Lesson 31

# 5z 105 y z z 0 5 z 2 5 z 6 prepare for

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Unformatted text preview: ( z ) − 5 z −1Y ( z ) + y [ −1] − 6 z −2Y ( z ) + z −1y [ −1] + y [ −2] = 3z −1 X ( z ) + 5z −2 X ( z ) X(z)=z/(z-1/2) Lesson 31 ) z-Transforms Z-transform of difference equation parameterize with initial conditions and X(z): ( )( Y ( z ) − 5 z −1Y ( z ) + 11 / 6 − 6 z −2Y ( z ) + z −1(11 / 6) + 37 / 36 = 3z −1 ( z /( z − 1/ 2)) + 5z −2 ( z /( z − 1/ 2)) z(3z 2 − 9.5z + 10.5) Y (z) = ( z − 0 .5 ) z 2 − 5 z + 6 ( ) Prepare for Heaviside expansion Lesson 31 ) z-Transforms Prepare for residue analysis » n=[3 -9.5 10.5]; » conv([1 -0.5],[1 -5 6]) %(z-0.5)(z2-9.5z+6) 1.0000 -5.5000 8.5000 -3.0000 » d=conv([1 -0.5],[1 -5 6]) d = 1.0000 -5.5000 8.5000 -3.0000 » [R,P,K]=residue(n,d) R= 3.6000 -2.3333 1.7333 P= 3.0000 # 2.0000 0.5000 K = » Lesson 31 Y ( z ) (3z 2 − 9.5z + 10.5) = z ( z − 0 .5 ) z 2 − 5 z + 6 ( ) Combined homogeneous and inhomogeneous solutions – total solution. 1.733 z 2.333 z 3. 6 z − + ( z − 0. 5 ) ( z − 2 ) ( z − 3 ) Y (z) = [ y [k ] = 1.73(0.5)k − 2.33(2)k + 3.6(3)k u[k ] Choice: Solve difference equations or z-transform equation. z-Transforms From Lesson 29 y [k ] − 0.6 y [k − 1] − 0.16 y [k − 2] = 5 x[k ] y [−1] = 0; y [−2] = 25 / 4; x[k ] = (4) −k u[k ] Author’s solution. y [k ] = 0.2(−0.2) k + 0.8(0.8) k Zero input − 1.26(4) − k + 0.444(−0.2) k + 5.8(0.8) k Zero state We worked this problem in class. The component parts can be generated using z-transforms to form the total solution. Lesson 31 z-Transforms Zero Input solution (z-domain) ( ) ( ) Y ( z ) − 0.6 z −1Y ( z ) + y [ −1] − 0.16 z −2Y ( z ) + z −1y [ −1] + y [ −2] = 0 ( ) ( ) Y ( z ) − 0.6 z −1Y ( z ) + 0 − 0.16 z −2Y ( z ) + z −10 + 25 / 4 = 0 10 10z 2 Y (z) = = 1 − 0.6z −1 − 0.16z −2 z 2 − 0.6z1 − 0.16 Y (z) 10 z 2 =3 z z − 0.6z 2 − 0.16z Lesson 31 (Prepare for Heaviside expansion) z-Transf...
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