Lesson_29 - Lesson 29 Challenge 28 Lesson 29 Introduction...

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Lesson 29 Lesson 29 Challenge 28 Lesson 29 – Introduction to Frequency Response Challenge 29
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  Lesson 29 Challenge 28 Given 0 , 2 0 , 0 1 , 1 0 1 1 = - = = - = d c b A
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  Lesson 29 Challenge 28 Given 0 , 2 0 , 0 1 , 1 0 1 1 = - = = - = d c b A ( 29 ( 29 [ ] = - - = + - + - + = + - = - 0 ) 1 /( 1 2 0 1 1 1 0 1 1 ) ( ) ( 1 z d b z z z z c d b A zI c z H T T = 0
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  Lesson 29 Challenge 28 Given 0 , 2 0 , 0 1 , 1 0 1 1 = - = = - = d c b A » A=[1 1; 0 -1]; b= [1; 0]; ct= [0, -2]; d=0; » [NUM,DEN]=SS2TF(A,b,ct,d); » NUM      0     0     0    {N(z)=0} » DEN      1     0    -1       {D(z)= z 2  -1}
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  Lesson 29 Challenge 28 Given 0 , 2 0 , 0 1 , 1 0 1 1 = - = = - = d c b A Changing the state model to: results in: » A=[1 1; 0 -1]; b= [1; 0]; ct= [-2, 0]; d=0; » [NUM,DEN]=SS2TF(A,b,ct,d); » NUM      0    -2    -2       {N(z)= -2z -2} » DEN      1     0    -1       { D(z)= z 2  -1} or H(z) = -2(z+1)/(z 2 -1)
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  Lesson 29 Lesson 29 Chapter 10:   Random thoughts on frequency response analysis and system analysis
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  Lesson 29 Lesson 29 How would one study the following 1 th  order ODE system?              dy(t)/dt + ay(t) = b v(t) Simulation diagram shown below. v(t) b x(t)   =       y(t) dx(t)/dt
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  Lesson 29 Lesson 29 State model:   Assign x 1 (t)=x(t) [dx 1 (t)/dt]=  A  x 1 (t) +  b v(t) = [a] x 1 (t) + [b]v(t) Output model :  y(t) = 
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This note was uploaded on 08/21/2010 for the course EEL 3135 taught by Professor ? during the Spring '08 term at University of Florida.

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Lesson_29 - Lesson 29 Challenge 28 Lesson 29 Introduction...

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