ppt on variable entered map method(VEM) in digital electronics useful for GTU students

1. SY CE 1 BATCH B
170410107027
DHANANJAYSINH JHALA

2. CONTENTS
• Why VEM is used?
• Example 1 & verification
• Grouping in VEM
• Example 2 & verification

3. WHY VEM IS USED ?
• Due to the difficulty in managing K map exceeding4 variables like 5
or 6 variables we have the technique called map entered
variables[VEM].
• As we all know for n variables a K map has 2n variables.
• To allow a smaller map to handle a larger number of variables that is
to reduce the number of squares the concept of variable entered
mapping was introduced.
• This is done by choosing one variable as map entered variable and
thus including it in the k map along with the zeros, ones and the don't
care conditions.

4. EXAMPLE 1:
A B C Y
0 0 0 1
0 0 1 1
0 1 0 0
0 1 1 0
1 0 0 1
1 0 1 1
1 1 0 0
1 1 1 1
𝑌 = 𝑓 𝐴, 𝐵, 𝐶 = 𝑚𝑖(0,1,4,5,7)
000
000
000
MEV
0
1
C
1
Truth table considering C as MEV
is
A B Y
0 0 1
0 1 0
1 0 1
1 1 C

5. MEV K-map is
1 0
1 C
B
A
1
0
0 1

6. GROUPING IN VEM
Considering our example, you can group C and C’s, 1 and C’s, 1 and
C’s but your cannot group C and C′
s.
Also either group 1 with C and C, or group 1 again i.e.

7. So grouping in our MEV K-map is as shown and equation is Y = 𝐵 +
𝐴 𝐶
1 0
1 C
B
A
0
10
1

8. VERIFICATION USING K-MAP
𝑌 = 𝑓 𝐴, 𝐵, 𝐶 = 𝑚𝑖(0,1,4,5,7)
A B C Y
0 0 0 1
0 0 1 1
0 1 0 0
0 1 1 0
1 0 0 1
1 0 1 1
1 1 0 0
1 1 1 1
1 1 0 1
1 1 0 0
BC
A 00 01 11 10
1
0
Y = 𝐵 + 𝐴 𝐶, verified!

9. EXAMPLE
2:𝑌 = 𝑓 𝐴, 𝐵, 𝐶, 𝐷 = 𝑚𝑖(0,1,4,5,6,7,11,15)
A B C D Y
0 0 0 0 1
0 0 0 1 1
0 0 1 0 0
0 0 1 1 0
0 1 0 0 1
0 1 0 1 1
0 1 1 0 1
0 1 1 1 1
1 0 0 0 0
1 0 0 1 0
1 0 1 0 0
1 0 1 1 1
1 1 0 0 0
1 1 0 1 0
1 1 1 0 0
1 1 1 1 1
00
0000
00
0000
0
1
1
1
D
0
D
0
Truth table considering C as MEV
isA B C Y
0 0 0 1
0 0 1 0
0 1 0 1
0 1 1 1
1 0 0 0
1 0 1 D
1 1 0 0
1 1 1 D
MEV

10. MEV K-map is
1 0 1 1
0 D D 0
BC
A 00 01 11 10
1
0
So grouping in our MEV K-map is as shown and equation is Y =𝐴𝐶𝐷 +
𝐴 𝐶 + 𝐴𝐵

11. VERIFICATION USING K-
MAP
1 1 0 0
1 1 1 1
0 0 1 0
0 0 1 0
BC
A
B
00 01 11 10
01
00
Y =𝐴𝐶𝐷 + 𝐴 𝐶 + 𝐴𝐵,
verified!
A B C D Y
0 0 0 0 1
0 0 0 1 1
0 0 1 0 0
0 0 1 1 0
0 1 0 0 1
0 1 0 1 1
0 1 1 0 1
0 1 1 1 1
1 0 0 0 0
1 0 0 1 0
1 0 1 0 0
1 0 1 1 1
1 1 0 0 0
1 1 0 1 0
1 1 1 0 0
1 1 1 1 1
11
10
𝑌 = 𝑓 𝐴, 𝐵, 𝐶, 𝐷 = 𝑚𝑖(0,1,4,5,6,7,11,15)