Tutorial Set 1 A-B Solutions

Tutorial Set 1 A-B Solutions - EE2006: Digital Design...

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EE2006: Digital Design 1 EE2006 Solutions for Number Systems 1. (a) N N N N NN 1110 1100 1111 1001 0101 1010 . 59 = FCE A (b) 0 5 0 5 16 5 7 80 87 16 87 8 1392 1400 16 ...... 76 . 16 36 . 36 . 16 96 . 96 . 16 56 . 56 . 16 16 . 5 F 8 2 = × = × = × = × () 16 .... 5 28 . 578 F = (c) N N N N N 7 0 4 3 5 111 000 . 100 011 101 8 07 . 534 = (d) 0 1 0 1 2 1 0 2 2 2 2 1 4 5 2 5 0 10 10 2 10 0 20 20 2 20 1 40 41 2 41 1 82 83 2 83 0 166 166 2 ...... 44 . 2 72 . 72 . 2 36 . 36 . 2 68 . 68 . 2 34 . 1 0 1 0 = × = × = × = × = (10100110.0101…) 2 (e) NNN N 3 2 6 4 0011 0010 0110 0100 4623 = (f) 6- 2 3 2 6 4 x + 2 - 6 = 4. x = 8. The number system is octal. 2. (a) No. locations = End-Start+1 = 0x4FFF – 0x3000+1 = 0x5000-0x3000 = 0x2000 NNNN 2000 0010000000000000 ⎯→ 8K 0010000000000000 ±²³² ´±²²³ ²²´ 0x2000 = 8K = 8 x 1024 =8192 locations.
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EE2006: Digital Design 2 (b) No. locations = End-Start+1 = 0x533F-0x4000+1 = 0x5340-0x4000 = 0x1340 NN 1340 0001001101000000 ⎯→ 512 256 64 832 4K 0001001101000000 ++= ±²³² ´±²³²´ 0x1340 = 4K+832 = 4928 locations. 3. 13K = 13K 0011010000000000 ±³´ ±²²³ ²²´ N NNN 3400 0011010000000000 End address = Start address + No. locations – 1 = 0x6000+0x3400-1 = 0x9400-1 = 0x93FF. 4. Decimal Sign Mag. 127 01111111 -0 10000000 -55 10110111 5.
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This note was uploaded on 01/08/2011 for the course EE 2006 taught by Professor Dr. during the Spring '10 term at National University of Singapore.

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Tutorial Set 1 A-B Solutions - EE2006: Digital Design...

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