Question

8. Design a PD controller for the system shown in Figure P9.2 to reduce the settling time by a factor of 4 while continuing to operate the system with 20.5% overshoot. Compare the performance of the compensated system to that of the uncompensated system. Summarize the results in a table similar to that in Example 9.7. R(s) + [K / (s(s + 8)(s + 25))] C(s) FIGURE P9.2

          8. Design a PD controller for the system shown in
Figure P9.2 to reduce the settling time by a factor of
4 while continuing to operate the system with 20.5%
overshoot. Compare the performance of the compensated
system to that of the uncompensated system. Summarize
the results in a table similar to that in Example 9.7.
R(s) + [K / (s(s + 8)(s + 25))] C(s)
FIGURE P9.2
        
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8. Design a PD controller for the system shown in
Figure P9.2 to reduce the settling time by a factor of
4 while continuing to operate the system with 20.5%
overshoot. Compare the performance of the compensated
system to that of the uncompensated system. Summarize
the results in a table similar to that in Example 9.7.
R(s) + [K / (s(s + 8)(s + 25))] C(s)
FIGURE P9.2

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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8. Design a PD controller for the system shown in Figure P9.2 to reduce the settling time by a factor of 4 while continuing to operate the system with 20.5% overshoot. Compare the performance of the compensated system to that of the uncompensated system. Summarize the results in a table similar to that in Example 9.7. R(s) K C(s) s(s + 8)(s + 25) FIGURE P9.2
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Transcript

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00:01 Here we need to use the matlap and find the solution.
00:06 So the percentage overshoot is 20%.
00:17 So percent os that is equal to e raised 2 minus pi e divided by under root 1 minus e squared multiplied by 100.
00:33 So 20 that is equal to e raised 2 minus pi e divided by under root 1 minus 1 minus 1 minus minus e square multiplied by 100.
00:44 So using log here we can say e rest minus pi e divided by under root 1 minus e square will be equal to log 0 .2.
00:59 So e that is equal to 0 .5123 under root 1 minus e square.
01:09 So e that is equal to 0 .22.
01:16 Minus 0 .262 is square.
01:22 So here e will be equal to 0 .456...
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