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process control

process control - Spring 2003 10.450 Process Dynamics...

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10.450 Process Dynamics, Operations, and Control Spring 2003 Exam 1 from the 10.450 Policies: "The grade for either test may be raised by reworking the test out of class and turning it in the next class meeting. Final test grade will then be 2/3 in -class and 1/3 at-home. Up to 5 bonus points will be added for creativity in the presentation of the at-home portion." (1) (25%) Represent this block diagram as an equation in the Laplace domain. The equation should be arranged so that a dependent variable is expressed as the sum of independent variables, each of these multiplied by some transfer function. (Note in the diagram that y is subtracted from x sp .) If you could choose G f to be anything you wanted, how could you use it to improve the response of y to the disturbance x d ? G f G d G m G c x d x sp y - Build the equations by starting at the output signal y and moving backwards through the diagram: y = G d x d + G m x m = G d x d + G m ( G f x d + G c e ) = G d x d + G m ( G f x d + G c ( x sp - y )) = G d x d + G m G f x d + G m G c x sp - G m G c y G d + G m G f G G m c y = x 1 + G G x d + 1 + G G sp m c m c If we know the disturbance and manipulated variable transfer functions, G d and G m , perhaps we can create a device G f to be -G m /G d . In this case, the denominator of the closed loop disturbance transfer function will be zero, making response variable y independent of disturbances! (2) (20%) An old 18.03 exam says
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10.450 Process Dynamics, Operations, and Control Spring 2003 Exam 1 u ( t - 2) = 3 u ( t - 8) - 1 y - 2 dy at t = 0, y = 2 2 2 dt where u(t-a) is the unit step function (0 for t < a; 1 thereafter). Write the solution y(t), and sketch the plot. State the time at which the maximum absolute value of y occurs. First rewrite in standard form to clarify the structure of the problem. 4 dy + y = - 2 u ( t - 2) + 3 u ( t - 8) y (0) = 2 dt This is a first-order system that will react immediately to its initial condition and be subsequently disturbed by two step changes. by-the-book Laplace transform solution 4 ( sy ( s ) - y (0) ) + y ( s ) = - 2 e - 2 s + 3 e - 8 s s s ( 4 s + 1 ) y ( s ) = 8 + - 2 e - 2 s + 3 e - 8 s s s 8 - 2 - 2 s 3 - 8 s e e y ( s ) = 4 s + 1 + s ( 4 s + 1 ) + s ( 4 s + 1 ) The denominator product is easily resolved by partial fraction expansion 1 A B 1 - 4 = + = + s ( 4 s + 1 ) s 4 s + 1 s 4 s + 1 Substituting y ( s ) = 8 +� - 2 + 8 e - 2 s +� 3 + - 12 e - 8 s 4 s + 1 Ł s 4 s + 1 ł Ł s 4 s + 1 ł On being inverted, the terms in parentheses will be delayed in time by the exponential functions in s.
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