Digital Circuits
Midterm Solutions
1. Mix of Questions (40 points)
Answer the following 8 questions. Simple reasoning will be enough. Each of them worth
5 points.
(a) Covert the binary number 1011.01(2) to decimal.
(b) Add binary numbers 101.011(2) + 11.101(2) (Your answer should be a binary number
as well).
(c) Determine the 2’s complement of 10011000.
(d) Construct a truth table of X = ĀB + AB C̄ + ĀC̄ + AB̄C.
(e) Convert X = AB(CD + EF ) to sum-of-product (SOP) form.
(f) Implement a logic circuit for X = AD + B C̄.
(g) For the full-adder with input A = 0, B = 1, Cin = 1, determine the outputs Σ and
Cout .
(h) Design a simple decoder that detects the presence of the binary code 0110.
Solution: Mix of Questions
(a) 23 + 21 + 20 + 2−2 = 11.25.
(b) 1001.
(c) 1’s complement of the given number is 01100111. We need to add 1 after that which
gives us 01101000.
(d) X = ĀB + AB C̄ + ĀC̄ + AB̄C.
A B C X
0 0 0 1
0 0 1 0
0 1 0 1
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 0
(e) We have
X =AB(C̄D + EF ) (1)
=AB + C̄D + EF (2)
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=Ā + B̄ + C̄D · EF (3)
=Ā + B̄ + (C + D̄)(Ē + F̄ ) (4)
=Ā + B̄ + C Ē + D̄Ē + C F̄ + D̄F̄ . (5)
(f) Figure 1 shows the logic circuit.
Figure 1: Problem 1(f).
(g) Σ = 0 and Cout = 1.
(h) Figure 2 shows the decoder.
Figure 2: Problem 1(h).
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2. Karnaugh Map (30 points)
Let X = (A + BC)(AC + D) + ĀB C̄ D̄.
(a) Develop a truth table of X. (10 points)
(Hint: You can simplify the formula first)
(b) Use a Karnaugh map to reduce X to a minimum SOP form.(10 points)
(c) Use a Karnaugh map to reduce X to a minimum POS form.(10 points)
Solution: Karnaugh Map.
(a) We can simplify X first.
X =(A + BC)(AC + D) + ĀB C̄ D̄ (6)
=A + BC + AC + D + ĀB C̄ D̄ (7)
=Ā(B̄ + C) + (Ā + C̄)D + ĀB C̄ D̄ (8)
=ĀB̄ + ĀC + ĀD + C̄D + ĀB C̄ D̄. (9)
This provides
A B C D X
0 0 0 0 1
0 0 0 1 1
0 0 1 0 1
0 0 1 1 1
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 1
1 0 1 0 0
1 0 1 1 0
1 1 0 0 0
1 1 0 1 1
1 1 1 0 0
1 1 1 1 0
(b) It is not hard to show that X = Ā + C̄D.
(c) It is not hard to show that X = (Ā + D)(Ā + C̄).
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Figure 3: Problem 2.
3. Functions of Combinational Logic (30 points)
(a) For the multiplexer in Figure 4, input states are given by D0 = 1, D1 = 1, D2 =
0, D3 = 0. Then, determine the output waveform when the data-select inputs are
sequenced as shown by the waveforms in Figure 5. (15 points)
Figure 4: Multiplexer.
Figure 5: Data-Select Input Waveforms.
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(b) Suppose you have two 4-bit odd parity generators as decribed in Figure 6. This
4-bit odd parity generators outputs 1 if even numbers of inputs are 1 and outputs
0 if odd numbers of inputs are 1. For example, the output is P = 1 if A = 1, B =
0, C = 1, D = 0, and the output is P = 0 if A = 1, B = 0, C = 1, D = 1. Construct
8-bit odd parity generator using two 4-bit odd parity generators with one additional
logic gate. (15 points)
(Hint: Again, 8-bit odd parity generator outputs 1 if even number of inputs are 1
and outputs 0 if odd numbers of inputs are 1)
Figure 6: Data-Select Input Waveforms.
Solution: Functions of Combinational Logic
(a) Figure 7 shows the output waveform. Y = 1 if and only if S1 = 0, S0 = 0 or
S1 = 0, S0 = 1. In other words, Y = S¯1 .
Figure 7: Problem 3(a).
(b) Figure 8 shows the 8-bit odd parity generators using two 4-bit odd parity generator
and one XNOR gate. This is because the output should be 1 if and only if two 4-bit
odd parity generators have the same output.
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Figure 8: Problem 3(b) 8-bit Odd Parity Generator.
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