hey, igot questino from digital design, it\'s not from thebook it\'s from teache

Question

hey, igot questino from digital design, it's not from thebook it's from teacher and this is the problem: Convert the first letter of your last name to ASCII. Design acircuit to
detect that sequence. Example: GATLIN > G >1000111. Recommendations:
Spend a lot of time developing your state diagram. You can brutforce it,
but it may make for a complicated circuit.
Use any flip-flop you want, but the D may be the easiest hey, igot questino from digital design, it's not from thebook it's from teacher and this is the problem: Convert the first letter of your last name to ASCII. Design acircuit to
detect that sequence. Example: GATLIN > G >1000111. Recommendations:
Spend a lot of time developing your state diagram. You can brutforce it,
but it may make for a complicated circuit.
Use any flip-flop you want, but the D may be the easiest

Explanation / Answer

one of my friend gave me this  introduction: Your last name starts with A, if I remembercorrectly, and the ASCII number
for "A" is 1000001.

I'm not sure what he means by "detect that sequence," but we couldsay,
for instance, that if that sequence is entered, it will light anLED.

So, one way to do that is to make a state machine with 9 states (1for
each number and an extra one for errors and another one for theoutput
LED).

For example, you start out in State 0. If a "1" is encountered, youwill
go to State 1. From there, if a "0" is encountered, you willgo to State
2, and so on. When you reach State 7, an LED willlight. If you make an
error anywhere in the middle of the sequence, you'll be sent backto State
0. When you reach State 8 (1000), (You get there by enteringthe final
"1" while in State 7), an LED will light up. Any input, a "0"or a "1" in
State 8 will send you back to State 0. (You can change this ifyou'd
like).

With that in mind, you can decide that you'll need 4 (2^4 > 9> 2^3)
flip-flops and can build a transition table. It wouldactually be much
easier if you could eliminate one state because then you would onlyneed
three flip flops, but since this is practice it will help you moreto do
it this way.

I attached a possible transition table using D Flip Flops. DFF's are
easy to use because the output is the same as the input. Ialso did two
of the Karnaugh maps. You will need to do two more todetermine the
expression for each flip flop.

After you've determined that, you can make a final logic gate foryour
LED. Since you only want the LED on when you enter state 8(1000), you
will make a 4-input AND gate with inputs wx'y'z' and an output thatgoes
to an LED.
Your last name starts with A, if I remembercorrectly, and the ASCII number
for "A" is 1000001.

I'm not sure what he means by "detect that sequence," but we couldsay,
for instance, that if that sequence is entered, it will light anLED.

So, one way to do that is to make a state machine with 9 states (1for
each number and an extra one for errors and another one for theoutput
LED).

For example, you start out in State 0. If a "1" is encountered, youwill
go to State 1. From there, if a "0" is encountered, you willgo to State
2, and so on. When you reach State 7, an LED willlight. If you make an
error anywhere in the middle of the sequence, you'll be sent backto State
0. When you reach State 8 (1000), (You get there by enteringthe final
"1" while in State 7), an LED will light up. Any input, a "0"or a "1" in
State 8 will send you back to State 0. (You can change this ifyou'd
like).

With that in mind, you can decide that you'll need 4 (2^4 > 9> 2^3)
flip-flops and can build a transition table. It wouldactually be much
easier if you could eliminate one state because then you would onlyneed
three flip flops, but since this is practice it will help you moreto do
it this way.

I attached a possible transition table using D Flip Flops. DFF's are
easy to use because the output is the same as the input. Ialso did two
of the Karnaugh maps. You will need to do two more todetermine the
expression for each flip flop.

After you've determined that, you can make a final logic gate foryour
LED. Since you only want the LED on when you enter state 8(1000), you
will make a 4-input AND gate with inputs wx'y'z' and an output thatgoes
to an LED.

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