Unformatted text preview: 3. The zero input response is given by the solution of the homogeneous equation of the system. Then, 1n y (n) = C1 + C2 (2)n 3 Now we use the initial conditions y (−1) = α and y (−2) = β to solve for C1 and C2 . 1 3C1 + C2 = α 2 1 9C1 + C2 = β 4 Then C1 =
1 15 (2β − α) and C2 = 4 (3α − β ). 5 – If we want only λ1 = 1 to be present, then we choose C2 = 0 or 3α = β . 3 – If we want only λ2 = 2 to be present, then we choose C1 = 0 or α = 2β . Problems from Chapter #8 8.1 Solving a diﬀerence equation in 4 diﬀerent ways. Method 1. Using the recursion: 1 3 +1= 2 2 13 +1= 22 17 +1= 24 yc (0) = yc (1) = yc (2) = . . . yc (n) = That is, 7 4 15 8 1 2
n+1 2(n+2) − 1 =2− 2(n+1) 1 2
(n+1) yc (n) = 2 − , n≥0 Method 2. Using yc (n) = yzs (n) + yzi (n) where yzs (n) = yp (n) + yh (n). Solving for yp (n): Since the input is u(n), yp (n) = ku(n). Now substituting yp (n) = ku(n) into the diﬀerence equation, we get 1 ku(n) = ku(n − 1) + u(n) 2 Evaluating the above equation at n ≥ 1, we ﬁnd k = 2. Therefore, yp (n) = 2u(n) for n ≥ 1. Solving for yh (n) of the relaxed system: 1n λ − 1 = 0. Hence the mode is 1 . yh (n) = C1 2 . 2 2 n Therefore yzs (n) = C1 1 + 2u(n) for n ≥ 1. Now, we need to solve for the coeﬃcient, 2 C1 using y (0) = 1 = C1 + 2 Therefore, we get yzs (n) = − For yzi (n): yzi (−1) = 2C1 = 1 → C1 = 1 1n , n≥0 yzi (n) = 22 Since the complete solution is yc (n) = yzs (n) + yzi (n), yc (n) = 1 2
n 1 2 n + 2, n≥0 1 2 Therefore, +2− 1 2
n+1 3 2 1 2 n n≥0 yc (n) = 2 − n≥0 Method 3. yc (n) = yp (n) + yh (n). We already have from Method 2. yp (n) = 2u(n) for n ≥ 1 and yh (n) = C1 1 2
n we just need to compute y (0) by substituting the initial condition in to the diﬀerence equation. yc (0) = 1 + 1 = 3 2 2 Then, yc (0) = 2 + C = 3 . Therefore, 2 yc (n) = 2 − 1 2 1 2
n n≥0 Method 4. yc (n) = yzs (n) + yzi (n) where yzs (n) = hrelax (n) ∗ x(n) Substituting x(n) = δ (n) and the relaxed initial condition, y (−1) = 0, we get h(0) = 1. Therefore 1n u(n) h(n) = 2 Then, yzs (n) = hrelax (n) ∗ x(n) n ≥ 0
∞ =
k=−∞ 1 2 k n u(k )u(n − k ) =
k=0 1 2 k = 1− 1 1 n+1 2 −1 2 Therefore, using yzi (n) from method 2, 1 yc (n) = 2 1 − ( )n+1 2 yc (n) = 2 − 8.7. (1) The characteristic equation is 3 1 λ2 − λ + 4 8 1 1 λ− λ− 2 4 Then the modes of the equation are λ1 =
n + 1 2 1 2 n 1 2 n+1 , n≥0 =0 =0 1 1 and λ2 = . 2 4 (2) All solutions to the homogeneous equations are yh (n) = C1 1 2 + C2 1 4
n for all n (3) The particular solution is given below for the following three cases • For x(n) = u(n), yp (n) = Ku(n). Substituting yp (n) = Ku(n) into the diﬀerence equation, we get 1 3 Ku(n) − Ku(n − 1) + Ku(n − 2) = u(n) 4 8 Evaluating the above equation at n ≥ 2, we ﬁnd K = 8 yp (n) = , 3 n≥2 8 . Therefore, 3 • For x(n) = nu(n), yp (n) = [K1 n + K2 ] u(n). Substituting yp (n) into the diﬀerence equation, we get [K1 n + K2 ] u(n) − For n ≥ 2, 3 1 3 K1 n + K1 + K2 = n 8 2 8 Equating the coeﬃcients of diﬀerent powers of n on both sides we get 8 K1 = , 3 Therefore, the particular solution is yp (n) = 8 32 n− , 3 9 n≥2 K2 = − 32 9 3 1 [K1 (n − 1) + K2 ] u(n − 1) + [K1 (n − 2) + K2 ] u(n − 2) = nu(n) 4 8 ...
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 Spring '08
 Walker
 difference equation

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