Professor Fearing EE C128 / ME C134 Problem Set 4 Solution Fall 2010 Jansen Sheng and Wenjie Chen, UC Berkeley. control input. error Controller D(s)

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1 Professor Fearing EE C18 / ME C13 Problem Set Solution Fall 1 Jansen Sheng and Wenjie Chen, UC Berkeley reference input r(t) + Σ error e(t) Controller D(s) grid 8 pixels control input u(t) plant G(s) output 1. ( pts) For each of the systems with open loop transfer function D(s)G(s) given below: i) Apply root locus rules 1 to, specifying asymptotes, angle(s) of arrival, and angle(s) of departure. ii) Determine the frequency of any imaginary axis crossing. iii) Hand sketch the closed loop root locus with respect to k. iv) Specify the range of k for stability. v)verifyyourrootlocususingmatlab. Turninhandsketchesandthe MATLABresultsonthesamescales. y(t) Solution: Part a) k s(s +s+1) Rule1: Thereis no zero andthereare poles at and 1± 11 (or( ± 1 1)/( 1)) so there are 3 branches Rule : On the real axis, there are poles between and 1 11 and between and Rule3: n m = 3 = 3sothereareasymptoteswithangle φ = (k +1)π/(n m) = π/3, π, andπ/3 rad These asymptotes intersect at the point s = α = a1 b1 n m = ( )/3 = 8 Rule : The angle of departure for s = is φ = π +() (+) = π rad The angle of departure for s = is φ = π +() (π +) = rad The angle of departure for s = 1 11 is φ = π +() (π +π) = π rad For imaginaryaxis crossings,we can plug in jω for s into the characteristicequation s(s +s+1)+k = and solve for ω and K (both real): (jω)((jω) +(jω)+1)+k = (1) jω 3 ω +1jω+K = () K ω =, ω 3 +1ω = (3) ω =,K = or ω = ±1,K = () For stability, we need to have the poles on the open left half plane so < K < Part b) k(s+) s(s +s+1) Rule1: Thereisa zero at - andthereare poles at and 1± 11 (or( ± 1 1)/( 1)) so there are 3 branches Rule : On the real axis, there are poles between 1 11 and and between and Rule3: n m = 3 1 = sothereareasymptoteswithangle φ = (k +1)π/(n m) = π/ and 3π/ rad These asymptotes intersect at the point s = α = ( )/(n m) = 11

2 Real Axis Real Axis Root locus plot for Problem 1-a) Root locus plot for Problem 1-b) Rule : The angle of departure for s = is φ = π +() (+) = π rad The angle of departure for s = is φ = π +(π) (π +) = π rad The angle of departure for s = 1 11 is φ = π +(π) (π +π) = rad The angle of arrival for s = is φ = π +() (π ++) = rad For imaginary axis crossings, we can plug in jω for s into the characteristic equation s(s +s +1)+ K(s+) = and solve for ω and K (both real): (jω)((jω) +(jω)+1)+k(jω+) = () jω 3 ω +1jω+Kjω+K = (6) K ω =, ω 3 +1ω+Kω = (7) K = 1ω, ω 3 +1ω+1ω 3 = 11ω 3 +1ω = (8) For stability, we need to have the poles on the open left half plane so K > ω =,K = () Real Axis Real 1 Axis.. Root locus plot for Problem 1-c) Root locus plot for Problem 1-d) Part c) k s(s +s+)

3 Rule 1: There is no zero and there are poles at and ±j ( or ( ± 1 )/( 1)) so there are 3 branches Rule : On the real axis, there are poles between and Rule3: n m = 3 = 3sothereareasymptoteswithangle φ = (k +1)π/(n m) = π/3, π, andπ/3 rad These asymptotes intersect at the point s = α = (a 1 b 1 ))/(n m) = ( )/3 = /3 Rule : The angle of departure for s = is φ = π +() (arctan( 1/)+arctan(1/)) = π rad The angle of departure for s = +j is φ = π +() (π/+π arctan(1/)) = rad The angle of departure for s = j is φ = π +() ( π/+π +arctan(1/)) = rad For imaginary axis crossings, we can plug in jω for s into the characteristic equation s(s +s+)+k = and solve for ω and K (both real): (jω)((jω) +(jω)+)+k = (1) jω 3 ω +jω+k = (11) K ω =, ω 3 ω = (1) ω =,K = or ω = ±,K = (13) For stability, we need to have the poles on the open left half plane so < K < Part d) k(s+) s(s +s+) Rule 1: There is a zero at - and there are poles at and ±j ( or ( ± 1 )/( 1)) so there are 3 branches Rule : On the real axis, there are poles between and Rule 3: n m = 3 1 = so there are asymptotes with angle φ = (k +1)π/(n m) = π/ and 3π/ rad These asymptotes intersect at the point s = α = ( )/() = 1 Rule : The angle of departure for s = is φ = π +() (arctan( 1/)+arctan(1/)) = π rad The angle of departure for s = +j is φ = π +(π/) (π/+π arctan(1/)) =.636 rad The angle of departure for s = j is φ = π+( π/) ( π/+π +arctan(1/)) =.636 rad The angle of arrival for s = is φ = π+() (π π/+π/) = rad For imaginary axis crossings, we can plug in jω for s into the characteristic equation s(s + s +) + K(s+) = and solve for ω and K (both real): (jω)((jω) +(jω)+)+k(jω +) = (1) jω 3 ω +jω+kjω +K = (1) K ω =, ω 3 ω+kω = (16) K = ω, ω 3 ω +ω = (17) For stability, we need to have the poles on the open left half plane so K > ω =,K = (18) Part e) k(s+) s (s+3) 3

4 Rule 1: There are two zeros at - and there are two poles at and two poles at 3 so there are branches Rule : On the real axis, there are poles only at and -3 Rule 3: n m = = so there are asymptotes with angle φ = (k +1)π/(n m) = π/ and 3π/ These asymptotes intersect at the point s = α = (6 )/ = 1 Rule : The angle of departure for s = is φ = (π +(+) ( )+πl)/ = π/ and π/ The angle of departure for s = 3 is φ = (π +( π π) ( π π)+πl)/ = π/ and π/ The angle of arrival for s = is φ = (π +() ( π π +π +π)+πl)/ = π/ and π/ Forimaginaryaxiscrossings,we canplug in jω forsinto the characteristicequation s (s+3) 3 +K(s+) = and solve for ω and K (both real): (jω) ((jω) +6(jω)+)+K((jω) +jω +) = (1) ω 6jω 3 ω Kω +Kjω+K = () ω ω Kω +K =, 6ω 3 +Kω = (1) K = 6ω, ω ω 6/ω +6ω = () ω 3ω 6ω = ω 6ω = (3) For stability, we need to have the poles on the open left half plane, so K > ω =,K = () Real 1. Axis 1.. Figure 1: Root locus plot for Problem 1-e). ( pts) Routh and root locus Given D(s)G(s) = k(s +s ) s(s+1)(s+6)(s +s+1) use Routh s criteria to find the range of gain for which the system is stable, and hand sketch the root locus

5 to confirm your calculation. Solution: a. Routh s criteria The closed loop characteristic equation for this system is 1+ 1+D(s)G(s) = () k(s +s ) s(s+1)(s+6)(s +s+1) = (6) s +s +1s 3 +(1+k)s +(k +6)s k = (7) The Routh array for this characteristic equation is s 1 1 k +6 s 1+k k s 3 a 1 = 1 (1+k) = k+17 a = (k+6)+k = 11k+ s b 1 = a1(1+k) a a 1 = k +k+7 k+17 b = a1( k) a 1 = k s 1 c 1 = b1a a1b b 1 = k3 16k +7k b 1a 1 s d 1 = c1b c 1 = b = k For stability, we need to have a 1 >,b 1 >,c 1 >,d 1 >, which yields k +17 > k < 17 k +k+7 > 3.88< k < 8.88 k 3 16k +7k > k < 1.,or 1.677< k <.7181 k > k < (8) () which is 1.677< k <. b. Root locus From characteristic equation (7), we know k should be negative for stability. Thus, we need to apply the rules for negative root locus (FPE6e P68). RULE1. Thelocushas branches,startingfrom s =, 1, 1, 1, 6 andendingat s =,1 and. RULE. The real axis segment 6 < s < or 1 < s < or 1 < s is part of the locus. RULE3. The 3 asymptotesarecenteredatα = +1 = 8 3 withtheangles φ l = 36o (l 1) = o,1 o, o. RULE. The departure from s = is at φ dep1 = (18 o +) = 18 o. The departures from s = 1 are at φ depl = 1 3 [(18o +) (18 o +) 36 o (l 1)] =,1 o, o. The departure from s = 6 is at φ dep3 = (18 o +18 o ) (18 o ) = 36 o =. The arrival at s = is at ψ arr1 = (18 o +) 18 o = o = 18 o. The arrival at s = 1 is at ψ arr = =. From the above rules, the root locus can be sketched as in the following figure. Remember that the gains on this root locus are negative. (To plot this negative root locus in MATLAB, you can use command rlocus(tf([1 1 -],[ ]),-1:.1:), where -1:.1: specifies the values for k). To get the range of k for stability, we need to calculate the imaginary axis crossing point. Thus, we can plug in jω for s into the characteristic equation (7) and solve for ω (real) and k (negative real): (jω) +(jω) +1(jω) 3 +(1+k)(jω) +(k +6)(jω) k = (3) j(ω 1ω 3 +(k +6)ω)+ω (1+k)ω k = (31) ω 1ω 3 +(k +6)ω =, ω (1+k)ω k = (3) ω =.86rad/sec,k = 1.63 or ω = rad/sec,k = (33) For stability, we need to have the poles on the open left half plane, so 1.63 < k <.

6 1 8 6 System: untitled1 Gain: 1.63 Pole:. +.i Damping:.3 Overshoot (%): 1 Frequency (rad/sec): Real Axis Figure : Root locus plot for Problem 3. ( pts) Root locus The open loop transfer function for a system is given by: D(s)G(s) = c+16s c+s a) Determine the characteristic equation for the closed loop system. b) Sketch the root locus with respect to positive values of c, showing direction in which c increases on the locus. Solution: a. The characteristic equation for the closed loop system is b. From (36), we have s 1+D(s)G(s) = (3) 1+ c+16s c+s s = (3) s 3 +cs +1s+c = (36) s 3 +cs +1s+c = s 3 +1s+c(s +) = (37) 1+c s + s 3 +1s = (38) s Thus, the root locus can be sketched with respect to the open loop transfer function + s 3 +1s and the positive gain c. In this case, the open loop poles are s =,±1j, and the open loop zeros are s = ±3j. RULE 1. The locus has 3 branches, starting from s =,±1j and ending at s = ±3j and. RULE. The real axis segment s is part of the locus. RULE 3. The only 1 asymptote is centered at α = 3 = with the angle φ = 18o. RULE. The departure from s = is at φ 1 = 18 o = 18 o. 6

7 The departure from s = 1j is at φ = ( o o ) ( o o ) 18 o 36 o = o = 18 o. The departure from s = 1j is at φ 3 = ( o + o ) ( o + o ) 18 o 36 o = o = 18 o. The arrival at s = 3j is at ψ 1 = ( o o o ) ( o )+18 o = 18 o. The arrival at s = 3j is at ψ = ( o + o o ) ( o )+18 o +36 = = 18 o. RULE. From Rules 1-, we can expect that a closed loop pole with multiplicity will occur on the negative real axis. By setting = s 3 +cs +1s+c = (s+a) (s+b), we obtain ab+a = 1 a+b = c a b = c Solving above, we get a = 1.3,b = 1.3 or a = 3.,b = Thus, the locus arrives at s = 1.3 at o and 7 o, and departs at o and 18 o. The locus arrives at s = 3. at o and 18 o, and departs at o and 7 o. From the above rules, the root locus is sketched as in the following figure. The arrows in the figure show direction in which c increases on the locus. 1 1 System: sys Gain:. Pole: e 7i Damping: 1 Overshoot (%): Frequency (rad/sec): Real Axis Figure 3: Root locus plot for Problem 3 7