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lecture9_small - State-space systems Solving the...

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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 ( t ) is defined as L- 1 braceleftbig ( sI- A )- 1 bracerightbig ), x ( t ) = ( t ) x (0) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright + integraldisplay t ( t- ) Bu ( ) d. bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright zero-input solution convolution of ( t ) and B u ( t ) We can also use this equation for an arbitrary initial time, x ( t ) = ( t- t ) x ( t ) + integraldisplay t t ( t- ) B u ( ) d. 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) = ( t ) x (0) . ( 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, ( t ) = L- 1 braceleftbig ( sI- A )- 1 bracerightbig = L- 1 braceleftbigg 1 s + bracerightbigg = e- t (this looks just like the impulse reponse) So we can calculate , x ( t ) = e- t x (0) + integraldisplay t e- ( t- ) u ( ) d. Step response: Zero initial condition ( x (0) = 0). As u ( t ) = 1 for t 0, y ( t ) = x ( t ) = e- t 0 + integraldisplay t e- ( t- ) d, = e- t integraldisplay t e d, = e- t [e | = t- e | =0 ] = e- t ( e t- 1 ) = 1- e- t . Roy Smith: ECE 147b 9 : 4 State-space systems A simple example: A first order system ( > 0). s + a27 a27 u ( s ) y ( s ) Take the initial conditions to be zero. The differential equation is, dy ( t ) dt + y ( t ) = u ( t ) . Define the state as, x ( t ) = y ( t ), then, dx ( t ) dt =- x ( t ) + u ( t ) , y ( t ) = x ( t ) or dx ( t ) dt = bracketleftBig- bracketrightBig x ( t ) + bracketleftBig bracketrightBig u ( t ) , y ( t ) = bracketleftBig 1 bracketrightBig x ( t ) + bracketleftBig bracketrightBig u ( t ) . So the state-space representation is: A =- , B = , C = 1 and D = 0....
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lecture9_small - State-space systems Solving the...

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