1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

348 CHAPTER 9 • z -TRANSFORMS AND APPLICATIONS

Difference Equation Model

Exponential growth or decay

y[n+ l] = (1 + r)y[n]

Newton's law of cooling

y[n + 1] = aL + (1 - a)y[n]

Repeated dosage drug level

y[n+ 1] = ay[n] + b

Value of an annuity due

Solution

y[n] = Yo(l + r)n

y[n] = Yo(l -ar + L (1 - (1 - a)n)

y[n] = Yoan + o.,,,~ 11 b

y[n + 1) = (1 + r)(y[n) + P) where y[O] = O y[n] = (l+r)~+n_l p - p

Table 9.3 Some examples of first-order linear difference equat ions.

9.1.7 M ethods for Solving First-Order Difference

Equations

Consider the first-order linear constant coefficient difference equation (LCCDE):

y[n + l] - ay[n) = x [n ) with the initial condition y[O) =YO·

Trial solution method

First, solve the homogeneous equation Yn[n+ 1)-ay,.[n) = 0 and get Y1i[n] = c1an.
Then use a trial solution that is appropriate for the sequence x[n] on the right
side of the equation and solve to obtain a particular solution yp[n]. Then the
general solution is

y[n) = Y1i[n) + Yp[n).

The shortcoming of this method is that an extensive list of appropriate trial solu-
tions must be available. (Details can be found in difference equations textbooks.)
vVe will emphasize techniques that use the z-transform.

z-transform method

(i) Use the time forward property 3[y[n + l)] = z(Y(z) -yo). Take the

z-transform of each term and get

z(Y(z) - yo) -aY(z) = X(z).

(ii) Solve the equation in (i) for Y(z).