6 z 1 016 z 2 1 1 4 z 1 dconv1 0251 06 016 d

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Unformatted text preview: orms » n=[1 0 0]; » d=[1 -0.6 -0.16 0]; » [R,P,K]=residue(n,d) R= 0.8000 0.2000 0 P= 0.8000 -0.2000 0 K= Zero input solution Y ( z ) = 0 .2 [ z z + 0.8 z + 0 .2 z − 0 .8 y [k ] = 0.2(−0.2) k + 0.8(0.8) k u[k ] Notice: Eigenvalues provided by the system (only). No contribution from the input. Lesson 31 z-Transforms Zero state solution (z-domain) ( ) ( ) Y ( z ) − 0.6 z −1Y ( z ) − 0.16 z −2Y ( z ) = 5(1 /(1 − 1 / 4 z −1 )) Y ( z )(1 − 0.6 z −1 − 0.16 z − 2 ) = 5(1 /(1 − 1 / 4 z −1 )) Y ( z ) = 5 /((1 − 0.6 z −1 − 0.16 z − 2 )(1 − 1 / 4 z −1 )) d=conv([1 -0.25],[1 -0.6 -0.16]) d = 1.0000 -0.8500 -0.0100 0.0400 Y ( z) = 5 1 − 0.85 z −1 − 0.01z − 2 − 0.04 z −3 ( 5z 3 Y ( z) = 3 z − 0.85 z 2 − 0.01z + 0.04 Y ( z) 5z 2 =3 z z − 0.85 z 2 − 0.01z + 0.04 Lesson 31 % Multiply 2 polynomials ) (Prepare for Heaviside expansion) z-Transforms » n=[5 0 0]; » d=conv([1 -0.25],[1 -0.6 -0.16]) d = 1.0000 -0.8500 -0.0100 0.0400 » [R,P,K]=residue(n,d) R= 5.8182 Zero state solution -1.2626 z z z 0.4444 Y ( z ) = 0.444 + 5.8182 + −1.26 P= z + 0 .2 z − 0.8 z − 0.25 0.8000 0.2500 y [k ] = 0.444( −0.2)k + 5.8(0.9)k − 1.26(0.25)k u[k ] -0.2000 K= Notice: Eigenvalues provided by both the system and input signal (i.e., convolution). [ Lesson 31 Stability Lesson 31 Stability Conditionally Stable Stable s­domain z­domain Unstable Lesson 31 Stability 7th order discrete-time lowpass filter. Stable Poles |H(ejω)| Lesson 31 Continuous to Discrete A first look. A study of mapping continuous-time systems (h(t) or H(s)) into their discretetime equivalent (h[k] or H(z)). H(s) H(z)? Lesson 31 Continuous to Discrete Accepted z-transform forms: • • • • • • • Impulse invariant (standard z-transform) Step invariant Zero order hold Finite difference Matched transform z-transform Bilinear z-transform others … You will only be responsible to the standard z-transform. Le...
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