()()(06.0)(9995.0)1( 31312877 k d k xk xk xk xk x −+++=+
)(00002106.0)(0019789.0)(022.0)(98.0)(000008778.0)1( 3109878 k uk xk xk xk xk x −−++=+
)(00001053.0)(0009894.0)(999.0)(00000438.0)1(310979k uk xk xk xk x +++−=+
)(0002053.0)(999794.0)(00008555.0)1( 310710 k uk xk xk x ++−=+
)()(0044422.0)(0044422.0)1(114111
k xk xk xk x +−=+
)()(0044422.0)(0044422.0)1(127112 k xk xk xk x +−=+
)()(0044422.0)(0044422.0)1(137413 k xk xk xk x +−=+
)()(01.0)(01.0)(00425.0)1(141211114
k xk xk xk xk x +++=+
)()(01.0)(01.0)(00425.0)1( 151311415 k xk xk xk xk x ++−=+
)()(01.0)(01.0)(00425.0)1(161312716 k xk xk xk xk x +−−=+
)(3517.0)(0663.0)(9337.0)(4309.0
)(0483.0)(3826.0)(2976.0)(8709.0)(2863.0)(0564.0
)(0129.0)(0777.0)(0717.0)(2017.0)(1107.1)(5992.0)1(
16151413
121110987
6543211
k xk xk xk x
k xk xk xk xk xk x
k xk xk xk xk xk xk u
−+−−
−++−−+
−−−−−−=+
)(3517.0)(9337.0)(0663.0)(0483.0
)(4309.0)(3826.0)(2976.0)(8709.0)(2863.0)(0564.0
)(2017.0)(1107.1)(5992.0)(0129.0)(0777.0)(0717.0)1(
16151413
121110987
6543212
k xk xk xk x
k xk xk xk xk xk x
k xk xk xk xk xk xk u
−−+−
−−+−−+
−−−−−−=+
)(8675.0)(3517.0)(3517.0
)(3183.0)(3183.0)(0975.8)(3587.10)(5975.0)(1983.0
)(0005.0)(003.0)(0027.0)(0005.0)(003.0)(0027.0)1(
161514
131210987
6543213
k xk xk x
k xk xk xk xk xk x
k xk xk xk xk xk xk u
−++
−−−−−−
+++++=+
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33..55 CCOONNCCLLUUDDIINNGG RREEMMAARRKKSS
The models of interconnected power systems comprising of areas of different
characteristics have been developed with integral as well as optimal control
strategies. The optimal controllers have been designed and the control equations in
continuous time have been obtained for all the power system models under
consideration. All these models have been studied for system stability and it has
been ensured that they are asymptotically stable with parameters values as given in
Appendix – I, after applying optimal control strategy. Also, for all the power system
models under consideration, the discrete time equations for system states as well as
control inputs have been obtained for both integral control and optimal control
strategies. These equations have been used for development of ANN controllers,
which has been discussed in Chapter 5.