Tutorial Answers

If the system already exists, different scenarios can be tried using he model before tests being carried out on the real system. Alternatively if the system does not exist, the model can be used to help decide on the final design of a system. Often there are constraints on the design that need to be investigated e. G. Constraints on cost, space, etc. Modification to systems once they exist can be expensive hence it is important to try and get design of systems ‘right-first-time’ and this is where modeling and simulation can be useful engineering tools.

A proportional relationship for a component is here considered to be an unchanging relationship (and is often referred to in modeling terms as a constitutive or physical relationship).

These are the natural physical laws which the individual components of the system obey e. g. For an electrical system, the relationship between voltage and current and in the special case of an ideal resistor – Ohm’s Law FRR.

Current Law : algebraic summation of all currents flowing into a junction of a network is zero.

Kerchiefs Voltage Law: algebraic summation of all voltages acting around a loop of a circuit is zero.

Analogies: This is analogous to Kerchiefs voltage law, particularly if one treats the inertia acceleration as an equivalent force.

Note: In order to model a mechanical system, the usual practice is to form a free body diagram around each inertia (mass) component.

One will then end up with a set of simultaneous differential equations, the solution of which dictates the dynamics and constitutes the system model. In the case where there are no mass components, then ensure a force balance at selected points in the system. That is the net force acting on any point must be zero, I. E. Multiple Choice: Mechanical Systems MI . G) MM. (a,b)

Find the Lovelace transform of the following signals:

1. Students should use MENTAL to check their working here, e. G. Ray the command.
2. Use partial fractions, a lookup table and inverse Lovelace to find the underlying signals with the following transforms. Students should use MAT to check their working, e. G.
3. What is the final value for signals with the following transforms? Use the Pit but note that: (I) there is no final value if the signal is divergent which is the case for 5th (obvious from negative sign) and (it) for convergent signals, the final value must be zero if there is no integrator. Hence only 2nd and 6th have a non-zero values which must be 4 and 0. 5 respectively.
4. Which of the following transforms has the fastest settling time? What are the settling times to within 5% of steady-state? Time constants are negative inverses of poles. One can estimate time to 5% error as approximately three times slowest time constant (exact for 1st order but no strict generalization when many poles due to uncertainty about partial fractions). Time constant is the negative inverse of the pole. So pole at -0. 25 gives T=4, etc.
5. Sketch the poles and zeros of the following transforms on an Regard diagram. By marking the LAP and RAP clearly, hence determine which represent stable and unstable behavior.

Students should use mental to check their working for his, for example, doing 4th as follows will produce a fugue with poles marked in Y and zeros in ‘o’: Systems are stable if and only if all the poles are in the LAP – the origin is counted as being in the LAP. The positions of the zeros do not affect stability.

1. The inverse Lovelace transform of a transfer function is called the “impulse response function”. If a system has an impulse response function given by g(t) t(l-sin(t)). Compute its transfer function, G(s).
2. Use Lovelace methods to solve the following ODE equations.
3. Give examples of type O, type 1 and type 2 systems. Has does this affect the expected behavior? Bookwork
4. Which of the following transforms for 1st order ODES has the highest gain? What are the gains? What are the time constants?

Determine and sketch the step responses for each of these. Gains are 4, 3, 1. 5 and 1. 125 respectively. Time constants are 4, 0. 2, 1. 25, 0. 5 respectively. As these are 1st order, sketching step response follows same procedures as tutorial 5,6.

1. Bookwork – read some control text books to broaden your views on the uses and potential of control.
2. This is also straight from the notes but your understanding will also be improved by some wider reading. Don’t Just stick to your main discipline, but look at examples from chemical, aerospace, automotive, medical, electrical, biological, etc.
3. Straightforward application of the Pit. 2nd set has an integrator and hence the offset is known to be zero. Otherwise, use formula. Confirm this with MENTAL, I. E. Plot is seen to settle at 0. 52
4. The 1st part is taken direct from the lecture slides so not repeated here. The closed-loop time constant and rise time are: Time constant +AKA), closed-loop game = AKA/(I+AKA), where A=4/5, -r=o. 2 Hence 0. 2/(1 +K/5)O. 8 which gives K >4+3. K or 0. K>4 or K>
5. Confirm this using MENTAL, ii. Use G=TFH(4,[1 %% plot in a figure It is clear that the time closed loop pole polynomial is (s+ [1 +AKA]/T) and hence the pole is in the LAP for all positive K which implies closed-loop stability. Discussion of large K is bookwork – read some text books.
6.  This question is designed to make a student think and experiment. To meet specifications, the closed-loop is given as Clearly the steady-state gain is unity as expected so the offset requirement is met. The closed-loop poles are determined from the roots of the denominator and we want the poles to be to the left of -2. 5 ‘e. (s+2. 5) is equivalent to (0. As+1). Both roots can be placed at 2. 5 if In the future students will recognize that lower values of K will give a slower pole and higher values of K will give rise to oscillation.
7. Standard question. Form closed-loop transfer function and find characteristic polynomial for all 3 cases.

You will need to do the partial fractions for all 3 and sketch, but you can use MENTAL to check your answers. E. G form the three closed-loop transfer functions and then type feedback(GO,GO,GO) to see all 3 together. N.B.: 63 is seem 2 content. Clearly Just proportional is fastest, but gives a large offset. GIG is smooth (2 real poles) ND no offset. But poles are well spaced so this is conservative. 63 has similar response time to GIG (same slowest time constant), but has complex poles and thus oscillation.

Conclusion

Typical exam type question outline answer

a) Let the internal temperature be given as T degrees. The rate of heat supplied is given as: The heat loss OHIO(T+50) Hence the temperature is given by: In steady-state we desire T=20 which implies that

b) If the external temperature drops by 10 degrees, then the model becomes: which implies the new steady-state temperature will be 6 degrees! The time constant is clearly 1000 sec. Students should sketch a graph showing the temperature moving from 20 to 6 with the appropriate time constant.

c) If the heat input from the passengers is increased, the model becomes In the case, the change in temperature is negligible which suggests that for this scenario the key factor is the external temperature and heaters rather than any heat coming from the passengers.

d) Clearly the open-loop choice of voltage does not maintain the temperature correctly in general and so some control is needed. It is known that the correct steady-state can only be achieved in the presence of uncertainty if integral action is included.

The steady-state error too change in desired temperature is given by because K(O) is infinite, irrespective of changes in the gain of G or disturbances such as changes in external temperature! Students should first put the equations for the model and integral control law into Lovelace transforms about the steady-state: Hence The closed-loop transfer function is given as Students should validate that the time constants are reasonable and that the closed- loop is stable! The time constants are given from the roots of the closed-loop denominator. Students should note that these are similar to the original time constant and thus satisfactory.