The so-called cruise control in a car has the task of keeping the speed constant. This...
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The so-called cruise control in a car has the task of keeping the speed constant. This can be comfortable on long trips. You set the desired speed and the system takes over. We have linearized a system so that it is described by two equations. Firstly, the work point is defined and in balance based on the non-linear system: Ua + va = Cy Then the deviation from this is defined in a linear equation: Au + Av = DaAy Here Da is a 'constant' that varies with the work point, and which you found a value for earlier. We will use a regulator with feedback plus nominal input: u = Kp(r - y) + Unom „where unom is the nominal input: the power we think we need in this point of work. It may be okay to just insert: Unom = ua With the same working point y, for both r and y we get: Au = K,(Ar – Ay) This inserted in the linearized process equation gives: Kp(Ar - Ay) + Av = DaAy Based on this information, solve the following problems: a) Set up the equation that expresses speed as a function of working point, disturbances and desired speed. Use deviation variables. Enter your numbers in at the end. Use K, = 1000 b) The working point is calculated for a situation where the car drives horizontally, but now assume it goes downhill. The motor power is given by the regulator. In addition, the downhill| provides a traction aid at A = 500 N. Calculate the effect this has on the speed. First Ay, then the real speed. Compare with the unregulated system. The so-called cruise control in a car has the task of keeping the speed constant. This can be comfortable on long trips. You set the desired speed and the system takes over. We have linearized a system so that it is described by two equations. Firstly, the work point is defined and in balance based on the non-linear system: Ua + va = Cy Then the deviation from this is defined in a linear equation: Au + Av = DaAy Here Da is a 'constant' that varies with the work point, and which you found a value for earlier. We will use a regulator with feedback plus nominal input: u = Kp(r - y) + Unom „where unom is the nominal input: the power we think we need in this point of work. It may be okay to just insert: Unom = ua With the same working point y, for both r and y we get: Au = K,(Ar – Ay) This inserted in the linearized process equation gives: Kp(Ar - Ay) + Av = DaAy Based on this information, solve the following problems: a) Set up the equation that expresses speed as a function of working point, disturbances and desired speed. Use deviation variables. Enter your numbers in at the end. Use K, = 1000 b) The working point is calculated for a situation where the car drives horizontally, but now assume it goes downhill. The motor power is given by the regulator. In addition, the downhill| provides a traction aid at A = 500 N. Calculate the effect this has on the speed. First Ay, then the real speed. Compare with the unregulated system.
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