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14.3 Input-Output (I/O) Equation
Another form in which a system model may be represented in
an
input-output (I/O) equation, which is a single differential
equation in terms of the system input, system output, and
their
(time) derivatives
nm
ububububyayayay mmmm
nn
nn
++++=++++
)(
)(
!
)(
)( ("-##)
In $quation ("-##), ),,#,( niai = and ),,,!( mkbk = are
constant
coefficients (for linear system, y is the system output, and u
is
the input%) &or a dynamic system involving many
generali'ed
coordinates, one often finds it etremely difficult or
impossibleto obtain the input-output equation directly from the
governing
equations% his is mainly because, in most cases, the
generali'ed
coordinates are coupled through the governing differential
equations%
Strategy
he idea is to ta*e the +aplace transform of each
differential
equation in the system model, assuming 'ero initial
conditions%onsequently, a set of algebraic equations in terms of
the
transfer functions of the coordinates will be obtained% hen
the
unwanted variables may be eliminated to produce a single
equation in terms of the +aplace of the desired coordinate
and
input% ltimately, this equation is transformed into time
domain
and interpreted as a differential equation in the form of
$quation
("-##)%
Example 14.9
onsider the mechanical system of $ample "%. with the
equation of motion
)(tfkxxbx =++
where the applied force )(tf represents the system input%
Obtainthe input-output equation assuming that the system output is
the
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displacement, )(tx of the bloc*, and assuming 'ero initial
conditions%
Solution
y assumption, the output is xy= and the input is fu = %
0irect
substitution of these into the equation of motion results in
ukyyby =++
which is in the general form of $quation ("-##) with
,!,,,# !# ===== bmkaban
where m is the order of the right-hand side of $quation
("-##)%
herefore, the governing equation is already in the desired
form,
and no further analysis is needed%
Example 14.10
Obtain the input-output equation for the mechanical system
of
$ample "%", where the input and output are )(tf and )( tx
respectively%
Solution
&rom previous results, the equations of motion are
!)()( #### =+
xxbxxkxbxm ("-1a)
)()()( ###### tfxxkxxbxm =++
("-1b)
he input-output equation must be a differential equation in
terms of )(),( txtf , and their time derivatives% a*ing the
+aplace transform of $quations ("-1a and b) results in
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)()2()(3)2()(3)(
!)2()(3)2()(3)()()(
#####
#
#
####
#
sFsXsXkssXssXbsXsm
ssXssXbsXsXksXkssXbsXsm
=++
=++
("-#.)
ollect li*e terms in $quation ("-#.) to obtain
)()()()()(
!)()()(2)(3
###
#
###
#####
#
sFsXksbsmsXksb
sXksbsXkksbbsm
=++++
=+++++
or, in matri form,
=
+++
+++++
)(
!
)(
)(
)(
)()(
#
##
#
###
####
#
sFsX
sX
ksbsmksb
ksbkksbbsm
ecause the output is x , directly solve the above for )( sX
via
ramer4s rule, as follows
######
####
#
##
#
#
##
)(
)()(
)(
)(!
)(
ksbsmksb
ksbkksbbsm
ksbsmsF
ksb
sX
+++
+++++
++
+
=
#
####
#
###
#
##
)()2()(3
)()()(
ksbksbsmkksbbsm
sFksbsX
+++++++
+= ("-#")
Algebraic manipulation of $quation ("-#") yields
)F(s)ks(b
sXkkskbkbsmkkbbkmsbbmbmsmm
##
###
#
####
.
###
"
#
)()(2)(3)2(3
+=
++++++++++
In time domain, representing the system4s output and input,
this
equation then reads
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fkfbxkkxkbkb
xmkkbbkmxbbmbmxmm
#####
########
)(
2)(3)2(3
+=+++
+++++++
("-#5)
which is the system4s input-output equation% As epected,
equation ("-#5) is a differential equation relating input f
,
output x , and their derivatives, and is precisely in the
general
form of $quation ("-##)
Problems
14.5onsider a dynamic system with input )(tf and output x ,
whose state-variable equations are
)2(#.3#
##
#
tfxxx
xx
+=
=
0irectly from these equations, determine the input-output
equation%
14. In 6roblem "%5, assume that #x is the output% &ind
theinput-output equation%
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14.4 !rans"er #un$tion
Once again, consider a linear, time-invariant (constant
coefficients) system described by
nm
ububububyayayay mmmm
nn
nn
++++=++++
)(
)(
!
)(
)( ("-#7)
in which u and y denote the system input and output,
respectively% &urthermore, assume that initial conditions
are all
'ero8 i%e%, )!(!)!( )( == muu and )!(!)!( )( === nyy % a*e
+aplace transforms of both sides of $quation ("-#7) to
obtain
)()()()
!
sUbsbsbsYasasas mmm
nn
nn +++=++++
("-#9)
hen, assuming 'ero initial conditions, the transfer function
is
defined as the ratio of the +aplace transform of the output
and
+aplace transform of the input% &rom equation ("-#9),
the
transfer function is then determined to be in the form of a
rational function%
n
n
nn
m
m
asasas
bsbsmb
sU
sYsG
++++
+++==
!
)(
)()(
("-#1)
:ecall that each pair of system input and output corresponds
to
an input-output equation% he same is true here in the sense
that
corresponding to each pair of input and output, there eists
a
single transfer function% In general, for a ;I;O system with
p
inputs and q outputs there are a total of pq transfer
functions%
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{ } { })()()()( sUsGLsYLty == ("-#=)
Example 14.11
onsider the single-degree-of-freedom mechanical system
studied in $ample "%.% Assume it to be sub>ect to 'ero
initial
conditions% ?uppose x is the output and )(tf is the input%
0etermine the transfer function%
Solution
a*ing the +aplace transform of the equation of motion,
assuming 'ero initial conditions, results in
)()()( # sFsXkbsms =++
so that the system transfer function is obtained as
kbsmssF
sX
++=
#
)(
)(
Example 14.12In the system shown in &igure "%" , )(tx and
)(ty denote the
output and input, respectively%
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Solution
a*ing the +aplace transform of both sides of the equation of
motion and collecting li*e terms, one obtains
)()()()( # sYkcssXkcsms +=++
onsequently, the transfer function is
kcsms
kcs
sY
sXsG
++
+==
#)(
)()(
Example 14.13
onsider the two-degree-of-freedom mechanical system in&igure
"%5 ?ub>ected to 'ero initial conditions% he equations
of motion are given as
)(
)(
#####
##
tfkxkxxcxcxm
tfkxkxxcxcxm
=++
=++
("-.!)
where x and #x are system outputs, and f and #f are system
inputs% 0etermine the transfer function matri%
&igure "%5 wo-degree-of-freedom mechanical system%
Solution
ecause there are two inputs and two outputs, there are four
transfer functions, denoted by )(and),(),(),( #### sGsGsGsG
%
onsequently, the transfer matri is formed as
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=
)()(
)()()(
###
#
sGsG
sGsGs%
;ore specifically, these transfer functions are defined as
follows
!)(#
#
##
!)(#
#
#
!)(#
#
!)(#
)(
)()(,
)(
)()(
)(
)()(,
)(
)()(
==
==
==
==
sFsF
sFsF
sF
sXsG
sF
sXsG
sF
sXsG
sF
sXsG
("-.)
$press the equations of motion, $quations ("-.!), in second-
order matri form as
=
+
+
#
#
#
#
#
!
!
f
f
x
x
kk
kk
x
x
cc
cc
x
x
m
m("-.#)
a*ing the +aplace transform of both sides of $quation ("-.#)
after setting initial conditions to 'ero, one obtains
=
+++
+++
)(
)(
)(
)(
##)(
)(
#
#
#
sF
sF
sX
sX
kcssmkcs
kcskcssm("-..)
@et, use ramer4s rule to solve for )( sX % his requires
replacing the first column of the coefficient matri by the
vector
on the right-hand side, and
)(
)(
)(
)(
)(
)()()()(
)(
)()(
)(
)(
#
#
#
#
#
#
#
##
sF
s
kcssF
s
kcssm
s
sFkcssFkcssm
kcssmsF
kcssF
ssX
++
++=
++++=
++
+
=
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where )(s denotes the determinant of the ## coefficient
matri in $quation ("-..), and is defined as
##
#
#
#
#
#
)())(()(
)()( kcskcssmkcssm
kcssmkcs
kcskcssms +++++=
+++
+++=
?imilarly, solve $quation ("-..) for )(# sX using ramer4s
rule%
his time the second column of the coefficient matri is
replaced by the vector on the right-hand side, and
)()(
)()()()(
)(
)(
)( #
#
#
#
# sFs
kcssF
s
kcssm
sFkcs
sFkcssm
ssX
++
++=
+
++
=
?ubsequently, all four transfer functions, defined through
the
relations in $quation ("-.), can be obtained as
)()(
)()(
#
#
!)(
#
s
kcssm
sF
sXsG
sF
++==
=
)()(
)()(
!)(#
#
s
kcs
sF
sXsG
sF
+==
=
)()(
)()(
!)(
##
#
s
kcs
sF
sXsG
sF
+==
=
)()(
)()(
#
!)(#
###
s
kcssm
sF
sXsG
sF
++==
=
hen, constitute the transfer matri, )(s% , as defined
earlier%
Example 14.14
0etermine the transfer matri for the electrical circuit of
$ample "%1% Assume that q and #q are system outputs, and
that initial conditions are 'ero%
Solution
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he governing equations of this system are given by $quation
("-#!) and may be epressed in the standard second-order
matri form as
=
+
+
!/!
!/
!
!
#
#
#
#
#
e
Q
Q
C
C
q
q
RR
RR
q
q
L
L
+aplace transformation of the governing equations, ta*ing
into
account 'ero initial conditions, yields
=
++
++
!
)(
)(
)(
/
/
#
#
#
#
#
sE
sQ
sQ
CRssLRs
RsCRssL
sing ramer4s rule, solve for )( sQ and )(# sQ separately to
obtain
)(
/
)(
)()(
)()(
/
/!
)(
)(
)(
#
#
#
#
#
#
#
#
#
s
CRssL
sE
sQsG
sEs
CRssL
CRssL
RssE
ssQ
++==
++=
++
=
)()(
)()(
)()(!
)(/
)(
)(
##
#
#
s
Rs
sE
sQsG
sEs
Rs
Rs
sECRssL
ssQ
==
=
++
=
where#
##
#
#
#
)/)(/()( sRCRssLCRssLs ++++=
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Insert this into the epressions of )( sG and )(# sG and
simplify
to obtain
)()()()( ##
##.
##"
##
##
##
+++++++
++
= sCCRsCLCLsLLCRCsCCLL
CsCRCsCCL
sG
and
)()()()(
#
#
##
.
##
"
##
#
#+++++++
=sCCRsCLCLsLLCRCsCCLL
sCRCsG
onsequently, the transfer matri is defined as
=)(
)()(
#
sG
sGs%
&elation bet'een state-spa$e "orm an trans"er "un$tion
:egardless of the type of representation for the
mathematical
model of a dynamic system, similar information about the
system may be etracted% ;ore specifically, given that the
state-
space form of a system model is available, its transfer
function
or transfer matri can be determined using state, input,
output,and direct transmission matrices% o this end, we consider
two
separate cases single input-single output (?I?O) systems,
and
multiple input-multiple output (;I;O) systems%
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Single input-single output (SISO) systems
onsider a dynamic system with a single input and a single
output with a state-space representation
Duy
u
+=
+=
)*
+,**("-.")
and a transfer function
)(
)()(
sU
sYsG = ("-.5)
Assume 'ero initial state8 i%e%, !)!( n=* % +aplace
transformation
of state and output equations yields
)()()()()()()()()( sUsssUsssUsss +,I-+-,I+,-- ==+=
("-.7a)
)()()( sDUssY +=)-
("-.7b)
?ubstitution of )(s- from $quation ("-.7a) into $quation ("-
.7b) results in
)(2)(3)()()()( sUDssDUsUssY +=+= 1+,I)+,I)
hus, the transfer function defined by $quation ("-.5) may be
epressed in terms of the state, input, output, and direct
transmission matrices as
DssU
sYsG +== +,I) )(
)(
)()( ("-.9)
:ecall that
)(ad>
)( ,I,I
,I
= s
ss
Insert into $quation ("-.9) and simplify to obtain
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,I,I
,I+,I)
,I
+,I)
=
+
=+
=
s
sN
s
DssD
s
ssG
)()(ad>)(ad>)( ("-.1)
in which )(sN is an nth-degree polynomial in s%
Example 14.15
0etermine the transfer function for the
single-degree-of-freedom
mechanical system of $ample "%., using its state-space form%
Solution
he state-space form for this system was determined in $ample
"%9 to be
Duy
u
+=
+=
)*
+,**
where
[ ] !,!),(,/
!,
//
!,
#
===
=
=
= Dtfu
mmbmkx
x)+,*
6rior to substitution into $quation ("-.9), we note that
+
=
)/(/
)(
mbsmk
ss ,I
+++=
mbsmk
s
mkmbsss
//
/)/(
)(
,I
onsequently, $quation ("-.9) yields
[ ] [ ]
kbsmsmmkmbss
mbsm
m
mkmbssmmbsmk
s
mkmbsssG
++=
++=
+++=
+++=
#
/)/(
)/(/
/!
/)/(
/
!
//
/)/(
!)(
ultiple input-multiple output (IO) systems
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he formulation leading to $quation ("-.9) was based on the
assumption of a single input and single output that caused
)(sG
and D to be scalars ()% Bhen the system has multiple inputs
and outputs, )(sG and D etend to the transfer matri, )(s% ,
and the direct transmission matri, / , respectively% In
treating
these systems, there eist two possible scenarios (a) a
specific
transfer function, or a few selected transfer functions, are
desired8 or (b) the entire transfer matri is sought% In
either
situation, what turns out to be significant consideration is
the
ad>ustment of si'es of the matrices +, , and in $quation
("-
.9)% 6roper modification of these matrices leads to the
desired
transfer function or transfer matri%
Example 14.16
?uppose a dynamic system has the following governing
equations
####
##
uxxxxx
uxxxxx
=++
=++
("-.=)
where uand u#denote the system inputs, and xand x#represent
the outputs% 0etermine the transfer function X(s)/U(s), via
a
modification of $quation ("-.9)%
Solution
here are four state variables,
#".##,,,
==== xxxxxxxx
which result in the following state-variable equations
#".#"
".#.
"#
.
uxxxxx
uxxxxx
xx
xx
++=
+++=
=
=
onsequently, the state equation is+u,** +=
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where
=
=
=
=#
"
.
#
,
!
!
!!
!!
,
!!!
!!!
,u
u
x
x
x
x
u+,*
ecause G(s)CX(s)/U(s) is the transfer function to be
determined, xmust be chosen as the output and uas the input%
In other words, the problem at hand reduces to a single
input-
single output system, and may be treated as before% Dowever,
in
doing so, certain matrices must be ad>usted properly% ecause
uis the input, the input matri +, which was originally "#,
reduces to " matri +, which is the first column of +% his is
because the elements of the first column of +correspond to
u%
&urthermore, since xis the output, the output equation
reads
[ ]!!!where, == )*)y
Bith this information available, we now apply $quation
("-.9)
as
DssU
sXsG +==
)()(
)()( +,I) ("-"!)
with
=
!
!
!
1+ and !=D
alculate (sI-,) -and insert into $quation ("-"!) to find the
desired transfer function
DssG +=
)()( +,I)
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[ ]
)##(
!
!
!
)(
)(
)(
)(
)##(
!!!
##
#
####
####
##
##
##
++++=
+++
+++
++++
++++
++=
sss
ss
sssssss
sssssss
ssssss
ssssss
sss
Example 14.17
:eferring to the system of $ample "%7, determine the
transfer matri %(s)%
Solution
his problem falls under category (b), discussed earlier% Be
aresee*ing the transfer matri associated with inputs uand u#,
and
outputs xand x#8 hence, %(s) is ##%
++
++
++
+++
+
++
++
=+=
)##(
)##(
)##(
)##(
)()(
##
#
##
####
#
sss
ss
sss
s
sss
s
sss
ss
ss /+,I)%
Problems
14.0he equation of motion of a rotational mechanical system
is derived as
ioo !o"# =++
where oi !! and denote angular displacements, and are
systeminput and output, respectively% 6arameters E, , and F are
constants% Assuming that the system is sub>ected to 'ero
initial
conditions, determine the transfer function )(/)( ss io %
14.he governing equations of an electromechanical system
can be shown to be
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!#
=+
=++
i"$t$#
%Ri$t
$iL
Dere, i and denote the current and the angular velocity,
respectively, and are system outputs, and applied voltage %
is
the input% 6arameters E, +, :, , F, and F#are constants%
&ind
the two possible transfer functions, represented by
)(
)()(and
)(
)()(
# s&
ssG
s&
s'sG
==
and determine the transfer matri%(s)%
14.2onsider the system of 6roblem "%9% sing a suitable set
of state variables, epress the equation of motion in the form
of
state equation% 0etermine the transfer function via the
state,
input, and output matrices%
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14.5 State-spa$e representation "rom te input-output
equation
?uppose the input u and output y of a dynamic system are
related through the input-output equation, as
nm
ububububyayayay mmmm
nn
nn
++++=++++
)(
)(
!
)(
)( ("-")
hen, ta*ing the +aplace transformation of this equation and
assuming 'ero initial conditions, the corresponding transfer
function is obtained as
nnn
n
nn
asas
bsbsb
sU
sY
+++
+++=
!
)(
)(("-"#)
:ewrite the transfer function, defining ((s), as
++++++==
n
nnn
n
asasbsbsnb
sU
s
s
sY
sU
sY
!
)(
)(
)(
)(
)(
)(
)(("-".)
Interpretation of the newly constructed transfer functions in
thetime domain yields
bbbtybsbsnbs
sYn
nn
n
n +++=+++= )(
)(
!
!)(
)(
)(("-"")
and
u)a)a)asassU
s(n
nn
n
nn =+++
+++=
)(
)(
)(
)(("-"5)
$quation ("-"5) represents an nth-order differential
equation
and hence, n initial conditions are required for a complete
solution8 that is, )!(,),!(),!( )(
n))) %
he state variables are then chosen as
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)(
#
=
=
=
nn )x
)x
)x
("-"7)
he corresponding n state-variable equations may then be
obtained as before% he first n* of these equations are
merely
automatic relations between the state variables, and the last
one
is generated using equation ("-"5)
uxaxaxa)x
xx
xx
n
nn
nn +==
=
=
#
)(
.#
#
("-"9)
herefore, epressing the state-variable equations, $quation
("-
"9), in matri form, the state equation is given as
u+,** +=
where
=
=
=
!
!
!
,
!!!
!!!
!!!
,
#
#
+,*
aaaax
x
x
nn
n
("-"1)
in which the state matri , is referred to as the lowercompanion
matri% he system output y is given by $quation
("-""), that can be epressed in terms of the state variables
as
!
)(
)(
! xbxbxb)b)b)by nnnnnn +++=+++=
("-"=)
?ubstituting the last equation in $quation ("-"9) for nx
in
$quation ("-"=), we have
#!)( xbxbuxaxaxaby
nnnnn ++++=
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:earranging and collect li*e terms to obtain
ubxbabxbabxbabynnnnn !!#!!
)()()( +++++++=
("-5!)
As a result, rewriting $quation ("-5!) using the matri
notation, the output equation is given as
Duy +=)*
where [ ] !!!! , bDbabbabbab nnnn =+++= ) ("-5)
$quations ("-"1) and ("-5) constitute the system4s state-
space form%
Example 14.18
A dynamic system is described by its transfer function, as
#
)(
)(# ++
=sssU
sY
0etermine the state-space form%
Solution
;anipulation of the transfer function results in
uyyy =++
#
which is the system4s input-output equation% omparison
reveals
that
,,#,# # ==== baan
Dence, defining the state variables, and the resulting state
vector
as
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=
==
yx
yx
#
*
the state-space form is
[ ]*
**
!
!
#
!
=
+
=
y
u
Example 14.19
Obtain the state-space representation for the input-output
equation below
uuyyyy +=+++
#.#" ("-5#)
Solution
he system4s transfer function, assuming 'ero initial
conditions,
is
.#"#
)()(
#. ++++=
ssss
sUsY
.#"
)(
)(
#)(
)(
#. +++=
+=
ssssU
s
ss
sY
and in time domain,
u))))
))y
=+++
+=
.#"
#
0efining the state variables and the corresponding state
vector
as
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=
=
=
=
)x
)x
)x
.
#
*
the state-variable equations are determined and epressed in
matri form to give the state equation, as
uxxxx
xx
xx
+=
=
=
.#.
.#
#
"#.
u
+
=
!
!
"#.
!!
!!
**
he output equation is
### xxy +=+=
[ ]*!#=y
Problems
14.1 0etermine the state-space representation of the
input-output equation
uuuyyy #.# ++=++
14.11he transfer function of a dynamic system is given as
).(
)(
)(
+
+=
ss
s
sU
sY
() 0etermine the input-output equation and, subsequently,
find the state-space form%
(#) sing the state, input, and output matrices obtained in
part
(a), find the transfer function% Is this in agreement with
the
transfer function provided originallyG
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