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Here is a trace of the call f(5) to the recursive factorial function defined in Example 9.2:

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For n = 2, F2 = Fn = Fn 1 + Fn 2 = F(2) 1 + F(2) 2 = F1 + F0 = 1 + 0 = 1. For n = 3, F3 = Fn = Fn 1 + Fn 2 = F(3) 1 + F(3) 2 = F2 + F1 = 1 + 1 = 2. For n = 4, F4 = Fn = Fn 1 + Fn 2 = F(4) 1 + F(4) 2 = F3 + F2 = 2 + 1 = 3. For n = 5, F5 = Fn = Fn 1 + Fn 2 = F(5) 1 + F(5) 2 = F4 + F3 = 3 + 2 = 5. For n = 6, F6 = Fn = Fn 1 + Fn 2 = F(6) 1 + F(6) 2 = F5 + F4 = 5 + 3 = 8. For n = 7, F7 = Fn = Fn 1 + Fn 2 = F(7) 1 + F(7) 2 = F6 + F5 = 8 + 5 = 13.

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Find the coefficients h(k) of the smallest-order filter that satisfies the following conditions:

2. 3.

The call originates in the main() function, passing 5 to the f() function. There, the value of the parameter n is 5, so it calls f(4), passing 4 to the f() function. There the value of the parameter n is 4, so it calls f(3), passing 3 to the f() function. This process continues (recursively) until the call f(1) is made from within the call f(2). There, the value of the parameter n is 1, so it returns 1 immediately, without making any more calls. Then the call f(2) returns 2*1 = 2 to the call f(3). Then the call f(3) returns 3*2 = 6 to the call f(4). Then the call f(4) returns 4*6 = 24 to the call f(5). Finally, the call f(5) returns the value 120 to main().

1 2 3 4 5 6

The filter has (generalized) linear phase. It completely rejects a sinusoid of frequency wo = n/3. The magnitude of the frequency response is equal to 1 at w = 0 and w = n.

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The trace in Example 9.7 shows that the call f(n) to the recursive implementation of the factorial function will generate n 1 recursive calls. This is clearly very inefficient compared to the iterative implementation shown in Example 9.3. EXAMPLE 9.8 Tracing the Recursive Fibonacci Function

To reject a sinusoid of frequency @ = 7r/3, the system function must have a pair of zeros on the unit circle at z = e*jnl3. Therefore, H(z) must contain a (linear phase) factor of the form

public static int fib(int n) { if (n < 2) { return n; // basis } return fib(n-1) + fib(n-2); // recursive part }

+F2)

The Fibonacci function (Example 9.6) is more heavily recursive than the factorial function (Example 9.2) because it includes two recursive calls. The consequences can be seen from the trace of the call

Thus, no value for A will allow us to simultaneously satisfy both unit magnitude constraints, and it is necessary to add another linear phase term to H (2). To minimize the order of the filter, we will pick a factor of the form

Here is a simple test driver for the Fibonacci method:

+ z - ~ )( 1. + B Z - ' + z-')

fib(5), shown in Figure 9.2 on page 168. The call originates in the main() function, passing 5 to the fib() function. There, the value of the parameter n is 5, so it calls fib(4) and fib(3), passing 4 and 3,

1 2 3 4 5

Now, with two free parameters, A and B , we should be able to satisfy the magnitude constraints. With

~ ( e ~ " ' ) )= 3A(2 - B ) = I ~,,

respectively. Each of these calls then makes two more recursive calls, continuing down to the basis calls

public static void main(String[] args) { for (int n = 0; n < 16; n++) { System.out.println("fib(" + n + ") = " + fib(n)); } }

is sufficient, but not necessary, for h(n) to be the unit sample response of a system with generalized linear phase.

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