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The converse of a statement is formed by interchanging the hypothesis and conclusion. Hence to form the converse of an if-then statement, interchange the if and then clauses. In the case of the subject-predicate form, interchange the subject and the predicate. Thus, the converse of triangles are polygons is polygons are triangles. Also, the converse of if a metal is heated, then it expands is if a metal expands, then it is heated. Note in each of these cases that the statement is true but its converse need not necessarily be true.

_ (b)

See Fig. 8-37.

PRINCIPLE 1:

Write the transfer function H s whose pole-zero plot is given in Fig. 8-38. s2 50s 400 s2 40s 2000

and the Laplace-domain transfer function is T s V2 R = R1 R2 Kb 2 R1 R2 VS sReq C2 1 s C 1 R1 R2

The converse of a true statement is not necessarily true. The converse of a definition is always true.

Ans:

From T s , it is apparent that the circuit forms a low-pass lter with low-frequency gain T 0 R2 = R1 R2 and a corner frequency at !1 1= 1 1=Req C2 . Its Bode plot is sketched in Fig. 8-16(b).

PRINCIPLE 2:

CHAP. 8]

Thus, the converse of the definition a triangle is a polygon of three sides is a polygon of three sides is a triangle. Both the definition and its converse are true.

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For the ampli er of Fig. 3-10, assume that CC ! 1; hre hoe 0, and Ri 0. The bypass capacitor CE cannot be neglected. Find expressions for (a) the current-gain ratio Ai s , (b) the current-gain ratio at low frequencies, and (c) the midfrequency current-gain ratio. (d) Determine the low-frequency cuto point, and sketch the asymptotic Bode plot (Mdb only).

Fig. 8-36 8.38

2.11 Determining the hypothesis and conclusion in subject-predicate form Determine the hypothesis and conclusion of each statement. Solutions Hypothesis (subject) Perpendiculars Complements of the same angle An equilateral triangle A right triangle A triangle

The pole-zero plot in Fig. 8-39 shows a pole at s 0 and zeros at s 50 j50. method to evaluate the transfer function at the test point j100. Ans: H j100 223:6 26:578

(a) The small-signal low-frequency equivalent circuit is given in Fig. 8-4. By current division for Laplacedomain quantities, Ib RB I RB hie ZE i 1

Statements (a) Perpendiculars form right angles. (b) Complements of the same angle are congruent. (c) An equilateral triangle is equiangular. (d) A right triangle has only one right angle. (e) A triangle is not a quadrilateral.

Fig. 8-38

(2) (3)

Conclusion (predicate) form right angles are congruent is equiangular has only one right angle is not a quadrilateral

Fig. 8-39

Substitution of (1) into (3) gives the current-gain ratio as hfe RB IL RC Ii RC RL RB hie ZE Using (2) in (4) and rearranging leads to the desired current-gain ratio: hfe RC RB sRE CE 1 IL RC RL RE RB hie Ai s R C R hie Ii 1 s E E B RE RB hie (b) The low-frequency current-gain ratio follows from letting s ! 0 in (5): Ai 0 lim (c) hfe RC RB IL Ii RC RL RE RB hie 6 4

and L 0:1 H in the other. First, apply an excitation, Ii s , and obtain the natural frequency from the denominator of the network function. Try di erent locations for applying the current source. Second, insert a voltage source, Vi s , and obtain the natural frequency. Ans: 300 Np/s in all cases

2.12 Determining the hypothesis and conclusion in if-then form Determine the hypothesis and conclusion of each statement. Solutions Hypothesis (if clause) Conclusion (then clause) If a line bisects an angle then it divides the angle into two congruent parts If it is an obtuse triangle (then) a triangle has an obtuse angle If a student is sick (then) she should not go to school If he wishes to pass (then) a student must study regularly

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