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x.plus(y) returns the Fraction that represents x + y x.times(n) returns the Fraction that represents n*x x.times(y) returns the Fraction that represents x*y x.reciprocal() returns the Fraction that represents 1/x x.value() returns the numerical value of x

Therefore, we need to find the frequency response of the system. In Example 2.2.3, it was shown that the unit sample response of an ideal low-pass filter,

Most programming languages have types for integers and real (decimal) numbers, but not for fractions. Such numbers can be implemented as objects. Here is a design for a fraction type:

sin nw, h i @ )= 7 n Because h(n) = 2hl (n - 1) with w, = 1712,an expression may be derived for H ( e J W ) terms of H , (ej") as follows: in

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UML diagrams can be used to specify ADTs simply by omitting the state information. The Fraction ADT defined in Example 1.1 is shown as a UML diagram in Figure 1.4. ADTs can be used in pseudocode to implement algorithms independently of any specific programming language. This is illustrated in Example 1.2.

Therefore,

ADT: Fraction plus(Fraction): Fraction times(Integer): Fraction times(Fraction): Fraction reciprocal(): Fraction value(): Real

h ( n ) = S(n) - c d ( n - I )

Node clone = new Node(list.data); Node p=list, q=clone; for (int i=m+1; i<n; i++) { if (p.next == null) { throw new IllegalArgumentException(); } q.next = new Node(p.next.data); p = p.next; q = q.next; } return clone; }

The phase, on the other hand, is H[(eJW) gh( w ) = tan-' ---- = tanp' HR(eiw)

This ADT specifies five operations. Note that the times() operation is overloaded. Note that the ADT uses generic terms for types: Integer instead of int, and Real instead of double. That is because it is supposed to be independent of any specific programming language. In general, a complete ADT would also include documentation that explains exactly how each operation should behave. For example,

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Finally, the group delay may be found by differentiating the phase (see Prob. 2.19). Alternatively, we may note that because this system is the inverse of the one considered in Example 2.2.2, the phase and the group delay are simply the negative of those found in the example. Therefore, we have

void append(Node list1, Node list2) { if (list1 == null) { throw new IllegalStateException(); } while (list1.next != null) { list1 = list1.next; } list1.next = list2; } Node concat(Node list1, Node list2) { Node list3 = new Node(0), p=list1, q=list3; while (p != null) { q.next = new Node(p.data); p = p.next; q = q.next; } p = list2; while (p != null) { q.next = new Node(p.data); p = p.next; q = q.next; } return list3.next; // discard dummy head node } void set(Node list, int i, int x) { if (i < 0) { throw new IllegalArgumentException(); } for (int j=0; j<i; j++) { if (list == null) { throw new IllegalStateException(); } list = list.next; } list.data = x; return; } int void get(Node list, int i) { // returns the value of element number i; Node p = list; int j = 0; while (j < i && p != null) { ++j; p = p.next; }

Note that the magnitude is constant for all o,and the phase is -n/2 for 0 < o < n and n/2 for -IT < w < 0. Find the unit sample response of this system.

CHAP. 3]

The unit sample response may be found by integration:

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