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barcode generator for ssrs The solution to Eq. (1.13)may be found by assuming a solution of the form in Software
The solution to Eq. (1.13)may be found by assuming a solution of the form Code128 Scanner In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Making ANSI/AIM Code 128 In None Using Barcode drawer for Software Control to generate, create ANSI/AIM Code 128 image in Software applications. Substituting this solution into Eq. (1.13) we obtain the polynomial equation
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For the particular solution, it is necessary to find the sequence yp(n) that satisfies the difference equation for the given x(n). In general, this requires some creativity and insight. However, for many of the typical inputs that we are interested in, the solution will have the same form as the input. Table 12 lists the particular solution for some commonly encountered inputs. For example, if x ( n ) = a n u ( n ) , the particular solution will be of the form provided a is not a root of the characteristic equation. The constant C is found by substituting the solution into the difference equation. Note that for x ( n ) = CS(n) the particular solution is zero. Because x ( n ) = 0 for n > 0, the unit sample only affects the initial condition of y(n). Table 12 The Particular Solution to an LCCDE for Several Different Inputs Term in x(n) Particular Solution
Cln C2 Clan C I cos(nm) C2sin(nm) C I cos(n@) C2sin(nm) Cian cos(nm) C2ansin(nm) None
EXAMPLE 1.5.1 Let us find the solution to the difference equation
y(n)  0.25y(n  2) = x(n) for x ( n ) = u(n) assuming initial conditions of y( 1) = 1 and y(2) = 0. We begin by finding the particular solution. From Table 12 we see that for x(n) = u(n) y,(n) = C I Substituting this solution into the difference equation we find C I 0.25CI = 1 I 4 1  0.25 3 To find the homogeneous solution, we set yh(n) = zn, which gives the characteristic polynomial In order for this to hold, we must have cl= z2  0.25 = 0 Therefore, the homogeneous solution has the form yh(n) = A1(0.5)" Thus, the total solution is y(n) = $ + A2(0.5)" + Al(0.5)" + A2(0.5)" SIGNALS AND SYSTEMS
[CHAP. 1
The constants A, and A2 must now be found so that the total solution satisfies the given initial conditions, y(1) = 1 and y(2) = 0. Because the solution given in Eq. (1.17) only applies for n 0,we must derive an equivalent set of initial conditions for y(0) and y(1). Evaluating Eq. (1.16) at n = 0 and n = 1. we have Substituting these derived initial conditions into Eq. (1.17) we have
Solving for A and A2 we find A, = I Thus, the solution is y(n) =  (0.5)"" A2=d
+ d(0.5)" Although we have focused thus far on linear difference equations with constant coefficients, not all systems and not all difference equations of interest are linear, and not all have constant coefficients. A system that computes a running average of a signal x ( n ) over the interval [0, n ] , for example, is defined by This system may be represented by a difference equation that has timevarying coefficients: Although more complicated and difficult to solve, nonlinear difference equations or difference equations with timevarying coefficients are important and arise frequently in many applications. Solved Problems
DiscreteTime Signals
Determine whether or not the signals below are periodic and, for each signal that is periodic, determine fundamental period.

