Matlab code snippet syms a n declares the symbolic

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Unformatted text preview: MATLAB Exponential: x[k]=bk u[k], |b|<1 MATLAB code snippet » syms b n % declares the symbolic variables » x= b^n; » X=ztrans(x); » X z/b/(z/b­1) Result: X(z)=(z/b) / ((z/b)­1) = z / (z­b). Lesson 31 MATLAB Sinusoid x[k] = sin(ak) u[k]. MATLAB code snippet » syms a n % declares the symbolic variables » x=sin(a*n); » X=ztrans(x); »X z*sin(a)/(z^2-2*z*cos(a)+1) X(z) = sin(a)z /(z2 – 2cos(a) +1), |b|<1 There are times when interpreting the outcome becomes challenging. Lesson 31 Properties See Table 5.2 Property Time-series z-transform Scaling Additively αx[k] x1[k]+x2[k] αX(z) X1(z)+X2(z) Linearity αx1[k]+βx2[k] αX1(z)+βX2(z) Shift (Delay) Reversal Complex Modulation Complex Multiply Ramping Reciprocal Decay others ………………. x[k–1] x[–k] ejθk x[k] wk x[k] kx[k] (1/k)x[k] z-1X(z)-x[0] X(1/z) X(e–jθz) X(z/w) -zdX(z)/dz Lesson 31 − ∫ X (ζ ) dζ ζ z-Transforms Initial valued theorem x[0]= X(z)|z→∞ Final valued theorem x[∞ ]= (z-1)X(z)|z→1 Provided that X(z) has all its poles interior to the unit circle with the exception being a single pole at z=1, Suppose x[k]=u[k], X(z)=z/(z-1), x[∞ ]= (z-1)X(z)|z→1= z|z→1=1 Lesson 31 z-Transforms Property Illustration Ramping kx[k] ⇔ -zdX(z)/dz Suppose x[k]=u[k] and X(z) = z/(z+1) Then kx[k] ⇔ -zdX(z)/dz = z d 1 z z z − 1 = −z = − z − z − 1 ( z − 1) 2 ( z − 1) 2 dz Check » syms n a f F » f=n; F=ztrans(f); » 'F= ', pretty(F) F= z -------2 Lesson 31 (z - 1) z-Transforms Property Illustration Time scaling x[k]=aku[k] x[Nk] ⇔ X(z)=z/(z-a) ⇔ X(zN) y[k]=x[k/2] ⇔ z2/(z2+a) = 1/(1-az-2) y[k]=1+az-2+a2z-4+… Lesson 31 z-Transforms How can the z-transform be used to study linear discrete-time signals and systems. Reconsider discrete-time system previously studied. Lesson 31 z-Transforms Author’s example using MATLAB extensions (Lesson 29). y [k + 2] − 5 y [k + 1] − 6 y [k ] = 3 x[k + 1] + 5 x [k ] y [ −1] = 11/ 6; y [ −2] = 37 / 36; x[k ] = (2)−k u[k ] = (1/ 2)k u[k ] Perform a z-transform of the difference equation. Shift (Delay) Theorem: x[k–1] ⇔ z-1X(z)-x[0] ( )( Y...
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