Equivalent POG block schemes
• Let us consider the following inductor connected in series:
V1 V2
I1 I2
I
L
Three mathematically equivalent POG block schemes can be used:
V1
I
-
?
1 s
?
1 L
?
-
φ
V2
I
a) Initial condition φ0
V1
I
-
1 L
1 s
?
?
?
-
V2
I
b) Initial condition I0
V1
I
-
1 L s
?
?
-
V2
I
c) Zero initial condition.
• For linear systems it is always possible to switch the position of two casca- de connected linear blocks without modifying the input-output dynamic behavior of the considered system.
• If the three block schemes were implemented in Simulink, their initial conditions would be φ = φ0, I = I0 and I = 0, respectively.
Connecting POG block schemes
• Let us consider the following two electric circuits and POG schemes:
Va VA
I1 L IA
R1
Va
I1
-
? 1 s
?
1 L
?
-
φ
-
R1 6
6
-
P VA
IA
VB V2
IB Ib
C R2
VB
IB P
-
1 R2
?
?
-
-
6 1 s 6
1 C
6
-
V2
Ib
Va
I1
-
? 1 s
?
1 L
?
-
φ
-
R1 6
6
-
P
-
1 R2
?
?
-
-
6 1 s 6
1 C
6
-
V2
Ib
• Two POG block schemes can be “directly connected” only if:
1) the two power sections “are oriented in the same way”;
2) the two power sections “share the same positive power flow”.
• The following two POG schemes “CANNOT be directly connected” be- cause “they are NOT oriented in the same way”:
Va
I1
-
? 1 s
?
1 L
?
-
φ
1 R1
?
?
- -
P VA
IA
VB
IB P
-
1 R2
?
?
-
-
6 1 s 6
1 C
6
-
V2
Ib
• The following two POG schemes “CANNOT be directly connected” be- cause “they do NOT share the same positive power flow”:
Va
I1
(a) Pa
-
? 1 s
?
1 L
?
-
φ
-
R1 6
6
-
P VA
IA
(A)
VB
IB
(B) P
-
1 R2
?
?
-
-
6 1 s 6
1 C
6
-
V2
Ib
Pb
(b)
• The two above POG schemes “CAN be directly connected” along sections (a)-(b) because “they share the same orientation and positive power flow”:
VB
IB (B)
P
-
1 R2
?
?
-
-
6 1 s 6
1 C
6
-
Pa = Pb
(a)=(b)
-
? 1 s
?
1 L
?
-
φ
-
R1 6
6
-
P VA
IA (A)
Algebraic loop
• Let us consider the following electric circuit:
Va Vr1 V2
I1 Ib
Ir2 L
C R1
R2
In the corresponding POG block scheme is present an “algebraic loop”:
Va
I1
-
?
1 s
?
1 L
?
-
φ
-
R1
6
6
Vr1
-
-
1 R2
?
?
Ir2
-
-
6
1 s
6
1 C
6
-
V2
Ib The corresponding POG state space model can still be determined:
L 0 0 C
˙I1 V˙2
=
−R∆1 −RR21∆
R1
R2∆ −R21∆
| {z }
A˜
I1 V2
+ 1 0 0 −1
Va
Ib
(⋄)
where ∆ is the determinant of the POG scheme without the dynamic blocks:
∆ = 1 + R1 R2.
The coefficient Aij of matric A is the “transfer gain” (computed using the
“Mason formula”) of the path that links the j-th state variable xj with the input of the i-th integrator.
• The “algebraic loop” can also be eliminated “graphically” as follows:
Va
I1
-
?
1 s
?
1 L
?
-
φ
-
R1
6
6
Vr1
-
-
1 R2
?
?
Ir2
-
-
6
1 s
6
1 C
6
-
V2
Ib
Va
I1
-
?
1 s
?
1 L
?
-
Vr1
φ
R1
∆
6
6
- - - R1
R2∆ -
R1
R2∆
1 R2∆
?
?
- -
Ir2
-
6
1 s
6
1 C
6
-
V2
Ib
• The same result can also be obtained using the following “mathematical”
approach. Add a “spurious” physical element Cs to the original system
Va Vs V2
I1 Ib
Ir1 L
Cs C R1
R2
such to eliminate the algebraic loop in the corresponding POG scheme:
Va
I1
-
?
1 s
?
1 L
?
-
φ
1 R1
?
?
Ir1
- -
-
6
1 s
6
1 Cs
6
Vs
-
-
1 R2
?
?
Ir2
-
-
6
1 s
6
1 C
6
-
V2
Ib
The corresponding POG state space model can now be written easily:
L 0 0
0 Cs 0 0 0 C
˙I1 V˙s
V˙2
=
0 −1 0
1 −R11 − R12 R12 0 R1
2 −R12
I1 Vs
V2
+
1 0 0 0 0 −1
Va Ib
When Cs = 0 one obtains the following constraint:
I1 +
− 1
R1 − 1 R2
Vs + 1
R2V2 = 0 which can be rewritten as follows:
Vs = R1R2 R1 + R2
I1 + R1 R1 + R2
V2
Let us consider the following “congruent” state space transformation:
x = Tx where T =
1 0
R1R2
R1+R2
R1
R1+R2
0 1
The transformed and reduced matrices have the following structure:
x= Tx =
I1 V2
, L= TTLT=
L 0 0 C
A = TTAT =
"
−RR11+RR22 −R1R+R1 2
R1
R1+R2 −R1+R1 2
#
, B= TTB=1 0 0 −1
.
One can easily verify that the obtained reduced system is equal to the system (⋄) obtained applying the Mason formula to solve the algebraic loop. One can easily verify that the two matrices A and ˜A are equal:
A=
"
−RR11+RR22 −R1R+R1 2
R1
R1+R2 −R1+R1 2
#
=
−1+RR11 R2
−R R1
2(1+R1R2) R1
R2(1+R1R2) − 1
R2(1+R1R2)
=
"
−R∆1 −RR21∆
R1
R2∆ −R12∆
#
= ˜A
• Example. Electric circuit.
Let us consider the following electric circuit:
Va V3 V4
L1 L2
Ra
Rb
C3 C4 Ib
I1 I2
VKL1 VKL2 VKL3
CKL1 CKL2 CKL3
The corresponding P.O.G. block scheme is:
Va
VKL1 -
? 1 s
?φ1
1 L1
?
I1
-
CKL1
-
Ra
6
6
-
VKL2 -
? 1 s
?φ2
1 L2
?
I2
-
CKL2
-
6 1 s
6Q3
1 C3
6
V3
-
VKL3 -
1 Rb
?
?
-
CKL3
-
6 1 s
6Q4
1 C4
6
V4
- Ib
The POG state space dynamic model has the following structure:
L1 0 0 0 0 L2 0 0 0 0 C3 0 0 0 0 C4
| {z }
L
˙I1
˙I2 V˙3 V˙4
|{z}˙x
=
−Ra Ra 0 0
Ra −Ra −1 0 0 1 −R1b R1b
0 0 R1
b −R1b
| {z }
A
I1 I2 V3 V4
|{z}x +
1 0 0 0 0 0 0 −1
| {z } B
"
Va
Ib
#
|{z}u
The connection coefficients appear within matrix A only when one type of energy (i.e. magnetic energy in L) converts to another different type of energy (i.e. electrostatic energy in C).
A dissipative parameter (i.e. Ra and Rb) appears 4 times in a symmetric way within matrix A when the corresponding physical element connect two dynamic elements of the same type.
• Example. A chain of carriages and snubbers:
The symbolic scheme:
b1 b2 b3
K1 K3 K3
˙x1 ˙x2 ˙x3
M1 M2 M3
F
0
The corresponding POG block scheme:
F -
1 M1s
?
?
˙x1
-
R1 F1
-
6 6
b1
K1
s
6 6
- -
1 M2s
?
?
˙x2
-
R2 F2
-
6 6
b2
K2
s
6 6
- -
1 M3s
?
?
˙x3
-
R3 F3
-
6 6
b3
K3
s
6 6
-
0 The POG state space dynamic model is:
M1 1 K1
M2 1 K2
M3 1 K3
¨ x1
R˙1
¨ x2
R˙2
¨ x3
R˙3
=
−b1 −1 b1 0 0 0
1 0 −1 0 0 0
b1 1 −b1−b2 −1 b2 0
0 0 1 0 −1 0
0 0 b2 1 −b2−b3 −1
0 0 0 0 1 0
˙x1
R1
˙x2 R2
˙x3 R3
+
1 0 0 0 0 0
F
• The “across” dynamic elements De connected “in parallel” can be graphi- cally represented only by the following particular POG scheme:
Electrical Mech. Tras. Mech. Rot. Hydraulic
Capacitor Mass Inertia Hyd. Capacitor
ElementDe
V V
I1 I2
C
x˙ x˙
F1 F2
M
ω ω
τ1 τ2
J
P P
Q1 Q2
Ci
POGscheme
I1
V
-
? 1 s
?Q
1 C
?
-
I2
V
F1
x˙
-
? 1 s
?p
1 M
?
-
F2
x˙
τ1
ω
-
? 1 s
?pr
1 J
?
-
τ2
ω
Q1
P
-
? 1 s
?V
1 Ci
?
-
Q2
P
• The “through” dynamic elements Df connected “in series” can be graphi- cally represented by only one particular POG scheme:
Electrical Mech. Tras. Mech. Rot. Hydraulic Inductor Spring Rot. Spring Hyd. Inductor
ElementDf
V1 V2
I L I
x˙1 x˙2
F E F
ω1 ω2
τ τ
Er
P1 P2
Q Li Q
POGscheme
V1
I
-
? 1 s
?φ
1 L
?
-
V2
I
x˙1
F
-
? 1 s
?x
1 E
?
-
x˙2
F
ω1
τ
-
? 1 s
?θ
1 Er
?
-
ω2
τ
P1
Q
-
? 1 s
?φi
1 Li
?
-
P2
Q
• The “across” dynamic elements De connected “in series” can be graphi- cally represented by the following two different POG schemes:
Electrical Mech. Tras. Mech. Rot. Hydraulic
Capacitor Mass Inertia Hyd. Capacitor
ElementDe
V1 V2
I C I
x˙1 x˙2
F M F
ω1 ω2
τ J τ
P1 P2
Q Ci Q
POGscheme1
V1
I
6 1 s
6Q
1 C
6
- -
V2
I
x˙1
F
6 1 s
6p
1 M
6
- -
x˙2
F
ω1
τ
6 1 s
6pr
1 J
6
- -
ω2
τ
P1
Q
6 1 s
6V
1 Ci
6
- -
P2
Q
POGscheme2
V1
I
- -
6 1 s
6Q
1 C
6
V2
I
x˙1
F
- -
6 1 s
6p
1 M
6
x˙2
F
ω1
τ
- -
6 1 s
6pr
1 J
6
ω2
τ
P1
Q
- -
6 1 s
6V
1 Ci
6
P2
Q
• The two POG schemes have opposite input/output power variables. The choice between the two POG block schemes depends on
• The “through” dynamic elements Df connected “in parallel” can be gra- phically represented by the following two different POG schemes:
Electrical Mech. Tras. Mech. Rot. Hydraulic Inductor Spring Rot. Spring Hyd. Inductor
ElementDf
V V
I1 I2
L x˙ x˙
F1 F2
E ω ω
τ1 τ2
Er P P
Q1 Q2
Li
POGscheme1
I1
V
6 1 s
6Q
1 L
6
- -
I2
V
F
x˙
6 1 s
6p
1 E
6
- -
F
x˙
τ1
ω
6 1 s
6pr
1 Er
6
- -
τ2
ω
Q1
P
6 1 s
6V
1 Li
6
- -
Q2
P
POGscheme2
I1
V
- -
6 1 s
6Q
1 L
6
I2
V
F1
x˙
- -
6 1 s
6p
1 E
6
F2
x˙
τ1
ω
- -
6 1 s
6pr
1 Er
6
τ2
ω
Q1
P
- -
6 1 s
6V
1 Li
6
Q2
P
• The two POG schemes have opposite input/output power variables. The choice between the two POG block schemes depends on
• “Across-variable” generators:
Eletttrico Mecc. Tras. Mecc. Rot. Idraulico
Generator
V1
V1
I1
x˙e
x˙e
F1
ω1
ω1
τ1
P1
P1
Q1
POG
V1
I1 P
˙xe
F1 P
ω1
τ1 P
P1
Q1 P
• “Through-variable” generators:
Eletttrico Mecc. Tras. Mecc. Rot. Idraulico
Generator V
1
I1
I1
x˙e
F1
F1
ω1
τ1
τ1
P1 Q1
Q1
POG
V1
I1 P
˙xe
F1 P
ω1
τ1 P
P1
Q1 P
• Example. Let us consider the following electric circuit:
Va Vr1 Vr2 Vr3
I1 I2 I3
Ib
L1 L2 L3
R1 R2 R3
The corresponding POG block scheme:
Va -
?
1 s
?
1 L1
?
I1
-
| {z }
-
R1
6
6
Vr1
-
-
?
1 s
?
1 L2
?
I2
-
-
R2
6
6
Vr2
-
-
?
1 s
?
1 L3
?
I3
-
-
R3
6
6
Vr3
- Ib
The POG state space dynamic model is:
L1 0 0 0 L2 0 0 0 L3
˙I1
˙I2
˙I3
=
−R1 R1 0
R1 −R1−R2 R2 0 R2 −R2−R3
I1 I2 I3
+
1 0 0 0 0 R3
Va Ib
I1 Vr3
=
"
1 0 0
0 0 R3
#
x + 0 0 0 −R3
Va Ib
Note that, in this case, the power matrix A is symmetric because the system is characterized only by one type of stored energy, in this case the magnetic energy stored in the three inductances L1, L2 and L3.
• Example. Let us consider the following electric circuit:
Va V1 V2 V3
Vr1 Vr2 Vr3
Ir1 Ir2 Ir3
I1
I1 I2 I3
Ib
L1 L2 L3
R1 R2 R3
The corresponding POG block scheme:
Va - -
R1 6
6
Vr1
V1
Ir1
| {z }
- -
1 sL1
?
?
I1
- -
R2 6
6
Vr2
V2
Ir2
- -
1 sL2
?
?
I2
- -
R3 6
6
Vr3
V3
Ir3
- -
1 sL3
?
?
I3
Ib
The POG state space dynamic model is:
L1 0 0 0 L2 0 0 0 L3
˙I1
˙I2
˙I3
=
−R1 −R1 −R1
−R1 −R1−R2 −R1−R2
−R1 −R1−R2 −R1−R2−R3
I1
I2
I3
+
1 −R1 1 −R1−R2 1 −R1−R2−R3
Va Ib
Ir1 V3
=
"
1 1 1
−R1 −R1−R2 −R1−R2−R3
# x+
0 1
1 −R1−R2−R3
Va Ib
The power matrix A is symmetric because the system is characterized only by one type of stored energy.
• Example. Let us consider the following electric circuit:
Va Vc Vd Vb
Vc1 Vl1
Vl2
Vl3 Ic1
I1
I2
I3
Ib C1
L1
L2
L3
R2
The corresponding POG block scheme:
Va
Vl1
-
1 s L1
?
?
I1
-
Vc
1 s C1
6
6
Vc1
- -
1 s L2
?
?
I2
- -
Vl2
Ir2
R2 6
6
Vr2
- -
I3
Vd
sL3 6
6
Vl3
- -
Vb
−Ib
• The system CANNOT be modeled using the “integral causality”: the inductance L3 is graphically represented using “derivative causality”.
• From the physical scheme it is evident that the state variables I1, I2 and I3 are linked by the following constraint:
I1 = I2 + I3
• The system can be modeled adding a spurious dynamic element C2 in parallel connection between elements R2 and L3:
Va Vc Vb
Vc1 Vl1
Vl2
Vl3 Vr2
Vc2 Ic1
Ic2
Ir2 I1
I2
I3
Ib
C1
C2
L1
L2
L3
R2
The new POG block scheme is:
Va
I1
Vl1
-
1 s L1
?
?
I1
-
Vc
1 s C1
6
6
Vc1
- -
1 s L2
?
?
I2
- -
Vl2
Ir2
R2 6
6
Vr2
- -
-
1 s C2
6
6
Vc2
-
Vl3
-
1 sL3
?
?
I3
-
Vb
−Ib
The corresponding POG state space dynamic model is:
L1
C1
L2
C2
L3
˙I1
V˙c1
˙I2
V˙c2
˙I3
=
−R2 −1 R2 −1 0
1 0 0 0 0
R2 0 −R2 1 0
1 0 −1 0 −1
0 0 0 1 0
I1
Vc1 I2
Vc2 I3
+
1 0 0 0 0 0 0 0 0 −1
Va Vb
I1
−Ib
=
1 0 0 0 0 0 0 0 0 1
x
When C2 = 0from the forth equation one obtains the following constraint:
I1 − I2 − I3 = 0 ⇒ I3 = I1 − I2
Let us consider the following state space “congruent” transformation:
x = Tx ⇔
I1 Vc1
I2 Vc2
I
=
1 0 0 0 1 0 0 0 1 0 0 0 1 0 −1
I1 Vc1
I2
Using the congruent transformation x = Tx one obtains the following transformed and reduced POG dynamic system:
L1 + L3 −L3 0 C1 0
−L3 L2 + L3
˙I1
V˙c1
˙I2
=
−R2 −1 R2
1 0 0
R2 0 −R2
I1
Vc1 I2
+
1 −1 0 0 0 1
Va Vb
I1
−Ib
=
1 0 0 1 0 −1
x
• The state variables Vc2 and I3 are no more present because C2 = 0 and the constraint I3 = I1 − I2.
• The transformed system is NO more decoupled because matrix L = TTLT is not diagonal. The reduced system has still the structure of a POG dynamic model. Matrix L, for example, is still symmetric and definite positive.
• Calculations using Matlab:
-- Matlab commands ---
syms L1 L2 L3 C1 C2 R2 LT =
C2=0;
LM=diag([L1 C1 L2 C2 L3]); [ L1 + L3, 0, -L3]
AM=[-R2 -1 R2 -1 0; [ 0, C1, 0]
1 0 0 0 0; [ -L3, 0, L2 + L3]
R2 0 -R2 1 0;
1 0 -1 0 -1; AT =
0 0 0 1 0];
BM=[1 0; 0 0; 0 0; 0 0; 0 -1]; [ -R2, -1, R2]
CM=[1 0 0 0 0; [ 1, 0, 0]
0 0 0 0 1]; [ R2, 0, -R2]
T1=sym([ 1 0 0;
0 1 0; BT =
0 0 1;
0 0 0; [ 1, -1]
1 0 -1]); [ 0, 0]
LT=T1.’*LM*T1 [ 0, 1]
%%
AT=simplify(T1.’*AM*T1) CT =
BT=T1.’*BM
CT=CM*T1 [ 1, 0, 0]
[ 1, 0, -1]
---
• Example. Tank with variable section.
Q input volume flow rate P input pressure
V volume of liquid in the tank z height of the liquid
a(z) area of the liquid at height z A area of the base section α slope of the liquid area p0 pressure per unit of height
6
z α
a(z)
Q A
• Two equivalent POG block schemes can be used:
Q
P Φ−1(V )
V
-
?
1 s
?
p0
√A2+2αV−A α
?
-
0
⇔
Q
P
P˙
-
p2 0 Ap0+αP
1 s
?
?
?
-
[Ψ(P )]−1 0
• The area of the liquid a(z) at height z and the volume of liquid V are:
a(z) = A + αz, V = Z z
0
a(z)dz = A z + α z2 2
• Since z = pP0, functions V = Φ(P ) and P =Φ−1(V ) are defined as follows:
V = Φ(P ) = A P
p0 + α P2
2p20 , P = Φ−1(V ) = p0
√A2 + 2α V − A
α .
• Functions Ψ(P ) and [Ψ(P )]−1 can be obtained as follows:
Ψ(P ) = ∂Φ(P )
∂P = A
p0 + α P
p20 , [Ψ(P )]−1 = p20 Ap0 + αP
• The energy Es stored in the system is:
Es = Z V
0
Φ−1(V ) dV = p0 3α2
h(A2 + 2α V )32 − 3α A V − A3i .
