Solutions for Test 1 - Introduction to Computer Engineering | ECE 2030, Exams of Electrical and Electronics Engineering

Download Solutions for Test 1 - Introduction to Computer Engineering | ECE 2030 and more Exams Electrical and Electronics Engineering in PDF only on Docsity! Georgia Institute of Technology ECE 2030D: Introduction to Computer Engineering Instructor: Mike Reid Fall Semester 2000 Test 1 Friday, September 15, 2000 Name: SOLUTIONS Directions: © One 8 1/2” x 11” page of notes is allowed (front and back). ¢ Calculators are not allowed. © Verify that you have 6 pages. . In order to get partial credit, you must show all of your work. Please make your answers clear and obvious. If] can’t determine your answer or I can’t read your handwriting, then you won't get credit. ¢ Please ask me if you need clarification with a problem. Good Luck! Problem Points Grade Tast 4 Dr afin huctd A + 1 25 25 2 15 15 3 18 is Jo-too 21 4 10 10 5 is 15 Bo's (7 6 20 20 Total 100 100 Fo 'g 13 Homework 3 2 60's 8 High loo (S people) < 60 $ Lu 42 Averwe 82  Problem 1. (25 points) No partial credit was given for any of these 5 problems. la. (5 points) If a binary code were assigned to uniquely identify 35 distinct objects, what would be the minimum number of bits required? Since 25 = 32 and 2° = 64 > a minimum of 6 bits are required 1b. (5 points) Find the simplest expression for f from the following logic diagram. AB+B=B(A+1)=B(1)=B f= B 1c, (5 points) Find the simplest expression for f from the following logic diagram. AA+1=A+1=1 f=__1 ld. (5 points) Find the simplest expression for f from the following logic diagram. (A+A)A=(DA=A f= A le. (5 points) The following timing diagrams could be from which of the following functions? Circle only one answer. Only one answer is correct. (a) f= XY x ¥ f=XY (b) f= K+Y 0 0 0 (o) f= XY 0 1 0 (a) f= X. Y > Answer 1 0 1 ()f=X+Y 1 1 0 (f= X+¥ Problem 5. (15 points) Find the simplified sum-of-products (SOP) expressions for the following Boolean expressions using Boolean Algebra techniques. a. (5 points) F = ABA+A Fo AB +A+A DeMorgan’s Theorem. F=AB+1 Identity: ¥+X=1 Fe=l Identity: 4+1=1 Alternatively, F=(At+ B)A+A DeMorgan’s Theorem F=AA+AB+A Distributive Theorem F=A+ABt+A Identity: XX=X F=AQ+B)+A4 Distributive Theorem F=AQ)+4 Identity, X+1=1 F=At+tA Identity: X-1=X F= Identity: X¥+X =] b. (5 points) F= B(A+C)+B F=B(A+C)- B DeMorgan’s Theorem F=(B+A+OB DeMorgan’s Theorem, Identity: ¥ =X F =(B+AC)B DeMorgan’s Theorem F =BB+ABC Distributive Property F =0+ ABC Identity: XX =0 F = ABC Identity: X+0=X Alternatively, F=AB+BC+B DeMorgan’s Theorem Fea (AB)(BC)(B) DeMorgan’s Theorem F =(A+BYBt+ C)B DeMorgan’s Theorem, Identity: y¥=X F=(At+ B)BC Distributive Property, Identity: XX =0 F = ABC +BBC Distributive Property F = ABC +0 Identity: XX =0 F = ABC Identity: ¥ +0 =X c. (5 points) F = ABC + A+B+C(D+ AB) A “A+C The AA term simplifies the equation to (X)(0)+C. Therefore, F = C, since (X)(0)= 0. No partial credit was given for this problem. Problem 6. (20 points) 5a. (5 points) Find the simplified sum-of-products (SOP) expression for the function, f(W,X,Y,Z), from the K-map below. Ye - ie 0 10 u a 7WX a 0 0 10 00 01 11 5b. (5 points) Find the simplified product-of-sum (POS) expression for the function, f(W,X,Y,Z), from the K-map below. Ye - Wy 10 Xe 10 1 ot To XY mie} tt since F=XZ+XV > f= XZ+XY = (KZ XV)=(X+ZX+¥) 5c. (10 points) Using a 4-variable K-map, find the simplified sum-of-products (SOP) expression for the following function: AW,XY,Z)= WY -Z + WXYZ +WeX-Z +X¥-Z Yr Wx 00) ol ul 10 00 1 r 0 0 Ci ol 1 0 0 0 li 0 0 0 GT 10 1 @) 0 0 fin XK z XR