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Equation (I. l o ) is a special case of what is known as a linear constant coeficient difference equation, or LCCDE. The general form of a LCCDE is

A recursive function that prints permutations:

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where the coefficients a ( k ) and h ( k ) are constants that define the system. If the difference equation has one or more terms a ( k ) that are nonzero, the difference equation is said to be recursive. On the other hand, if all of the coefficients a ( k ) are equal to zero, the difference equation is said to be nonrecursive. Thus, Eq. ( 1 . l o ) is an example of a first-order recursive difference equation, whereas Eq. ( 1 . 9 ) is an infinite-order nonrecursive difference equation. Difference equations provide a method for computing the response of a system, y ( n ) , to an arbitrary input x ( n ) . Before these equations may be solved, however, it is necessary to specify a set of initial conditions. For example, with an input x ( n ) that begins at time n = 0 , the solution to Eq. ( 1 . 1 1 )at time n = 0 depends on the

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int ackermann(int m, int n) { if (m == 0) { return n + 1; } else if (n == 0) { return ackermann(m - 1, 1); } else { return ackermann(m - 1, ackermann(m, n - 1)); } // basis // basis // recursion

values of y ( - l ) , . . . , y ( - p ) . Therefore, these initial conditions must be specified before the solution for n 2 0 may be found. When these initial conditions are zero, the system is said to be in initial rest. For an LSI system that is described by a difference equation, the unit sample response, h(n), is found by solving the difference equation for x ( n ) = 6(n) assuming initial rest. For a nonrecursive system, a ( k ) = 0, the difference equation becomes

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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)); } }

and the output is simply a weighted sum of the current and past input values. As a result, the unit sample response is simply

Consider the relationship c(8,3) = 56 = 35 + 21 = c(7,3) + c(7,2) from the expansion of (x + 1)8: (x + 1)8= (x + 1)(x + 1)7 = (x + 1)(x7 + 7x6 + 21x5 + 35x4 + 35x3 + 21x2 + 7x + 1) = x8 + 7x7 + 21x6 + 35x5 + 35x4 + 21x3 + 7x2 + x + x7 + 7x6 + 21x5 + 35x4 + 35x3 + 21x2 + 7x + 1 8 + 8x7 + 28x6 + 56x5 + 70x4 + 56x3 + 28x2 + 7x + 1 =x The coefficient c(8,3) is for the x5 term, which is 35x5 + 21x5 = 56x5. The sum 35x5 + 21x5 came from (x)(35x4) and (1)(21x5). So those coefficients are 35 = c(7,3) and 21 = c(7,2). The general proof is based upon the same argument: c(n,k) is the coefficient of the term xk in the expansion of (x + 1)n. Since (x + 1)n = (x + 1)(x + 1)n 1, that term comes from the sum (x)(c(n 1, k 1) x k 1) + (1)(c(n 1, k)x k) = (c(n 1, k 1) + c(n 1, k))x k Therefore c(n, k) = c(n 1, k 1) + c(n 1, k). Figure 9.7 shows the trace of the call gcd(616, 231):

Thus, h(n) is finite in length and the system is referred to as a fmite-length impulse response (FIR) system. However, if a ( k ) # 0, the unit sample response is, in general, infinite in length and the system is referred to as an infinite-length impulse response (IIR) system. For example, if

Iterative implementation of the Fibonacci function:

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