Figure 1: Unity Feedback System
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1 MEM 355 Sample Midterm Problem Stability 1 a) I the following ytem table? Solution: G() = Pole: -1, -2, -2, i, i 1 ( + 1)( )( + 2) 2 A you can ee, all pole are on the LHP. Thu, the ytem given above i table. 1b) Unity feedback ytem i a hown below Figure 1: Unity Feedback Sytem Cloed loop tranfer function: Y() R() = K()G p() 1 + K()G p ()
2 For what range of value of i the cloed loop ytem table? Characteritic equation; 1 + K()G p () = K ( 1 2 )2 ( + 1) = 0 Reorganize characteritic equation; ( + 1) + K( 1 2 )2 = 0 (1 + K) 2 + (1 K) + K 4 = 0 2 (1 + K) K 4 1 (1 K) 0 0 K 4 Table 1: Routh-Hurwitz Stability Criteria Hence, 0 < K < 1 When proportional Controller K i greater than 0 and le than 1, cloed loop ytem depicted in Figure 1 i table!
3 Ultimate State Error 2) What i the velocity error contant for the ytem? ( 2 + 2) G() = 5 ( + 2)( ) You may look at lecture note (Lecture 3: Ultimate State Error), or textbook Chapter 7.3, pg.350 (Nie, Control Sytem Engineering, 6th ed.) velocity error contant K v lim G() ( 2 + 2) K v lim 5 ( + 2)( ) = 10 2 = 5 3) For a unity feedback ytem, compute the teady-tate error in repone to: a) a unit tep and b) a unit ramp for a unity feedback ytem with the following forward loop tranfer function a) ( + 1)( 2 + 4) G() = 10 ( )( ) poition error contant K p lim G() velocity error contant K v lim G() K p lim 10 ( + 1)( 2 + 4) ( )( ) = 40 3 ( + 1)( 2 + 4) K v lim 10 ( )( ) = 0 Then, poition and velocity contant can be plugged in teady tate equation a follow.
4 For Step input > teady tate error: e( ) = 1 1+K p = 3 43 For Ramp input > teady tate error: e( ) = 1 K v = b) ( + 3) G() = 2 ( ) poition error contant K p lim G() velocity error contant K v lim G() K p lim 2 ( + 3) ( ) = ( + 3) K v lim 2 ( ) = 6 Then, poition and velocity contant can be plugged in teady tate equation a follow. For unity Step input ( 1 1 ) > teady tate error: e( ) = = 0 1+K p For unity Ramp input ( 1 2) > teady tate error: e( ) = 1 K v = 1 6
5 Solution#1: A we can ee, forward loop TF i G() = K ( + 5) velocity error contant K v lim G() K v lim K ( + 5) = K 5 We know that, For unity Ramp input ( 1 2) > teady tate error: e( ) = 1 K v = Therefore, we can redefine teady tate error For an arbitrary Ramp input, which i ( ) > teady tate error: e( ) = = K v K v = 10 K = 5 K v Then, K = 50
6 Solution #2: (Alternative olution) Error TF ----> E() = 1 1+G() R() In quetion, ramp input i given a R() = Steady tate error for ramp input > Thu, K = 50 e( ) = lim e( ) = lim E() = lim 1 K 1 + ( + 5) = G() R() Characteritic equation: K 1 + ( + 5) = 0 ( + 5) + K = K = K K Table 1: Routh-Hurwitz Stability Criteria Sytem given in quetion i table if K>0 Therefore, ytem i table for K=50!!!
7 Solution #1: Solution #2: (Alternative olution) Forward loop TF i G() = K( 2) ( + 1)( + 10) Error TF ----> E() = 1 1+G() R() In quetion, ramp input i given a R() = 20 Steady tate error for ramp input > e( ) = lim E() = lim 1 1+G() R()
8 e( ) = lim 1 K( 2) 1 + ( + 1)( + 10) 20 e( ) = lim ( + 1)( + 10) 20 ( + 1)( + 10) + K( 2) = K = K = Characteritic equation: K( 2) 1 + ( + 1)( + 10) = 0 ( + 1)( + 10) + K( 2) = (11 + K) K = K 1 (11 + K) K Sytem given in quetion i table if 11 < K < 5 Table 1: Routh-Hurwitz Stability Criteria And, we found previouly that K = Hence, the ytem i NOT table with K =
9 Solution: Let find forward loop TF to find error contant: a = K 2( ) J 2 ( K 1( ) G c ) a = K 1K 2 (( )) 2 ( 2 J + K K 2 ) G = a 1 = K 1 K 2 (( )) 2 2 ( 2 J + K K 2 )
10 After forward loop G i found, we can find error contant poition error contant K p lim G() velocity error contant K v lim G() accelarition error contant K v lim 2 G() K p lim K 1 K 2 (( )) 2 2 ( 2 J + K K 2 ) = K v lim K 1 K 2 (( )) 2 2 ( 2 J + K K 2 ) = K a lim 2 K 1 K 2 (( )) 2 2 ( 2 J + K K 2 ) = 0.01K 1 Error Contant Steady State Error K p = K p = 0 K v = 1 K v = 0 K a = 0.01K 1 1 K a = 100 K 1
11 Root Locu Solution: Forward loop TF i G() = K (2 + 1)( + 2) 2 ( + 1)( + 4)( + 5) Step#1: Pole: 0,0, 1, 4, 5 zero: 2, i, i Step#2: There are 5 branche becaue there are 5 pole! Step#3: Aymptote angle θ l = (2l + 1) π where l = 0,1,2 m n 1 m n
12 n: # of pole m: # of zero θ l = (2l + 1) 5 3 π θ l = π 2, 3π 2 for l = 0,1 Step#4: interection of aymptote with real axi (finite pole) (finite zero) σ = #pole #zero ( 1 4 5) ( i i) σ = 5 3 σ = Step 5: Break away-in point on the Root Loci 10 ( 1) 5 3 The breakaway point on the root loci of 1 + KG = 0 = 9 2 Satify, K = 1 G dk d = 0 If we want to imply the equation above, we can write dk = 0 d d d (1) = 0 G If define G() = N() D() where G() i open loop TF. d d (1) = d G d (D() d(n()) ) = D() N() d N() d(d()) =0 d N() = ( 2 + 1)( + 2) D() = 2 ( + 1)( + 4)( + 5)
13 ( ) = 0 There are a lot of olution but we know that ome of them are not on the loci. When we olve the equation above, we will ay that Break away point 0 (it i obviou from tep#1 ) and ( we peak thi becaue we know that break away point mut be between -4 and -5 in thi problem) G= zpk([-2, i, i],-[0,0,1,4,5],1) % Step2: Look at pole/zero of ytem [p,z]=pzmap(g) % Step3: Look aymptote angle i=1; for k=0:length(p)-length(z)-1 theta(i)=(2*k+1)*pi/( length(p)- length(z)) % angle for aymptote i=i+1; end theta=theta*180/pi % [deg] % theta= pi/2,3pi/2 % Step4: look interection of aymptote imga=(um(p)-um(z))/( length(p)- length(z)) % igma= = -7/3 in thi examle % Step5: Breakawway or break in if exit! ym N=((+2)*^2-+1); D=^2*(+1)*(+4)*(+5); DK=D*diff(N,)-N*diff(D,)
14 Breakaway=double(olve(DK==0,)) % Breakaway point are 0 and %Step6: Look at root loci rlocu(g) We have 5 branche and all are ymmetrical with real axi. Break away point are at 0 and at Aymptote interect with real axi at -4.5 and aymptote angle are 90 and 270 degree! Not to forget that, draw root locu tarting from pole to zero!
15 Solution: H i contant and aume that it i 0! Let find forward loop TF. G() K( ) a = ( 1 a) a = K( )0.04 ( ) G= 10 4 K(+0.01) 2 a = (+0.04) Pole 0,0, 0.04 Zero 0.01 σ θ l π 2, 3π 2 Break-away point 0 ( other are not on loci)
16 %Forward Loop TF G= zpk([-0.01],-[0,0,0.04],4e-6) % Step2: Look at pole/zero of ytem [p,z]=pzmap(g) % Step3: Look aymptote angle i=1; for k=0:length(p)-length(z)-1 theta(i)=(2*k+1)*pi/( length(p)- length(z)) % angle for aymptote i=i+1; end theta=theta*180/pi % [deg] % theta= pi/2,3pi/2 % Step4: look interection of aymptote imga=(um(p)-um(z))/( length(p)- length(z)) % igma= % Step5: Breakawway or break in if exit! ym N=(4e-06)*(+0.01); D=(^2)*(+0.04); DK=D*diff(N,)-N*diff(D,) Breakaway=double(olve(DK==0,)) % Breakaway point 0 %Step6: Look at root loci rlocu(g)
17
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