![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.](https://image.slidesharecdn.com/deppt27ce1-181004132429/85/variable-entered-map-digital-electronics-3-320.jpg)
variable entered map digital electronics
- 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)