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Unformatted text preview: DISCRETETIME e
SIGNALS
_ AND SYSTEMS Nasir Ahmed Department of Electrical Engineering
Kansas State University
Manhattan, Kansas T. Natarajan ARCO Oil and Gas Company
Dallas, Texas A PrenticeHall Company @ RESTON PUBLISHING COMPANY, INC.
Reston, Virginia Elemnts
ofDi/i‘érence
Equations In this chapter we present an elementary discussion of linear difference
equations with constant coefﬁcients. Our motivation for doing so is that
such difference equations will be used in Chapter 5 to describe and
analyze discretetime (DT) systems. Two methods for solving this class
of difference equations will be included in this chapter, while a third
method will be discussed in Chapter 3. 2.1 INTRODUCTORY REMARKS The notion of linear difference equations with constant coefficients is
best introduced by means of the simple resistive network that is shown
in Fig. 2.11, where V(n) denotes the voltage at the nth node, for —2 S
n S 3. We wish to describe this network by means of a difference equa
tion. To this end, we consider a typical section (below Fig. 2.11) of this
network, where 11, 12, and I 3 denote currents leaving the node n — 1.
Application of Kirchhoff’s current law to node n — 1 leads to the equa
tion I] + 12 + 13 = 0
Substituting for I 1, 12, and I 3 in the preceding equation, we obtain V(n—l)——V(n)+V(n—1)—V(n—2)+V(n—l)——O= 1 1 l 0 70 71 2.2 Solution of Difference Equations V(—2) V(—1) V(O) J— 1 1
89 volts :— 1 W1) W2) V(n) V(3) V(n — 2) which simpliﬁes to yield
V(n) — 3V(n — 1) + V(n — 2) = 0, 0 s n s 3 (2.1.1) Equation (2.11) is the desired difference equation that describes the
network in Fig. 2.11 in terms of its node voltages. We observe that it
is a secondorder difference equation since the voltage at node n [i.e.,
V(n)] is expressed as a linear combination of the voltages at two previous
node voltages V(n  l) and V(n —2). 2.2 SOLUTION OF DIFFERENCE EQUATIONS A logical question that arises at this point is how one can solve (2.11)
to obtain V(n). Since (2.11) represents a secondorder difference equa
tion, we would require two known voltages, say V(2) and V(— 1), to
obtain the rest. To illustrate, V(2) = 89 volts
and V(— 1) = 34 volts Then a simple procedure for obtaining the remaining V(n) for O S n <
3 would be a recursive method, since (2.11) implies that V(n) = 3V(n — 1) — V(n — 2), O s n s 3 (2.21) 72 ELEMENTS OF DIFFERENCE EQUATIONS With n = O, 1, 2, and 3, (2.21) yields the desired voltages to be as
follows: V(O) = 3V(—1) — V(—2) = 13 volts
V(l) = 3V(0) — V(—1) = 5 volts
V(2) = 3V(l)  V(O) = 2 volts and V(3) = 3V(2) — V(l) = 1 volt We shall refer to the preceding scheme as the recursive method
for solving difference equations. It is observed that, although this method
yields each V(n) in a simple recursive manner, it does not provide a
closedform solution, that is, a solution which yields V(rz) without having
to ﬁrst compute V(O), V(l), . . . , V(rt — 1). If a closedform solution is
desired, one can solve difference equations using the method of unde
termined coefﬁcients, which parallels the classical method of solving
linear differential equations with constant coefﬁcients. Method‘of Undetermined Coefficients We illustrate this method via examples. Suppose we seek the general
solution of the secondorder difference equation y(n) — gym — 1) + gym — 2) = 5‘", n 2 0 (222) with initial conditions y(—2) = 25 and y(—— 1) = 6. In (2 .22), y(rt) may be interpreted as the response (output) of a
DT system to the input (forcing) function 5‘" for rt 2 0, where rt is a
time index. It is apparent that (222) is a secondorder difference equa
tion since it expresses the output y(rt) at time rt as a linear combination
of two previous outputs y(n — 1) and y(rt — 2). The general (or closedform) solution y(n) of (2 .22) is obtained in
three steps that are similar to those used for solving secondorder dif
ferential equations. They are as follows: Obtain the complementaiy solution yc(n) in terms of two arbitrary con
stants cl and c2. Obtain the particular solution y,,(rz), and write
y(n) = yam) + yp(n) = f(cl,02) + yp(n) (2.23) where yc(n) = f(cl, c2) implies that yc(rz) is a function of Cl and c2. 73 2.2 Solution of Difference Equations 3. Solve for CI and 02 in (2.23) using two given initial conditions. In what follows, we elaborate on the preceding steps. STEP 1. We assume that the complementary solution yc(n) has
the form yc(n) = cla’f + c2613 (2.24) where the ai are real constants.
' Next substitute y(n) = a“ in the homogeneous equation to get n i n—l .1 11—2  y
a 6a + 6a — 0 (2.25) Dividing both sides of (2.25) by a"‘2, we obtain 5a +1
2 — — — =
a 6a +6 0
1
°r (a ‘ 5X 3) = 0
which yields the characteristic roots 1 1
a1 = 5 and a2 = 3 Thus the complementary solution is
yc(n) = 012‘" + 023‘" where cl and cz are arbitrary constants. STEP 2. The particular solution yp(n) is assumed to be yp(n) = 035 —n since the forcing function is 5 "; see (2. 2 2).
Substitution of y(n) = yp(n)= C35 " in (2. 2 2) leads to 5 1
C3[5_"  (8)5—01—1) + <g>5_(""2)] = 5‘" 74 ELEMENTS OF DIFFERENCE EQUATIONS Dividing both sides of this equation by 5'", we obtain 5 l 2 _
C3[l  (6)5 + (6)5] —l which implies that 03 = 1. Thus y(n) yam) + yp(n) (2.26) c12‘" + 023“" + 5‘” STEP 3. Since the initial conditions are
y(——2) = 25 and y(—1)= 6
(2.26) yields the simultaneous equations 4C1 + 9C2 = 0 (22_7) and 201 + 302 = 1 Solving (2.2—7) for CI and c2, we obtain 3 2
01:5 and Cz=§ Thus the desired general solution is given by (2.26) to be
3 _ 2 _ _
y(n) = 5(2 ") — 3(3 ") + 5 ", n 2 0 (2.28) As mentioned earlier, y(n) can be interpreted as the output of a
DT system when it is subjected to the exponential input (forcing func tion) 5‘”, which is the righthand side of the given difference equation
in (2.22). RULES FOR CHOOSING PARTICULAR SOLUTIONS. As is the case
with the solution of differential equations, there are a set of rules one
must follow to form appropriate particular solutions while solving dif
ference equations, as summarized in Table 2.21. For example, the form
of the particular solution related to the difference equation in (2.22)
was 035‘", which agrees with line 3 of Table 2.21. We will illustrate
the use of this table by means of more examples. 75 2.2 Solution of Difference Equations Table 2.21 Rules for Choosing Particular Solutions Terms in forcing function Choice of particular solution* 1. A constant c; c is a constant co + cm + c2142 +    + cm"; 2. b n"‘ b is a constant
1 ’ 1 the c, are constants 3. bzdi"; b2 and d are constants Proportional to di"
4. 17; cos (mo) b3 and b4 are c1 sin (nw) + 0; cos (mu)
5. [)4 sin (mo) constants *If a term in any of the particular solutions in this column is a part of the
complementary solution, it is necessary to modify the corresponding choice
by multiplying it by 11 before using it. If such a term appears r times in the
complementary solution, the corresponding choice must be multiplied by 11’. Example 2.21: Solve the secondorder difference equation y<n> —— gym — 1) + gym — 2)=1+ 3", n a 0 (2.29) with the initial conditions y(—2) = 0 and y(— 1) = 2.
Solution: The solution consists of three steps. STEP 1. Assume the complementary solution as yc(n) = cla’f +
czag. Substituting y(n) = a" in the homogeneous counterpart of (2.29),
we obtain the characteristic equation 3 1
2 — — — =
a 2 a + 2 0
the roots of which are a1 = g and a2 = 1.
Thus
yc(n) = c12”" + 021" = c12"1 + c2 (2.210) STEP 2. To choose an appropriate particular solution, we refer
to Table 2.21. From the given forcing function and lines 1 and 3 of Table
2.2—1, it follows that a choice for the particular solution is 03 + c43‘“.
However, we observe that this choice for the particular solution and
yc(n) in (2.210) have common terms, each of which is a constant; that 76 ELEMENTS OF DIFFERENCE EQUATIONS is, 03 and 02, respectively. Thus in accordance with the footnote of Table
2.21, we modify the choice 03 + c43‘" to obtain yp(n) = Cs" + 043‘“ (2211) Next, substitution of yp(n) in (2 .21 1) into (2.29) leads to cn+c3‘"——§cn+g —2 3‘"
3 4 2 3 203 204
1 9 _
+2c3n—c3+§c43‘”=3 "+1 (2.212) From (2.212) it follows that 1
2C3 = 1
9 9 ‘71 —— —H
and c4[1— 2 + §]3 — 3 which results in Thus, combining (2.210) and (2211), we get
y(n) = 012*" + c; + 2n + 3"" (2.213) STEP 3. To evaluate cl and c2 in (2213), the given initial con
ditions are used; that is, y(32) = 0 and y(— 1) = 2. This leads to the
simultaneous equations 4C1 + CZ = —5
2c1 + 02 = 1
Solving, we obtain c1 = —3 and 02 = 7, which yields the desired solution
as
y(n) = (—3)2'" + 7 + 2n + 3—", n 2 0 Example 2.22: Find the general solution of the ﬁrstorder difference
equation 77 2.2 Solution of Difference Equations y(n) — 0.9y(n — 1) = 0.5 + (O.9)"“, n 2 0 (2.214)
with y(—1) = 5. Solution:
STEP 1. Substituting y(n) = a" in the homogeneous equation y(n) ~ 0.9y(n — 1) = 0
we obtain yc(n) = c1(0.9)” (2.215) since we are dealing with a ﬁrstorder difference equation. STEP 2. From the forcing function in (2.214), the complementary
solution in (2.215), and lines 1 and 3 of Table 2.21, it follows that yp(n) = czn(0.9)" + c3
Substitution of y(n) = yp(n) in (2.214) results in
c3 + czn(0.9)" — 0.902(n — 1)(O.9)"‘l —— 0.9c3 = 0.5 + (0.9)"—1 which leads to 0.1c3 = 0.5
and (0.9)"c; = (0.9)11—1
Thus we have
10 c3 = 5 and cz = —9— which implies that
10
yp(n) = 3 n(0.9)" + 5 (2.216) Combining (2.215) and (2.216), we get y(n) = 01(09)" + 199n(0.9)" + 5 (2.217) 78 ELEMENTS OF DIFFERENCE EQUATIONS STEP 3. From (2.217) and the initial condition y(— 1) = 5, it 10 . . .
follows that (:1 = —9. Hence the deSIred solut10n can be wr1tten as y(n) = (n + l)(0.9)"—1 + 5, n 2 0 Example 2.23: Find the general solution of the secondorder difference
equation y(n) — l.8y(n — 1) + 0.81y(n — 2) = 2‘", n 2 0 (2.218) Leave the answer in terms of unknown constants, Which one can eval
uate if the initial conditions are given. Solution: STEP 1. With y(n) = a" substituted into the homogeneous coun
terpart of (2218), we obtain a2  1.8a + 0.81 = 0
which results in the repeated roots
611 = a2 = 0.9 Thus, as in the case of differential equations, we consider the comple
mentary solution to be yc(n) = 01(09)“ + Czn(0.9)" (2.219) STEP 2. From the given forcing function in (2218), yc(n) in (2 .2
19), and line 3 of Table 2.21, it is clear that Mn) = c32‘" (2.220)
Substitution of (2220) in (2218) leads to €3[1  (3X6) + (3)(24)]2‘" = 2‘" which yields c3 = 15. Thus the desired solution is given by (2219) and 16
(2220) to be 3’01) = 01(09)” + cm(0.9)" + 6%)2—11 79 2.2 Solution of Difference Equations where c1 and cz can be evaluated if two initial conditions are speciﬁed. Example 2.24: Find the particular solution for the ﬁrstorder difference
equation n77 y(n) — O.5y(n — 1) = sin (7), n 2 0 (2.221) Solution: Since the forcing function is sinusoidal, we refer to line 5 of
Table 2.2—1 and choose a particular solution of the form YMT 7’27T y,,(n) = 01 sin (7) + 0; cos (5) (2.222) Substitution of y(n) = yp(n) in (2221) leads to . 1177 1117 , n — 1 7r
c1 Sin (—2—) + cz cos (7) — 0.501 Sln [L2—)] — 0.5c2 cos [Gt—3.1.17] = sin (112—77) (2.223) We now use the following identities: sin [—(n — D77] sin ("—71 — I) = —cos (y)
' 2 2 2 2 (2.2—24) cos [#01 _ 1”] cos ("—7 — 3—7) = sin (:13)
2 2 2 2 Substituting (2224) in (2223), we obtain (Cl — 0502) sin (”777) + (0.501 + c2) cos (2271) = sin (fl—271) which yields the simultaneous equations cl — 0.5c2 = 1 (2.245)
0.501 + cz = 0 30 ELEMENTS OF DIFFERENCE EQUATIONS The solution of (2225) yields cl = g and 02 = :Z. Hence the desired result is given by (2222) to be (VD—ﬂs'n 11—77 —2 "—77 n>0
y” ‘5‘ 2 5“” 2' " Example 2.25: In Example 2.23, suppose the forcing function is (0.9)" instead of 2‘", n 2 0. What would be the appropriate choice for the
particular solution? Solution: The complementary solution is given by (2219) to be
yc(n) = c1(0.9)“ + c2n(0.9)" Since the forcing function is (0.9)", line 3 of Table 2.21 implies that a
choice for the particular solution is c3(0.9)". However, since this choice
and the preceding yc(n) have a term in common, we must modify our
choice according to the footnote of Table 2.21 to obtain c3n(0.9)". But
this choice again has a term in common with yc(n). Thus we apply to
the footnote of Table 2.21 once again to obtain yp(n) = c3n2(0.9)" (2 .226) which has no more terms in common with yc(n). Hence yp(n) in (2226) is the appropriate choice for the particular
solution for the difference equation in (2218) when the forcing function
is (0.9)"; that is, y(n) — 1.8y(n — 1) + O.81y(n — 2) = (0.9)", n 2 0 2.3 SUMMARY Our treatment of linear difference equations with constant coefﬁcients
in this chapter was conﬁned to ﬁrst and secondorder difference equa
tions. Higherorder difference equations of this type will be considered
in Chapters 3 and 5. Although our interest in such difference equations
is restricted to DT systems as they relate to electrical engineering, they
have a variety of applications in diverse areas such as economics, psy
chology, and sociology. The interested reader may refer to [3] for more
details. ...
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