3 Mass – 4 Spring System Control Design
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EGR 326 HW 11 Solutions April 28, 2017 3 Mass – 4 Spring System Control DesignTo apply the design criteria given: PO = 3% and ts = 3s, use the expressions on pages 240 – 241 in the text. Using Matlab to perform these calculations results in: dampR = 0.7448 natlF = 1.7902 pole1 = -1.3333 + 1.1946i pole2 = -1.3333 - 1.1946i These are also written out on EGR paper, on the following page. Note that you do not actually need to calculate any feedback gain matrix for the generic models. You only need to calculate the feedback gain matrix for your actual 6th order system model. This is to say, you need to: Determine the eigenvalues for a generic 2nd order system to behave according to the specified criteria Determine the 4 additional eigenvalues required in order to model a generic 6th order system that behaves as close as possible to the 2nd order system you defined above in part (1) Force your actual 6th order system model to have the eigenvalues defined in part (2) above. To do this, you need to apply a closed-loop state feedback controller, which is defined by the matrix K. So only at this point do you need A and B matrices (from your actual 6th order MSD model) in order to use them with the ‘place()’ command. parts (a), (b) and (c) The Matlab script below is used to define the transfer functions for the two generic systems – the 2nd order and augmented 6th order systems – that have the two performance characteristics given in the problem statement: 3% overshoot and a settling time of 3s. This follows the discussion on page 238, which moves from the state space model, to the transfer function representation. Either form for representing the systems is fine. The two transfer functions are found to be: Transfer function: 3.204
--------------------- s^2 + 2.666 s + 3.204 Transfer function: 1.533e05
----------------------------------------------------------------------- s^6 + 62.0 s^5 + 147 s^4 + 1.6e04 s^3 + 8.6e04 s^2 + 1.69e05 s + 1.5e05 The step responses of each are shown below.  
fig. 1: Step response for generic 2nd order system fig. 2: Step response for generic 6th order system These figures show that the 2nd order system achieves the desired performance criteria with PO = 3% and settling time = 3.23s. The 6th order system also has PO = 3%, with a longer settling time of 3.5 seconds. This slower response (also seen in the rise time increasing from 1.27s to 1.31s) is expected as a result of the dynamic response from the 4 additional eigenvalues (see discussion on pp 247-248). Additional comparison of these two step responses is shown below in figure 3. 
fig. 3: Step response of 2nd and 6th order systems compared Download 96.76 Kb. Share with your friends:
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