Question Solved1 Answer Problem 5. Consider a discrete time system, the block diagram of which is shown below. a) Obtain the state and output equations with the state variables as defined in the figure. What will be the system matrix, i.e. A-matrix, if the system is transformed to modified canonical form? b) Also obtain the transfer function from \( U(z) \) to \( Y(z) \). Obtain Problem 5. Consider a discrete time system, the block diagram of which is shown below. a) Obtain the state and output equations with the state variables as defined in the figure. What will be the system matrix, i.e. A-matrix, if the system is transformed to modified canonical form? b) Also obtain the transfer function from \( U(z) \) to \( Y(z) \). Obtain the poles and zeros of the transfer function.

QVJIV5 The Asker · Electrical Engineering

Transcribed Image Text: Problem 5. Consider a discrete time system, the block diagram of which is shown below. a) Obtain the state and output equations with the state variables as defined in the figure. What will be the system matrix, i.e. A-matrix, if the system is transformed to modified canonical form? b) Also obtain the transfer function from \( U(z) \) to \( Y(z) \). Obtain the poles and zeros of the transfer function.
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Transcribed Image Text: Problem 5. Consider a discrete time system, the block diagram of which is shown below. a) Obtain the state and output equations with the state variables as defined in the figure. What will be the system matrix, i.e. A-matrix, if the system is transformed to modified canonical form? b) Also obtain the transfer function from \( U(z) \) to \( Y(z) \). Obtain the poles and zeros of the transfer function.
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the state equation and the output equation are given in the below picture system matrix A = 0 1 0 0 0 1 sol The conanical form is:. By comparing with standard CCF{:[" ire, "(y(z))/(u(z))=(b_(0)z^(3)+b_(1)z^(2)+cdotsb_(n))/(z^(n)+a_(1)z^(n-1)+cdotsa_(n))],[{::.quadb_(0)=0,b_(1)=1,b_(0)=0,b_(3)=1,quad}TF=(z^(2)+1)/(z^(3)-1)],[a_(1)=0quada_(2)=0","a_(3)=01]:}n-3{:[:.[[x_(2)(k+1)],[x_(2)(k+1)],[x_(3)(k+1)]]=[[0,1,0],[0,0,1],[H,-0,0]][[x_(1)(k)],[x_(2)(k)],[x_(3)(k)]]+[[0],[0],[1]]u(k)],[y[(k)=[[1-1.0,0-0,1-0.0]]*[[x_(1)(k)],[x_(2)(epsi)],[x_(3)(k)]]+0.0(k):}],[[[x_(1)(k+1)],[x_(2)(k+1)],[{:x ... See the full answer