The content of a memory fragment of the Basic Computer is given as: / Le contenu d'un fragment de mmoire de l'ordinateur de base est donn comme suit: Adresse/Address A18 A19 A1A A1B A1C A1D A1E A1F Contenu/Content 8A1C 9A1D AA1F DA1E 0A1F 0A1E 0A1D 0A1C Using Table 5.6 fill out the Blank # boxes below (as defined in the next table), with the values of the corresponding registers (PC, AR, IR, DR and AC) at the end of every clock cycle (TO, T1, T2, ....T5). All values are in hexadecimal. / l'aide du tableau 5.6, remplissez les cases vides ci-dessous (comme dfini dans le tableau suivant), avec les valeurs des registres correspondants la fin de chaque cycle d'horloge (TO, T1, T2, ....T5). Toutes les valeurs sont en hexadcimal / PC AR IR DR AC Initiale A1A 123 1234 1234 1234 A1A Box Box 1234 1234 #6 #13 T1 Box Box #1 Box 1234 1234 #7 #14 T2 T3 T4 T5 Box #2 Box #3 Box #4 Box #8 Box # 10 Box #11 Box 1234 1234 #15 Box 1234 1234 #16 Box #17 Box Box Box #5 # 12 # 18 Box #19 Box #21 Box Box #20 #23 Instructions: The values (in hex) which registers may contain are enlisted in the right column below, each being encoded with a decimal code from 1 to 20. The left column represents the empty # boxes (#1 to #23) of the table above. So, to complete your table, you must match each empty box (from the left column below) with the correct value (in hex) from the right column, by specifying its unique number (from 1 < < < Box #12 Box #21 Box #8 Box #18 Box #23 Box #13 Box #14 Box #19 Box #20 Rox #10 1. 2. 3. A1A 5. 4. A1B 6. A18 7. A19 8. A1C A1E 123 1234 9. 8A1C ..... < < Box #6 Box #1 Box #4 Box #2 Box #3 Box #7 16. 1C51 17. 0214 18. Other/Autre chose 19. A1F 20. A1D V Box #10 Box #17 Box #11 Box #15 Box #5 Box #16 Box #6 9. 8A1C 10. 9A1D 11. AA1F 12. DA1E 13. 0A1C 14. 0A1D 15. 0A1F 16. 1C51 17 0214