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ME5659_LecNotes_StateSpace

# ME5659_LecNotes_StateSpace - ECE4510 Feedback Control...

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ECE4510: Feedback Control Systems. 10–1 STATE-SPACE MODELS 1. What are they? 2. Why use them? 3. How do we formulate them? 4. How are they related to the transfer functions we have used already? What are They? Representation of the dynamics of an N th-order system as a first-order differential equation in an N -vector called the STATE. N first-order equations. Classic example: 2 nd-order E.O.M. m f ( t ) y ( t ) k b m .. y ( t ) = f ( t ) - b . y ( t ) - ky ( t ) .. y ( t ) = f ( t ) - b . y ( t ) - ky ( t ) m . Define a state vector E x ( t ) = " y ( t ) . y ( t ) # Lecture notes prepared by Dr. Gregory L. Plett

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ECE4510, STATE-SPACE MODELS 10–2 then , . E x ( t ) = " . y ( t ) .. y ( t ) # = . y ( t ) - k m y ( t ) - b m . y ( t ) + 1 m f ( t ) . We can write this in the form . E x ( t ) = A E x ( t ) + B f ( t ) , where A and B are constant matrices. A = , B = . Complete the picture by setting y ( t ) as a function of E x ( t ). The general form is: y ( t ) = C E x ( t ) + Df ( t ) where C and D are constant matrices. C = h i , D = h i Fundamental form for linear state-space model: . E x ( t ) = A E x ( t ) + Bu ( t ) y ( t ) = C E x ( t ) + Du ( t ). where u ( t ) is the input, E x ( t ) is the “state”, A , B , C , D are constant matrices. We usually assume that E x ( t ) is a vector, so simplify notation by simply using x ( t ) . Why Use Them? Transfer functions provide: u G ( s ) y . That is, only an input-output mapping. State variables provide easy access to what is going on inside the system (homogeneous dynamics). Convenient way to express E.O.M. The matrix format is great for computers. Lecture notes prepared by Dr. Gregory L. Plett
ECE4510, STATE-SPACE MODELS 10–3 Allows new analysis and synthesis tools. GREAT for multi-input, multi-output systems. These are very hard to work with transfer functions. How do we Formulate Them? A variety of ways. e.g. , from E.O.M. Also from transfer functions. Three cases: 1] Transfer function is only made up of poles. G ( s ) = 1 s 3 + a 1 s 2 + a 2 s + a 3 = Y ( s ) U ( s ) ... y ( t ) + a 1 .. y ( t ) + a 2 . y ( t ) + a 3 y ( t ) = u ( t ) Choose output and derivatives as the state. x ( t ) = h .. y ( t ) . y ( t ) y ( t ) i T . Then . x ( t ) = ... y ( t ) .. y ( t ) . y ( t ) = - a 1 - a 2 - a 3 1 0 0 0 1 0 .. y ( t ) . y ( t ) y ( t ) + 1 0 0 u ( t ) y ( t ) = h 0 0 1 i x ( t ) + h 0 i u ( t ). 2] Transfer function has poles and zeros, but is strictly proper. G ( s ) = b 1 s 2 + b 2 s + b 3 s 3 + a 1 s 2 + a 2 s + a 3 = Y ( s ) U ( s ) Lecture notes prepared by Dr. Gregory L. Plett

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ECE4510, STATE-SPACE MODELS 10–4 Break up transfer function into two parts. V ( s ) U ( s ) contains all of the poles of Y ( s ) U ( s ) . Then, Y ( s ) = [ b 1 s 2 + b 2 s + b 3 ] V ( s ) . Or, y ( t ) = b 1 .. v( t ) + b 2 . v( t ) + b 3 v( t ) . But, V ( s ) [ s 3 + a 1 s 2 + a 2 s + a 3 ] = U ( s ) , or, ... v( t ) + a 1 .. v( t ) + a 2 . v( t ) + a 3 v( t ) = u ( t ) . The representation for this is the same as in Case [1]. Let x ( t ) = h .. v( t ) . v( t ) v( t ) i T . Then . x ( t ) = ... v( t ) .. v( t ) . v( t ) = - a 1 - a 2 - a 3 1 0 0 0 1 0 .. v( t ) . v( t ) v( t ) + 1 0 0 u ( t ) represents the dynamics of v( t ) . All that remains is to couple in the zeros of the system.
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ME5659_LecNotes_StateSpace - ECE4510 Feedback Control...

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