.1
‫ 𝑛[𝛿 ∶ 𝑤𝑜𝑛𝑘 𝑒𝑤 (الف‬− 𝑛0 ] ∗ 𝑥 [𝑛] = 𝑥 [𝑛 − 𝑛0 ] => 𝛿[𝑛 + 1] ∗ 𝑥 [𝑛] = 𝑥 [𝑛 + 1],
𝛿 [𝑛 − 3] ∗ 𝑥[𝑛] = 𝑥[𝑛 − 3] => 𝑦[𝑛] = 𝑥 [𝑛 + 1] − 𝑥[𝑛 − 3] = 𝛿 [𝑛 + 1] +
2𝛿[𝑛] − 𝛿 [𝑛 − 1] − 𝛿 [𝑛 − 3] − 2𝛿[𝑛 − 4] + 𝛿 [𝑛 − 5]
‫ ]𝑛[ 𝛿 = ]𝑛[ 𝑥 (ب‬+ 2𝛿[𝑛 − 1] − 𝛿 [𝑛 − 2] => 𝑥[𝑛 − 1] = 𝛿 [𝑛 − 1] + 2𝛿[𝑛 − 2] −
𝛿 [𝑛 − 3] => 𝑦[𝑛] = 𝑥[𝑛 − 1] ∗ ℎ[𝑛] = 𝛿 [𝑛] + 2𝛿[𝑛 − 1] − 𝛿[𝑛 − 2] − 𝛿 [𝑛 − 4]
−2𝛿[𝑛 − 5] + 𝛿 [𝑛 − 6]
‫ (ج‬ℎ[𝑛] = 𝛿 [𝑛 + 1] − 𝛿 [𝑛 − 3] => ℎ[𝑛 + 2] = 𝛿 [𝑛 + 3] − 𝛿 [𝑛 − 1] => 𝑦[𝑛] =
𝑥 [𝑛 + 3] − 𝑥 [𝑛 − 1] = 𝛿 [𝑛 + 3] + 2𝛿[𝑛 + 2] − 𝛿[𝑛 + 1] − 𝛿 [𝑛 − 1] −
2 𝛿[𝑛 − 2] + 𝛿 [𝑛 − 3]
.2
𝑦[𝑛] = 𝑢[𝑛 + 3] − 𝑢[𝑛 − 2]: −3 ≤ 𝑛 ≤ 1 => 𝑦[𝑛] = 1, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 𝑦[𝑛] = 0
=> 𝑀 = 𝑛 + 3, 𝑁 = 𝑛 − 1
𝑁
.3
𝑤𝑒 𝑘𝑛𝑜𝑤 ∶ 𝛿 [𝑛 − 𝑛0 ] ∗ 𝑥[𝑛] = 𝑥[𝑛 − 𝑛0 ] => ℎ[𝑛] = ∑ 𝛿 [𝑛 − 𝑘] =>
𝑁
𝑁
𝑘=0
𝑦[𝑛] = 𝑥 [𝑛] ∗ ℎ[𝑛] = ∑ 𝑥 [𝑛 − 𝑘] => 𝑦[4] = ∑ 𝑥[4 − 𝑘] = 5 => 𝑘𝑚𝑎𝑥 = 4
𝑘=0
4
𝑘=0
=> 𝑁 = 4 => 𝑦[14] = ∑ 𝑥[14 − 𝑘] = 0
𝑘=0
‫ (الف‬ℎ(𝑡) = 𝑒−(1−2𝑗) 𝑡 𝑢(𝑡) = 𝑒 −𝑡 (𝑐𝑜𝑠2𝑡 + 𝑗𝑠𝑖𝑛2𝑡)𝑢(𝑡) => 𝑠𝑡𝑎𝑏𝑙𝑒
‫ (ب‬ℎ(𝑡) = 𝑒 −𝑡 cos (2𝑡) 𝑢(𝑡) => 𝑠𝑡𝑎𝑏𝑙𝑒
𝜋
‫ (ج‬ℎ[𝑛] = 𝑛𝑐𝑜𝑠 ( 𝑛) 𝑢[𝑛] => 𝑢𝑛𝑠𝑡𝑎𝑏𝑙𝑒
4
‫ (د‬ℎ[𝑛] = 3𝑛𝑢[−𝑛 + 2] => 𝑠𝑡𝑎𝑏𝑙𝑒
𝑥 (𝑡) = 𝑒( −1+3𝑗) 𝑡 𝑢(𝑡) = 𝑒 −𝑡 (𝑐𝑜𝑠3𝑡 + 𝑗𝑠𝑖𝑛3𝑡)𝑢(𝑡) =>
=> 𝑦(𝑡) = 𝑐1 𝑒−𝑡 𝑐𝑜𝑠3𝑡 + 𝑐2 𝑗𝑒−𝑡 𝑠𝑖𝑛3𝑡 =>
.4
𝑑
𝑦(𝑡) + 4𝑦(𝑡) = 𝑥(𝑡)
𝑑𝑡
𝑑
𝑦(𝑡) = −𝑐1 𝑒−𝑡 𝑐𝑜𝑠3𝑡 − 3𝑐1 𝑠𝑖𝑛3𝑡
𝑑𝑡
.5
‫>= 𝑡‪−𝑐2 𝑗𝑒−𝑡 𝑠𝑖𝑛3𝑡 + 3𝑐2𝑗𝑒−𝑡 𝑐𝑜𝑠3𝑡 + (4𝑦(𝑡) =)4𝑐1𝑒−𝑡 𝑐𝑜𝑠3𝑡 + 4𝑐2 𝑗𝑒−𝑡 𝑠𝑖𝑛3‬‬
‫𝐼 ‪𝑒−𝑡 𝑐𝑜𝑠3𝑡(−𝑐1 + 3𝑗𝑐2 + 4𝑐1) = 𝑒 −𝑡 𝑐𝑜𝑠3𝑡 => 3𝑐1 + 3𝑗𝑐2 = 1‬‬
‫𝐼𝐼 ‪𝑒−𝑡 𝑠𝑖𝑛3𝑡(−3𝑐1 − 𝑗𝑐2 + 4𝑗𝑐2) = 𝑒 −𝑡 𝑠𝑖𝑛3𝑡 => −3𝑐1 + 3𝑗𝑐2 = 1‬‬
‫𝑗‪1 𝑗 −‬‬
‫‪1‬‬
‫= × = ‪𝐼, 𝐼𝐼 => 6𝑗𝑐2 = 1 => 𝑐2‬‬
‫>= = ‪=> 𝑐1‬‬
‫𝑗 𝑗‪6‬‬
‫‪6‬‬
‫‪6‬‬
‫𝑡‪1 −‬‬
‫𝑡‪1 −‬‬
‫)𝑡(𝑢 𝑡‪𝑦𝑝 (𝑡) = ( 𝑒 𝑐𝑜𝑠3𝑡 − 𝑗𝑒 𝑠𝑖𝑛3𝑡)𝑢(𝑡), 𝑦ℎ (𝑡) = 𝑒 −4‬‬
‫‪6‬‬
‫‪6‬‬
‫)𝑡( ‪=> 𝑦(𝑡) = 𝑦𝑝 (𝑡) + 𝑦ℎ‬‬
‫‪.6‬‬
‫‪1‬‬
‫‪1‬‬
‫‪1‬‬
‫= ]‪𝑦[−1] = 𝑦[−2] + 𝑥 [−1] = 0 + 1 = 1 => 𝑦[0] = × 1 + 0 = => 𝑦[1‬‬
‫‪4‬‬
‫‪4‬‬
‫‪4‬‬
‫‪1 1‬‬
‫‪1‬‬
‫‪1‬‬
‫]𝑛[𝑢 ‪× + 0 = ( )2 => ℎ[𝑛] = ( )𝑛+1‬‬
‫‪4 4‬‬
‫‪4‬‬
‫‪4‬‬
‫‪ .7‬الف)‬
‫‪1‬‬
‫𝛼‬
‫‪𝑦[𝑛] = 𝛼𝑦[𝑛 − 1] + 𝛽𝑤[𝑛] => 𝑤[𝑛] = 𝑦[𝑛] − 𝑦[𝑛 − 1],‬‬
‫𝛽‬
‫𝛽‬
‫‪1‬‬
‫𝛼‬
‫‪1‬‬
‫‪1‬‬
‫‪𝑤[𝑛 − 1] = 𝑦[𝑛 − 1] − 𝑦[𝑛 − 2] => 𝑤[𝑛] − 𝑤[𝑛 − 1] = 𝑦[𝑛] −‬‬
‫𝛽‬
‫𝛽‬
‫‪2‬‬
‫𝛽‬
‫𝛼‬
‫‪1‬‬
‫𝛼‬
‫‪1‬‬
‫‪𝑦[𝑛 − 1 ] −‬‬
‫‪𝑦[𝑛 − 1 ] +‬‬
‫‪𝑦[𝑛 − 2] = 𝑥[𝑛] => 𝑦[𝑛] = (𝛼 + ) 𝑦[𝑛 − 1] −‬‬
‫𝛽‬
‫𝛽‪2‬‬
‫𝛽‪2‬‬
‫‪2‬‬
‫𝛼‬
‫‪1‬‬
‫‪𝑦[𝑛 − 2] + 𝛽𝑥[𝑛] => 𝛼 = , 𝛽 = 1‬‬
‫‪2‬‬
‫‪4‬‬
‫ب)‬
‫‪1‬‬
‫‪1‬‬
‫= ]𝑛[ ‪ℎ1[𝑛] = ( )𝑛𝑢[𝑛], ℎ2[𝑛] = ( )𝑛 𝑢[𝑛] => ℎ[𝑛] = ℎ1[𝑛] ∗ ℎ2‬‬
‫‪2‬‬
‫‪4‬‬
‫∞‬
‫𝑛‬
‫𝑛‬
‫𝑘‪1 𝑘 1 𝑛−‬‬
‫𝑘‪1 𝑘 1 𝑛−‬‬
‫‪1‬‬
‫) ( ) (∑‬
‫) ( ) (∑ = ]𝑘 ‪𝑢[𝑛 −‬‬
‫= )𝑘‪= ∑( )2(𝑛−‬‬
‫‪2 4‬‬
‫‪2 4‬‬
‫‪2‬‬
‫‪𝑘=0‬‬
‫‪𝑘=0‬‬
‫‪ .8‬قسمت الف عکس سمت راست و قسمت ب عکس سمت چپ است‪.‬‬
‫‪𝑘=0‬‬
‫𝑛 ‪1‬‬
‫𝑛 ‪1‬‬
‫]𝑛[𝑢 ] ) ( ‪[2 ( ) −‬‬
‫‪2‬‬
‫‪4‬‬
.9
∞
∞
‫∫ (الف‬−∞ 𝑢0 (𝑡) cos(𝑡) 𝑑𝑡 = ∫−∞ 𝛿 (𝑡) cos(𝑡) 𝑑𝑡 = cos(0) = 1
5
‫∫ (ب‬0 𝑠𝑖𝑛(2𝜋𝑡) 𝛿 (𝑡 + 3)𝑑𝑡 = sin(2𝜋(5 + 3)) − sin(2𝜋(0 + 3)) = 0
)‫ الف‬.10
𝑟 2 + 3𝑟 + 2 = 0 => (𝑟 + 1)(𝑟 + 2) = 0 => 𝑟 = −1, −2 => 𝑦(𝑡) = 𝑐1𝑒−𝑡 +
𝑐2 𝑒 −2𝑡 => 𝑦(0) = 𝑐1 + 𝑐2 = 0 => 𝑦̇ (0) = −𝑐1 − 2𝑐2 = 2 => 𝑐1 = −𝑐2 =>
−𝑐2 = 2 => 𝑐2 = −2 => 𝑐1 = 2 => 𝑦(𝑡) = 2𝑒−𝑡 − 2𝑒−2𝑡
)‫ب‬
𝑟 3 + 𝑟 2 − 𝑟 − 1 = 0 => (𝑟 + 1)(𝑟 2 − 1) = (𝑟 + 1)2 (𝑟 − 1) =>
𝑦(𝑡) = 𝑐1 𝑒 −𝑡 + 𝑐2𝑡𝑒−𝑡 + 𝑐3 𝑒 𝑡 => 𝑦(0) = 𝑐1 + 𝑐2 + 𝑐3 = 1
𝑦̇ (0) = −𝑐1 − 𝑐2 + 𝑐3 = 1, 𝑦̈ (0) = 𝑐1 + 𝑐3 − 2𝑐2 = −2 =>
2𝑐3 = 2 => 𝑐3 = 1 => 𝑐1 + 𝑐2 = 0,2𝑐2 − 𝑐1 = 3 => 𝑐2 = 1, 𝑐1 = −1 =>
𝑦(𝑡) = −𝑒 −𝑡 + 𝑡𝑒 −𝑡 + 𝑒𝑡