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String hexadecimal(int n) { if (n < 16) { return Integer.toString(n%16) } return hexadecimal(n/16) + s; // recursion }

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)); } }

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(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:

A recursive function that prints permutations:

A recursive function that converts decimal to binary:

CHAP. I]

If the power in x(n) is P = 5, find the power in the odd part, x,(n), of x(n).

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void print(String str) { print("",str); } void print(String left, String right) { int n = right.length(); if (n == 0) { return; } else if (n == 1) System.out.println(left+right); return; } StringBuilder buf = new StringBuilder(right); for (int i = 0; i < n; i++) { char temp = s.charAt(i); s.setCharAt(i, s.charAt(0)); s.setCharAt(0, temp); print(left+temp, s.substring(1, n)); } }

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; }

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,

Iterative implementation of the Fibonacci function:

A recursive function that converts decimal to hexadecimal:

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Another method for performing convolutions, which we call the slide rule method, is particularly convenient when both x ( n ) and h ( n ) are finite in length and short in duration. The steps involved in the slide rule method are as follows: Write the values of x ( k ) along the top of a piece of paper, and the values of h ( - k ) along the top of another piece of paper as illustrated in Fig. 1-7. Line up the two sequence values x ( 0 ) and h(O), multiply each pair of numbers, and add the products to form the value of y(0). Slide the paper with the time-reversed sequence h ( k ) to the right by one, multiply each pair of numbers, sum the products to find the value y ( l ) , and repeat for all shifts to the right by n > 0. Do the same, shifting the time-reversed sequence to the left, to find the values of y ( n ) for n i0 .

The convolution sum expresses the output of a linear shift-invariant system in terms of a linear combination of the input values x ( n ) . For example, a system that has a unit sample response h ( n ) = a n u ( n )is described by the equation

int fib(int n) { if (n < 2) { return n; } int f0 = 0, f1 = 1, f = f0+f1; for (int i = 2; i < n; i++) { f0 = f1; f1 = f; f = f0 + f1; } return f; }

String hexadecimal(int n) { if (n < 16) { return Integer.toString(n%16) } return hexadecimal(n/16) + s; // recursion }

Although this equation allows one to compute the output y ( n ) for an arbitrary input x ( n ) , from a computational point of view this representation is not very efficient. In some cases it may be possible to more efficiently express the output in terms of past values of the output in addition to the current and past values of the input. The previous system, for example, may be described more concisely as follows:

The Ackermann function:

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