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# sol2 - ECE 351 Systems I Feb 2 2009 OSU Winter 2009...

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Unformatted text preview: ECE 351, Systems I Feb. 2, 2009 OSU, Winter 2009 Solutions - Problem Set 2 Problem 1 (a) The system diagram has the following difference equation: y ( n ) =- a 1 y ( n- 1)- a 2 y ( n- 2)- a 3 y ( n- 3) + b x ( n ) (b) The two figures provided are equivalent system descriptions. The question is asking for the one on the right hand side, but both will be accepted. Σ b 1 b b 2 b 3 D D D [ n ] y [ n ] x Σ b 1 b b 2 b 3 Σ a 2 a 3 a 1 [ n ] x [ n ] y-- - D D D a 2 a 3 a 1 D D D-- - Σ + + + + + + + + + + Problem 2 (b) Using Euler’s approximation for ¨ y and ˙ y we obtain: y (( n + 2)∆)- 2 y (( n + 1)∆) + y ( n ∆) ∆ 2 + y (( n + 1)∆)- y ( n ∆) ∆ + 4 . 25 y ( n ∆) ≈ . Defining the sequence y [ n ] = y (( n + 2)∆) and rearranging the equation we obtain: y [ n ] + (- 2 + ∆) y [ n- 1] + (1- ∆ + 4 . 25∆ 2 ) y [ n- 2] = 0 . The initial conditions required are y [- 2] and y [- 1]. In the question, y [- 2] = y (0) = 2 is given and y [- 1] can be obtained by solving the equation 1 = ˙ y (0) ≈ y (∆)- y (0) ∆ = y [- 1]- y [- 2] ∆ , which gives y [- 1] = ∆ + 2. (c) Use the recur function in Matlab with the inputs and initial conditions a, b, x, x , y from part (b). The code and plots are given below. Note that, in place of ∆, we shall use T which is the notation used in the textbook. For ∆ = 0 . 1 1 n=[2:100];T=0.1; a=[-2+T 1-T+4.25*(T^2)]; b=[0]; x= zeros(1,length(n)); x0=;y0=[2 T+2]; % initial conditions...
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sol2 - ECE 351 Systems I Feb 2 2009 OSU Winter 2009...

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