Searching for the third pole on the RL at each of the corresponding gains 7.36, 12.79 and 39.64. Note that the third pole cannot be complex as the CLTF is third order, ie. the third pole must be on the real axis. The poles are at s = -9.25, -8.6 and 1.8 respectively. Root L ocus 10 8 6 4 System:System: sys sys Gain: 7.4 G ain: 12.8 Pole: -9.25 Po le: -8.61 Dampin g: Da1mping: 1 OvershoOotversh (%): oot 0 (%): 0 Fre quenFreque cy (ra d/se ncyc): (rad/sec): 9 .2 5 8.61 Imag Axis 2 System: sys Ga in: 3 9.2 Pole: -1.8 Damping: 1 Ove rsho ot (%): 0 Frequen cy (ra d/sec): 1.8 0 -2 -4 -6 -8 -10 -1 0 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 Real A xis Using Ts = 4 ςωn ; Tp = π ωd and the real and imaginary parts of the dominant pole. The velocity error constant is sK ( s + 1.5) K (1.5) K v = lim sG (s ) = lim = s→ 0 s →0 s( s + 1)( s + 10) (1)(10) Subst. various values of K gives the K v values in the Table below. 1 Second- and third-order responses a. Case 2; b. Case 3 Cases 1 and 2 have the third pole far away from the complex pair. However, there is no approx. pole zero cancellation. Case 3, the third pole is closer to the zero, so a 2nd order approx can be considered valid. The plots are relatively close. A step input is used to show the second order dynamics and validity of the second order approximation. We will now re-evaluate for the third pole analytically knowing the gain at the corresponding dominant pair. As an example, Case 3 will be used. In case 3, K = 39.64. Using the magnitude criterion assuming the third pole p exists somewhere between the pole at − 10 and the zero at − 1.5. The point in question is p. L1 = 10 − p L2 = p − 1 L3 = p L4 = p − 1.5 (10 − p )( p − 1) p = 39.64 ( p − 1.5) Solving for p gives K = p = −1.795 which is the third pole. The same can be done for the finding the other third poles. Note the RL does NOT exist on portions of the real axis where the sum of poles and zeros to the left is even. 2 The closed loop TF is using K = 39.64 C ( s) 39.64 s + 59.64 = R ( S ) s3 + 11s 2 + 49.64 s + 59.64 1 (discuss Matlab) s2 The figure shows that there is a steady state error e( ∞) of (4.5 − 4.33) = 0.17. Since R ( s) = NOTE THAT THIS IS = 1 Kv For Case 3, ( K = 39.64 ) K v = 5.9 e (∞) = 1 = 0.17 which matches the Matlab plot. 5.9 Ste p Re spo n se 6 5 Syst em: sys Time (se c): 4. 5 Amp lit ud e : 4 .3 3 Amp litude 4 3 2 1 0 0 1 2 3 4 5 6 Time (se c) Example Given a unity feedback system that has a forward TF K G (s ) = , do the following (s + 2)(s + 4)( s + 6) (a) Obtain a RL using Matlab (b) Using a 2nd order approx. determine the value of K to give a 10% overshoot for a unit step input (c) Determine settling and peak times (d) The natural frequency (e) The SSE for the value of K . (f) Determine the validity of the 2nd order approx. 3 Repeat part (b) using the magnitude criterion. This is to be done at the point − 2.028 + j 2.768 From the poles L1 = (2.028 − 2) 2 + (2.768)2 = 2.76814 L2 = (2.028 − 4) 2 + (2.768) 2 = 3.399 L3 = (2.028 − 6) 2 + (2.768) 2 = 4.841 There are no zeros K= L1 L2 L3 = 45.55 1 (d) ωn = 2.0282 + 2.7682 = 3.432 (e) The system is Type 0, the position error constant is K 45.55 = = 0.949 2 *4 *6 48 1 Therefore, e(∞) = = 0.51 1+ K p lim G( s ) = K p = s→ 0 (f) In this case, K = 45.55. Using the magnitude criterion assuming the third pole p exists somewhere to the left of the pole at − 6. The point in question is p. L1 = p − 6 L2 = p − 4 L3 = p − 2 4 ( p − 6)( p − 4)( p − 2) = 45.55 1 Solving for p gives p = −7.943 which is the third pole. K= Since this third close loop pole is NOT 5 times or more the magnitude of the real part of the closed loop dominant pole, the second approx. NOT valid. Roo t L o cu s 10 8 6 4 S yste m: sys Ga in : 4 4. 6 P o le: -7 .9 2 Da mp in g: 1 Oversh oo t (%): 0 Fre qu e ncy (ra d/ se c): 7 .9 2 Ima g Axis 2 0 -2 -4 -6 -8 -1 0 -1 6 -1 4 -1 2 -1 0 -8 -6 -4 -2 0 2 Re al A xis 5

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