Please solve for system 3 and 4 PLEASE DO SYSTEM 3 AND 4 IDE

Please solve for system 3 and 4

PLEASE DO SYSTEM 3 AND 4, IDENTIFY EACH OF THE PLOTS THAT CORRELATE. This is my second time posting this question. I need a detailed response.

For each system, we have plotted the Bode and Nyquist plots of the 6/ and the closed-loop step loop gain L(s), the sensitivity function S(s) = response. Correctly identify which plots belong to the same system. Briefly justify your answers. The systems are .System 1. Loop gain too small . System 2. Bad robustness (small gain/phase margin) .System 3. Good response (low bandwidth) .System 4. Good response (high bandwidth) Give vour answers in the form shown below System 1 Bd Nyqi ), Sensivity), Step-response (Justification your Teason Bode Diagram Bode Diagam 50 135 10 10 Frequency Oaseac) Frequency (radlsec) Bode Diagram Bode Diagam 50 60 -180 -180 10 10 Frequency (adbc) Frequency (radisec)

Solution

The first thing to be realized is that Bode plots shown in Figures b) and d) are exactly the same, then, there could\'nt be a comparison between bandwidths.

SYSTEM 3: Bode(a) Nyquist (d) Sensitivity (d) Step-Response(c) Justification:

It can be seen that the Bode plot a) has a lower cut-off frequency than the c)\'s. It can also be seen in the Phase Bode plot that its phase margin is about to 45º(Lower than the phase margin shown in the Bode Plot shown in Figure c), which belongs to the remaining system, 4). Then, system 3 is nearest to the inestability than system 4, therefore, its Nyquist diagram must be d, which it is closest to the s=-1 value. Because of the Bode plot (Phase response) it can be estimated that system 3 refers to a compensated or superior order system, then, its sensitivity Bode plot should be d), which presents differents slope values along its frequency response, while the other sensitivity Bode plot only shows a -20 dB/decade slope. Since the system shows a good response but a low bandwidht, its step response should be c), where the output shape is quite aceptable but the time settling is high (slow system\'s response means low bandwidth).

SYSTEM 4: Bode(c) Nyquist (b) Sensitivity (c) Step-Response(b) Justification:

It can be seen that the Bode plot c) has a higher cut-off frequency than the a)\'s. It can also be seen in the Phase Bode plot that its phase margin is about to 90º(Higher than the phase margin shown in the Bode Plot shown in Figure a), which belongs to system 3). Then, system 4 is farther to the inestability than system 3, therefore, its Nyquist diagram must be b, which it is farthest to the s=-1 value. Its sensitivity Bode plot should be c), which only shows a -20 dB/decade slope. Since the system shows a good response and a high bandwidht, its step response should be b), where the output shape is good and the time settling is low (fast system\'s response means high bandwidth).