Assignment 3 Solutions 2009

2 1 2 2 2

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Unformatted text preview: = ¶F s s V1 ¶T ss V1 ¶Q ss r pV1 c 1 b = 1 ¶f 2 (T - T ) ¶f F ¶f - Fs = 1s 2 s ; b = 2 = s ; b = 2 = 2 3 ¶F ss V ¶T ss V ¶T ss V 2 1 2 2 2 By subtracting the steady­state equation and defining deviat ion variables (like F = F - F ), we obtain the fo llowing equat ions: s dT 1 = a F + a 2 T + a 3Q 1 1 dt d 2 T = b F + b T + b T 1 21 32 dt Taking the Laplace transformat ion, sT ( s ) = a F ( s + a 2 T ( s ) + a 3Q ( s ) ) 1 1 1 s 2 ( s ) = b1 F ( s + b2 T ( s + b3T ( s ) T ) 1 ) 2 and rearranging, a 3 a 1 F ( ) + s Q ( ) s s - a 2 s - a 2 b b T ( s = 1 F ( s + 2 T ( s ) 2) 1) s - b s - b 3 3 T1 ( s = ) Replacing the first equation into the second equation and rearranging again, T2 ( s = ) ù b b é a a 3 1 F ( s + 2 ê 1 F ( s + ) ) Q ( s )ú s - b s - b ë s - a 2 s - a 2 3 3 û T ( s = 2) b é a ù b a 3 1 1 2 ) ) ê1 + s - a ú F ( s + s - b s - a Q ( s s - b ë 3 2 û 3 2 The transfer funct ions can now be ident ified, T ( s = g TF F ( s + g TQ Q ( s ) ) 2) T (...
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