process control

process control - Spring 2003 10.450 Process Dynamics,...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/01/2011 for the course EE 223 taught by Professor Thw during the Spring '09 term at MIT.

Page1 / 7

process control - Spring 2003 10.450 Process Dynamics,...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online