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Unformatted text preview: ME 132 Solutions # 8 1 Sensitivity Analysis It is pretty simple to analyze the block diagram and determine that y ( t ) = AC 1 + ABC- AD u ( t ) We can now compute the sensitivities: S a = ∂y ∂A · A y = 1 1 + ABC- AD S c = ∂y ∂C · C y = 1- AD 1 + ABC- AD 2 Model Properties Many of you had difficulty with this problem, particularly part (c). (a) This model is memoryless, nonlinear, time-invariant, causal. (b) This model is memoryless, nonlinear (because of the affine term b ), time-invariant, causal. (c) This model is dynamic, linear, time-varying, non-causal. It is time-varying because of the following. If we apply the pulse input u a ( t ) = 1 for | t | ≤ 1, we get the output y a ( t ) = 1 for | t | ≤ 1. If we delay this input by one second and use u b ( t ) = 1 for 0 ≤ t ≤ 2 we do not get a one second delayed version of y a . You can verify that y b ( t ) = 1 for- 2 ≤ t ≤ 0. Plot u a ,u b ,y a ,y b to better understand my solution. It is not causal because y (- 4) depends on u (4). This also shows the model is dynamic. (d) This model is dynamic, linear, time-invariant, non-causal. 2 4 6 8 10-4-2 2 4 Time (sec) (a) (b) (c) 3 Homogeneous Solutions (a) The system has only one pole which is at- 1, and so the form of the homogeneous solution is...
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This note was uploaded on 10/20/2010 for the course ME 132 taught by Professor Tomizuka during the Spring '08 term at Berkeley.
- Spring '08