I need help with a 4 bit bcd up/down counter that counts 1001 to 0000 ascendant and 0000 to 1001 for descendant
A->E, Q3->D, Q2->C, Q1->B, Q0->A
J3= Q3'(A'Q2Q1Q0+AQ2'Q1'Q0') = D'(E'CBA+EC'B'A')
K3= A'Q0+AQ0' = E'A+EA'
J2= Q2'(A'Q3'Q1Q0+AQ3Q1'Q0') = C'(E'D'BA+EDB'A')
K2= A'Q1Q0+AQ1'Q0' = E'BA+EB'A'
J1= A'Q3'Q1'Q0+AQ3'Q2Q1'Q0'+AQ3Q2'Q1'Q0' = E'D'B'A+ED'CB'A'+EDC'B'A'
K1= A'Q3Q1Q0+AQ3'Q1Q0' = E'DBA+ED'BA'
How did you find the problem?I think I found the answer, cause k1 = E'D'BA+ED'BA'
Not sure what your asking for.Is there any aplication for Kmaps for those kind of ecuations?
This is my T. tableNot sure what your asking for.
If you had populated your truth table with the don't care states and transferred all entries to Kmaps, the equation for K1 would have simplified to E'A+EA'. No amount of boolean algebra would have allowed you to get there from E'D'BA+ED'BA' = D'B(E'A+EA').
EDIT: This is the Kmap for KB from my design. I used Z as the control and the counter counted up with Z=1 and down with Z=0.
View attachment 148643
I think what they teach in schools now is to use 2 4x4 Kmaps to solve 5 variables. With the technique I use, I can solve for 6 variables in one Kmap.
I take it that you haven't studied Kmaps yet? That's one of the problems with Homework Help. We don't know what you've studied or what you're currently studying, so we either have to ask a lot of questions, or make assumptions.My teacher told me to do it like that
The only equation I agree with is K1, but the groupings you made in the Kmap don't agree with the answer. To simplify to 2 variables, you needed a grouping of 8. You show two groups of 4:Not sure if i made them correct, cause is my first time doing Kmaps this way.
I think the preferred method is to use 2 4x4 maps.I have, but don´t know how to do Kmaps for 5 variables.
How are you eliminating Q3'? There's no identity/rule that allows it.A'Q3'Q0+AQ3'Q0' is the answer but next simp. is AQ0'+A'Q0
I´m learning by myself to this, I am newbie.Not sure what your asking for.
If you had populated your truth table with the don't care states and transferred all entries to Kmaps, the equation for K1 would have simplified to E'A+EA'. No amount of boolean algebra would have allowed you to get there from E'D'BA+ED'BA' = D'B(E'A+EA').
EDIT: This is the Kmap for KB from my design. I used Z as the control and the counter counted up with Z=1 and down with Z=0.
View attachment 148643
I think what they teach in schools now is to use 2 4x4 Kmaps to solve 5 variables. With the technique I use, I can solve for 6 variables in one Kmap.
And I am wrong from that sorry I was reading my first ec.How are you eliminating Q3'? There's no identity/rule that allows it.
Your truth table has 20 rows, it should have 32.Can you tell me how did you get that equation, from what truth table did you get that, and how that i didn´t transfer all the data.
From the data in your Kmap, and the groupings, you can't get the answer you wrote down for K1.And I am wrong from that sorry I was reading my first ec.
Either you misunderstood your teacher, or the teacher is wrong. With 5 variables, a fully populated truth table would have 32 rows:The teacher told me that my truth table should have just 20 rows as I have it. Is it wrong?
by Jake Hertz
by Jake Hertz