6.003 TA Office Hour NotesSpring 2010Alison [email protected] of you had questions about the solution to the difference equation of the formy[n] = αy[n − 1] + β (1)You can use recursion to solve for y[n]y[n] = αy[n − 1] + β (2)= α(αy[n − 2] + β) + β = α2y[n − 2] + αβ + β (3)= α(α2y[n − 3] + αβ + β) + β = α3y[n − 3] + α2β + αβ + β (4)= αny[0] + αn−1β + αn−2β + ... + αβ + β (5)= αny[0] + βn−1Xk=0αk(6)= αny[0] + β1 − αn1 − α(7)= αny[0] −β1 − α+β1 − α(8)(9)Notice this is of the formy[n] = Aαn+ B (10)In HW1 Problem 7 some of you did something similar to the calculation above using atable and noticing a pattern (this was a bit simpler since β = 1 and y[0] = 1 in the HW),or you might have assumed the answer would be of this form, using boundary conditionsto solve for A and B (with |α| < 1). These produce equivalent results.y∞= αy∞+ β (11)y∞=β1 − α(12)y[0] = A + B (13)y∞=β1 − α= B (14)A = y[0] −β1 − α(15)So again,y[n] = αny[0] −β1 − α+β1 − α(16)Yet another way to think about this solution parallels the homogeneous/inhomogeneoussolution of a differential equation, with y[n] = y[0]αnthe homogeneous solution, i.e. thesolution of the difference equationy[n] = αy[n − 1]
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