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which says that the power in x ( n ) is equal to the sum of the powers in its even and odd parts. Evaluating the power in the even part of x ( n ) , we find
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Therefore, with P = 5 we have
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) (ynl I + 2 ) ( f )2n = f : =:
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P, = 5 - P,
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Consider the sequence Find the numerical value of
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Compute the power in x(n),
If x(n) is input to a time-varying system defined by y(n) = nx(n), find the power in the output signal (i.e., evaluate the sum)
This is a direct application of the geometric series
With the substitution of -n for n we have
Therefore, it follows from the geometric series that
SIGNALS AND SYSTEMS
( h ) To find the power in x ( n ) we must evaluate the sum
[CHAP. 1
Replacing n by -n and using the geometric series, this sum becomes
(c) Finally, to find the power in y ( n ) = n x ( n )we must evaluate the sum
In Table 1- I there is a closed-form expression for the sum
However, we may derive a closed-form expression for this sum as follow^.^ Differentibut not for C:,n2an. ating both sides of Eq. (1.19) with respect to a , we have
Therefore. we have the sum
Using this expression to evaluate Eq. (1.18).we find
Express the sequence
.r(n) =
as a sum of scaled and shifted unit steps.
n=O n=l
n=2 else
In this problem, we would like to perform a signal decomposition, expressing x ( n ) as a sum of scaled and shifted unit steps. There are several ways to derive this decomposition. One way is to express x ( n ) as a sum of weighted and shifted unit samples, x ( n ) = S(n) 2S(n - I) 3S(n - 2)
and use the fact that a unit sample may be written as the difference of two steps as follows:
Therefore,
x ( n ) = u(n) - u(n - I )
+ 2[u(n- I) - u(n - 2)] + 3[u(n- 2 ) - u(n - 3)]
which gives the desired decomposition:
"his
method is very useful and should be remembered
CHAP. I]
SIGNALS AND SYSTEMS
Another way to derive this decomposition more directly is as follows. First, we note that the decomposition should begin with a unit step, which generates a value of I at index n = 0.Because x(n) increases to a value of 2 at n = 1, we must add a delayed unit step u(n - 1). At n = 2, x(n) again increases in amplitude by 1, so we add the delayed unit step u(n - 2). At this point, we have
Thus, all that remains is to bring the sequence back to zero for n > 3. This may be done by subtracting the delayed unit step 3u(n - 3), which produces the same decomposition as before.
Discrete-Time Systems
For each of the systems below, x ( n ) is the input and y(n) is the output. Determine which systems are homogeneous, which systems are additive, and which are linear.
(a) If the system is homogeneous, y(n) = T[cx(n)] = cT[x(n)] for any input x(n) and for all complex constants c. The system y(n) = log(x(n)) is not homogeneous because the response of the system to xl(n) = cx(n) is
which is not equal to c log(x(n)). For the system to be additive, if yl(n) and y2(n) are the responses to the inputs and xz(n), respectively, the response to x(n) = xl(n) x2(n) must be y(n) = yl(n) y2(n). For this system we have T[xl(n) + x h N = log[x~(n) x2(n)l # log[x~(n)l+log[x2(n)l
Therefore, the system is not additive. Finally, because the system is neither additive nor homogeneous, the system is nonlinear. (b) Note that if y(n) is the response to x(n).
the response to xl(n) = cx(n) is
which is not the same as y1 (n). Therefore, this system is not homogeneous. Similarly, note that the response to x(n) = x,(n) x2(n) is
which is not equal to yl(n)
+ y2(n). Therefore, this system is not additive and, as a result, is nonlinear.
SIGNALS AND SYSTEMS
[CHAP. 1
This system is homogeneous, because the response of the system to xl(n) = cx(n) is
The system is clearly, however, not additive and therefore is nonlinear. Let y,(n) and yz(n) be the responses of the system to the inputs x,(n) and x2(n), respectively. The response to the input x(n) = axl(n) bxz(n)
y(n) = x(n) sin
(y) =
[axl(n)
+ bx2(n)] sin
Thus, it follows that this system is linear and, therefore, additive and homogeneous. Because the real part of the sum of two numbers is the sum of the real parts, if y,(n) is the response of the system toxl(n), and yz(n) is the response to x2(n), the response to y(n) = yl(n) yz(n) is
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