Question

demonstrate the following, in order

1. Write 1001 to address 7.
2. Write 0101 to address 3.
3. Read from address 7.
4. Write 0000 to address 7 E. Read from address 3
   U2 1 2.5 V 2.5 V 2.5 V 2.5 V X4 X3 X1 X2 Key = A U3 0 Key = B U4 O Key = C U5 1 Key =D U1 DO 00 D1 01 D2 02 D3 03 D4 04 D5 05 D6 06 D7 07 U6 1 BO Key = 4 U7 0 Key = 3 U8 1 Key = 2 U9 0 Key = 1 A11 A10 A9 A8 A7 A6 A5 A4 A3 A2 A1 АО WE CS U10 0 Key = W 2K8RAM VCC 5.0V
   alll i got

Answer 1

as hot address seven be 2007 in the memory locations ro ranging from 2001 from 2001 thisugua 2007 To cestte 1001 org to ceddsess 7. g MORE Memory Data Register- MDR g MAR - Memory Address Register - MAR 2007 2007 Manesy 2A 2001 ТА 3€ Memory 2001 Zet2 ID 2007 - Before Execution 9 2007 & After Execution 2 Before enection 2D was stored at addsess 7 but after execution of Memory write operation was ust then to addsess z. 100 or g

or similar to past (a) let 2003 be addsess 3. To wsite 0101 or 5 to adds ese 3. MDR MOR MAR [2003] MAR 2003 AB Memory 2eet 2002 AB Memasy) 2001 2002 2003 5c 2003 5 Before Execution After Enection Before execution, 3k mas stossed at 2003 but after execution of Memory Operation 5 was usitlen to address 3 of 2003 MDR 3D MAR 2007) Memory GATool MDR Tok MAR 2009 Momory 2001 BA - 2007 Before Execution Memory 9 2007 Affer enocation Read Opesation

dy Writing oooo to address 7 뇌 similar to past (a). In this case, after the eneoetion of Memory write operation the address 7 will coutežu 0000. Bane es Reading from addsess 3 is again the as done for reading address 7 in past (c). In this case, however, the address 3 will read OLOL