6.3 Transfer Func
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Transcript of 6.3 Transfer Func
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8/12/2019 6.3 Transfer Func
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6.3 State Space and the Transfer Function
The basic state equations are
( ) 0( ), 0 ,t= =&x Ax + Bf x x (6.3.1 a)
( ) ( )t t= +y Cx Df . (6.3.1 b)We have since established the solution:
( ) ( ) ( ) ( ) ( )1 1
0s s s s
= + X I A x I A BF . (6.3.2)
To establish a transfer function, we disreard initial conditions, so !q. (6.3.2) beco"es
( ) ( ) ( )1
s s s= X I A BF .
We also have deter"ined the transition "atri# in the $a%lace do"ain
( ) ( ) 1
s s = I A , (6.3.3)
so
( ) ( ) ( )s s s
= X BF
&ow ' tae !q. (6.3.1 b) into the $a%lace do"ain
( ) ( ) ( )s s s= +Y CX DF ,which ' can write as
( ) ( ) ( ) ( ) ( ) ( )s s s s s s= + = + Y C BF DF C B D F .o we conclude that the transfer function is
( ) ( )s s= + C B D . (6.3.*)
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!xa"p#e 6.3.$
ind the transfer function usin state s%ace "ethods of the 2ndorder s-ste" that we
looed at %reviousl-:
( ) ( )( ) ( )
2
23 2
d y t dy t y t f t
dt dt + + = .
olution
We found that .
( ) 0( ), 0 ,t= = =&x Ax + Bf x x y Cx
[ ] [ ]0 1 0
, , 1 0 , 02 3 1
= = = =
A B C D
[ ]( ) ( )
1 3 11
21 2
ss
ss s
+
= + + I % A .
o
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( )( ) ( )
[ ]
( ) ( )
1 121 11
11 21 12
21
11 21 12 1 1
101
0 0
10
R Cs a s as a s a a a
a
s a s a a a R C
=
=
C B
&ow our transfer functionis
( ) ( )( ) ( )
[ ]2111 21 12 1 1
10 0 1
as s
s a s a a a R C
= + = +
C B D .
ince we are onl- interested in the in%ut due to inAV ,
( ) ( )( ) ( )
21
1 1 11 21 12
1 as s
R C s a s a a a= + =
C B D
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!xercises
6.3.1 !#%lain how -ou would esti"ate a transfer function for the s-ste" of %roble"
6.2.1.
6.3.2 onsider a linear continuous+ti"e s-ste" with described b-
( ) ( )( )
( )( )
2
26 2
d y t dy t df t y t f t
dt dt dt + + = +
(a) !#%ress this is state s%ace usin the %hase variable canonical for".
(b) ind the $a%lace do"ain transition "atri#.
(c) ind the s-ste" transfer function usin the state s%ace for"ula. 4oes it aree withwhat -ou would et b- tain the oriinal equation into the $a%lace do"ain5
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