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Unformatted text preview: of (2), we have the following three equations: 4 3 K D = 2 !" n + b 1 + 4 3 K P = 2 b n + " n 2 4 3 K I = b n 2 (3) Since we desire ! = 0.5 and = 1 "# n = 1 , this gives: n = 1 = 1 0.5 = 2 . Substitution of this into equations (3) gives: 4 3 K D = 2 n + b = 2 + b # K D = 3 4 2 + b ( ) 1 + 4 3 K P = 2 b n + n 2 = 2 b + 4 # K P = 3 4 2 b + 3 ( ) 4 3 K I = b n 2 = 4 b # K I = 3 b What should we use for “b”? Recall that b is the third (real) pole of the CLCE. We simply need to choose b such that we are insured that the design poles above are DOMINANT. That is, we need to choose b such that its corresponding time constant is less than our design poles. In other words, we need b > n = 1 ....
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This note was uploaded on 12/23/2011 for the course ME 375 taught by Professor Meckle during the Fall '10 term at Purdue.
 Fall '10
 Meckle

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