lecture9_small

# lecture9_small - State-space systems Solving the...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: State-space systems Solving the state-space differential equation (Laplace approach) Now consider the case where there is an input. dx ( t ) dt = A x ( t ) + B u ( t ) . Taking Laplace transforms gives, sx ( s )- x (0) = A x ( s ) + B u ( s ) , which means that, x ( s ) = ( sI- A )- 1 x (0) + ( sI- A )- 1 B u ( s ) . Now the inverse Laplace gives (recall that Φ( t ) is defined as L- 1 braceleftbig ( sI- A )- 1 bracerightbig ), x ( t ) = Φ( t ) x (0) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright + integraldisplay t Φ( t- τ ) Bu ( τ ) dτ. bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright zero-input solution convolution of Φ( t ) and B u ( t ) We can also use this equation for an arbitrary initial time, x ( t ) = Φ( t- t ) x ( t ) + integraldisplay t t Φ( t- τ ) B u ( τ ) dτ. Roy Smith: ECE 147b 9 : 2 State-space systems Solving the state-space differential equation (Laplace approach) Consider a continuous time state-space representation. dx ( t ) dt = A x ( t ) + B u ( t ) , y ( t ) = C x ( t ) + D u ( t ) . To begin, look at the zero-input solution dx ( t ) dt = A x ( t ) . Taking unilateral Laplace transforms gives, sx ( s )- x (0) = A x ( s ) where x (0) = x ( t ) at t = 0 . This gives, ( sI- A ) x ( s ) = x (0) , or x ( s ) = ( sI- A )- 1 x (0) . Taking an inverse Laplace transform gives, x ( t ) = L- 1 braceleftbig ( sI- A )- 1 bracerightbig x (0) = Φ( t ) x (0) . Φ( t ), is also known as the “State Transition Matrix” . Roy Smith: ECE 147b 9 : 1 State-space systems Example (continued) Now look at the State Transition Matrix, Φ( t ) = L- 1 braceleftbig ( sI- A )- 1 bracerightbig = L- 1 braceleftbigg 1 s + α bracerightbigg = e- αt (this looks just like the impulse reponse) So we can calculate , x ( t ) = e- αt x (0) + integraldisplay t e- α ( t- τ ) α u ( τ ) dτ. Step response: Zero initial condition ( x (0) = 0). As u ( t ) = 1 for t ≥ 0, y ( t ) = x ( t ) = e- αt 0 + integraldisplay t e- α ( t- τ ) α dτ, = e- αt integraldisplay t e ατ α dτ, = e- αt [e ατ | τ = t- e ατ | τ =0 ] = e- αt ( e αt- 1 ) = 1- e- αt . Roy Smith: ECE 147b 9 : 4 State-space systems A simple example: A first order system ( α > 0). α s + α a27 a27 u ( s ) y ( s ) Take the initial conditions to be zero. The differential equation is, dy ( t ) dt + α y ( t ) = α u ( t ) . Define the state as, x ( t ) = y ( t ), then, dx ( t ) dt =- α x ( t ) + αu ( t ) , y ( t ) = x ( t ) or dx ( t ) dt = bracketleftBig- α bracketrightBig x ( t ) + bracketleftBig α bracketrightBig u ( t ) , y ( t ) = bracketleftBig 1 bracketrightBig x ( t ) + bracketleftBig bracketrightBig u ( t ) . So the state-space representation is: A =- α , B = α , C = 1 and D = 0....
View Full Document

## This note was uploaded on 04/06/2010 for the course ECE 145 taught by Professor Rodwell during the Spring '07 term at UCSB.

### Page1 / 11

lecture9_small - State-space systems Solving the...

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

View Full Document
Ask a homework question - tutors are online