EECS C128/ME C134 Fall 2011 Problem Set 7 Solutions. 24(s + 4) (s + 2)(s + 6)(s + 8)
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1 EECS C28/ME C34 Fall 2 Problem Set 7 Solutions. (25 pts) Lag compensation. For open loop plant G(s) with unity gain feedback, G(s) 24(s + 4) (s + 2)(s + 6)(s + 8) a) Use MATLAB to find a lag compensator D(s) k lag α s+/t s+/αt to obtain phase margin of 45 and static error less than % for a step input. Note that D(j) k l ag and G(j) Use the Final Value Theorem to express the steady state error for a unit step: with appropriate gain k lag so as SSE lim s s s D(s)G(s) + D(s)G(s) D()G() + D()G() k lag k lag + SSE <. k lag > 99 From here we can use trial and error. MATLAB s sisotool is a good choice. Alternately, we can follow the procedure in Nise section.3. The form of the lag compensator we are using has a zero at T and a pole at αt. For a lag compensator, the pole has to be to the right of the zero. In order for both the pole and the zero to be in the left half plane, and in the right order, T > and α >. Following the Nise procedure:. K lag 99 for the steady state error target, as described. (We were asked to make SSE less than %, which means k lag > 99, but we can proceed with exactly 99). 4 Bode Diagram 2. Make the Bode plot for G(s) with the new k lag gain. Look for the frequency where the phase margin is 5 2 higher than our target of 45. The Bode plot shown here shows that the phase margin is 57 at.4 rad/s, and the system gain is 24 db at that point. Magnitude (db) Phase (deg) System: untitled Frequency (rad/s):.4 Magnitude (db): 24 System: untitled Frequency (rad/s):.4 Phase (deg): Frequency (rad/s)
2 5 3. We need the lag compensator to have a gain of -24 db at.4 rad/s; putting the zero one decade below this frequency means it should be at.4 rad/s. (Round that off to rad/s and have the zero at s ). The lag compensator starts at db at DC; since one pole causes -2dB/dec drop, and we need to go from db to -24dB, the pole must be decades below the zero With the lag compensator in the form presented, the gain doesn t need to be re-corrected. The zero at - and the pole at -.63 translates to T, α 5.8. There are many different values for (T, α) that will satisfy the phase margin requirement. The following figure shows a contour plot of the phase margins achieved for k lag 99 and different values of z and α. The heaviest line is 45. (z T is the zero location). phase margin z α So, for example, the plot shows that for T.4, α 2, the phase margin is approximately 5. Any values on or to the right of the heavy line should work. (Note that this plot would be different if k lag were different. Hopefully most answers will use k lag close to 99).
3 b) Use a MATLAB Bode plot to verify gain and phase margin for the lag compensated system. 5 Bode Diagram Gm Inf db (at Inf rad/s), Pm 52 deg (at.5 rad/s) Magnitude (db) Phase (deg) MATLAB s margin command makes a Bode plot and annotates it with phase and gain margin information. Here we see the system for k lag 99, T, α 5.8. Note that the crossover is almost exactly where we designed it, and the phase margin is better than 45 (and consistent with the contour plot of phase margins) Frequency (rad/s) c) Using MATLAB, plot the steady state error for a step input for the lag compensated system..4 Step Response.2 Amplitude Time (seconds). 2. Figure :. This plot uses the same values as (b). Here we see an oscillatory transient that dies out after about second, and a slow convergence to the steady state value of Same ω c, but choosing zero at -3 and pole at -.5. This meets phase margin spec, but slightly faster pole and zero give a bit nicer quicker settling.
4 d) Using MATLAB, plot the steady state error for a step input for 99G(s) with unity gain feedback, and compare to the response in c)..8 Step Response.6 Amplitude While the systems in (c) and (d) are converging to the same value in the steady state, the transient behavior is drastically different. The overshoot for the 99G(s) system is much larger. Comparing the time scales, we see that the oscillations have more or less died out at second for both systems, but the 99G(s) system oscillates faster Time (seconds) 2. (2 pts) State Space For the following transfer function, (s + )(s + 4)(s + ) U(s) (s + 3)(s + 2) 2 give state space description (with state variable x) in the following forms (see Nise 2.4 and Fig 5.3): Note that the solutions can be verified using the ss2tf MATLAB command a) phase variable form (Nise 2.4). Part (b) was done first. Since phase variable form is equivilant to controllable canonical with the variables renamed x x 2 x 3 x 3 x 2 x, we can write the matrices: A B C D
5 b) controllable canonical form (Nise p. 265) U(s) s3 + 5s s + 4 s 3 + 7s 2 + 6s + 2 As in Problem 3 from this homework and example 3.4 from Nise, the denominator gives A The numerator gives B y 4x x 2 + 5x + ẋ 4x x 2 + 5x + ( 7x 6x 2 2x 3 + u) 28x x 2 + 8x + u C D c) cascade form (Nise p. 265) Doing some block diagram manipulation: U(s) s + s + 3 s + 4 s + 2 s + s + 2 U 2 (s) U(s) s + s + 3, U (s) U 2 (s) s + 4 s + 2, Following the signals through, and substituting: U (s) s + s + 2
6 ẋ 3 u 3x 3 ẋ 2 ẋ 3 + x 3 2x 2 (u 3x 3 ) + x 3 2x 2 u 2x 3 2x 2 ẋ ẋ 2 + 4x 2 2x (u 2x 3 2x 2 ) + 4x 2 2x u 2x 3 + 2x 2 2x Y ẋ + x u 2x 3 + 2x 2 2x + x u 2x 3 + 2x 2 + 8x in matrix form: A B C D d) parallel form (Nise p. 265) Partial Fraction Expansion: s 3 + 5s s + 4 s 3 + 7s 2 + 6s s s (s + 2) 2 + From p of Nise (and considering the pass-thru term): 3 4 A 2, B 2 6 C 22/6, D Alternatively, could push the scaling around, and use: B T C ( pts) State space Convert the following differential equation to state space in controllable canonical form (with state variable x), where u(t) is the input and y(t) is the output. d 2 y dt 2 + 6dy + 8y u + 3du dt dt + u 2d2 dt 2
7 Start with a Laplace Transform of the Differential Equation (ignoring initial conditions): s 2 + 6s + 8 U(s) + 3sU(s) + 2s 2 U(s) U(s) 2s2 + 3s + s 2 + 6s + 8 Splitting the system into two for the numerator and denominator: U(s) X(s) X(s) U(s) (2s 2 + 3s + ) s 2 + 6s + 8 Following example 3.4 for phase variable form, A and B are given by: ẋ ẋ 2 From 2s 2 + 3s +, we get C and D: ẋ Ax + Bu 8 6 x x 2 + y(t) x + 3x 2 + 2ẋ 2 x + 3x 2 + 2( 8x 6x 2 + u) 5x 9x 2 + 2u x u x 2 y Cx + Du Putting into Controller Canonical Form (or at least what Nise calls it): 6 8 A B C 9 5 D 2 u 4. (2 pts) State space to transfer function Given x x ẋ Ax+Bu x 2 x 2 + x 3 x 3 u(t) and y x x 2 x 3 +u(t)
8 a) Draw a block diagram for this system with input u(t) and output y(t) using integrators, addition junctions, and static gains. b) Starting from the block diagram, determine the transfer function U(s).
9 Note this can be verified using tf2ss U(s) s3 + 3s 2 + 3s + 2 s 3 + 6s 2 + 2s + 9
10 5. (25 pts) State space solutions Given the following 5 6 ẋ Ax + Bu x + u(t), y x x() Solve for the state transition matrix e At, x(t), and y(t), where u(t) is the unit step. (Hint: use Laplace transform.) Take the Laplace transform of the unforced (u ) system: L {ẋ} L {Ax} sx(s) x() AX(s) (si A) X(s) x() X(s) (si A) x() x(t) L { (si A) } x() This process is described in section 4. of Nise. Since the ODE solution is unique, we know e At L { (si A) } We need to invert a 2 2 matrix and find the inverse Laplace transform for each element. (si A) s s s 6 s(s + 5) + 6 s + 5 s { L (si A) } { L L { s 6 s+5 Now it s just a whole bunch of Partial Fraction Expansions: L { 2 s s+3 L { s+2 + s+3 e At 2e 2t + 3e 3t e 2t e 3t } } } } L { L { 6 s+5 L { 6 s s+3 } } { L 3 s s+3 6e 2t + 6e 3t 3e 2t 2e 3t } }
11 To find x(t) and y(t) by Laplace transform, start with: then substitute in (si A) from above: X(s) si A X(s) (si A) (x() + BU(s)) + s Now with inverse Laplace transforms, e 2t + 2e 3t x(t) e 2t 2 3 e 3t s+ s+ s y(t) Alternatively, x(t) and y(t) can also be found by convolution. Referring to eqn. 4.9 in Nise: x(t) e At x() + t 6s + s s+3 2(s+2) + 2 3(s+3) x(t) e 2t + 2e 3t e A(t τ) Bu(τ) dτ Note that u(τ) is over the integration (assuming t > ). e A(t τ) 2e B 2t + 3e 3t e 2t e 3t t t e A(t τ) B dτ 2e 2(t τ) + 3e 3(t τ) dτ t e 2(t τ) e 3(t τ) dτ ( e 2t ) + ( e 3t ) 2 ( e 2t ) 3 ( e 3t ) e 2t e 3t 6 2 e 2t + 3 e 3t e A(t τ) 2e x() 2t + 3e 3t e 2t e 3t 2e x(t) 2t + 3e 3t e e 2t e 3t + 2t e 3t 6 2 e 2t + 3 e 3t e 2t + 2e 3t e 2t 2 3 e 3t y(t) e 2t + 2e 3t Note that these expressions for x(t) and y(t) are only valid for t > due to the unit step input.
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