Chapter 10, Problem 1SE

To determine

To write: The Boolean expression and truth table for the given circuit.

Answer to Problem 1SE

The expression of the circuit is $Y=\underset{\_}{\left(x\prime \vee y\right)\wedge \left(x\prime \vee z\prime \right)}$.

Explanation of Solution

Suppose, that the output of the first AND gate is $y_1$ whose inputs are $x,z$. So, the output of the AND gate is $y_1=\left(x\wedge z\right)$.

And suppose the output of the first OR gate is $y_2$ which output is, input of the first AND gate. So, the output of the NOT gate is $y_2=\left(x\wedge z\right)\prime$

Let the output of the OR gate be $y_3$ whose inputs are $x\prime ,y$ Then, the output of the OR gate is $y_3=\left(x\prime \vee y\right)$.

If the output of the second AND gate is $Y$ whose inputs are the output of the NOT gate and OR gate.

So, the output of the AND gate is $Y=\left(y_3\wedge y_2\right)$.

Thus, the output of the circuit is as shown below.

$\begin{array}{c}Y=\left(y_3\wedge y_2\right)\\ =\left[\left(x\prime \vee y\right)\wedge \left(x\wedge z\right)\prime \right]\\ =\left(x\prime \vee y\right)\wedge \left(x\prime \vee z\prime \right)\end{array}$

Therefore, the expression of the circuit is $Y=\underset{\_}{\left(x\prime \vee y\right)\wedge \left(x\prime \vee z\prime \right)}$.

Now, the truth table of the expression of $Y=\left(x\prime \vee y\right)\wedge \left(x\prime \vee z\prime \right)$ is as shown below.

$x$	$x\prime$	$y$	$z$	$z\prime$	$Y$
0	1	0	0	1	1
0	1	0	1	0	1
0	1	1	0	1	1
0	1	1	1	0	1
1	0	0	0	1	0
1	0	0	1	0	0
1	0	1	0	1	1
1	0	1	1	0	0

Table 1

Hence, the truth table 1 for $Y=\underset{\_}{\left(x\prime \vee y\right)\wedge \left(x\prime \vee z\prime \right)}$ is obtained.