429-2009-HW3key

429-2009-HW3key - Systems Bioengineering III: Homework 3...

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Unformatted text preview: Systems Bioengineering III: Homework 3 Key Rahul Karnik October 5, 2009 1 Assuming the system has equilibrated, we know that at t = 0- both dx 1 dt and dx 2 dt are zero. Setting dx 1 dt = 0 in the first ODE: 0 = k 1 S- α 1 x 1 (0- ) x 1 (0- ) = k 1 S α 1 Similarly, setting dx 2 dt = 0 in the second ODE: 0 = k 2 x 1 (0- )- α 2 x 2 (0- ) x 2 (0- ) = k 2 x 1 (0- ) α 2 = k 1 k 2 S α 1 α 2 We can now use these values as initial conditions in the Laplace transform for modeling the system’s behavior after t = 0. 1 2 S ( t ) = S 1 cos( ωt- θ ) ˜ S ( s ) = Z ∞ S 1 cos( ωt- θ ) e- st dt = S 1 Z ∞ cos( ωt )cos( θ ) e- st dt + S 1 Z ∞ sin( ωt )sin( θ ) e- st dt = S 1 cos θ Z ∞ cos( ωt ) dt + S 1 sin θ Z ∞ sin( ωt ) dt = S 1 s cos θ s 2 + ω 2 + S 1 ωsinθ s 2 + ω 2 = S 1 ( s cos θ + ω sin θ ) s 2 + ω 2 We can use this expression for ˜ S ( s ) to find ˜ x 1 ( s ). ˜ x 1 ( s ) = 1 s + α 1 k 1 ˜ S ( s ) + k 1 S α 1 = 1 s + α 1 k 1 S 1 ( s cos θ + ω sin θ ) s 2 + ω 2 + k 1 S α 1 x 1 ( t ) = Z iω- iω e st 1 s + α 1 k 1 S 1 ( s cos θ + ω sin θ ) s 2 + ω 2 + k 1 S α 1 ds = k 1 S 1 cos θ- α 1 e- α 1 t α 2 1 + ω 2 + iωe iωt ( α 1 + iω )(2 iω ) +- iωe- iωt ( α 1- iω )(- 2 iω ) + k 1 S 1 ω e- α 1 t α 2 1 + ω 2 + e iωt ( α 1 + iω )(2 iω ) + e- iωt ( α 1- iω )(- 2 iω ) + k 1 S e- α 1 t α 1 (ignoring transient terms) = k 1 S 1 cos θ 2 e iωt α 1 + iω + e- iωt α 1- iω + k 1 S 1 sin θ 2 i e iωt α 1 + iω- e- iωt α 1- iω using r 1 = p ω 2 + α 2 , φ 1 = tan- 1 ω α 1 = k 1 S 1 cos θ 2 e iωt r 1 e iφ 1 + e- iωt r 1 e- iφ 1 + k 1 S 1 sin θ 2 i e iωt r 1 e iφ 1- e- iωt r 1 e- iφ 1 = k 1 S 1 cos θ r 1 e i ( ωt- φ 1 ) + e- i ( ωt- φ 1 ) 2 + k 1 S 1 sin θ r 1 e i ( ωt- φ 1 )- e- i ( ωt- φ 1 ) 2 i = k 1 S 1 r 1 [cos θ cos( ωt- φ 1 ) + sin θ sin( ωt- φ 1 )] = k 1 S 1 r 1 cos( ωt- θ- φ 1 ) = k 1 S 1 p ω 2 + α 2 1 ! cos( ωt- θ- tan- 1 ω α 1 ) So the theoretical values for gain and phase shift for x 1 are: 2 A x 1 = k 1 p ω 2 + α 2 1 φ x 1 = tan- 1 ω α 1 We can use the expression for ˜ x 1 ( s ) to find ˜ x 2 ( s ). ˜ x 2 ( s ) = 1 s + α 2 k 2 ˜ x 1 ( s ) + k 1 k 2 S α 1 α 2 = 1 s + α 2 1 s + α 1 k 1 k 2 S 1 ( s cos θ + ω sin θ ) s 2 + ω 2 + k 1 k 2 S α 1 + k 1 k 2 S α 1 α 2 Following a procedure very similar to that shown above for x 1 ( t ), we get: x 2 ( t ) = k 1 k 2 S 1 r 1 r 2 cos( ωt- θ- φ 1- φ 2 ) = k 1 k 2 S 1 p ω 2 + α 2 1 p ω 2 + α 2 2 !...
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This note was uploaded on 03/30/2010 for the course SBE 580.429 taught by Professor Joelbader during the Fall '09 term at Johns Hopkins.

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429-2009-HW3key - Systems Bioengineering III: Homework 3...

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