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Transfer Function
• Introduce a new concept: Transfer Function
It’s governing equation:
Input, r(t) Output, c(t)System
)()()()()()(0101 trB
dttdrB
dttrdBtcA
dttdcA
dttcdA m
m
mn
n
n +++=+++ LL
Transfer Function– Laplace Transfer ( assuming zero initial
condition)
– Rearrange:
( ) ( ) )()( 0101 sRBsBsBsCAsAsA mm
nn +++=+++ LL
01
01)()()(
AsAsABsBsBsG
sRsC
nn
mm
++++++
==L
L
Definition of the Transfer Function!
Transfer Function
The Transfer function G(s) is a property of the system elements only, and is not dependent on the excitation and initial conditions. In addition, transfer functions can be used to represent both closed-loop and open-loop systems
Transfer Function
• Block diagram
Input, r(t) SystemG(s)
Output, c(t)
Input, R(s) Output, C(s)
Transfer Function
• Block diagram
• Laplace Transform of the output
Input, r(t) SystemG(s)
Output, c(t)
Input, R(s) Output, C(s)
)()()( sRsGsC =
Transfer Function of Systems• Cascaded system
G1(s)E1(s)
G2(s)E2(s)
G3(s)E3(s)
G4(s)E4(s) E5(s)
)()()()(/)()()()()()()()()()()()(
432115
445
334
223
112
sGsGsGsGEEsEsGsEsEsGsEsEsGsEsEsGsE
=====
Transfer Function of Systems
• Cascaded system• Single-loop feedback system
Input R(s) SystemG(s)
Output, C(s)+−
FeedbackH(s)
E(s)
B(s)
Transfer Function of Single-loop Feedback System
• Use the definition of transfer function:
• Solve for C(s)/R(s))()()(
)()()()()()(
sEsGsCsBsRsEsCsHsB
=−=
=
)()(1)(
)()(
sHsGsG
sRsC
+=
Transfer Function of Single-loop Feedback System
)(1
)()(1)(
)()(
1)()( sHsHsGsG
sRsC
sHsG
≈+
=>>
Independent of G(s) !
Transfer Function of Single-loop feedback system
• Characteristic equation (denominator)
• Solving for error
0)()(1 =+ sHsG
)()(1)(
)()()()(
sHsGsG
sRsEsGsC
+=
=
Transfer Function of Single-loop feedback system
Therefore
)()(11
)()(
sHsGsRsE
+=
)(1
)(11
)()(
1)(1)( sGsGsRsE
sGsH
≈+
=>>=
Error should be very small, if G(s) is large !
Transfer Function of Single-loop feedback system
• Block diagram transformation– Table 2.5– Figure 2.11– Example p2.23
Signal-Flow Graphs and Mason’s Theorem
A signal-flow graph is a topological representation of a set of linear equations having
niyayn
jjiji ,,2,1 L==∑
435
314
23
4212
hygyyfyeyy
dyycybyayy
+=+=
=++=
Signal-Flow Graph
– A source is node having only outgoing branches y1
– A sink is a node having only incoming branches y5
y1 y2 y3 y4 y5a
b
c
d
e
f
g
h
Signal-Flow Graph
– A path is a group of connected branches having the same sense of direction (eh b)
– Forward paths are paths which originate from a source and terminate at a sink and along which no node is encountered more than once (eh adfh b)
y1 y2 y3 y4 y5a
b
c
d
e
f
g
h
Signal-Flow Graph
y1 y2 y3 y4 y5a
b
c
d
e
f
g
h
– Feedback loop is a path originating from a node and terminating at the same node. In addition, a node cannot be encountered more than once (b dfc)
Signal-Flow Graph– Path gain is the product of the coefficients
associated with the branches along the path
– Loop gain is the product of the coefficients associated with the branches forming a feed back loop
Reduction of the Signal-Flow Graph
• Signal-Flow Graph Reduction– Addition–Multiplication– Feedback loops• Figure 2.13 d & e
• Mason’s theorem
∆
∆= ∑k kkGG
Reduction of the Signal-Flow Graph
• Example 1• Example 2• Example 3• Example 4
Single-loop feedback system
Input R(s) SystemG(s)
Output, C(s)+−
FeedbackH(s)
E(s)
B(s)
)()(1)(
)()(
sHsGsG
sRsC
+=
Apply Mason’s Theorem to Single-loop feedback system
[ ] )()(1)()(1 sHsGsHsG +=−−=∆
C(s)R(s)1 G(s)
-H(s)
)()(1 sGsGGA =⋅=
1=∆A
)()(1)(
)()(
sHsGsG
sRsCG
+==
Apply Mason’s Theorem to Single-loop feedback system
• Example 2• Example 3• Example 4
Review of Matrix Algebra• Identity Matrix (aii ＝1, aij ＝ 0)• Diagonal Matrix (aii≠0, aij ＝ 0)• Symmetric Matrix (aij ＝ aji )• Skew-Symmetric Matrix
(aii ＝ 0, aij ＝ －aji )• Zero Matrix• Adjoint Matrix (aij ← Aij )• Transpose (aij aji )↔
Review of Matrix Algebra
• Addition and subtraction• Multiplication by a scalar• Multiplication of two matrices• Inverse of a matrix• Differentiation of a matrix• Integration of a matrix
State Variable Method
Use a representation of the system dynamics that contain the system’s input-output relationship (similar to that of a transfer function) but in terms of n first-order differential equations to represent the nth order system
State Variable Method
• State Representation in Phase-Variable Canonical Form
)()()( tuBtxPtx +=&
where is the state vector, is its time derivative, is the input vector, is the state (companion) matrix, and is the input matrix
)(tu)(tx
PB
)(tx&
State Variable Method– Block Diagram of the Phase-Variable
Canonical Form (from Definition Equation)
)(txB L
P
Integrator++
)(tx&
)(tu
)(tc+
+
D
State Variable Method– Block Diagram of the Phase-Variable
Canonical Form (from Definition Equation)
)(txB L
P
Integrator++
)(tx&)(tu )(tc
State Variable Method
=
=
nmnn
m
m
nnnn
n
n
bbb
bbbbbb
B
ppp
pppppp
P
L
M
L
L
L
M
L
L
21
22221
11211
21
22221
11211
,
)()()()(
)()()()()()()()(
1111
212121212
111111111
tubtubtxptxpx
tubtubtxptxpxtubtubtxptxpx
mnmnnnnnn
mmnn
mmnn
LL&
M
LL&
LL&
++++=
++++=++++=
State Variable MethodSystem’s Output
where is the output vector, is the out-put matrix, is the coefficient matrix represents the direct transmission between input and output, in most case equal to zero.Therefore
)(tc
)()()( tuDtxLtc +=
LD
)()( txLtc =
State Variable Method
• Example 1 (Figure 2.28)• Example 2 (Figure 2.29)• Example 3
θ
-Fg
r
Rocket trajectory
v
State-Variable Diagram– Example
Dividing numerator and denominator by s3
Force terms in the numerator, pure integrators !
sssss
sUsCsP
8914
)()()( 23
2
++++
==
21
321
8914
)()()( −−
−−−
++++
==sssss
sUsCsP
State-Variable DiagramDefine the error node of the system
then
And
Draw diagramMore example ?
21 891)()( −− ++
=ss
sUsE
( ) )(4)( 321 sEssssC −−− ++=
)(8)(9)()( 21 sEssEssUsE −− −−=
State Transition Matrix
• Recall phase-variable canonical equation
• Laplace transfer
• Rearrange
)()()( tuBtxPtx +=&
)()()0()( sUBsXPxsXs +=− +
)()0()()( sUBxsXPsXs +=− +
State Transition Matrix
• Inverse Laplace transfer (the state transition equation)
• The state transition matrix is defined asτττ duBtxttx
t)()()0()()(
0−Φ+Φ= ∫+
[ ] [ ] )()0()( 11 sUBPIsxPIssX −+− −+−=
[ ]{ }11)( −− −=Φ PIst L
State Transition Matrix
• Properties of state transition matrix
)()(
)()()()()()(
)0(
1
011202
tt
tttttttt
I
−Φ=Φ
ΦΦ=+Φ
−Φ−Φ=−Φ
=Φ
−
ττ
State Transition Matrix
• For more general initial time, recall
• Rearrange and let t = t0
τττ duBtxttxt
)()()0()()(0
−Φ+Φ= ∫+
τττ duBtttxtxt
)()()()()()0( 0001
001 0 −ΦΦ−Φ= ∫−−+
Pay attention to the order of the terms in matrix multiplication !The—commutative law
State Transition Matrix
• Substitute back to the state transition equation
• Second term becomes (using the properties)
ττττττ duBtduBttt
txtttxtt
)()()()()()(
)()()()(
0000
00
0 −Φ+−Φ−ΦΦ−
−ΦΦ=
∫∫
τττττ duBtduBtttt
)()()()(000
0 −Φ+−Φ− ∫∫
State Transition Matrix
• and
• then
τττ duBttxtttxt
t)()()()()(
000 −Φ+−Φ= ∫
τττ
τττττ
duBt
duBtduBttt
t
tt
)()(
)()()()(
0
0
000
−Φ=
−Φ+−Φ−
∫∫∫
State Transition Matrix
• Example: an open loop system,
– Differential equation form is
– Therefore, define the state variables
2
1)()()(
ssUsCsP ==
)()( tutc =&&
)()( )()( 21 tctxtctx &==
State Transition Matrixthus
in the phase-variable canonical form
====
)()()( )()()(
2
21
tutctxtctxtx
&&&
&&
)()()( tuBtxPtx +=&
=
=
=
=
)()(
)( )()(
)( 10
0010
2
1
2
1
xxtx
txxxtx
txBP&
&&
State Transition Matrix
– The state transition matrix is,
[ ]{ }function stepunit theis )( where
)(0)(
)( 11
tUtUttU
PIst
=−=Φ −−L
[ ]
−=
−
=−
ss
sPIs0
10010
1001
[ ] [ ]
=
=−
=−
−=− −
s
ss
ss
s
ss
ss
PIsPIs
PIs1
11
21
00
1
01
01
adj 2
State Transition Matrix– Assume the initial conditions,
and u(0) = 0
=
=+
21
)0()0(
)0(2
1
xx
x
)0()()( +Φ= xttx
State Transition Matrix– Therefore, (notice there is an error in the
book)
– or
=
=
21
)(0)(
)()(
)(2
1
tUttU
txtx
tx
)(22)()(
2
1
tUxttUtx
=+=
Total Solution of the State Equation
• Example: a system describe by
• Determine the output c(t), given r(t) = sin t
• Initial conditions
)()()()(2)( trtrtctctc +=++ &&&&
0)0( 1)0( == cc &
Total Solution of the State Equation
• Determine the state transition matrix
• Determine the output c(t)
[ ]{ }11)( −− −=Φ PIst L
τττ duBtxttxt
)()()0()()(0
−Φ+Φ= ∫+
)()( txLtc =
Total Solution of the State Equation
• Determine the state transition matrix– Define the state variable
and we have
)()()()(
2
1
tctxtctx&=
=
)()()()(trtutrtu
&& ==
Total Solution of the State Equation
– First order differential equation representation of the system
– The phase-variable canonical form is,
)()()()(2)()()(
122
21
tututxtxtxtxtx
&&
&
++−==
( ))()()()( tutuBtxPtx && ++=
Total Solution of the State Equation
where
and
=
=
=
−−
=)()(
)( )()(
)( 10
21
10
2
1
2
1
xxtx
txxxtx
txBP&
&&
[ ]
+−
=
−−
−
=−
211
2110
1001
ss
sPIs
Total Solution of the State Equation
– Then
– Therefore, the state transition matrix is
[ ] [ ]
−
=+
−+
=
+−
−+
=−
−=−
++
+++
−
22
22
)1()1(1
)1(1
)1(2
21
)1(1
12
211
112
adj
ss
s
sss
ss
s
ss
ss
PIsPIs
PIs
[ ]{ }
−−+
=−=Φ−
−−−−
)1()1(
)( 11
tteteet
PIst t
tt
L
Total Solution of the State Equation
• Determine the output c(t)
Substitute x(t) into
result in
τττ duBtxttxt
)()()0()()(0
−Φ+Φ= ∫+
)()( txLtc =
τττ duBtLxtLtct
)()()0()()(0
−Φ+Φ= ∫+
Total Solution of the State Equation
from
and given
we have
)()()()(
2
1
tctxtctx&=
=
0)0( 1)0( == cc &
[ ]01=L
=
=
=+
01
)0()0(
)0()0(
)0(2
1
cc
xx
x&
Total Solution of the State Equation
at the same time, given
Therefore
and
ττττττ cossin)()()()( +=+=+ rruu &&
ttrtu sin)()( ==
=
10
B
Total Solution of the State Equation
Substitute all of them into c(t) we have,
[ ]
[ ]
τττ
ττ
ττ
ττ
d
eteeet
eteeet
tc
t
tt
tt
tt
tt
)cos(sin10
)1()1(
01
01
)1()1(
01)(
0 )()(
)()(
+
×
+−−+
+
−+
=
∫ −−−−
−−−−
−−
−−
Total Solution of the State Equation
On simplifying
check the initial conditions
[ ]tttee
dettetc
tt
t tt
cos21sin
21
23
)cos(sin)()1()(0
)(
−++=
+−++=
−−
−−− ∫ ττττ τ
0012110
23)0(
112100
23)0(
=+×+++−=
=×−++=
c
c
&
