ece3101_Homework9_SOLNS_F19.docx - B Olson ECE 3101 Solving Difference Equations 1 Solve the following difference equations for n0 a y n)−.8 y n−1

ece3101_Homework9_SOLNS_F19.docx - B Olson ECE 3101 Solving...

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B. Olson ECE 3101 Solving Difference Equations 1) Solve the following difference equations for n 0 a) y ( n ) .8 y ( n 1 ) = 0 , y(-1) = 25 y ( n ) .8 y ( n 1 ) = 0 (1) We assume y ( n ) = C λ n Substituting into (1) C λ n .8 C λ n 1 = 0 λ .8 = 0 λ = .8 y ( n ) = C ( .8 ) n (2) Applying IC to determine C . Working with equation (1) and IC y(-1) y ( 0 ) = .8 y ( 1 ) = y ( 0 ) = .8 ( 25 ) = 20 ( 3 ) Working with equation (2) y ( 0 ) = C ( 4 ) Combining (3) and (4) y ( n ) = 20 ( .8 ) n b) 8 y ( n ) 2 y ( n 1 ) y ( n 2 ) = 0 , y(-2) = 5, y(-1) = 10 8 y ( n ) 2 y ( n 1 ) y ( n 2 ) = 0 (1) We assume y ( n ) = C λ n Substituting into (1)
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8 C λ n 2 C λ n 1 C λ n 2 = 0 8 λ 2 2 λ 2 1 = 0 λ = 1 / 2, 1 / 4 y ( n ) = C 1 ( 1 4 ) n + C 2 ( 1 2 ) n (2) Applying ICs to determine C 1 and C 2. Working with equation (1) and ICs 8 y ( 0 ) = 2 y ( 1 ) + y ( 2 ) 8 y ( 0 ) = 2 ( 10 ) + ( 5 ) = 25 8 ( 3 ) Working with equation (2) y ( 0 ) = C 1 + C 2 ( 4 ) y ( 1 ) = C 1 ( 1 4 ) 1 + C 2 ( 1 2 ) 1 ( 5 ) Combining (3), (4) and (5) { C 1 5 8 ,C 2 15 4 }} y ( n ) = ( 5 8 )( 1 4 ) n + 15 ( 1 2 ) n + 2 c) 50 y ( n ) + 5 y ( n 1 ) y ( n 2 ) = 0 , y(-2) = 5, y(-1) = 2 Following the approach of the previous problem {{ λ 1 1 5 } , { λ 2 1 10 }} {{ y ( 0 ) 1 10 ,C 1 1 5 ,C 2 1 10 }} y ( n ) = ( 1 5 )( 1 5 ) n + ( 1 10 )( 1 10 ) n 2) Determine the total solution to the following difference equations for n 0 a) y ( n ) 0.8 y ( n 1 ) = 4 δ ( n ) , y (-1) =12.5 There is no particular solution. The influence of the delta function will be taken into consideration when applying the IC y ( n ) 0.8 y ( n 1 ) = 4 δ ( n ) (1)
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Determining the homogeneous solution which solves: y ( n ) 0.8 y ( n 1 ) = 0 y ( n ) = C 1 ( 0.8 ) n ( λ = 0.8 ) y ( 0 ) 0.8 y ( 1 ) = 4 δ ( 0 ) y ( 0 ) = 14 y ( 0 ) = C 1 ( 0.8 ) 0 C 1 = 14 y ( n ) = 14 ( 0.8 ) n b) y ( n ) 0.125 y ( n 1 ) = 7 u ( n ) , y (-1) = 6 y ( n ) 0.125 y ( n 1 ) = 7 u ( n ) (1) y ( n ) = y h ( n ) + y p ( n ) Determining the homogeneous solution which solves: y ( n ) 0.125 y ( n 1 ) = 0 y h ( n ) = C 1 ( 0.125 ) n ( λ = 0.125 ) Determining the particular solution: Assume that the particular solution has the following form: y p ( n ) = C 2 Substituting the particular solution into 1 C 2 0.125 C 2 = 7 → y p ( n ) = C 2 = 8 y ( n ) = y h ( n ) + y p ( n ) = C 1 ( 0.125 ) n + 8 (2) Using IC to solve for C 1 Solving for y(0) using equation 1: y ( 0 ) 0.125 y ( 1 ) = 7 u ( 0 ) y ( 0 ) 0.125 ( 6 ) = 7 → y ( 0 ) = 7.75 y ( 0 ) = C 1 ( 0.125 ) 0 + 8 = 7.75 →C 1 =− .25 y ( n ) =(− .25 ) ( 0.125 ) n + 8 c) y ( n ) + 0.25 y ( n 1 ) = nu ( n ) , y ( 1 ) = 5 y ( n ) + 0.25 y ( n 1 ) = nu ( n ) (1) y ( n ) = y h ( n ) + y p ( n )
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Determining the homogeneous solution which solves: y ( n ) + .25 y ( n 1 ) = 0 y h ( n ) = C 1 ( 0.25 ) n ( λ = 0.4 ) Determining the particular solution: Assume that the particular solution has the following form: y p ( n ) = C 2 n + C 3 Substituting the particular solution into (1) [ C 2 n + C 3 ] + ( .25 ) [ C 2 ( n 1 ) + C 3 ] = n Terms containing n: [ C 2 n ] + ( .25 ) [ C 2 ( n ) ] = n Terms not containing n: [ C 3 ] + ( .25 ) [ C 2 + C 3 ] = ¿ 0 Solving: → y p ( n ) = 0.16 + 0.8 n y ( n ) = y h ( n ) + y p ( n ) = C 1 ( 0.25 ) n + 0.16 + 0.8 n (2) Using the IC to solve for C 1 Solving for y(0) using equation 1: y ( 0 ) + 0.25 y ( 1 ) = 0 y ( 0 ) + 0.25 ( 5 ) = 0 y ( 0 ) =− 1.25 Solving for C1 using equation (2): y ( 0 ) = C 1 ( 0.25 ) 0 ++ 0.16 + 0.8 ( 0 ) =− 1.25 →C 1 =− 1.41 y ( n ) =− 1.41 ( 0.25 ) n + 0.16 + 0.8 n d) y ( n ) 0.4 y ( n 1 ) =( .3 ) n u ( n ) , y (-1) = -3 y ( n ) 0.4 y ( n 1 ) =( .3 ) n u ( n ) (1) y ( n ) = y h ( n ) + y p ( n ) Determining the homogeneous solution which solves: y ( n ) 0.4 y ( n 1 ) = 0 y h ( n ) = C 1 ( 0.4 ) n ( λ = 0.4 )
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Determining the particular solution: Assume that the particular solution has the following form: y p ( n ) = C 2 ( .3 ) n
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