Practice_Final_Answers

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ECE2277 Practice Final ECE2277A: Practice Final Exam (Answers) Professor: Arash Reyhani & John McLeod Based on Exam given 2020 12 13 Part I: Number Systems [12 marks] 1. Convert (10110100) 2 into unsigned decimal and hexadecimal. [2 marks] Decimal 180 Hexadecimal B4 2. Convert (1110110) 2 into the appropriate decimal number according to the signed-magnitude, signed-1’s complement, and signed-2’s complement number systems. [3 marks] Signed-Magnitude 10 Signed-1’s Complement -137 Signed-2’s Complement -138 3. Convert decimal (35) 10 into 2421 binary code. [2 marks] 2421 code 00111011 4. What is the maximum range of signed-2’s complement and unsigned decimal numbers that can be accom- modated in a 9 -bit binary number system? (Please include negative signs as appropriate.) [4 marks] System Minimum Value Maximum Value Signed-Decimal -256 255 Unsigned-Decimal 0 511 5. What is the minimum number of bits needed to encode (26) 10 distinct quantities? [1 marks] Number of bits 5 1

ECE2277 Practice Final Part II: Logic Operations [4 marks] 6. Which logic operation with inputs x and y is represented by the Boolean function f ( x, y ) = ∑ (1 , 2 , 3) ? [1 mark] (a) AND (b) OR (c) NAND (d) NOR (e) XOR (f) XNOR b 7. Which logic operation with inputs x and y is represented by the Boolean function f ( x, y ) = Q (1 , 2 , 3) ? [1 mark] (a) AND (b) OR (c) NAND (d) NOR (e) XOR (f) XNOR d 8. What is the canonical sum-of-minterms (SOM) representation the logic operation x NAND y ? Please enter your answer as comma-separated indices (i.e. 0 , 1 , 2 , 3 ). [1 mark] f ( x, y ) = X (0 , 1 , 2) 9. What is the canonical product-of-maxterms (POM) representation the logic operation x XOR y ? Please enter your answer as comma-separated indices (i.e. 0 , 1 , 2 , 3 ). [1 mark] f ( x, y ) = Y (0 , 3) 2

ECE2277 Practice Final Part III: Boolean Expressions [20 marks] 10. What Boolean expression is represented by this Karnaugh map? [4 marks] yz wx 00 01 11 10 00 01 11 10 0 x x x 1 1 x 0 1 1 1 1 0 0 0 1 (a) f ( w, x, y, z ) = wx ′ + wyz + w ′ xy ′ (b) f ( w, x, y, z ) = wxz ′ + w ′ y + wxy ′ + w ′ x ′ (c) f ( w, x, y, z ) = wx ′ + wz + w ′ xy ′ (d) f ( w, x, y, z ) = wx ′ + wyz + w ′ y ′ z a 11. Which of the following Boolean expressions is equivalent to: f ( x, y, z ) = x ′ z + xz ′ + yz ′ [4 marks] (a) f ( x, y, z ) = ( x + y + z ) ( x ′ + z ′ ) (b) f ( x, y, z ) = ( x + y + z ) ( x ′ + y ′ + z ′ ) (c) f ( x, y, z ) = ( x + y ) ( x ′ + z ′ ) (d) f ( x, y, z ) = ( x + z ′ ) ( x ′ + z ) ( y ′ + z ) a 12. What are the minterms of: f ( x, y, z ) = x ′ y ′ + xy + yz ′ [4 marks] (a) f ( x, y, z ) = ∑ (1 , 2 , 6 , 7) (b) f ( x, y, z ) = ∑ (3 , 4 , 5) (c) f ( x, y, z ) = ∑ (0 , 1 , 2 , 5 , 6 , 7) (d) f ( x, y, z ) = ∑ (0 , 1 , 2 , 6 , 7) d 3

ECE2277 Practice Final 13. Find the sum-of-products (SOP) simplification of: f ( w, x, y, z ) = X (0 , 1 , 2 , 5 , 8 , 10 , 11 , 13 , 15) , with don’t cares d ( w, x, y, z ) = X (3 , 4 , 7 , 14) . You may assume that all variables ( w, x, y, z ) and their complements ( w ′ , x ′ , y ′ , z ′ ) are available. [4 marks] yz wx 00 01 11 10 00 01 11 10 f ( w, x, y, z ) = wy + xz + w ′ z + x ′ z ′ 14. Find the product-of-sums (POS) simplification of: f ( w, x, y, z ) = X (0 , 2 , 4 , 7 , 8 , 9 , 10 , 11 , 14) , with don’t cares d ( w, x, y, z ) = X (1 , 5 , 6 , 12) . You may assume that all variables ( w, x, y, z ) and their complements ( w ′ , x ′ , y ′ , z ′ ) are available. [4 marks] yz wx 00 01 11 10 00 01 11 10 f ( w, x, y, z ) = ( w + x + z ′ ) ( w ′ + x ′ + z ′ ) 4

ECE2277 Practice Final Part IV: Combinational Circuits [18 marks] 15. The circuit diagram is shown below. The 3 -to- 8 decoder chip output is active high. You may assume it is always enabled. 2 2 x 2 1 y 2 0 z 3-to-8 Decoder 7 6 5 4 3 2 1 0 f ( x, y, z ) What function is implemented by this circuit? [2 marks] (a) f ( x, y, z ) = yz + x ′ y (b) f ( x, y, z ) = z ′ + y ′ (c) f ( x, y, z ) = y ′ (d) f ( x, y, z ) = xz ′ + y ′ d 16. Implement the Boolean expression: f ( x, y, z ) = x ′ z + yz ′ + y ′ z, using the following 4 × 1 multiplexer circuit. Set the data inputs D 0 , D 1 , D 2 , and D 3 to the appropriate input from z , z ′ , 1 , or 0 . You may assume the multiplexer is always enabled. [4 marks] f ( x, y, z ) 2 1 x 2 0 y 4 × 1 MUX D 3 D 2 D 1 D 0 Input Function D 0 = z D 1 = 1 D 2 = z D 3 = z ′ 5

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