1. Consider the closed loop system shown in the figure below. Select the appropriate option to implement the system shown in dotted lines using

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1 1. Consider the closed loop system shown in the figure below. Select the appropriate option to implement the system shown in dotted lines using op-amps a. b. c. d.

2 Solution: b) Explanation: The dotted lines represents the system with a negative feedback with a gain of 40. Hence a difference amplifier with a gain of 40 to input C(t) and a gain of 1 to r(t) is required. Thus, the equation is e(t) = r(t) 40 C(t) Solving the option b we get, Apply superposition theorem to solve for output, Vo = - 40 *C(t) +(1/41) (1 + 40) r(t) = - 40 C(t) + r(t) The output matches with the required equation. 2. Match the following. List I(Transfer function) P. K 1S+K 2 K 3 Q. K 1S 2 +K 2 + K 3 K 4 S R. K 1S+ K 2 K 3 S S. K 1S K 2 S List II(Transfer function) 1. P-Controller 2. PI-Controller 3. PD-Controller 4. PID-Controller a. P-3 Q-4 R-2 S-1 b. P-4 Q-3 R-1 S-2 c. P-3 Q-2 R-4 S-1 d. P-4 Q-1 R-2 S-3 3. What is the advantage of integral controller a. Improve the transient response b. Reduce the offset c. Eliminate the offset d. Reduce the settling time Integration is a continual summing. Integration of error over time, adds up the complete controller error history up to the present time, starting from when the controller was first switched. Integral action continually resets the bias value to eliminate offset as operating level changes.

3 So integral action eliminates offset. But takes long time to do it. Hence the option C. It eliminates not reduce. 4. Consider the response of a second order closed loop feedback system is oscillatory with high frequency. To reduce the oscillations, choose the appropriate option. a. Can be reduced by increasing the proportional action b. Can be reduced by increasing the integral action c. Can be reduced by increasing the derivative action d. Cannot be reduced Further increase of proportional gain and integral gain increases the oscillations. One way to decrease the oscillatory motion is by adding or increasing the derivative gain. Hence, without compromising on steady state offset it can decrease the oscillatory action Adding a derivative into the system, zero adds to the system. Thus, results in shift of root locus to left. Hence the stability increased as well as oscillations decreases Ans: d) 5. How do you improve the dynamics of a temperature control system which is very slow? a. A PI controller can be used b. An I controller can be used c. A PID controller with large I and a negligible D action can be used d. A PD controller can be used Explanation: The integral control action slows down the closed loop response of the process. The speed of response is increased by increasing the proportional gain of the system. However, increase in the proportional gain leads to overshoots in the process. These overshoots can be brought down by using a derivative controller Thus, to improve the dynamics of a sluggish system a PD controller can be used Ans: b) 6. Consider a closed loop feedback system with a proportional controller. If the response of the system shows an offset for a constant input, then to reduce the offset which of the following action is chosen. a. Increasing the gain b. Adding integral mode c. Adding derivative gain d. Decreasing the gain It is given that the system uses proportional controller. Adding an integral controller represents it as a PI controller. However, whether it is a proportional-integral controller or Simple integral controller it eliminates the offset (as discussed in Q3). Increasing a higher gain will not work in all the cases/ with higher order systems (second order or third order systems). Having a proportional controller, at one point after reaching to a certain P gain further increase will not have any effect on the offset. Hence such a case increase of gain will not have any influence on the offset. But the use of Integral controller in a closed loop control can eliminate the offset with fast response and with less settling time.

4 7. Consider the closed loop system shown in the figure below. To implement the system shown in dotted lines, what kind of op-amp can be used. Solution: a) a. Inverting Summing amplifier b. Differential amplifier c. Instrumentation amplifier d. None of the mentioned Explanation: The dotted lines represents the system with a negative feedback with a gain of 1. Hence a inverting summing amplifier with a gain of 1 to input C(t) and a gain of 1 to r(t) is required. Thus, the equation is e(t) = - r(t) C(t) This equation can be implemented using inverting summer amplifier. 8. A reset controller is also referred to as? a. Proportional b. Derivative c. Integral d. None of the mentioned Integral controller is also called as reset controller, due to their unique ability that they can return the controlled variable back to the exact set point. 9. Suppose we have to design an electronic PID controller with Op-Amps. What is the minimum number of Op-Amps required? a. 1 b. 2 c. 3 d. 6 Ans: a) or b)

5 The impedances are given by Finding the relation between the input voltage e and the output voltage u, we get Comparing the above equation with PID controller equation given below And writing the parameters as Since, output is negative, another op-amp in inverting configuration can be considered 10. Transient and Steady State response are important parameters of any controller. Which of the following gives excellent characteristics for both of these parameters? a. Proportional b. PD c. PI d. PID Ans: PID

LECTURE 2: PD, PID, and Feedback Compensation. ( ) = + We consider various settings for Zc when compensating the system with the following RL:

LECTURE 2: PD, PID, and Feedback Compensation. 2.1 Ideal Derivative Compensation (PD) Generally, we want to speed up the transient response (decrease Ts and Tp). If we are lucky then a system s desired