3. Can a series connection of two BIBO unstable systems ever be BIBO stable? What about a parallel connection and a negative feedback connection? Explain. Hint: Think pole-zero cancellation.

For a lead compensator whose zero is (-2,0) is used in a feedback control system to provide a phase lead of 45 .for desired pole location (S1=-2+3.3j), that represents the design specifications, this compensators pole is at

An operational-amplifier circuit that may be used as a controller is shown in Figure Qla. (a) R4 ww R1 R3 e Co Figure Qla R1, R2, R3 and R4 are resistors and C1 and C2 are capacitors. Assume both op-amps are ideal and their input/output voltages are ei, e and eo, as shown in the Figure. Using the impedance approach: (i) obtain the transfer function for each op-amp, i.e. E(s)/E:(s) and Eo(s)/E(s). (ii) hence obtain the complete transfer function for the system, i.e. Eo(s)/Ei(s). (iii) Determine D(s) = Eo(s)/E:(s) when R1 = R2 = R3 = R4 = 1 N and C1=1 F, C2= 2 F. %3D (b) Consider the closed-loop control system shown in Figure Qlb. 1 Σ K.D(s) (s +2) Figure Qlb K is the gain and D(s) is the transfer function obtained in part (a)-iii. (i) Plot open loop zeros and poles for the above system in the s-plane. [. Determine whether the point (-1+j) is a closed loop pole for any value of K, by considering the angle criterion at this point. (ii) (c) Briefly summarise the steps, which need to be…

QUESTION 3 Consider a unity negative feedback control system. Assume that the plant G(s) does not have an integrator and you are using a proportional compensator K. You want to increase the closed-loop step tracking performance. Currently, the steady- state tracking error of the closed-loop system with K=K_old to a unit step signal is 0.1. You would like this error to be reduced to 0.02 with a new proportional compensator K=K_new. How does the proportional gain have to change to make this possible? We assume that the closed-loop is stable for all values of the proportional gain. Do not guess as incorrect answers will incur a penalty. K_new = 6.33 K_old K_new = 5.44 K_old K_new = 5 K_old K_new = 4.66 K_old

Could you please go further into how to apply nodal analysis to an op-amp circuit? Thank you

Thanks