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Shown in the figure below is a plot of the sequence h(-k)R4(k).

CHAP. 9]

A recursive function that returns the power xn and makes no more than lgn recursive calls:

To evaluate y(O), we multiply this sequence by x ( k ) and sum the product from k = 0 to k = 3. The result is y(0) = 1. Next, to find the value of y( I), we evaluate the sum

A recursive function that determines whether a string is a palindrome:

+ 1)

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double pow(double x, int if (n == 0) { return 1.0; } double p = pow(x,n/2); if (n%2 == 0) { return p*p; } else { return x*p*p; } } n) { // basis

(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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boolean isPalindrome(String s) { int len = s.length(); if (len < 2) { return true; } else if (s.charAt(0) != s.charAt(len-1)) return false; } else if (len == 2) { return true; } else { return isPalindrome(s.substring(1,len-1)); } }

(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

// recursion (n even) // recursion (n odd)

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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 ) .

A recursive function that converts decimal to binary:

(4 Finally, y4(n) = x(n2 - 2n + 1) is formed by a nonlinear transformation of the time variable n. This sequence

A recursive function that returns the integer binary logarithm of n:

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

String binary(int n) { String s; if (n%2 == 0) { s = "0"; } else { s = "1"; } if (n < 2) { return s; } return binary(n/2) + s; }

The sequence y4(n) is sketched in Fig. 1-8(g).

CHAP. 9]

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 recursive function that converts decimal to hexadecimal:

A recursive function that determines whether a string is a palindrome:

(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:

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