# ssrs 2d barcode ( d ) y d n ) = x ( n 2 - 2n in Software Printing Code-128 in Software ( d ) y d n ) = x ( n 2 - 2n

( d ) y d n ) = x ( n 2 - 2n
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CHAP. 11
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(a) The sequence x(n), illustrated in Fig. 1-8(a), a linearly decreasing sequence that begins at index n = 0 and is ends at index n = 5. The first sequence that is to be sketched, yl(n) = x(4 - n), is found by shifting x(n) by four and time-reversing. Observe that at index n = 4, yl(n) is equal to x(0). Therefore, yl(n) has a value of 6 at n = 4 and decreases linearly to the left (decreasing values of n) until n = - 1, beyond which y (n) = 0.The sequence y (n) is shown in Fig. 1-8(b).
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Fig. 1-8. Performing signal manipulations.
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(b) The second sequence, y2(n) = x(2n - 3), is formed through the combination of time-shifting and downsampling. Therefore, y&~) may be plotted by first shifting x(n) to the right by three (delay) as shown in
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[CHAP. 1
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Fig. 1-8(c). The sequence y2(n) is then formed by down-sampling by a factor of 2 (i.e., keeping only the even index terms as indicated by the solid circles in Fig. 1-8(c)). A sketch of yn(n) is shown in Fig. I-8(d). (c) The third sequence, y3(n) = x(8 - 3n), is formed through a combination of tirne-shifting, down-sampling, and time-reversal. To sketch y3(n) we begin by plotting x(8 - n), which is formed by shifting x(n) to the left by eight (advance) and reversing in time as shown in Fig. 1 -8(e). Then, y3(n) is found by extracting every third sample of x(8 - n), as indicated by the solid circles, which is plotted in Fig. 1-8(f ) .
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(4 Finally, y4(n) = x(n2 - 2n + 1) is formed by a nonlinear transformation of the time variable n. This sequence
may be easily sketched by listing how the index n is mapped. First, note that if n 2 4 or n 5 -2, then n2 - 2n + 1 2 9 and, therefore, y4(n) = 0. For - I 5 n 5 3 we have
The sequence y4(n) is sketched in Fig. 1-8(g).
The notation ~ ( ( n ) is ~ ) used to define the sequence that is formed as follows: ~ ( ( n ) = ~ modulo N) ) x(n where (n modulo N) is the positive integer in the range [0, N - 11 that remains after dividing n by N. For example, ((3))g = 3, ((12))g = 4, and ((-6))d = 2. If x(n) = (i)%in(nn/2)u(n), make a sketch of (a) x((n))3 and (b)x((n - 2))3.
(a) We begin by noting that ((n))3, for any value of n, is always an integer in the range [O, 21. In fact, because ((n))3 = ((n 3k)h for any k ,
Therefore, x((n))3 is periodic with a period N = 3. It thus follows t h a t ~ ( ( n )is~ ) formed by periodically repeating the first three values of x(n) as illustrated in the figure below:
(b) The sequence x((n - 2))3 is also periodic with a period N = 3, except that the signal is shifted to the right by no = 2 compared to the periodic sequence in part (a). This sequence is shown in the figure below:
The power in a real-valued signal x(n) is defined as the sum of the squares of the sequence values:
Suppose that a sequence x(n) has an even part x,(n) equal to
CHAP. I]
SIGNALS AND SYSTEMS
If the power in x(n) is P = 5, find the power in the odd part, x,(n), of x(n).
This problem requires finding the relationship between the power in x ( n ) and the power in the even and odd parts. By definition, x ( n ) = x , ( n ) x,(n). Therefore,
Note that x,(n)x,(n) is the product of an even sequence and an odd sequence and, therefore, the product is odd. Because the sum for all n of an odd sequence is equal to zero,