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Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition 9 61 Copyright © Orchard Publications Solutions to End of Chapter Exercises We can express for all as where the coefficients of and are the residues that were found for and at .The coefficient is multiplied by to emphasize that this value exists only for and coefficient is multiplied by to emphasize that this value exists only for . Check with MATLAB: syms z n; Fz=(z^3+2*z^2+1)/(z*(z 1)*(z 0.5)); iztrans(Fz) ans = 2*charfcn[1](n)+6*charfcn[0](n)+8-13*(1/2)^n 7 . Multiplication of each term by yields The long division of the numerator by the denominator is shown below. Therefore, (1) Also, (2) f n [ ] n 0 f n [ ] 6 δ n [ ] 2 δ n 1 [ ] 8 13 0.5 ( ) n + + = δ n [ ] δ n 1 [ ] n 0 = n 1 = z 0 = 6 δ n [ ] n 0 = 2 δ n [ ] n 1 = z 3 F z ( ) z 1 z 2 z 3 + 1 z 1 z 2 4z 3 + + + ----------------------------------------------- z 2 z 1 + z 3 z 2 z 1 + + + ---------------------------------- = = z 1 + z 3 z 2 z 1 + + + z 1 2 z 3 z 4 + + Divisor Quotient z 2 z 1 + Dividend z 2 z 1 z 1 + + + 2 z 1 1st Remainder 2 2 z 1 2 z 2 2 z 3 z 1 2z 2 2z 3 + + 2nd Remainder F z ( ) z 1 2 z 3 z 4 + + = F z ( ) f n [ ] z n 0 f 0 [ ] f 1 [ ] z 1 f 2 [ ] z 2 f 3 [ ] z 3 f 4 [ ] z 4 + + + + + = =

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Chapter 9 Discrete Time Systems and the Z Transform 9 62 Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition Copyright © Orchard Publications Equating like terms on the right sides of (1) and (2) we obtain 8 . a. Taking the Z transform of both sides we obtain and thus b. Then, or (1) By substitution into (1) and multiplication by we obtain Recalling that and we obtain 9 . a. Taking the Z transform of both sides we obtain f 0 [ ] 0 = f 1 [ ] 1 = f 2 [ ] 0 = f 3 [ ] 2 = f 4 [ ] 1 = y n [ ] y n 1 [ ] Tx n 1 [ ] = Y z ( ) z 1 Y z ( ) Tz 1 X z ( ) = H z ( ) Y z ( ) X z ( ) ----------- Tz 1 1 z 1 --------------- T z 1 ----------- = = = x n [ ] e naT = X z ( ) z z e aT ------------------ = Y z ( ) H z ( ) X z ( ) T z 1 ----------- z z e aT ------------------ Tz z 1 ( ) z e aT ( ) -------------------------------------------- = = = Y z ( ) z ----------- T z 1 ( ) z e aT ( ) -------------------------------------------- r 1 z 1 ----------- r 2 z e aT ------------------ + = = r 1 T z e aT ------------------ z 1 = T 1 e aT ------------------ = = r 2 T z 1 ----------- z e aT = T 1 e aT ------------------ = = z Y z ( ) Tz 1 e aT ( ) z 1 ( ) ---------------------------------- Tz 1 e aT ( ) z e aT ( ) ---------------------------------- = z z 1 ----------- u 0 n [ ] z z a ---------- a n u 0 n [ ] y n [ ] T 1 e aT ------------------ Te naT 1 e aT ------------------ T 1 e aT ------------------ 1 e naT ( ) u 0 n [ ] = = y n [ ] y n 1 [ ] T 2 --- x n [ ] x n 1 [ ] + { } =