Question: By Multisim, please implement two mod-8 Johnson counters, a right-shift Johnson counter and a left-shift Johnson counter. Fot each counter, you can use the following

By Multisim, please implement two mod-8 Johnson counters, a right-shift Johnson counter and a left-shift Johnson counter. Fot each counter, you can use the following items: one 74195 (a 4-bit SRG, see Fig. 1), 4 HIEX displays (see Fig. 2) to show the counter sequences, and basic logic gates (if necessary). (a) Right-shift Johinson counter, i.e., the counting sequenec is [0]0]0[0], 0[0]0], [110]0, [1]IIO, , IIIII], [0]I]II], [0]0|I]], [0[0]0|I], and then back to [0]0][0]0]. (b) Leff-shift Johnson counter, i.e., the counting sequence is [0]0[0]0], [0]0[0]], Required items: (1) 74195 (4-bit SRG). (2) HEX display (note: connecting HIGH (LOW) to LSB in HEX display shows 1 (0)), By Multisim, please implement two mod-8 Johnson counters, a right-shift Johnson counter and a left-shift Johnson counter. Fot each counter, you can use the following items: one 74195 (a 4-bit SRG, see Fig. 1), 4 HIEX displays (see Fig. 2) to show the counter sequences, and basic logic gates (if necessary). (a) Right-shift Johinson counter, i.e., the counting sequenec is [0]0]0[0], 0[0]0], [110]0, [1]IIO, , IIIII], [0]I]II], [0]0|I]], [0[0]0|I], and then back to [0]0][0]0]. (b) Leff-shift Johnson counter, i.e., the counting sequence is [0]0[0]0], [0]0[0]], Required items: (1) 74195 (4-bit SRG). (2) HEX display (note: connecting HIGH (LOW) to LSB in HEX display shows 1 (0))
Step by Step Solution
There are 3 Steps involved in it
Get step-by-step solutions from verified subject matter experts
