Partial preview of the text
Download Module 7 Notes for maths and more Lecture notes Mathematics in PDF only on Docsity!
4.12 Solution of Difference Equation with Constan; Coefficients by Z-Transforms Step Knowledge : For solving a linear difference equation with constant Coefficient Using z-transforms, we use the following steps : (i) Take the z-transforms of both sides of the given difference formulae of shifting property and given conditions. (ii) Transpose all terms without F(z), to the right hand side, (iii) Divide the coefficient of F(z), getting F(z) as a function of z. (iv) Express this function in terms Of the z-transforms of known functions ang take the inverse z-transforms of both sides, which gives f(k), i.e., f, asa Function of k which is the required solution, equation Using the Example 1 Solve the difference equation 6 fea - fey + fr =0, withf(0) = 0, (1) =1, by Z-transform, Solution * The given difference equation is 6 Fata — fae +f, =0 Taking Z-transform of both sides of (1), we get 2 (6f,.2 ~fasa +f] =0 = 2(6h.2)-Z (fs) +Z(h) = 9 = 6 [27F(2) ~27F(0) - 261) } - [2F(2) ~ 2F(0)} + F(z) = 9 Scanned by CamScanner 
