ove using Karnaughmap that the two c Question

Question

ove using Karnaughmap that the two circuit diagram Fig (1) & Fig (2)shown below are two different simplification oof the functi

sameerosameer8890 · Answer

the question requires proving that the two circuit diagrams shown (fig 1 and fig 2) are different simplified forms of the same logical function by using karnaugh maps (k-maps). explanation: a karnaugh map is a graphical method for simplifying boolean expressions. it organizes truth values in a way that makes it easier to identify groups of 1s, which correspond to simplified product terms in the expression. to solve this problem:

derive the boolean expression for each circuit diagram. draw the karnaugh map for each boolean expression. group the adjacent 1s in each karnaugh map to simplify the expressions. compare the simplified expressions obtained from both circuits. if the simplified expressions match, it confirms that both circuit diagrams represent the same logical function, though in different simplified forms. thus, using karnaugh maps provides a clear way to verify that the two given circuits are logically equivalent.

haider370 · Answer

since you mentioned “fig (1) & fig (2) shown below” but no image is attached, i’ll explain how you would generally do this. you can later share the diagrams for a specific solution. ✅ general steps to prove two circuits are equivalent simplifications using karnaugh map: 🔹 step 1: determine the original boolean function if given as a truth table or a standard expression, extract the minterms (the rows where output = 1). for example, if the function is: f(a, b, c) = σ(1, 2, 3, 5) 🔹 step 2: draw the karnaugh map for 3 variables (a, b, c), your k-map will have 8 cells. fill in the cells based on the minterms. ab \ c010001011111001010 (example for f = σ(1, 2, 3, 5)) 🔹 step 3: simplify the boolean function using k-map group 1s into rectangles of 1, 2, or 4 cells. write the simplified boolean expressions. let’s say this gives you: figure 1: a'b + b'c figure 2: b(a' + c) both are valid simplifications of the same function, just written differently. 🔹 step 4: compare with the circuit diagrams figure 1 implements one expression figure 2 implements another if both match their respective simplifications, and you proved both are logically equivalent (either using boolean algebra or the k-map), then you’ve proven that both circuits represent different simplifications of the same function. ✅ final statement: using a karnaugh map, we identify that both circuits represent simplified forms of the same original boolean function. though the circuits differ in structure, they yield the same output for all input combinations, proving they are logically equivalent.

mimikht · Answer

objective
use a karnaugh map (k-map) to prove that the two circuit diagrams in fig (1) and fig (2) represent two different simplifications of the same boolean function.
step-by-step outline
1. define the boolean function
if not explicitly given, derive or assume the boolean expression from the truth table or circuit diagram in fig (1) and fig (2).let’s suppose the function is given by a truth table or original boolean expression:f(a, b, c, d) = some combination of minterms.
2. draw the karnaugh map
use a 2-variable (4-cell), 3-variable (8-cell), or 4-variable (16-cell) k-map depending on how many variables are involved.fill the k-map with 1’s based on the function’s minterms.
3. simplify using the k-map
group the adjacent 1’s into groups of 1, 2, 4, 8, etc. (in powers of 2).from these groups, derive the simplified boolean expression (first version).this expression should match one of the circuit diagrams &mdash; say, fig (1).
4. find an alternative simplification
sometimes, there’s more than one way to group the 1’s in a k-map.try a different grouping to get a different (but logically equivalent) expression.this second expression should correspond to fig (2).
5. conclude that both circuits are equivalent
show that both simplified expressions (from fig 1 and fig 2) yield the same output for all combinations of inputs &mdash; i.e., they are logically equivalent, even if the circuit designs differ.
✅ example
let’s assume a sample function:f(a, b, c) = σm(0,1,2,5,6,7)
step 1: k-map
ab \ c010011011011111001

JangoTheOfficial · Answer

proof of circuit equivalence via karnaugh map simplification
introduction this analysis demonstrates that two given combinational logic circuits (figure 1 and figure 2) represent distinct minimal forms of the same boolean function. the proof employs karnaugh map (k map) simplification and boolean algebra to verify functional equivalence.
theoretical background a k map is a graphical method for simplifying boolean expressions by grouping adjacent minterms. valid groupings must follow these rules:
1. groups must contain 1, 2, 4, 8, ... minterms (powers of 2).
2. adjacency includes wrap around edges (top to bottom, left to right).
3. each group should be as large as possible.
methodology
step 1: function extraction assume both circuits implement the 3-variable function: f(a,b,c) = σm(1,3,5,7) = ¬a·b·¬c + ¬a·b·c + a·¬b·c + a·b·c
step 2: k map construction the 3 variable k map (variables a vs. bc) is constructed as follows:
    bc→  00  01  11  10  
a=0 ↓ 0 1 1 0 a=1 ↓ 0 1 1 0
step 3: k-map simplification
option 1: vertical grouping all '1's align vertically under c=1 (columns 01 and 11). this yields: f = c
option 2: horizontal grouping (incorrect) grouping rows a=0 and a=1 horizontally for b=1 would suggest: f = b however, this contradicts the truth table when b=1 and c=0 (output should be 0).
step 4: verification the truth table confirms f=c is correct:
a b c | f 0 0 0 | 0 0 0 1 | 1 0 1 0 | 0 0 1 1 | 1 1 0 0 | 0 1 0 1 | 1 1 1 0 | 0 1 1 1 | 1
conclusion the only valid minimal form is f = c. while the horizontal grouping f = b appears plausible, it fails to satisfy all minterms. thus:
- figure 1 (implementing f = c) is correct.
- figure 2 (if implementing f = b) is incorrect.
discussion this example highlights two critical aspects of k map simplification:
1. proper adjacency identification is essential. here, vertical grouping was mandatory.
2. all possible groupings must be verified against the truth table.
references
1. mano, m. morris. digital design. pearson, 2018. (chapter 3: simplification methods)
2. wakerly, j. f. digital design principles and practices. prentice hall, 2017.

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