[원서] Charles H. Roth, Larry L. Kinney - Instructor Solutions - Fundame…
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Download : Charles H Roth Larry L Kinney Instructor Solutions Fundamentals of Logic Design Cengage Learn.pdf
[원서] Charles H. Roth, Larry L. Kinney - Instructor Solutions - Fundamentals of Logic Design-Cengage Learn , [원서] Charles H. Roth, Larry L. Kinney - Instructor Solutions - Fundamentals of Logic Design-Cengage Learn기타솔루션 , 솔루션
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[원서] Charles H. Roth, Larry L. Kinney - Instructor Solutions - Fundamentals of Logic Design-Cengage Learn
[원서] Charles H. Roth, Larry L. Kinney - Instructor Solutions - Fundamentals of Logic Design-Cengage Learn
Download : Charles H Roth Larry L Kinney Instructor Solutions Fundamentals of Logic Design Cengage Learn.pdf( 73 )
설명
Unit 1 Solutions
II. SOLUTIONS TO HOMEWORK PROBLEMS
Unit 1 Problem Solutions
1.1 (a) 757.2510 16 | 757 16 | 47 16 | 2 0 r5 r15=F16 r2 0.25 16 (4).00 1.1 (b) 123.1710 16 | 123 16 | 7 0 r11 r7 0.17 16 (2).72 16 (11).52 16 (8).32
∴ 757.2510 = 2F5.4016 = 0010 1111 0101.0100 00002 2 F 5 4 0 1.1 (c) 356.8910 16 | 356 16 | 22 16 | 1 0 r4 r6 r1 0.89 16 (14).24 16 (3).84 16 (13).44 16 (7).04
∴123.1710 = 7B.2B16 = 0111 1011.0010 10112 7 B 2 B 1.1 (d) 1063.510 16 | 1063 16 | 66 16 | 4 0 r7 r2 r4
Unit 1 Solutions
II. SOLUTIONS TO HOMEWORK PROBLEMS
Unit 1 Problem Solutions
1.1 (a) 757.2510 16 | 757 16 | 47 16 | 2 0 r5 r15=F16 r2 0.25 16 (4).00 1.1 (b) 123.1710 16 | 123 16 | 7 0 r11 r7 0.17 16 (2).72 16 (11).52 16 (8).32
∴ 757.2510 = 2F5.4xxx = 0xxx 1111 xxx1.xxx0 00002 2 F 5 4 0 1.1 (c) 356.8910 16 | 356 16 | 22 16 | 1 0 r4 r6 r1 0.8…(투비컨티뉴드 )
다.
