ET 438A Lecture 2 PT 2
ET 438A Lecture 2 PT 2
ET 438A Lecture 2 PT 2
Procedure: 1.) find the transfer function X(s) 2.) multiply X(s)by s 3.) takethe limitof sX(s)as S goesto zero 4.) resultis valueof x(t)whent = infinity x ( t) =
@
lim
{--
lim
S-> 0
s. X(s)
Initial Value Theorem - determines thevalue of thetime function whent=0without finding theinverse transform Procedure: 1.) findthetransfer function X(s) 2.) multiply X(s)by s 3.) takethe limit as S of sX(s) goesto infinity 4.) result is value of x(t)whent =0 lim x(t) = lim
t-.0 s - > oo
s.X(s)
et438a-6.MCD 15
x ( s ) = z ; to t
x(oi=l, , *r t X C t) = l ,n tt sx(s): *
T -70 S +oc l,^, 1 _ Bo E
t
s . (s ' * 2 .s r 1 0 1 )
), mi
KBoV
S--+o0' J;Zg+E
de^o^rruo$uc /"Ter-
fc,trq-> s
x(o).J:,1 ;r#, . S g
Example 2: Findthe finalvalue of thetransfer function X(s)above. /''""* x(t):- l''^rl' sx(s) V*t*-<F^*C -7c-t S-+ O
f r '''r 'lSy ( s \= l ,- ,t
5-> o
c5 S-)o
K (so?)
(ftnas*lorl '\'
a^.J Ls-)o
s-)o
_1, ^rf
s-?o
goB
sztZs{lol
Sz >o
l r * r- f _3 :J5-9o +l-siio t
. jg-
g
5: -
lDl
fh*
g {, ^*f vc,-[u1
X(ou) = B
et438a-6.MCD 16
fortheabove system Writethedifferential equations Laplace themusing and solve withrespect to position f(t; = F andthatthe mass Assume methods. transform x(0)=g surface. on a frictionless slides
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
l -6
s\ M ' s
Et438a-6.mcd 17
+ B.s , K)
=X ( s )
s ( 1 . s ' + 4. s + 5)
/.t
b:= 4 c::5
'
S 1 = -2 +i
2.a
s2 s 2 =-2-i
2-a
=X ( s )
5
F,^d A
Fre+tr)(so(b6 o1 J
r' ^' *l+ 'elJ
.rr<
r)-
r ,, (stx) s+ (,_{ )
\
.B
- c.
7 ffiA-Tre*("6)){= f.)) {
S:o
sr'
5
_l
Al
Sr
(z+()(z-tr)
18 Et438a.mcd
<--:
zt+ l
(= t )t \ .-(* \ t *[) o
L u ''t
r -
Rdi'
g* * pl e X
r'fffil=JLtj-@J/ FTie
(-l L 1 ' ---19 Et438a-6.mcd
('^ ( -r[i l iS
_t_ \i ' tI
Mechanical Sofution
\ +ry1 v
IN"Jocs 8.-1,o-".)tzeJJr"' n^
-(atj ) f f s I - ( z - * ) t
Et438a4.mcd 20
response system Plotof the mechanical + 2.sin(t)) x(t):: 1- e-2't.(cos(t) andsee decrease of system to damping -2 relates effects xt(t) : 1- e '2't'(cos(t) " 2'sin(t)) ..10 t ,=0 ,0.25
x (t)
x 1( t ) 0
21 Et438a.mcd
Transfer Functions
given InpuUoutput relationships fora mathematical model usually bytheratio of twopolynominals of thevariable s Definitions polynomial. Poles of thedenominator Values thatcause - roots transfer function magnitude to go to infinity. polynomial. Zeros - roots of thenumerator Values thatcause the transfer function to go to 0. eigenvalues responses of a system. Roots of the - Characteristic polynomial. denominator Alleigenvalues mustbe (natural negative for a system response) transient to decay out.
X's indicate location of pole. 0 is location of zero poleis to imaginary Closer response. roots axisstower Comptex pairs appear in conjugate
et438a-7.MCDI
Examples
R(') f;
--f
input 'rl
G(s) |->
x(')
o u tp u t
X ( s )=G(s ).R (s )
ffi=G(s)
f
v ; ( r ) =R . i * (t ) I I t/l
J
i ( t )d t
u i(t)
TakeLapface V i (s )= R .l (s ) *
V;(s)
Rn
*r(s)
= l(s)
Remember V o (s )= C .s ' l (s ) 1
n. ftn-.Voltage divider formuta
C. s
So
vo(s)==;.V;(s)=
C .s
R'C's* 1
'V;(s)
et438a-7.MCD 2
RCs+ 1
RC is timeconstant of system. System has 1 poleat -1lRC andno zeros Larger RCslower response
Transfer functions of OPAMPcircuits Practical Differentiatoractive highpass filter withdifinite low frequency cutoff. Take Laplace of
components andtreatlike impedances
V,C
I
^\
vo General gainformula
-z g (s ) V o (s )
-Rf
z 1 ( s )=Rf
- R1C's
Av(s)=
A u( s )=
v o(s)
1 R, * ' C .s
et438a-7.MCD3
%(')
Block fortransfer Algebra functionsCascaded blocks connected functions Series transfer - multiply Note: do notcancel common terms fromnumerator and
X, ' ( s )
l\ t
c 1( s )
tr 2 (si
X(s ) =G 2 (s ).X1 (s )
(2 )
et438a-7.MCD 4
v,,(t)
vo(t)
ct( s)=v.r(il
V 1( s )
V o(s)
of TakeLaplace
A v 1(s )=
c 1's
-,
1's
1
-1
v o(s)
Vi(s)
_1
- R1C 2 ' s
et438a-7.MCD5
v o(s)
V;(s ) ( R 1. c 2. s + 1 )(R i .C 1 s )
R 1C 2 ' s
Simplified form
parameters
Fi ,= 10000 C1 R 1 := 2 5000 Rf
0.1.10-6 1 00000
R i 'C 1= 0. 001
Rt 'C2 =0.005
Above areall timeconstants forthesystem Final transfer function Function has1 zeroat 0.005.s = s 0 andtwopoles =1 s 1) . ( 0 .0 0 1 .s )s = -1 1 0 .0 0 V;(s) ( 0. 001. * 1000 ands=0 Parallel Blocks functions - addtransfer
v o(s)
et438a-7.MCD 6