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Note that the function header is identical to that used in Example 9.2; only the body is different. This allows us to use the same test driver for both implementations. The output should be the same.

The only difference between a type I and a type I11 linear phase filter is that h(n)= h(N - n ) for a type I filter, and h(n) = -h(N - n ) for a type 111 filter. Therefore, h ( N / 2 ) = 0, and Eq. (5.20) is modified as follows:

CHAP. 14]

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where The only modification required in Prob. 5.15 to find the form of the frequency response for a type IV linear phase filter is to use the fact that h ( n ) is odd to rewrite Eq. (5.21) as follows:

BASIS AND RECURSIVE PARTS To work correctly, every recursive function must have a basis and a recursive part. The basis is what stops the recursion. The recursive part is where the function calls itself. EXAMPLE 9.4 The Basis and Recursive Parts of the Factorial Function

h(n) = fh*(N - n )

1 2 3 4 5 6

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In the Java method that implements the factorial function in Example 9.2, the basis and the recursive parts are labeled with comments. The recursive part invokes the method, passing a smaller value of n. So starting with a positive value like 5, the values on the successive invocations will be 4, 3, 2, 1, and 0. When 0 is passed, the basis executes, thereby stopping the recursion and beginning the chain of returns, returning 1, 1, 2, 6, 24, and finally 120.

If ~ ( e j " is the DTFT of h ( n ) , it follows from the delay property and the time-reversal property that the DTFT of ) h ( N - n ) is h(N - n ) Applying the conjugation property. we then have h*(N - n )

e lNw~(e-jW)

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

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

e 'NW~*(e'w)

- n).

Here is a simple test driver for the Fibonacci method:

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0 1 1 2 3 5 8 13 21 34 55 89 144 233 377

Now, let us consider the case in which h ( n ) is conjugate symmetric, h ( n ) = hl*(N H ( e j w ) = e-jNWf/*(el"')

) and, expressing H (el"') and H * ( e J Win terns of their magnitude and phase, we have H(eJw) = I ~ ( ~ j w ) l ~ ~ @ l t ( w ) and Therefore, it follows that

1 2 3 4 5

The Fibonacci numbers are 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . Each number after the second is the sum of the two preceding numbers. This is a naturally recursive definition: Fn = 0, if n = 0 1, if n = 1 F n 1 + F n 2 , if n > 1

el@h(w)

+2nk(o)

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

The first 15 values of the Fibonacci sequence are shown in Table 9.2. The first two values, F0 and F1, are defined by the first two parts of the definition: F0 = 0 (for n = 0) and F1 = 1 (for n = 1). These two parts form the basis of the recursion. All the other values are defined by the recursive part of the definition:

where A(eJW) a real-valued (in general bipolar) function of w. Thus, h(n) has linear phase. is For the case in which h(n) is conjugate antisymmetric, h(n) = -h*(N - n) Eq. (5.22) becomes

Therefore,

TRACING A RECURSIVE CALL Hand tracing the execution of a method usually helps clarify it. EXAMPLE 9.7 Tracing the Recursive Factorial Function

CHAP. 9]

+ rr + 2 ~ k ( w )

The relationship between the input and the output of an FIR system is as follows:

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