Unformatted text preview: Introduction to Feedback Control
Review of Laplace Transform
Signals: Signals are represented by functions of time t. For a signal f t, its Laplace transform is de ned by EE3530: F s := L f t = ut Z1
0 f te,st dt: , yt
 Systems: Systems process signals in a certain way.
 H For a linear timeinvariant system, it can be represented by transfer function de ned by Y s : Ly = H s = L ut t zero initial condition U s Thus input and output have a simpler relation in sdomain: Y s = H sU s
Laplace transform is a tool to analyze signals and systems. 1 Inverse Laplace transform: Given F s, f t is given by 1 Z +j 1 f t = 2j ,j1 F sest ds where is large enough to include all pk , poles of F s, on left of straight line s = . Thus f t can be computed using residue theorem: f t = X Respk F sest = X Respk fF sg epk t: Example 1: Suppose that the system is governed by
yt + Kyt = ut; y0 = 0; _
1 and ut = cos!t =Re e,j!t . Compute output response yt. Answer: Applying Laplace transform to 1 yields 2 1 3 s sY s + KY s = U s = Re 4 s + j! 5 = s2 + !2 : Hence we obtain a Y s = s2 + !2s s + K = s + K + s ,c j! + s +c j! with a, c, c residues at poles ,K and j! respectively. Question: How to compute residues? 2 Partial Fraction for Residue Calculation:
Case 1: Real and distinct poles: , 1 , , F s = Kss, pzss, pz2 s, pzm s n 1 2 c c c = c0 + 1 + 2 + + n ; where residues are calculated as
8
0 s,p 1 s,p 2 s , pn c = slim F s = : K; if n = m; !1 0; if n m; ck = s! pk s , pk F s; k 0: lim Example 2: Y s = sss+ 2ss+ 4 . + 1 + 3
Since n = 3
0 1 2 Using the formulas, we obtain 8 ,1 3 c1 = 2 4 = 3 ; c2 = ,1 + 21 + + 4 = , 2 ; 13 ,1, 3 ,3 1 c3 = ,3 + 23 + + 4 = , 6 : ,3, 1 m = 2, c = 0, and Y s = cs + s c 1 + s c 3 : + +
3 3 Using the table in page 735, we have 0 8 3 ,t 1 ,3t1 yt = @ 3 , 2 e , 6 e A 1t: Case 2: Distinct complex poles where the pole pi is complex, and pi is also a pole. In this case, c c F s = s ,i p + s ,i p + rest terms; i i and pi = + j , pi = , j . We have f t = 2jcije t cos t + 6 ci1t + L, rest terms :
1 Continuation of Example 1: By s + ! = s , j!s + j!,
2 2 a Y s = s + ! s s + K = s + K + s ,c j! + s +c j! with p = j!, and thus = 0, = !. Hence, s K ; a = s +! =, K +! s ,K 0:5 s = c = s + j!s + K K + j! :
2 2 2 2 = 2 2 As jcj = 0:5= K 2 + !2, 6 c = , tan,1!=K = , we have 0 1 K e,Kt + p 1 yt = @, K 2 + !2 cos!t + A 1t: 2 2 K +! p s=j! 4 Discussion of Example 1: We notice that
We further notice that s 1 T s = Y s = s + K : U
2 jT j!j = 1= K + ! = 2jcj; 6 T j! = , tan, !=K = :
2 1 p Let K 0. Then with ut = cos!t, the output response of the system approaches to yt = p 21 2 cos!t + = jT j!j cos!t + 6 T j!: K +! as t becomes large value by K 0. This is true for any linear timeinvariant system. tem T s that is stable. Then the output response yt eventually approaches to Fact 1: Suppose that input cos!t, or sin!t, is applied to sysyt = yst = jT j!j cos!t + 6 T j!; or yt = yst = jT j!j sin!t + 6 T j!; which is called steadystate response of the system. 5 Comments: 1 T j! is called frequency response of the system with jT j!j magnitude, and 6 T j! phase responses. 2
Suppose that the system is unknown. We can obtain frequency response data by sweeping the frequency to extrapolate the given transfer function. Hence even if the system is unknown, it can be identi ed through frequency response method under some mild conditions. Case 3: Multiple poles p1 = p2 = = p` = p: s s F s = K ,,z1s ,, z2 s , zm s p ` p`+1 s , pn c c = c0 + 1 + ` ` + rest terms: s,p s , p The residues for c1 c` can be calculated as 1 di `F s c`,i = slimp i! dsi s , p ! for i = 0; 1; 2; ; ` , 1. 6 Example 3: Find inverse Laplace transform for
1 2 2 2 3 c F s = s + s +s + 1 = s c 2 + s + 2 + s c 1 : 2 + + Clearly ` = 2. Applying the formula yields: d 2s + 33 1d c = 1! ds s + 2 F s s , = ds 4 s + 1 5 s , 2 3 d 2 5 2 = 41 + =, = ,2; ds s+1 s , s + 1 s , ,2 + 3 = ,1; c = slim s + 2 F s = s + 3 = !, s + 1 s , ,2 + 1 , +3 c = ,11+ 2 = 2:
3 1 2 = 2 = 2 = 2 2 = 2 2 2 2 = 2 3 2 Therefore, and its inverse transform is given by 2 1 2 F s = , s + 2 , s + 2 + s + 1 ;
2 f t = ,2t , te,2t + 2e,t 1t: ,2e 7 Example 4: Find inverse Laplace transform for Answer: We rst decompose
s +s+1 =
2 F s = ss +1s + 1
2 0 12 0 p 12 @s + 1 A + B 3 C @ A = s , 2 2 2 + 2 = s , Thus = ,1=2, and = 3=2 that leads to F s = ss , , j 1 s , + j c c2 c2 : = 1+ + s s, ,j s, +j We now obtain c1 = s2 +1s + 1 = 1; s=0 1 1 c2 = ss , 1 + j = = p 1 p3 2j + j j 3, 2 + j 2 s= +j o p p3 6 o 2 2 p = p 6 180 , tan,1 3=3 = 150 : = 3 ,3 + j 3 12 Using the previous formulas we obtain p p 1 0 3 2 3 ,0:5t @ A f t = B1 + 3 e cos 2 t + 150oC 1t: p , j s , + j : 8 Solving Di erential Equations:
An important application of Laplace transform is to solve di erential equations. Consider yt + 5y_ t + 4yt = 3; y0 = ; y0 = : _
Applying Laplace transform gives
2 2 s s + 5 + + 3 = s 3, ,4 3,4 ,4 s s + 5 + + 3 = 3 , 3 + 12 : 4 Y s = ss2 + 5s + 4 s s+1 s+4 Therefore, 0 3 3 , , 4 ,t 3 , 4 , 4 ,4t1 yt = @ 4 , e + e A 1t: 3 12 3 s + 5s + 4Y s = s + 5 + s + 3 s Y s , s , + 5sY s , + 4Y s = s 9 Properties of Laplace Transform page 95:
L f t + f t = F s + F s. Time delay: L f t , T = e,sT F s. s Timescaling: L f at = jaj F a . Shift in frequency: L e,atf t = F s + a. _t = sF s , f 0,, and L f m t Di erentiation: L f = smF s , sm, f 0, , sm, f_0, , , f m, 0,:
Superposition:
1 1 2 2 1 1 2 2 1 1 2 1 L R t f d = F ss . Convolution: L f t f t = F sF s. 1 Time product: L f tf t = j R ,jj1 F xF s , x dx. Multiplication by time: L tf t = ,F 0s. Integration:
0 1 2 1 2 1 2 1 + 2 1 2 Initial Value Theorem: f 0 = slim sF s. !1
Proof: By di erentiation in time,
Z1
0 + sF s = by taking s ! 1 that completes the proof. , _ ,st dt + f 0, = f 0++ Z 1 fe,st dt _ fe
0 + ,! f 0 + 10 f(t) t f(t1) 1 t f(2t) f(t) 2 4 1 2 f(t/2) 3 4 4
11 8 its poles on open left half plane. Final Value Theorem: tlim yt = slim sY s, if sY s has all !1 !
0 D.C. gain = nal value of the step response. Example from Homework 3.7e:
2s + 2s + 52 2s + 2s + 52 F s = s + 1s2 + 42 = s + 1s , 2j 2s + 2j 2 c + c 1 + c 1 + c2 + c 2 : = s + 1 s , 2j s + 2j s , 2j 2 s + 2j 2 According to the formulas, 2,1 + 2,1 + 52 32 c = ,12 + 42 = 25 = 1:28; 2 3 d 6 2s + 2s + 52 7 = 2:96496 ,102:465o; c1 = ds 4 s + 1s + 2j 2 5 s=2j 2s + 2s + 52 c2 = s + 1s + 2j 2 = 4:58536 ,154:83o: s=2j 12 The key part is to compute 2 3 c2 + c2 7 = c te2jt + c te,2jt 1t 5 4 L,1 6 s , 2j 2 s + 2j 2 2 2 2jt = 2Re c2te 1t " j 6 c2 e2jt 1t = 2Re jc2jte
" = 2Re jc2 1t = 2jc2jt cos2t + 6 c21t: jtej2t+6 c2 Let c2 = + j . Then we have the following gure: c 2 c2 Hence we nally have f t = = 1:28e + 14:8244 cos2t , 65:63o +9:1706t cos2t , 154:83o 1t: ,t ce + 2jc1j cos2t + 6 ,t c1 + 2tjc2j cos2t + 6 c 1t
2 13 Example from Homework 3.3c: Find
F s = L f t = L sin t=t : Answer: Let gt = tf t = sint. Then
2 F s = , s 1 1 ds + C = , tan, s + C: + What is C ? Using initial value theorem,
Z
1 2 Gs = L gt = s 1 1 = L tf t = , dF s : + ds Hence F s can be obtained as that yields C = =2. Indeed,
slim s !1 f 0 = slim sF s = slim s !1 !1 ! sint = slim s C , = tlim =1 !1 !+ t 2
+ 0 ,1 s C , tan 2 , tan,1s
2 ! Therefore, F s = , tan, s.
1 , tan, s = lim , s2 = 1: = slim !1 s!1 , 2 s s 2 1 1 1 1+ 1 14 Use of MATLAB Part I
Representation of transfer function: b0sm + b1sm,1 + bm : i T s = n a0s + a1sn,1 + an MATLAB command is num = b0 b1 bm ; den = a0 a1 an ; s , z1s , z2 s , zm ii T s = K : s , p1s , p2 s , pn MATLAB command is K = ; Z = z1 z2 zm ; P = p1 p2 pn ; Conversion between i and ii: MATLAB command is Z,P,K = tf2zpnum,den; num,den = zp2tfZ,P,K; Output response for transfer function of num, den : MATLAB command for impulse response is y = impulsenum,den; or y = impulsenum,den,T; 15 The MATLAB command for step response is similar: simply replace impulse by step in the previous commands. Plot of output response: MATLAB command for plotting signals is plotT,y Residue computation: MATLAB command for residues is R,P,K = residuenum,den; If there is no repeated poles, then R Rn R T s = s , 1 + s , 2 + + s , P n + K P 1 P 2 where R,P,K are MATLAB variables: R = R1 R2 P = P 1 P 2 Rn P n For multiple poles with P i = P i + 1 = P i + l , 1, Ri + 1 + l 1 T s = s , Pi + sR,iP i2 + + Rsi, P ,l + rest: i Attention should be paid to multiple poles, or poles very close to each other which may result in numerical problems. 16 ...
View
Full
Document
This note was uploaded on 02/03/2012 for the course EE 3530 taught by Professor Chen during the Fall '07 term at LSU.
 Fall '07
 Chen

Click to edit the document details