100
Discrete-Time Signals and Systems
Chap. 2
Example 2.43
Determ ine if the linear tim e-invariant recursive system described by the difference
equation given in (2.4.7) is stable.
Solution
i*(n)l <
Let us assume that the input signal x(n) is bounded in amplitude, that is,
< oc for all n > 0. From (2.4.8) we have
j v{n J | < |a',+l_y(—1)| + ^
akx(n - k) ,
n>0
If n is finite, the bound M v is finite and the output is bounded independently of the
value of a. However, as n -*• oo, the bound My remains finite only if |aj < 1 because
|a |B -»■ 0 as n -*• oc. Then M y = Ms j(\ - |o|).
Thus the system is stable only if \a\ < 1.
For the sim ple first-order system in E xam ple 2.4,3, w e w ere able to express
the con d ition for B IB O stability in term s o f the system param eter a. nam ely \a\ < 1.
W e should stress, how ever, that this task b ecom es m ore difficult for higher-order
system s. F ortunately, as we shall see in su bsequent chapters, other sim ple and
m ore efficient tech n iqu es exist for investigating the stability o f recursive system s.
2.4.3 Solution of Linear Constant-Coefficient Difference
Equations
G iven a linear con stan t-coefficien t d ifferen ce eq u ation as the in p u t-o u tp u t rela­
tionship describing a linear tim e-invariant system , our objective in this subsection
is to determ ine an explicit exp ression for the output y{n). T h e m eth od that is
d ev elo p ed is term ed the direct m e t h o d . A n alternative m eth od based on the ztransform is described in Chapter 3. For reasons that will b eco m e apparent later,
the z-transform approach is called the indirect m e th o d .
Basically, the goal is to determ ine the output y(n ), n > 0, o f the system given
a specific input x ( n ), n > 0, and a set o f initial con d ition s. T h e direct solution
m ethod assum es that the total solu tion is the sum o f tw o parts:
y ( n ) = y h(n) + }'r (n)
T he part y/,(n) is know n as the h o m o g e n e o u s or c o m p l e m e n ta r y solu tion , w hereas
yp(n) is called the particular solution.
The homogeneous solution of a difference equation. W e begin the
problem o f solving the linear con stan t-coefficien t d ifferen ce eq u ation given by
Sec. 2.4
Discrete-Time Systems Described by Difference Equations
101
(2.4.13) bv assu m in g th a t th e in p u t x (n ) = 0. T h u s w e will first o b ta in th e so lu tio n
to th e h o m o g e n e o u s di ff erence equati on
\
J 2 a ky ( n - k ) = 0
*=(>
(2.4.14)
T h e p ro c e d u re fo r solving a lin e a r c o n s tan t-co efficien t d ifferen ce e q u a tio n
d irectly is very sim ilar to th e p ro c e d u re fo r solving a iin e a r c o n stan t-co efficien t
d iffe re n tia l e q u a tio n . B asically, we assum e th a t th e so lu tio n is in th e fo rm o f an
e x p o n e n tia l, th a t is.
yi,{n) = a"
(2.4.15)
w h ere th e su b scrip t h o n \ {n) is used to d e n o te th e so lu tio n to th e h o m o g e n e o u s
d ifferen ce e q u a tio n . If w e su b stitu te this assu m ed so lu tio n in (2.4.14), w e o b tain
th e p o ly n o m ial e q u a tio n
=0
fc=U
or
k" A (k* + ci)k^ ! + a2k^ * + • • - +
q n —\ k
+ aw) = 0
(2.4.16)
T h e p o ly n o m ial in p a re n th e se s is called th e characteristic p o l y n o m i a l o f the
system . In g e n e ra l, it h as N roots, w hich we d e n o te as Xi. k 2.........k ^ . T h e ro o ts
can b e real o r co m p lex v alued. In p ra c tic e the co efficien ts a \ , a 2........ are usually
re a l. C o m p le x -v a lu e d ro o ts o ccu r as c o m p le x -c o n ju g a te p airs. S om e o f th e N ro o ts
m ay be id en tical, in w hich case w e have m u ltip Je -o rd e r ro o ts.
F o r th e m o m e n t, let us assum e th a t th e ro o ts a re d istin ct, th a t is, th e re are
n o m u ltip le -o rd e r ro o ts. T h e n th e m o st g e n eral so lu tio n to th e h o m o g e n e o u s
d iffe re n c e e q u a tio n in (2.4.14) is
yh(n) = C \ k \ + C2k 2 + • ■■+ C^ k ^ ,
(2.4.17)
w h ere C i. C 2.........C s are w eig h tin g coefficients.
T h e se co efficien ts are d e te rm in e d fro m th e in itial c o n d itio n s specified fo r th e
sy stem . Since th e in p u t jr(n) = 0. (2.4.17) can b e u sed to o b ta in th e z ero- i nput
response o f th e system . T h e follow ing ex am p les illu stra te th e p ro c e d u re .
Example 2.4.4
D eterm ine the hom ogeneous solution of the system described by the first-order dif­
ference equation
_v(n) + a\y(rt — 1) = x(n)
Solution
The assumed solution obtained by setting x(n) = 0 is
y*(n) = V
(2.4.18)
D iscrete-Time Signals and System s
102
Chap. 2
When we substitute this solution in (2.4.18), we obtain [with x(n) = 0]
X -f-12]An ^ = 0
A" *(A + ai) = 0
A = —O)
Therefore, the solution to the hom ogeneous difference equation is
= CA" = C ( - a ,)"
(2.4.19)
The zero-input response of the system can be determ ined from (2.4,18) and
(2.4.19). With x(n) = 0, (2.4.18) yields
v(0) =
-a, v ( - l )
On the other hand, from (2.4,19) we have
v*(0) = C
and hence the zero-input response of the system is
Vii(«) = (—£i)n+1 v{—1)
n >0
(2.4.20)
With a = —ai, this result is consistent with (2.4,11) for the first-order system, which
was obtained earlier by iteration of the difference equation.
Example 2AS
Determ ine the zero-input response of the system described by the homogeneous
second-order difference equation
y(«) - 3y(n - 1) - 4y(n - 2) = 0
(2.4.21)
Solution First we determ ine the solution to the homogeneous equation. We assume
the solution to be the exponential
yh(n) = X"
U pon substitution of this solution into (2.4.21), we obtain the characteristic equation
a"
- 3xn- ' - 4 r ~ 2 = o
A"-2(X2 —3A —4) = 0
Therefore, the roots are X = - 1 , 4, and the general form of the solution to the
homogeneous equation is
}7i(n) = CjX" + C 2X2
(2.4.22)
= C1( - i r + C2(4)"
The zero-input response of the system can be obtained from the homogenous
solution by evaluating the constants in (2.4.22), given the initial conditions y (—1) and
y ( ~ 2). From the difference equation in (2.4.21) we have
y(0) = 3y(—1) + 4y(—2)
y(l) = 3v(0) + 4_y(—1)
= 3 [3 y (-l)+ 4y(-2)] + 4y(-l)
= 13y(—1) + l 2y( —2)
Sec. 2.4
Discrete-Time Systems Described by Difference Equations
103
On the other hand, from (2.4.22) we obtain
v(0> = C) + C;
V( 1 I = -C l +4C;
By equating these two sets of relations, we have
C i + C ; = 3_v( —1) + 4y(—2)
- C , + 4 C ; = 13 v{ —1) + 12 y( —2)
The solution of these two equations is
Ci = —7 v( —1 ) + ? v ( - 2 )
C; = X .V (-l) + T.v( -2 )
Therefore, the zero-input response of the system is
v,i(n) = [—5 v( —1 >- i v(-2)](-l)''
(2.4.23)
n >0
+ [ x . v ( - l » + t.v(-2)](4>"
For example, if v{—2) = 0 and y ( - l ) = 5. then C| = —1, C: = 16. and hence
y-,i(H) = ( -1 ) " * 1 + (4)"+:
n >0
T h e se e x am p les illu strate the m e th o d fo r o b ta in in g the h o m o g e n e o u s so lu tio n
and th e z e ro -in p u t re sp o n se o f th e system w hen the ch a ra c te ristic e q u a tio n co n tain s
d istin ct ro o ts. O n th e o th e r h an d , if th e c h a ra c te ristic e q u a tio n co n ta in s m u ltip le
ro o ts, th e form o f th e so lu tio n given in (2.4.17) m ust be m odified. F o r ex am p le, if
ai is a ro o t o f m u ltip licity m, th en (2.4.17) b eco m es
\h(ii) = Ci A.'.' + C2/iX’! + C v;:a',' + ■■• + Cm>7n'~lX'!
(2.4.24)
+ Cm+\)"m+\ + ■• • + C^Kl
The particular solution of the difference equation. T h e p a rtic u la r so ­
lu tio n \ p (n) is re q u ire d to satisfy th e d ifferen ce e q u a tio n (2.4.13) fo r th e specific
in p u t signal x (n ). n > 0. In o th e r w ords, y p(n) is any so lu tio n satisfying
N
M
' Y ^ a ky p (n - k) — ' Y ^ b kx ( n - k)
A-0
*=0
an — 1
(2.4.25)
T o solve (2.4.25). w e assu m e fo r yp (n), a fo rm th at d e p e n d s on th e fo rm o f th e
in p u t j:(h ). T h e fo llow ing ex am p le illu strates the p ro c e d u re .
Example 2.4.6
D eterm ine the particular solution of the first-order difference equation
y(n) + fliy(n - 1) = x(n).
| f l i | <l
when the input x(n) is a unit step sequence, that is.
x(n) = u (n )
(2,4.26)
104
Discrete-Time Signals and System s
Chap. 2
Solution Since the input sequence x(n) is a constant for n > 0. the form of the solu­
tion that we assume is also a constant. Hence the assumed solution of the difference
equation to the forcing function x(n), called the particular solution of the difference
equation, is
vr (n) = Ku(n)
where A- is a scale factor determined so that (2.4.26) is satisfied. U pon substitution
of this assumed solution into (2.4.26). we obtain
Ku i n ) + d]Ku{,n — 1) = u ( n )
To determine K, we must evaluate this equation for any n > 1. where none of the
terms vanish. Thus
K -t- O] K — 1
1
K = -------l+^i
Therefore, the particular solution to the difference equation is
v (n) = --------u(n)
'’
1 +fii
(2.4.27)
In th is ex am p le, th e in p u t * (« ). n > 0. is a c o n s ta n t an d th e fo rm assu m ed
for th e p a rtic u la r so lu tio n is also a c o n sta n t. If x{n) is an e x p o n e n tia l, w e w ould
assu m e th a t th e p a rtic u ia r so lu tio n is also an e x p o n e n tia l. If x {n) w e re a sin u so id ,
th e n >■/,(«) w o uld also be a sinusoid. T h u s o u r assu m ed fo rm fo r th e p a rtic u la r
so lu tio n tak es th e basic fo rm of th e signal x{n). T a b le 2.1 p ro v id e s th e g en eral
fo rm o f th e p a rtic u la r so lu tio n for sev eral ty p es of excitatio n .
Exam ple 2.4.7
Determ ine the particuiar solution of the difference equation
v(n) = | y(n - 1) - £ v(w - 2) + x(n)
when the forcing function x(n) = 2". n > 0 and zero elsewhere.
TABLE 2.1 GENERAL FORM OF THE PARTICULAR
SOLUTION FOR SEVERAL TYPES OF INPUT
SIGNALS
Input Signal,
x(n)
Particular Solution,
vr (n)
A (constant)
AM"
AnM
K
KM"
Kan” + K ^ " - ' 1 + . . . + KM
A cos Wf>n
A sin a>on
Kj cos toon -I- K2 sin won
AnnM
An(Ki)nM+ K\ttM 1
Sec. 2.4
Discrete-Time Systems Described by Difference Equations
Solution
105
The form of the particular solution is
yp (n ) = K2n
n >0
Upon substitution of yP(n) into the difference equation, we obtain
K2"u( n) =
-
1)
- I K 2 n- 2u(n - 2 ) + 2 " « ( « )
To determ ine the value of K, we can evaluate this equation for any n > 2, where
none of the terms vanish. Thus we obtain
4A" = \ ( 2K) - i f f + 4
and hence K =
Therefore, the particular solution is
yn(n) = ^2"
n > 0
W e h av e n o w d e m o n s tra te d how to d e te rm in e th e tw o c o m p o n e n ts o f the
so lu tio n to a d ifferen ce e q u a tio n w ith c o n s ta n t coefficients. T h e se tw o c o m p o n e n ts
are th e h o m o g e n e o u s so lu tio n and th e p a rtic u la r so lu tio n . F ro m th e s e tw o co m ­
p o n e n ts, we c o n s tru c t the to tal so lu tio n fro m w hich w e can o b ta in th e ze ro -sta te
re sp o n se .
The total solution of the difference equation. T h e lin e a rity p ro p e rty o f
th e lin ear c o n stan t-co efficien t d ifferen ce e q u a tio n allow s u s to a d d th e h o m o g e ­
n e o u s so lu tio n an d the p a rtic u la r so lu tio n in o rd e r to o b ta in th e total solution. T h u s
v(«) =
y h(n)
+ \ p( n)
T h e r e s u lta n t sum v(n) co n tain s th e c o n s ta n t p a r a m e te r s {C/} e m b o d ie d in th e
h o m o g e n e o u s so lu tio n c o m p o n e n t >v,(n). T h e se c o n s ta n ts can be d e te rm in e d to
satisfy th e in itial co n d itio n s. T h e follow ing ex a m p le illu stra te s th e p ro c e d u re ,
Exam ple 2.4.8
D eterm ine the total solution y(/i), n > 0, to the difference equation
y(n) +a i y ( n - 1) = x(n)
(2.4.28)
when x( n) is a unit step sequence [i.e., x(n) = «(«)] and y (—1) is the initial condition.
Solution
From (2.4,19) of Example 2.4.4, the hom ogeneous solution is
y*(n) = C(-fli)"
and from (2.4.26) of Example 2.4.6, the particular solution is
Consequently, the total solution is
y(n) = C ( - a i)" + — -—
1 + fli
n >0
where the constant C is determ ined to satisfy the initial condition y ( - l ) .
(2.4.29)
106
Discrete-Time Signals and Systems
Chap. 2
In particular, suppose that we wish to obtain the zero-siate response of the
system described by the first-order difference equation in (2.4.28). T hen we set
y( —1) = 0. To evaluate C. we evaluate (2.4.28) at n = 0 obtaining
y ( 0 ) + Oi y ( — 1 ) =
1
y(0) = 1
On the other hand, (2.4.29) evaluated at n = 0 yields
v(0) = C 4- —i —
1 + ax
Consequently.
1
C + -------- = 1
1 + «i
ai
1 + ai
c
Substitution for C into (2.4.29) yields the zero-state response of the system
l-f-fli)" * '
Vi*(/i) = ---- —-----—
1 + «i
n >0
If we evaluate the param eter C in (2,4.29) under the condition that y( —1) ^ 0. the
total solution will include the zero-input response as well as the zero-state response
of the system. In this case (2.4.28) yields
y(0) + a]V( —1) = 1
y (0) = —fliy( —1) 4- 1
On the other hand. (2.4.29) yields
]
1 + U\
By equating these two relations, we obtain
C + — ^— = —ai v(—1) 4- 1
1 4 - a,
C = -g ] v( 1) +
- —
1 4- a ]
Finally, if we substitute this value of C into (2.4.29). we obtain
y(n) = (—a , r +1y(—1) + -—
=
+
^ —
1 + a)
n >0
(2.4.30)
Vzs( n)
W e o b se rv e th a t th e system re sp o n se as given by (2.4.30) is c o n s iste n t w ith
th e re sp o n se y ( n) given in (2.4.8) for th e first-o rd e r system (w ith a = - o i ) . w hich
was o b ta in e d by solv in g th e d ifferen ce e q u a tio n iterativ e ly . F u rth e rm o r e , we n o te
th a t th e v alu e o f th e c o n s ta n t C d e p e n d s b o th o n th e initial co n d itio n y (—1) an d
o n th e ex citatio n fu n ctio n . C o n se q u e n tly , th e v alue o f C in flu en ces b o th th e zeroin p u t re sp o n se a n d th e z e ro -sta te re sp o n se . O n th e o th e r h a n d , if w e w ish to
Sec. 2.4
Discrete-Time Systems Described by Difference Equations
107
obtain the zero-state resp onse only, w e sim ply solve for C under th e con d ition
that y ( - l ) = 0, as d em onstrated in E xam ple 2.4.8.
W e further ob serve that the particular solution to the d ifferen ce equation can
b e o b tain ed from the zero-state resp onse o f the system . In d eed , if |o]| < 1, which
is the con d ition for stability o f the system , as w ill be show n in S ection 2.4.4, the
lim iting valu e o f >’zs(«) as n approaches infinity, is the particular solu tion , that is,
1
» ( n ) = lim >zs(«) = --------n-*Oo
1+ai
Since this com p onent o f the system response d o es not go to zero as n approaches
infinity, it is usually called the steady-state re spons e o f the system . T his response
persists as lon g as the input persists. T he com p onent that d ies out as n approaches
infinity is called the transient re sponse o f the system .
Example 2.4.9
D eterm ine the response y(n), n > 0, of the system described by the second-order
difference equation
v(n) - 3y(n - 1) - 4v(n - 2) = x( n ) + 2x(n - 1)
(2.4.31)
when the input sequence is
x(n) =4"u(n)
Solution We have already determ ined the solution to the homogeneous difference
equation for this system in Example 2.4.5. From (2.4.22) we have
y*(n) = C , ( - l ) n + C 2(4)n
(2.4.32)
The particular solution to (2.4.31) is assumed to be an exponential sequence of the
same form as x(n). Normally, we could assume a solution of the form
yp(n) = K(4,)au(n)
However, we observe that » ( n ) is already contained in the hom ogeneous solution,
so that this particular solution is redundant. Instead, we select the particular solution
to be linearly independent of the terms contained in the homogeneous solution. In
fact, we treat this situation in the same manner as we have already treated multiple
roots in the characteristic equation. Thus we assume that
yp(n) = Kn{4)"u(n)
(2.4.33)
Upon substitution of (2.4.33) into (2.4.31), we obtain
Kn(4)"u(n) - 3 K ( n - l)(4 ),' - 1u(n - 1) - 4 K{n - 2)(4)n- 2u(n - 2)
= (4)"u(n) + 2(4)"-1 m(/i —1)
To determ ine K , we evaluate this equation for any n > 2, where none of the
unit step term s vanish. To simplify the arithmetic, we select n = 2, from which we
obtain K = | . Therefore,
yP{n) « |rt(4)n«(/l)
(2.4.34)
108
Discrete-Time Signals and System s
Chap. 2
The total solution to the difference equation is obtained by adding (2.4.32) to
(2.4.34). Thus
y(«) — C ](—1)" + C;(4)n + |w(4)"
n > 0
(2.4.35)
where the constants C\ and C2 are determined such that the initial conditions are
satisfied. To accomplish this, we return to (2.4.31), from which we obtain
v(0) = 3 y (- l) + 4y (—2) + 1
y (1) = 3v(0) + 4_v(-l) + 6
= 1 3 y (- l) + 12y(-2) + 9
O n the other hand, (2.4.35) evaluated at n = 0 and n = 1 yields
y(0) = Ci + C2
y (l) = -C] + 4C2 + f
W e can now equate these two sets of relations to obtain C\ and C 2. In so doing, we
have the response due to initial conditions y (—1) and y (—2) (the zero-input response),
and the zero-state or forced response.
Since we have already solved for the zero-input response in Exam ple 2.4.5. we
can simplify the computations above by setting v (—1) = y (—2) = 0. Then we have
C, + C; = 1
- C l + 4 C2 + f
=9
Hence C : = —^ and C 2 =
Finally, we have the zero-statc response to the forcing
function
= (4)"«(n) in the form
vB(n) =
+5<4)" +H 4)"
nZ0
(2.4.36)
The total response of the system, which includes the response to arbitrary initial
conditions, is the sum of (2.4.23) and (2.4.36).
2.4.4 The Impulse Response of a Linear Time-Invariant
Recursive System
T h e im p u lse re sp o n se o f a lin e a r tim e -in v a ria n t sy stem w as p re v io u sly defin ed as
th e re sp o n se o f th e sy stem to a u n it sa m p le e x citatio n [i.e., x (n ) = <5(n)]. In th e
case o f a recu rsiv e sy stem , h ( n ) is sim ply e q u a l to th e z e ro -s ta te re sp o n s e o f the
system w h en th e in p u t j:(n ) = <5(n) an d th e system is initially re la x e d .
F o r ex am p le, in th e sim ple first-o rd e r re c u rsiv e system g iv en in (2.4.7), th e
z e ro -sta te re sp o n se g iv en in (2.4.8), is
n
(2.4.37)
*=o
W ith jr(n) = i ( « ) is substituted into (2.4.37), we obtain
n
yzs(n) = Y 2 a k8(n - k)
=
an
n > 0
Sec. 2.4
Discrete-Time Systems Described by Difference Equations
109
H e n c e th e im p u lse re sp o n se o f th e first-o rd e r rec u rsiv e system d e sc rib e d by
(2.4.7) is
h(n) = a nu(n)
(2.4.38)
as in d ic a te d in S ectio n 2.4.2.
In th e g e n e ra l case o f an a rb itra ry , lin e a r tim e -in v a ria n t rec u rsiv e system , the
z e ro -s ta te re sp o n s e ex p re sse d in te rm s o f th e co n v o lu tio n su m m a tio n is
>’zs(n) — ^ 2 h ( k ) x ( n — k)
n > 0
(2.4.39)
i=0
W h e n th e in p u t is an im p u lse [i.e., * (« ) = <5(«)], (2.4.39) re d u c e s to
Vzs(n) = h(n)
(2.4.40)
N o w , let us c o n s id e r th e p ro b le m o f d e te rm in in g th e im p u lse re sp o n se h i n ) given a
lin e a r c o n s tan t-co efficien t d ifferen ce e q u a tio n d e sc rip tio n o f th e system . In term s
o f o u r d isc u ssio n in th e p re c e d in g su b sectio n , we h av e e s ta b lis h e d th e fact th a t the
to ta l re sp o n s e o f th e sy stem to any e x citatio n fu n ctio n co n sists o f th e sum o f tw o
so lu tio n s o f th e d ifferen ce e q u atio n : th e so lu tio n to the h o m o g e n e o u s e q u atio n
p lu s th e p a rtic u la r so lu tio n to th e e x c itatio n fu n ctio n . In th e case w h ere the exci­
ta tio n is an im p u lse, th e p a rtic u la r so lu tio n is z e ro , since x ( n ) = 0 for n > 0. th at is.
yp (n) = 0
C o n s e q u e n tly , th e re sp o n se o f th e system to an im p u lse co n sists only o f th e so lu ­
tio n to th e h o m o g e n e o u s e q u a tio n , w ith the ( Q ) p a r a m e te r s e v a lu a te d to satisfy
th e in itial c o n d itio n s d ic ta te d by th e im pulse. T h e follow ing ex am p le illu strates
th e p ro c e d u re fo r o b ta in in g h(n) given th e d ifferen ce e q u a tio n fo r the system .
Example 2.4.10
D eterm ine the impulse response h(n) for the system described by the second-order
difference equation
y(n ) — 3v(n — 1) — 4 y (« — 2) = x ( n ) + 2 x i n — 1)
(2.4.41)
Solution W e have already determ ined in Example 2.4.5 that the solution to the
hom ogeneous difference equation for this system is
^ ( n ) = C, (-1 )" + C2(4)"
n > 0
(2.4.42)
Since the particular solution is zero when x(n) = 6(n), the impulse response of the sys­
tem is simply given by (2.4.42), where C] and C2 must be evaluated to satisfy (2.4.41).
For n = 0 and n = 1, (2.4.41) yields
v(0) = 1
y (1) = 3 y (0 )+ 2 = 5
where we have imposed the conditions y (—1) = y (—2) = 0. since the system must be
relaxed. On the other hand, (2.4.42) evaluated at n = 0 and n = 1 yields
y (0) = C, + C2
y (1) = —Ci + 4C2
Discrete-Time Signals and Systems
110
Chap. 2
By solving these two sets of equations for C] and C2, we obtain
=
C: = 5
Therefore, the impulse response of the system is
h{n) = [—i ( —1)" + f (4)-}u(n)
W e m ak e th e o b se rv a tio n th a t b o th th e sim ple firs t-o rd e r recu rsiv e system
an d th e se c o n d -o rd e r recu rsiv e system h av e im pulse re sp o n se s th a t a re infinite in
d u ra tio n . In o th e r w o rds, b o th o f th e s e recu rsiv e system s a re IIR system s. In
fact, d u e to th e recu rsiv e n a tu re of th e system , an y recu rsiv e sy stem d e sc rib e d by
a lin e a r co n stan t-co efficien t d iffe re n c e e q u a tio n is an I IR system . T h e co n v erse
is n o t tru e, h o w ev er. T h a t is, n o t ev ery lin e a r tim e -in v a ria n t I I R sy stem can be
d e sc rib e d by a lin e a r c o n s tan t-co efficien t d ifferen ce e q u a tio n . In o th e r w ords,
recu rsiv e sy stem s d e sc rib e d by lin ear co n sta n t-c o e ffic ie n t d iffe re n c e e q u a tio n s are
a su bclass o f lin e a r tim e -in v a ria n t I I R system s.
T h e ex ten sio n o f th e a p p ro a c h th a t w e h av e d e m o n s tra te d fo r d e te rm in ­
ing th e im p u lse re sp o n se o f th e first- a n d s e c o n d -o rd e r system s, g en e ra liz e s in a
stra ig h tfo rw a rd m a n n e r. W h e n th e system is d e sc rib e d by an A 'th -o rd e r lin e a r
d ifferen ce e q u a tio n o f th e ty p e given in (2.4.13), th e so lu tio n o f th e h o m o g e n e o u s
e q u a tio n is
*=i
w h en th e ro o ts {a*} o f th e c h a ra c te ristic p o ly n o m ial are d istin ct. H e n c e th e im pulse
re sp o n se o f th e sy stem is id en tical in fo rm , th a t is.
(2.4.43)
w h ere th e p a ra m e te rs {Ctl are d e te rm in e d by se ttin g th e initial c o n d itio n s v ( —1) =
. . . = y ( - N ) = 0.
T h is fo rm o f h{n) allow s us to easily re la te th e stab ility of a system , d escrib ed
by an N th -o rd e r d iffe re n c e e q u a tio n , to th e values o f th e ro o ts o f th e ch a ra c te ristic
p o ly n o m ial. In d e e d , since B IB O sta b ility re q u ire s th a t th e im p u lse re sp o n s e be
ab so lu te ly su m m ab le, th e n , fo r a causal system , w e h av e
x
oc
N
N
oo
oo
£ > (* )! = £
Y ^ C kXnk < £ | C * | £ | A * | "
«=0
n
=0
n=0 Ik=-l
k=\
n=0
N ow if |
j < 1 fo r all k, th e n
an d h en ce
OC
IM*)[ < oo
Sec. 2.5
im plementation of Discrete-Time Systems
111
O n th e o th e r h an d , if o n e o r m o re o f th e
> 1, h{n) is n o lo n g e r ab so lu te ly
su m m ab le, an d c o n se q u e n tly , th e sy stem is u n sta b le . T h e re fo re , a n ecessa ry an d
sufficient c o n d itio n fo r th e sta b ility o f a causal IIR system d e sc rib e d by a lin ear
c o n s tan t-co efficien t d iffe re n c e e q u a tio n , is th a t al! ro o ts o f th e c h a ra c te ristic p o ly ­
n o m ial b e less th a n u n ity in m a g n itu d e. T h e re a d e r m ay verify th a t th is c o n d itio n
ca rrie s o v er to th e case w h ere th e system h as ro o ts o f m u ltip lic ity m.
2.5 IMPLEMENTATION OF DISCRETE-TIME SYSTEMS
O u r tre a tm e n t o f d isc re te -tim e sy stem s h as b e e n fo cu sed on th e tim e -d o m a in c h a r­
ac te riz a tio n an d analysis o f lin ear tim e -in v a ria n t system s d esc rib e d by c o n stan tco efficien t lin e a r d ifferen ce e q u a tio n s. A d d itio n a l an aly tical m e th o d s a re d e v e l­
o p e d in th e n ex t tw o c h a p te rs, w h ere we c h a ra c te riz e an d an aly ze L TI sy stem s in
th e fre q u e n c y d o m ain . T w o o th e r im p o rta n t to p ics th a t will b e tr e a te d la te r are
th e d esig n an d im p le m e n ta tio n o f th e se system s.
In p ractice, system design a n d im p le m e n ta tio n a re usually tr e a te d jo in tly
r a th e r th a n se p a ra te ly . O fte n , th e system design is d riv en by th e m e th o d o f
im p le m e n ta tio n an d by im p le m e n ta tio n c o n stra in ts, such as cost, h a rd w a re lim ­
itatio n s, size lim ita tio n s, a n d p o w e r re q u ire m e n ts . A t th is p o in t, we h av e n o t
as y e t d e v e lo p e d th e n ecessa ry analysis an d design tools to tr e a t such com plex
issues. H o w ev er, we h ave d e v e lo p e d sufficient b a c k g ro u n d to c o n sid e r so m e b a ­
sic im p le m e n ta tio n m e th o d s for re a liz a tio n s o f L TI sy stem s d escrib ed by lin ear
c o n s tan t-co efficien t d iffe re n c e e q u atio n s.
2.5.1 Structures for the Realization of Linear
Time-Invariant Systems
In th is su b sectio n we d esc rib e s tru c tu re s fo r th e re a liz a tio n o f system s d escrib ed
by lin e a r c o n s tan t-co efficien t d ifferen ce e q u a tio n s. A d d itio n a l stru c tu re s fo r th ese
sy stem s a re in tro d u c e d in C h a p te r 7.
A s a b eg in n in g , let us c o n sid e r th e first-o rd e r system
y(n) = —a iy (n — 1) + box ( n) + b \ x ( n - 1)
(2.5.1)
w hich is re a liz e d as in Fig. 2.32a. T h is re a liz a tio n uses s e p a ra te delays (m em o ry )
fo r b o th th e in p u t a n d o u tp u t signal sa m p le s a n d it is called a direct f o r m I structure.
N o te th a t th is sy stem can b e view ed as tw o lin e a r tim e -in v a ria n t system s in cascad e.
T h e first is a n o n re c u rsiv e , sy stem d e sc rib e d by th e e q u a tio n
u(n) = Z»oj;(n) + 6 j j ( n — 1)
(2.5.2)
w h e re a s th e se c o n d is a rec u rsiv e system d e sc rib e d by th e e q u a tio n
y ( n ) = - a i y ( n - 1) + t>{«)
(2.5.3)
H o w e v e r, as we h a v e se e n in S ectio n 2.3.4, if w e in te rc h a n g e th e o r d e r o f the
c a s cad ed lin e a r tim e -in v a ria n t system s, th e o v erall system re sp o n s e re m a in s th e