In the previous lecture we showed how to obtain the transfer function of a closed loop system. In addition, we derived the transfer function of a PID controller. In this lecture we will discuss the response of a closed loop system, final value theorem and tracking problem.
We will deal with block diagrams of the form shown below. Comparing this diagram with the one given in the previous lecture we note that the sensor block S(s) is missing here. The closed loop system below is called a unity feedback system since S(s) = 1. In practice this can be easily achieved by adding an amplifier to the measurement system in order to make the measured and actual outputs equal in value. The units of two outputs will be usually different.
From the previous lecture the transfer function of the above closed loop (feedback) system is (with S(s) = 1):
The transfer function from the input R to the error E is equal to
Response of the closed loop system
If we know the input signal r(t), then the response of the system due to the input signal can be found as follows
Examples of the input signal are given below
1. Unit step, 
2. Step, 
3. Ramp, 
4. Sine signal, 
We now state the final value theorem which will be needed in the discussion of the tracking problem.
Final Value Theorem
If F(s) is the Laplace transform of the function f(t) , then
Example 1:
Tracking Problem
Tracking means that the output follows the input (or reference) signal as time goes to infinity.
In order to check if tracking is achieved mathematically we follow one of the relations given below.
Let us give an example that demonstrates the ideas of this lecture.
Example 2: Consider the following closed loop system.
Let
(a) Find the response of the closed loop system, y(t) if
What is the percent overshoot.
(b) Is tracking achieved in part (a)?
(c) Repeat part (a) with 
(d) Repeat part (a) with 
In this case proportional control action only is used.
(e) Choose the proportional and integral gains that will produce a stable closed loop system, eliminate oscillations in the response when the input is a step function, and achieve tracking.
Solution:
(a) We first find the transfer function of the closed loop system
The roots of the denominator of the transfer function are -1+2j and -1-2j. The system is therefore stable (negative real parts) and there will be oscillations in the response (imaginary roots). Think of the denominator of the transfer function as det(sI-A). Now we find the output y(t)
Use partial fraction expansion
It should be an easy exercise to find A, B, and C. We have
The plot of the response of the system (y versus t ) is shown below. From the graph note that the maximum value of y is equal to 1.234 whereas the steady state (final) value of y is equal to 1. The overshoot is the quantity by which the output goes above its final value:
The percent overshoot is equal to
method 1:
Therefore, tracking is achieved. Note that since the output y(t) was found in part (a) we could use it to find the value of the output as time goes to infinity.
method 2: Check if the error goes to zero as time goes to infinity. Using the final value theorem we have
Therefore, tracking is achieved.
(c) We will repeat part (a) with the new values of proportional and integral gains
The roots of the denominator of the transfer function are at -2 and -3. Therefore, the system is stable and no oscillations will be present in the response. Let us now find the response of the closed loop system when it is given a unit step function
It is an easy exercise to find the values of A, B, and C. We have
The output plot is shown below. Note that as time goes to infinity
the output goes to the set point which is equal to 1. Therefore, tracking is achieved. Also note that there are no oscillations since all the roots of the transfer function were real. There is an overshoot due to the existence of two exponentials in the response.
(d) We will repeat part (a) with the new values of proportional and integral gains
The roots of the denominator of transfer function (characteristic polynomial) are at 0 and -2. This indicates that the system is unstable.
Let us now find the response of the closed loop system when it is given a unit step function
It is an easy exercise to find the values of A and B. We have
The output plot is shown below. Note that as time goes to infinity the output goes to the 0.5 which is NOT equal to 1 (desired signal). Therefore, tracking is NOT achieved. In this case proportional control action alone is not enough to satisfy the objective of tracking.
(e) The transfer function of the closed loop system is equal to (see above)
For stability the denominator of the transfer function should have roots with negative real parts (Think of the denominator as det(sI-A)). For no oscillations the roots should be pure real (no imaginary part). In part (b) of this problem we gave values for the proportional and integral gains that will satisfy the conditions of part (e). As an exercise try to find other values of the proportional and integral gains that will satisfy the conditions of part (e). In addition, try to select values of the gains that will also eliminate the overshoot in the system.