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PTFE4761_Problem set

# PTFE4761_Problem set - PTFE 4761 Final Practice...

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PTFE 4761 Final Practice Problems (problems from Spring 2007 Final Exam) 1) (note that here I leave T to be whatever value; it’s also okay to take a value of T, e.g. T = 1 to make the notation simpler) a) The difference equation for a discrete time step dt = T: ) ( ) ( ) ( ) ( 2 ) ( ) ( nT c T nT c T nT c nT y T nT y T nT y + - = + - . Multiplying with T and collecting terms: (1 + 2T) y(nT+T) – y(nT) = c(nT+T) + (T – 1) c(nT) (note that we have used a “forward” difference equation; a “backward” one, i.e. using y(nT) – y(nT-T) will also work) b) Taking the z transform of the equation above: ( using the relationship that if y(nT) Y(z) then y(nT+T) z Y(z), with y(0) = 0) (1+2T) z Y(z) – Y(z) = z C(z) + (T-1) C(z) [ (1+2T) z – 1 ] Y(z) = [ z + T – 1 ] C(z) ) ( 1 ) 2 1 ( 1 ) ( z C z T T z z Y - + - = then the transfer function 1 ) 2 1 ( 1 ) ( ) ( ) ( - + - = = z T T z z C z Y z G 2) b z z a z z G = 2 ) ( 2 and we apply a unit step input: c(nT) = 1 for n > 0, C(z) = 1 - z z Then Y(z) = G(z) C(z) = 1 2 2 - z z b z z a z I assume that “zero steady-state error” means that the output will be y(nT) = c(nT) = 1 for n →∞ . Using the final value theorem in the z-domain: y ss = b a b a b z z a z z z b z z a z z z z Y z z nT y z z z n + = + + = = - - = - = 3 1 2 1 1 2 lim 1 2 1 lim ) ( 1 lim ) ( lim 2 1 2 1 1 Since we want y ss to be 1: 1 + a = 3 + b a = 2 + b b) For dead beat process, the pole(s) of G(z) have to be 0. One way is for the denominator to be: (z 2 + 2z + b) = (z-0)(z-0) = z 2 . But this is not possible.

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PTFE4761_Problem set - PTFE 4761 Final Practice...

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