Control System Answer with an explanation would be highly appreciated.(a) Determine the location(s) of the (finite) open-loop zero(s). open-loop zero(s) at s = ?(b) Determine the location(s) of the open-loop pole(s). open-loop pole(s) at s = ? A negative unity-feedback LTI system has a forward element consisting of an amplifier with adjustablereal-valued gain K in cascade with a transfer function G(s), the loop gain KG(s) for some fixed K > 0known to generate the asymptotic Bode plot description shown in Figure 4.Magnitude (dB)(deg)Phase60402080-20-40-60-8090450-45-90-135-180-22510440 100 400 1000Frequency (rad/sec)Figure 4: Asymptotic Approximation of Bode Plot for KG(s) for some fixed K > 00.010.1 0.4 110000

Given data Magnitude and phase plot of the Bode plot is given in the question. From the above two graph, we have to find out the location of the finite open loop zero and pole of the transfer function.Transfer function From the magnitude plot of the graph, we can observe that the slope of the graph is changing at ω=4 rad/sec and ω=40 rad/sec. Therefore the corner frequencies of the system are ωc1=4 rad/sec and ωc2=40 rad/sec. Consider the below magnitude plot; From the above plot, we can write the transfer function OLTFas, Gs=Ksωc1+1ssωc2+1 Substitute ωc1=4 rad/sec and ωc2=40 rad/sec. Gs=Ks4+1ss40+1=Ks+44ss+4040=10Ks+4ss+40 Therefore the transfer function is Gs=10Ks+4ss+40.(a) From the transfer function, we can find the open-loop zero Open loop zero at, s=-4 (b) From the transfer function, we can find the open-loop pole Open loop pole at, s=0 and s=-4 Conclusion The Open loop zero at s=-4. The Open loop pole at s=0 and s=-4.