Question

Write the continuous time mathematical expression of P 2 ID controller for the case in which two proportional terms have completely different control gains denoted by ๐‘˜๐‘1 and ๐‘˜๐‘2 , respectively. Obtain the mathematical expression of digital version of P 2 ID controller by utilizing backward derivation and backward integration methods.

          Write the continuous time mathematical expression of P 2 ID controller for the case in which two proportional terms have completely different control gains denoted by ๐‘˜๐‘1 and ๐‘˜๐‘2 , respectively.
Obtain the mathematical expression of digital version of P 2 ID controller by utilizing backward derivation and backward integration methods.
        
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Elementary and Intermediate Algebra
Elementary and Intermediate Algebra
Alan S. Tussy, R. David Gustafson 5th Edition
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Write the continuous time mathematical expression of P 2 ID controller for the case in which two proportional terms have completely different control gains denoted by ๐‘˜๐‘1 and ๐‘˜๐‘2 , respectively. Obtain the mathematical expression of digital version of P 2 ID controller by utilizing backward derivation and backward integration methods.
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Transcript

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00:02 All right so let here we are given that solid the black diagram is gc um s which is uh kab s plus z divided by s and g p of s is is 4 divided by s plus 2 so uh this is the controller okay controller and this is the plant right this is the plant now uh your question is not your question is not so clear to me okay so i think this would be your question okay so now part this let we are given that one plus g h of s is 0 and 1 plus k up s plus 2 divide s and 4 divided by s plus 2 is equal to 0.
01:08 Sorry, this is z.
01:09 This is not 2.
01:11 This is z.
01:13 Now, the control polynomial is, this is the polynomial, which is s up, s plus 2, plus 4k of s plus z.
01:27 This is 0, and this is their polynomial, which is s squared times 2 plus 4k, and this is times s plus 4 k z which is equal to 0 okay now i here given that that the mp the maximum peak is less than equal to 5 percent and the settling time is less than equal to 2 second okay so let here uh let this is a 1 p divided by square root of 1 minus e 1 square this is less than equal to 0 .0 5 and minus e1p divided by square root of 1 minus e1 square is less than equal to 0 .05 and minus e1p divided by square root 1 minus e 1 square is less than equal to natural log of 0 .05 minus 2 .995...
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