ECE150-Homework 2 Solved

. Construct a truth table for the following:

F = (x + y)(x’ + z’)(y’ + z’)

 

x
y
z
x+y
x’ + z’
y’ + z’
F(x + y)(x’ + z’)(y’ + z’)
0
0
0
0
1
1
0
0
0
1
0
1
1
0
0
1
0
1
1
1
1
0
1
1
1
1
0
0
1
0
0
1
1
1
1
1
0
1
1
0
1
0
1
1
0
1
1
1
1
1
1
1
1
0
0
0
 
 

2. Using truth tables, show that: xz = (x+y)(x+y’)(x’+z)

 

               (x+y)(x+y’)(x’+z) truth table:

              

x
y
z
x’
y’
x+y
x+y’
x’+z
F(x+y)(x+y’)(x’+z)
F(xz)
0
0
0
1
1
0
1
1
0
0
0
0
1
1
1
0
1
1
0
0
0
1
0
1
0
1
0
1
0
0
0
1
1
1
0
1
0
1
0
0
1
0
0
0
1
1
1
0
0
0
1
0
1
0
1
1
1
1
1
1
1
1
0
0
0
1
1
0
0
0
1
1
1
0
0
1
1
1
1
1
 
 

3. The truth table for a Boolean expression is shown. Write the Boolean expression in sum-of-products

form.

 

F = x’y’z’ + x’y’z + x’yz’ + xy’z’ + xy’z

 
4. Draw the combinational circuit that directly implements the following Boolean expression:

F(x,y,z) = y’ + xy + y’z.

 

 

 
 

5. Consider the parity generator (even parity) shown in the truth table below. The parity bit Y is a function of Boolean variables A, B, and C. Represent this parity function in the following ways:

1. As a Boolean algebraic expression.

A’B’C + A’BC’ + AB’C’ + ABC

 

2. As a combinational logic diagram (logic circuit).

 

 

 
 

                                      


6.

Input
Output
A
B
X(K)
J
A(t+1)
B(t+1)
0
0
0
1
0
1
0
0
1
1
0
1
0
1
0
0
1
1
0
1
1
0
1
0
1
0
0
0
0
0
1
0
1
0
0
0
1
1
0
1
1
1
1
1
1
1
1
0
 
7.

 

 

 

8.1.

I2
I1
Z1
Z0
0
0
0
0
0
1
0
1
1
0
1
0
1
1
1
0
 
   2. Z1 = I2I’1 + I2I1

       Z0 = I’2I1
   3.

 

 
9. 1. Jack Kilby

    2. He received the Nobel Prize in Physics in 2000 for his part in the invention of the integrated circuit.