Initialand FinalValueTheorems
FinalValueTheorem- determines
value
the steady-state
of the systemresponsewithoutfindingthe inverse
transform.
Procedure:
1.) find the transferfunctionX(s)
2.) multiplyX(s)by s
3.) takethe limitof sX(s)as S
goesto zero
4.) resultis valueof x(t)whent =
infinity
x ( t) =
lim
{--
@
lim
S->
s. X(s)
0
InitialValueTheorem
- determines
thevalueof thetime
functionwhent=0withoutfindingtheinversetransform
Procedure:
1.) findthetransferfunctionX(s)
2.) multiply
X(s)by s
3.) takethe limitof sX(s)as S
goesto infinity
4.) resultis valueof x(t)whent =0
lim x(t) = lim
t-.0
et438a-6.MCD
15
s - > oo
s.X(s)
Example
1: Findthe initialvalueof thetransfer
function
x ( s ) = z ; to t
i
s . (s ' * 2 .s r 1 0 1)
t
x(oi=l,
, *r t X C t) = l ,n tt sx(s):
*
T -70
l,^, 1 _
S +oc
KBoV
S- + oo V( S. + ZSt t ot )
Lar - c 1€rua- |, ue= , d s n, ok c -
Bo E
S--+o0' J;Zg+E
), mi
de^o^rruo$uc/"Ter-
fc,trq-> s
x(o).J:,1
;r#, . S g
Example
2: Findthe finalvalueof thetransferfunction
X(s)above.
/''""* x(t):- l''^rl'sx(s)
V*t*-<F^*C -7c-t
f r '''r 'lSy ( s \= l ,- ,t
s-)o
5-> o
c5
S-)o
Sz >o
K (so?)
_1, ^rf
goB
(ftnas*lorl
'\'
s-?o
sztZs{lol
a^.J
l r * r- f _3 :J5-9o
fh*
S-+ O
Ls-)o
. jg-
lDl
+l-siio t
g
{, ^*f vc,-[u1
X(ou) = B
F
et438a-6.MCD
16
g
5: -
UsingLaplace
for solvingmechanical
systems
fortheabovesystem
Writethedifferential
equations
withrespectto positionand solvethemusingLaplace
methods.Assumef(t; = F andthatthe mass
transform
surface.x(0)=g
slideson a frictionless
f(t) =M.
d2
x ( t) + B h x ( t ) * K ' x ( t )
dt2
r{
d 2 x ( t) + 8 .: -x(t
)*K. x(t )
F =M '
dt
dt2
of bothsides
TakeLaplace
transform
l -6
:-=M.sz.X(s) B.s.X(s). K.X(s)
"
S
Solveforthe positionX(s)
t_ 2
s\ M ' s
Et438a-6.mcd
17
\
+ B.s , K)
=X ( s )
LetM = 1,F = 5, B = 4 andK= 5. Solvethis
using
Laplace
andpartialfractionexpansion.
= X(s)
/.t
Usequadraticformula
to factordenominator
s ( 1 . s ' + 4. s + 5)
a
b:= 4 c::5
'
s2
2.a
2-a
s 2 =-2-i
S 1 = -2 +i
formof function
Factored
5
=X ( s )
s . ( s + (2 r j ))'(s r (2
))
Usepartialfractionexpansion
,
.B
5
s
r)-
- c.
r
,, (stx) s+ (,_{ )
Fre+tr)(so(b6o1 J
\
.rr<
F,^d A
sr'
r' ^' *l+ 'elJ
Al
_l
|
r
jrl-*--l
,
r
--^
s+(z+11 s+(z-6) I , =o
8s
7
ffiA-Tre*("6)){=
f.)){
S:o
<--:
5
(z+()(z-tr)
Et438a€.mcd18
Sr
zt+ l
I
7
I
Continued
systemsolution:
Mechancical
5
S
;
:
:
(= t )t
\ . - ( *\ t *[)o
-' L =
--1i -L(-z+
j)(.i)
\- '( 't
r
('^ (
-r[i l
iS
r -
_t_
\i ' tI
Rdi'
g* * pl e X
,,.qr,_1es
Conr,\
-+:'L--
r'fffil=JLtj-@J/ FTie
L
u
''t
19
Et438a-6.mcd
(-l
L
1
'
----
Mechanical
Sofution
€ \ +ry1 € v
-(atj ) f f s I - ( z - * ) t
zFTt.f, s :,i :z+^+(t
e*+ l _ z - ? l *e \
8.-1,o-".)tzeJJr"' n^IN"Jocs
Et438a4.mcd20
systemresponse
Plotof the mechanical
+ 2.sin(t))
x(t):: 1- e-2't.(cos(t)
-2 relatesto dampingof systemdecreaseandsee
effects
xt(t) : 1- e '2't'(cos(t)" 2'sin(t))
t ,=0 ,0.25..10
x (t)
x 1( t )
0
Et438a€.mcd21
TransferFunctions
InpuUoutput
relationships
fora mathematical
modelusuallygiven
bytheratioof twopolynominals
of thevariables
Definitions
polynomial.
Poles- rootsof thedenominator
Valuesthatcause
transferfunctionmagnitude
to go to infinity.
Zeros- rootsof thenumeratorpolynomial.
Valuesthatcausethe
transfer
function
to go to 0.
eigenvalues
responses
of a system.Rootsof the
- Characteristic
polynomial.
denominator
Alleigenvalues
mustbe
(naturalresponse)
negative
for a systemtransient
to decayout.
right holf
plone
le ft h olf
p rqn e
X's indicate
location
of pole. 0 is location
of zero
Closerpoleis to imaginary
roots
axisstowerresponse.
Comptex
pairs
appearin conjugate
et438a-7.MCDI
Examples
R(') f;
--f
input -
G(s)
|->
o u tp u t
'rl
Transfer
Function
is a "gain"as
a function
of
x(')
X ( s )=G(s ).R (s )
ffi=G(s)
S
PassiveLowpassfifter- intergrator
f
v ; ( r ) =R . i *( tI) I
t/l
i ( t )d t
J
TakeLapface
u i(t)
V i (s )= R .l (s ) *
V;(s)
Rn
1
= l(s)
C. s
1
Remember V o (s )= C .s ' l (s )
1
n.
So
vo(s)==;.V;(s)=
ftn-.-
C .s
et438a-7.MCD2
*r(s)
'V;(s)
R'C's* 1
Voltage
divider
formuta
Drawas a blockdiagram
RC is timeconstant
of
system.Systemhas
1 poleat -1lRCandno
zeros
LargerRCslowerresponse
1
RCs+
1
Transfer
functions
of OPAMPcircuits
Practical
Differentiatoractivehighpassfilterwithdifinitelow
frequency
cutoff.
Take Laplaceof
components
andtreatlike
impedances
V,C
I
vo Generalgainformula
^\
-z g (s ) V o (s )
=vi l t
Au (t)=
=,(r)
z ; ( s ) = R i .a :
z 1 ( s )=Rf
SimplifyAu
A u( s )=
et438a-7.MCD3
- R1C's
-Rf
Av(s)=
v o(s)
R ;'C ' s n 1 V; (s)
1
R, *
' C .s
Transferfunction
has1 zeroat s=0
and1 poleat
s = -1lRiC
Transferfunctionfor PracticalDifferentiator
-RtCs
R'Cs+
1
%(')
BlockAlgebrafortransferfunctionsCascaded
blocks
functions
Seriesconnected
transfer
- multiply
Note:do notcancelcommontermsfromnumerator
and
X, ' ( s )
l\
c 1( s )
X r ( s )=R ( s ) ' G1 ( s ) ( 1)
X ( s )=G1( s ) ' c 2 ( s ) ' R ( s)
t
tr 2 (si
X(s ) =G 2 (s ).X1 (s )
2(s )
\ G
5+=G
1(s
)
R (s )
'
(1)into(2)andsimplifyto getoverallgain
Substitute
et438a-7.MCD4
(2 )
ExamplewithOP AMPs
0.1 uF
\J1
v,,(t)
vo(t)
Firststage-integrator Secondstage-practicaldifferentiator
V o(s)
TakeLaplaceof
Forstage1
_z
1(s)
z i ( s )=R i z
A v 1( s )= ;i u
1(s)=c
-1
V 1( s )
ct( s)=v.r(il
G z(s)=
v
V 1G)
'
anduse
::Tff.ents
generalgainformula
c 1's
A v 1(s )=
1's
-1
-,
SimplifyAu1to get G1(s) G 1(s)=Ri'c
1€
- RgC 2's
examPle
Gz(s)fromPrevious
= G 2 (s )
A y (s )=
1
R1' C2' s t
v o(s)
Vi(s)
_1
- R1C 2 ' s
R;' C1 's R1'C2's+ 1
et438a-7.MCD5
signscancel
Negative
v o(s)
R 1C 2 ' s
-
Simplifiedform
V;(s ) ( R 1. c 2. s + 1 )(R i .C 1 s )
Plugin givenvaluesfor the component
symbolsand compute
parameters
Fi ,= 10000 C1
R 1 := 2 5000
0.1.10-6
Rf
R i 'C 1= 0. 001
1 00000
C2,=0.05.10-6
R 1'C 2= 0.001
Rt 'C2 =0.005
Aboveareall timeconstants
forthesystem
Finaltransferfunction
Function
has1 zeroat
0.005.s
=
s 0 andtwopoles
V;(s) ( 0. 001. *s 1) . ( 0 .0 0 1 .s )s = -1 1 0 .0 0=11 0 0 0
ands=0
Parallel
Blocks- addtransfer
functions
v o(s)
et438a-7.MCD6