barcode generator for ssrs The solution to Eq. (1.13)may be found by assuming a solution of the form in Software

Printer Code-128 in Software The solution to Eq. (1.13)may be found by assuming a solution of the form

The solution to Eq. (1.13)may be found by assuming a solution of the form
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Substituting this solution into Eq. (1.13) we obtain the polynomial equation
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The polynomial in braces is called the characteristic polynomial. Because it is of degree p , it will have p roots, which may be either real or complex. If the coefficients a ( k ) are real-valued, these roots will occur in complexconjugate pairs (i.e., for each complex root z , there will be another that is equal to z f ) . If the p roots z i are distinct, zi # z k fork # i , the general solution to the homogeneous difference equation is
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CHAP. 11
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SIGNALS AND SYSTEMS
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where the constants Ak are chosen to satisfy the initial conditions. For repeated roots, the solution must be modified as follows. If z I is a root of multiplicity m with the remaining p - m roots distinct, the homogeneous solution becomes
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yh(n) = ( A I
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+ A2n + . . . + ~ , n ~ - ' ) z ;+
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(1.15)
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k=m+l
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 1-2 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 1-2 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 1-2 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 time-varying coefficients:
Although more complicated and difficult to solve, nonlinear difference equations or difference equations with time-varying coefficients are important and arise frequently in many applications.
Solved Problems
Discrete-Time Signals
Determine whether or not the signals below are periodic and, for each signal that is periodic, determine fundamental period.
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