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Returning to the example in the previous paragraph, we expect only one Bill Smith, and if there are two or more, we will find a way to make the different Bill Smith entities unique We will assign a key to each entity of the student entity type such that we can distinguish each student Having selected the entity types to include in the database, the data modeler then specifies the relationships among the entities For instance, we mentioned previously the advisor/advisee relationship between professors and students One of the important decisions to make is whether the relationship will be 1:1 (one-to-one), 1:N (one-to-many), or N:M (many-to-many) These ratios are called cardinality ratios In the case of the advisor/advisee relationship, the designer might decide the relationship is 1:N, with 1 advisor advising N students.

CHAP. 11]

For this triangular waveform, under the assumption that the ac component of iL ows through C and the dc component of iL ows through RL , I2 1 V 1 D Ts I 2 2 max 2Lc

However, I2 V2 =RL which can be equated to the above expression for I2 . Rearrangement of the result gives Lc RL 1 D Ts RL 1 D 2fs 2 10:6

By extending the AbstractCollection class, the BinaryTree class automatically inherits these methods that are defined by using the iterator() and size() methods:

On the other hand, if the school assigns multiple advisors to each student (for instance, one for the student s major field and one for student life questions), the relationship could be defined as N:M, multiple advisors for each student, and multiple students for each advisor Another pair of decisions related to the cardinality ratio of a relationship is the specification of minimum cardinalities Must a student have an advisor If so, then the minimum cardinality on the professor side of the advisor/advisee relationship must be 1 If not, then the minimum cardinality on the professor side of the relationship will be 0; a student entity may exist who is not associated with any advisor Likewise, must every professor be an advisor If so, then the minimum cardinality on the student side of the advisor/advisee relationship must be 1.

Example 10.2. A buck converter having a switching frequency of 25 kHz is to be operated with a duty cycle such that 0:1 D 1. The load is described by RL 5 . Determine the value of critical inductance L Lc so that current iL is continuous. The critical inductance must be determined for the minimum value of duty cycle. By (10.6), Lc RL 1 D 5 1 0:1 90 H 2fs 2 25 103

isEmpty() contains(Object object) toArray() toArray(Object[] objects) toString() add(Object object) addAll(Collection collection) clear() containsAll(Collection collection) remove(Object object) removeAll(Collection collection) retainAll(Collection collection) However, the mutating methods will throw an UnsupportedOperationException because they invoke other methods that are not implemented, namely the add() and the Iterator.remove() methods. public public public public public public public public public public public public boolean boolean Object[] Object[] String boolean boolean void boolean boolean boolean boolean

The boost converter SMPS circuit of Fig. 10-4 produces an average value output voltage V2 hv2 i > V1 . When the ideal transistor Q is ON, vQ 0. Conversely, when Q is OFF continuity of current through inductor L requires that ideal diode D be in the forward conducting state. With vD 0; vQ V2 . Thus, vQ is a rectangular pulse with a delay of DTs and duration 1 D Ts . If capacitor C is large, reasonable approximations are that the time-varying component of iD ows through C and that the voltage across the load resistor RL is constant. Due to periodic switching of Q, voltage vL and current iC are periodic once initial transients die out.

If not, then the minimum cardinality on the student side will be 0; a professor entity can exist with no associated advisees Other relationships might be 1:1 Imagine an entity type called Parking Permit, and that the policy is to allow each student one and only one parking permit The relationship between student and parking permit could be called parks/permit-to-park, and the relationship is 1:1 The minimum cardinality on the student side would probably be 1, since otherwise it would mean the database tracks parking permits that are not issued to anyone The minimum cardinality on the parking permit side would probably be 0, since some students probably will not have cars to park A many-to-many, N:M, relationship would exist between students and courses We could call this relationship takes/is-taken-by Each student will take many courses, and many students will take each course.

This example builds the same tree as the one in Example 11.21 and then tests the contains() method on it and its subtrees:

CHAP. 10]

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