数字逻辑设计及应用教学课件：2和3章作业评讲

1、Chapter 2 homework,2.4 321076543218 =(11010001)(00011111)(01011000)(11010001)2=(321)(037)(130)(321)8,2.11 decimal -42 signed-magnitude 10101010 twos-complement 11010110 ones-complement 11010101,2.22,2.35 001010, 011100, 101110, 111000. 2.36 When the LSB changes from 0 to 1, theres no problem. When t

2、he LSB changes from 1 to 0, so does the next bit (and possibly others), guaranteeing a problem. So, half the boundaries are bad, a total of 2n-1 for an n-bit encoding disc.,E X E R C I S E S O L U T I O N S,Chapter 3,3.1 The “probably” cases may cause damage to the gate if sustained. (a) 0 (b) 0 (c)

3、1 (d) probably 0 (e) undefined (f) probably 0 (g) 1 (h) probably 1,3-11 (A.B)+(C+D),3.12 (A+B).(C.D),3.20,The first answer for each parameter below assumes commercial operation and that the device is used with the maximum allowable (TTL) load. The number in parentheses, if any, indicates the value o

4、btained under a lesser but specified (CMOS) load. VOHmin 3.84V (4.4V ) VIHmin 3.15V VILmax 1.35V VOLmax 0.33V (0.1V) IImax 1mA IOLmax 4 mA (20 mA) IOHmax -4 mA (-20 mA),3.23,Current is positive if it flows into a node. Therefore, an output with negative current is sourcing current.,3-56,4.12 Each pr

5、oduct term of a canonical sum has n literals, regardless of whether or not the canonical sum happens to be a minimal sum. However, if the canonical sum is a minimal sum, there can be no other minimal sum. Another minimal sum would have to have the same number of product terms, each with n literals. But since n literal product terms (minterms) correspond exactly to the 1s of the logic function, such a minimal sum would