Find the recursive system is both stable and causal y[n] • −y[n − r = = + 1] with 1 and is +ve  x[n] r 0 1 = , a pole is at = and is greater than one, the pole is  H(z) z • outside the unit circle and hence system can not be both causal and stable system Ex3: Stable and causal system Find the impulse response of a stable and causal system described by a difference equation 1 1 − + = + 4 1] 8 2] 4 1] y[n] y[n y[n − − − −2x[n] Solution: find the from the difference equation and then z-transform find the impulse response Ex3: Stable and causal system is • z-transform kM=0 bkz−k kN=0 = akz−k H(z) 45 + −2 z−1 = H(z) 41 + 1 z−1 − 81 z−2 Find the poles of and Write the denominator in product form H(z) • Ex3: Stable and causal system We get •
• Find the recursive system is both stable and causal y[n] −y[n − 1] = x[n] with  = 1 + r and r is +ve • H(z) = 1 , a pole is at z =  and is greater than one, the pole is outside the unit circle and hence system can not be both causal and stable system Ex3: Stable and causal system Find the impulse response of a stable and causal system described by a difference equation y[n] + 1 y[n − 1] − 1 y[n − 2] = −2x[n] + − 1] 4 8 4 Solution: find the z-transform from the difference equation and then find the impulse response Ex3: Stable and causal system • z-transform is H(z) = kM=0 bkz−k kN=0 akz−k H(z) = −2 + 45 z−1 1 + 41 z−1 − 81 z−2 • Find the poles of H(z) and Write the denominator in product form Ex3: Stable and causal system • We get 0
U8-L47_Solved-Examples-on-Causality-stability-and-Inverse-systems.docx