Tutorial-Memory-System-Answer

Tutorial-Memory-System-Answer - Tutorial Memory System 1(i...

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1 Tutorial : Memory System 1. (i) ROM (4M x 16-bit) : 16M/(4Mx2 ) = 2 SRAM (2M x 16-bit) : 16M/(2Mx2) = 4 SRAM (1M x 16-bit): 8M/(1Mx2) = 4 (ii) Address lines Device nGCS 24 23 22 21 20 19 03 02 01 ROM 0 0(000) 0 0 X X X X X X X ROM 1 0(000) 0 1 X X X X X X X RAM1_0 6(110) 0 0 0 X X X X X X RAM1_1 6(110) 0 0 1 X X X X X X RAM1_2 6(110) 0 1 0 X X X X X X RAM1_3 6(100) 0 1 1 X X X X X X RAM2_0 7(111) 0 0 0 0 X X X X X RAM2_1 7(111) 0 0 0 1 X X X X X RAM2_2 7(111) 0 0 1 0 X X X X X RAM2_3 7(111) 0 0 1 1 X X X X X
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2 (iii) $0000 0000 $007F FFFF ROM0 4M 16 8 Mbytes 16 Mbytes $0800 0000 $0FFF FFFF ROM1 4M 16 8 Mbytes $0C00 0000 $0C3F FFFF RAM1_0 2M 16 4 Mbytes 16 Mbytes $0C40 0000 $0C7F FFFF RAM1_1 2M 16 4 Mbytes $0C80 0000 $0CBF FFFF RAM1_ 2 2M 16 4 Mbytes $0CC0 0000 $0CFF FFFF RAM1_ 3 2M 16 4 Mbytes
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Unformatted text preview: 16  4 Mbytes  $0E00 0000 $0E1F FFFF RAM2_0 1M  16  2 Mbytes      8 Mbytes     $0E20 0000 $0E3F FFFF RAM2_1 1M  16  2 Mbytes  $0E40 0000 $0E5F FFFF RAM2_ 2 1M  16  2 Mbytes  $0E60 0000 $0E7F FFFF RAM2_ 3 1M  16  2 Mbytes  3 (iv) (v) ROM 0: 23 24 A A nGCS0 ROM 1: A A nGCS0 23 24 RAM1_0: 22 23 24 A A A nGCS6 RAM1_1: 22 23 24 A A A nGCS6 RAM1_2: 22 23 24 A A A nGCS6 RAM1_3: 22 23 24 A A A nGCS6 RAM2_0: 21 22 23 24 A A A A nGCS7 RAM2_1: 21 22 23 24 A A A A nGCS7 RAM2_2: 21 22 23 24 A A A A nGCS7 RAM2_3: 21 22 23 24 A A A A nGCS7 i/p 1 A23 CS_ROM1 CS_ROM0 A24 nGSC7 E1 E2 1 3 2 1 A24 nGSC6 A23 A22 CS_RAM1_3 E1 E2 CS_RAM1_2 CS_RAM1_1 CS_RAM1_0 2 (v) 1 3 2 1 A23 A24 nGSC7 A22 A21 CS_RAM2_3 E1 E2 CS_RAM2_2 CS_RAM2_1 CS_RAM2_0 3 1 3 2 1 nGSC7 A22 A23 CS_0 A24 2 CS_7 CS_6 CS_5 CS_4 CS_3 CS_2 CS_1 7 6 5 4...
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