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Fig. 2-28
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In Fig. 2-28 the output of the unit delay element is y [ n - 11. Thus, from Fig. 2-28 we see that
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which is the required first-order linear difference equation.
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2.40. The discrete-time system shown in Fig. 2-29 consists of two unit delay elements and two scalar multipliers. Write a difference equation that relates the output y[n] and the input x [ n ] .
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CHAP. 2 1
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LINEAR TIME-INVARIANT SYSTEMS
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In Fig. 2-29 the output of the first (from the right) unit delay element is y[n - 11 and the output of the second (from the right) unit delay element is y[n - 21. Thus, from Fig. 2-29 we see that y[n]=a,y[n-l]+a,y[n-2]+x[n] or y [ n ] - a , y [ n - 1 ] - a , y [ n - 21 = x [ n ] (2.144) (2.145)
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which is the required second-order linear difference equation. Note that, in general, the order of a discrete-time LTI system consisting of the interconnection of unit delay elements and scalar multipliers is equal to the number of unit delay elements in the system.
2.41. Consider the discrete-time system in Fig. 2-30. Write a difference equation that relates the output y [ n ] and the input x [ n ] .
Unit delay
11 Fig. 2-30
Let the input to the unit delay element be q[nl. Then from Fig. 2-30 we see that q [ n ]= 2 q [ n - 1 1 + x [ n ] ~ [ n ] [ n I + 3 q [ n - 11 =s 9 [ n ] =f y [ n ]+ Changing n to ( n - 1) in Eq. (2.147a), we have q [ n - 1 1 = f y [ n - 11 + i x [ n - 1 1 Thus, equating Eq. (2.1476) and (Eq. (2.147c), we have ; y [ n ] - f x [ n ] = f y [ n - 1 1 + + x [ n - 11 Multiplying both sides of the above equation by 5 and rearranging terms, we obtain y [ n ] - 2 y [ n - 1 1 = x [ n ]+ 3 x [ n - 1 1 which is the required difference equation. (2.148) (2.147~)
( 2.146a)
(2.1466)
Solving Eqs. (2.146a) and (2.146b) for q [ n ]and q [ n - 11 in terms of x [ n ]and y[n],we obtain
(2.147a) (2.147b)
q [ n - 1 1 = i y [ n ]- i x [ n ]
2.42. Consider a discrete-time system whose input x [ n ] and output y [ n ] are related by
y [ n ] - a y [ n - 11 = x [ n ]
(2.149)
LINEAR TIME-INVARIANT SYSTEMS
[CHAP. 2
where a is a constant. Find y [ n ] with the auxiliary condition y [ - l] = y -
, and
(2.150)
x [ n ] =Kbnu[n]
yLn1 =ypLn1
+yhLn1
where y,[n] is the particular solution satisfying Eq. (2.149) and y h [ n ] is the homogeneous solution which satisfies
y [ n ] -ay[n
- 11 = O
Assume that
y,[n] = Abn
Substituting Eq. (2.1521 into Eq. (2.1491, we obtain
Abn - a A b n - ' = K b n
from which we obtain A
= Kb/(b
- a), and
To obtain yh["], we assume
y,[n]
= Bzn
Substituting this into Eq. (2.151) gives from which we have z = a and
~ ~ [=Ban n ]
Combining yp[n] and yh[n],we get
y [ n ] =Ban + b"+l b-a K n20
In order to determine B in Eq. (2.155) we need the value of y[O]. Setting n and (2.1501, we have
Y [ O ] -ay[-11 =y[O] - a y - , =x[O] = K
in Eqs. (2.149)
or Setting n
y[O] = K + a y - ,
in Eq. (2.155), we obtain
y[O] = B +Kb b-a
Therefore, equating Eqs. (2.156) and (2.1571, we have
from which we obtain
B =ay-, - K b-a
Hence, Eq. (2.155) becomes
CHAP. 21
LINEAR TIME-INVARIANT SYSTEMS
For n < 0, we have x [ n ] = 0, and Eq. (2.149) becomes Eq. (2.151). Hence.
y [ n ] =Ban
(2.159)
From the auxiliary condition y[ - 11 = y - we have
y [ - 11 = y - ,
= BU-'
from which we obtain B = y - ,a. Thus,
Combining Eqs. (2.158) and (2.160), y [ n ] can be expressed as
Note that as in the continuous-time case (Probs. 2.21 and 2.221, the system described by Eq. (2.149) is not linear if y [ - 11 # 0. The system is causal and time-invariant if it is initially at rest, that is, y [ - 11 = 0. Note also that Eq. (2.149) can be solved recursively (see Prob. 2.43).
2.43. Consider the discrete-time system in Prob. 2.42. Find the output y [ n ] when x [ n ] = K S [ n ] and y [ - l ] = y - , = a .
We can solve Eq. (2.149) for successive values of y [ n ] for n r 0 as follows: rearrange Eq. (2.149) as
y[n]=ay[n - 11 +x[n]
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