Exam 1
Chapters 1, 2, & 3.1
Sep 17th 2026
The exam will be administered in class during your corresponding section. Those with
time extensions through disability services will receive their exams at the designated
testing facility.
In the Course Documents section of Blackboard, you will find a list of the Properties
of a Switching Algebra. You should familiarize yourself with that list. It will be
available in this format during the exam.
You must show your work for each question. Answers given without proper accompanying work will not be considered and will be marked with zero points. Illegible
answers will be marked with a zero as well, as described in your syllabus.
As per our syllabus, without documentation from an appropriate provider make-up
exams will not be allowed. In the event of an appropriate excuse stating a reasonable
cause for not taking the exam in the provided time, your final exam grade will be used
replace this one. Valid exam excuses are subject to more rigor than other submissions.
Do not submit an excuse that is not on the legal SC State list with the expectation it
will be accepted. Verify first.
All problems are inspired by lecture material, solved problems from our book, and homework.
• Chapter 1: Introduction
– 1.2–1.2.1: Conversion between number systems/radix, especially binary, decimal and hexadecimal.
Example Problem & Response:
Convert 01102 from 4-bit unsigned binary to decimal.
0 × 23 + 1 × 22 + 1 × 21 + 0 × 20 = 610
or
1 × 22 + 1 × 21 = 610
or
2
2 + 21 = 610
– 1.2.2–1.2.4: Two’s complement encoding of a negative binary integer and addition of signed and
unsigned binary values.
Example Problems & Response:
∗ Convert the following 6-bit signed binary to decimal
· 101000 = -(010111 + 000001) = -011000 (from two’s complement)
· −0110002 = −(1 × 24 + 1 × 23 ) = −2410
· −710 = −0001112 = 111000 + 000001 = 1110012
· Proof of correctness: 1110012 + 0001112 = 0000002
Exam 1
Chapters 1, 2, & 3.1
Page 2 of 3
∗ Perform the following 6-bit signed binary arithmetic. Convert your 6-bit signed result to
decimal:
0000012 - 0110012
= 0000012 + −0110012 (to signed binary)
= 000001 + (100110 + 000001) (to two’s complement 6-bit binary)
= 000001 + 100111 (signed addition)
000001
+100111 (6-bit binary addition)
101000 = -(010111 + 000001) = -011000 (from two’s complement)
−0110002 = −(1 × 24 + 1 × 23 ) = −2410
– 1.2.5: Binary Coded Decimal (BCD)
∗ BCD was excluded for brevity. I want no more than eight questions.
• Chapter 2: Combinational Systems
– 2.1: Development of Truth Tables for binary expressions, especially SOP expressions.
See: Page 80, Problems 1.a–d.
– 2.3: Implementation of algebraic functions with AND, OR, and NOT gates.
See: Page 87, Problems 11.i.a and 11.i.b
– 2.4: The Complement
See: Page 87, Problems 12.a–c
– 2.5: Conversion from truth tables to minterms, minterms to truth tables, and either to or from
algebraic expressions.
See: Page 88, Problems 13.a–b.
– 2.7: Minimization of algebraic expressions.
Example Problem & Response:
Minimize a’b’c’ + a’b’c + ab’c’ to 2 Terms and 4 Literals
a’b’c’ + a’b’c + a’b’c’ + ab’c’ [P6a]
a’b’ + a’b’c’ + ab’c’ [P9a]
a’b’ + b’c’ [P9a]
See: Page 91, Problems 18.a–f. Page 84, Problems 8.a–c. Page 86, Problems 9.a–b.
– 2.8: Conversions between SOPs and POSs and vice versa. Remember, you are not minimizing,
you are converting.
See: Pages 93–95, Problems 20.a–b and 21.
Cont.
Exam 1
Chapters 1, 2, & 3.1
Page 3 of 3
• Chapter 3.1: Introduction to the Karnaugh Map
– 3.1: Plot the cover/one-implicants of an algebraic expression or minterms on a K-map.
See: Page 162, Problems 1.a–c, Especially 1.c.
∗ Ensure you can complete a three or four input K-map for any partially minimized SoP.
The End.