Solving For X: Variable Values In Logical Equations
Hey guys! Let's dive into how to find the values of X for different logical equations. This is super important in digital logic and computer science, so let's break it down step by step. We'll look at several examples, making sure you understand the process. So, grab your thinking caps, and let's get started!
(a) X = (A+B)C + B
In this first equation, X = (A+B)C + B, we need to figure out all the possible values of X based on the variables A, B, and C. To do this, we'll use a truth table. Truth tables are our best friends when dealing with logical expressions because they let us systematically evaluate every possible combination of inputs. For variables A, B, and C, we have 2^3 = 8 possible combinations. Let’s construct the truth table:
| A | B | C | A+B | (A+B)C | (A+B)C + B | X |
|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 | 0 | 0 |
| 0 | 1 | 0 | 1 | 0 | 1 | 1 |
| 0 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 | 1 | 1 | 1 |
| 1 | 1 | 0 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 |
Breaking it Down:
- A, B, C Columns: These show all the combinations of inputs (0 and 1) for the variables.
- A+B Column: This is the logical OR of A and B. It’s 1 if either A or B (or both) are 1.
- (A+B)C Column: This is the logical AND of (A+B) and C. It’s 1 only if both (A+B) and C are 1.
- (A+B)C + B Column: This is the logical OR of (A+B)C and B. It’s 1 if either (A+B)C or B (or both) are 1.
- X Column: This is the final value of X based on the equation. As you can see, X is 1 in most cases except when B is 0 and C is 0, or when A is 0, B is 0 and C is 1.
So, by looking at the truth table, we've nailed down exactly when X will be 0 or 1 for every possible input. Cool, right?
(b) X = \overline{(A+B)}C
Next up, let's tackle the equation X = \overline{(A+B)}C. This one involves a NOT operation (the overline), so it's crucial to get that negation right. Again, we'll roll out the truth table method. We've got A, B, and C, so 8 combinations to consider. Let’s see how it shakes out:
| A | B | C | A+B | \overline{(A+B)} | \overline{(A+B)}C | X |
|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 1 | 0 | 0 |
| 0 | 0 | 1 | 0 | 1 | 1 | 1 |
| 0 | 1 | 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 | 0 | 0 | 0 |
| 1 | 1 | 0 | 1 | 0 | 0 | 0 |
| 1 | 1 | 1 | 1 | 0 | 0 | 0 |
Let's break it down:
- A, B, C Columns: All input combinations.
- A+B Column: The logical OR of A and B.
- \overline{(A+B)} Column: The logical NOT of (A+B). This is 1 only when both A and B are 0.
- \overline(A+B)}C Column and C. It’s 1 only when \overline{(A+B)} and C are both 1.
- X Column: The final value of X. Notice that X is 1 only when A and B are 0, and C is 1.
So, in this case, X is only high (1) in a very specific condition. Understanding these negations is key, guys!
(c) X = A\overline{B}C + AB
Now, let’s jump into the equation X = A\overline{B}C + AB. This one's a bit more complex with both AND, OR, and NOT operations. But fear not! Our trusty truth table will guide us through. We still have three variables, A, B, and C, so let’s map out the possibilities:
| A | B | C | \overline{B} | A\overline{B}C | AB | A\overline{B}C + AB | X |
|---|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 |
| 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 |
| 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 |
Here’s the breakdown:
- A, B, C Columns: Input combinations as always.
- \overline{B} Column: The logical NOT of B.
- A\overlineB}C Column, and C. This is 1 only when A and C are 1, and B is 0.
- AB Column: The logical AND of A and B. This is 1 only when both A and B are 1.
- A\overlineB}C + AB ColumnC and AB. This is 1 if either A\overline{B}C or AB (or both) are 1.
- X Column: The final value of X. We see X is 1 when A is 1 and either B is 0 and C is 1, or A and B are both 1.
This equation has some interesting behavior. By breaking it down, we can clearly see the conditions that make X true (1).
(e) X = (A+B)(\overline{A}+B)
Let's move on to X = (A+B)(\overline{A}+B). This equation combines OR and AND operations along with a NOT, giving us a slightly different flavor. Again, our trusty truth table will come to the rescue. With just two variables, A and B, we only have 2^2 = 4 combinations to deal with, making it a bit simpler:
| A | B | A+B | \overline{A} | \overline{A}+B | (A+B)(\overline{A}+B) | X |
|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 0 | 0 | 0 |
| 1 | 1 | 1 | 0 | 1 | 1 | 1 |
Here’s what's happening:
- A, B Columns: Input combinations for A and B.
- A+B Column: The logical OR of A and B.
- \overline{A} Column: The logical NOT of A.
- \overlineA}+B Column and B.
- (A+B)(\overlineA}+B) Column+B).
- X Column: The final value of X. Looking at the truth table, X is 1 when B is 1, regardless of A’s value.
This equation highlights how different logical operations interact. Understanding these interactions is key to simplifying and optimizing logic circuits.
(f) X = (A+BC)(\overline{B}+\overline{C})
Last but not least, we have X = (A+BC)(\overline{B}+\overline{C}). This equation looks complex, but we’ll break it down using our truth table method. We've got A, B, and C, so 8 combinations to consider. Let’s get to it:
| A | B | C | BC | A+BC | \overline{B} | \overline{C} | \overline{B}+\overline{C} | (A+BC)(\overline{B}+\overline{C}) | X |
|---|---|---|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 |
| 0 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
Let’s dissect this one:
- A, B, C Columns: All input combinations.
- BC Column: The logical AND of B and C.
- A+BC Column: The logical OR of A and BC.
- \overline{B} Column: The logical NOT of B.
- \overline{C} Column: The logical NOT of C.
- \overlineB}+\overline{C} Column and \overline{C}.
- (A+BC)(\overlineB}+\overline{C}) Column+\overline{C}).
- X Column: The final value of X. We observe that X is 1 when A is 1 and B and C are not both 1.
This equation showcases how multiple logical operations combine to create specific conditions for X. Pretty neat, huh?
Wrapping Up
Alright, guys, we’ve walked through several examples of finding values of X for different logical equations. The key takeaway here is the power of truth tables. They let us systematically evaluate every possible scenario and see exactly how the variables interact. Remember, breaking down complex equations into smaller, manageable parts makes the whole process much easier. Keep practicing, and you’ll become a logic equation whiz in no time!