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ECE2030A Introduction to Computer Engineering Fall 2008 Homework Assignment # Assigned 09/08/08 Due in the first 5 min in class 09/15/ No late turn-in accepted
- (10%) Determine the numbers for all the question marks (?) shown in the following equations. Note that all these numbers are “ signed ” numbers and are represented by two’s complement using 2 bytes. Please show how you derive your answers step-by-step. You will not receive any credit if you only write down your answers with no derivation. 1.1. (-183) 10 = (?) 16 Note that the number is negative. First convert (183) 10 to base 2. 183 = 128 + 32 + 16 + 4 + 2 + 1 = 2^7 + 2^5 + 2^4 + 2^2 + 2^1 + 2^0 (183) 10 = (0000 0000 1011 0111) 2 [Note: leading zeros to ensure 2 bytes]
To get the negative complement : (1111 1111 0100 1000) 2 And then add 1. (-183) 10 (1111 1111 0100 1001) 2
Now convert to base 16 (-183) 10 = (FF49) 16
Convert to base 10 (0125) 7 = 0(7^3 ) + 1(7^2 ) + 2(7^1 ) + 5(7^0 ) = 0 + 49 + 14 + 5 (0125) 7 = (68) 10
Now, convert to base 9 (68) 10 = 7(9^1 ) + 5(9^0 )
1.3. (FFED) 16 = (?) 10
Convert to base 2 (FFED) 16 = (1111 1111 1110 1101) 2
Notice that the leading (sign) bit is a one; therefore the answer will be negative. Apply two’s complement and add one.
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Now convert from base 2 to base 10. = - (2^4 + 2^1 + 2^0 ) 10 = - (16 + 2 + 1 ) 10
(FFED) 16 = (-19) 10
Simple binary to hex conversion
(0111 0010 0000 1110 1111 1010 0110 1100) 2 = (720EFA6C) 16
Convert the digits of the right hand side one place at a time to base 10. (142)? = (1(?^2 ) + 4(?^1 ) +2(?^0 )) 10
Now solve for? 0 = (?^2 ) + 4(?) -
Using quadratic equation:
㎘4 ㎙ 㒓4⡰^ ㎘ 4䙦1䙧䙦㎘77䙧
Therefore? = -11, or 7. However, base must be positive.
- (20%) Gary tried to solve the equation 5x^2 – 50x + 125 = 0 for his algebra homework assignment. He came up with a solution x=5. All of a sudden, a flying saucer from Vega landed in his backyard and a little green man coming out from the craft told him their solution of this equation in fact is x=8 based on their number system. We (or Gray) use base-10 on earth, what are the possible base(s) these outer space visitors use in their number system?
5X^2 −50X+ 125 = 0
Let s = the unknown base. (5)s(8^2 ) 10 – (50)s(8) 10 + (125)s = 0;
Convert the coefficients one term at a time to base 10. (5)s = 5(s^0 ) = (5) 10 (50)s = 5(s^1 ) + 0(s^0 ) = (5s) 10 (125)s = 1(s^2 ) + 2(s^1 ) + 5(s^0 ) = (s^2 +2s + 5) 10
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Method 2. Start with PDN.
F 㐄 䙲A · C ㎗ B · D䙳 · 㐵E ㎗ D㐹 ㎗ C
㐄 䙲A · C ㎗ B · D䙳 · 㐵E ㎗ D㐹 · C
㐄 㐩䙲A · C ㎗ B · D䙳 ㎗ 㐵E ㎗ D㐹㐳 · C
㐄 䙲A · C ㎗ B · D ㎗ E ㎗ D䙳 · C
㐄 䙦A · C ㎗ B ㎗ D ㎗ E䙧 · C
Remember to invert the entire Boolean equation first.
Then draw the pmos in complementary form.
The final result regardless of the method used:
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F =B⋅D⋅E⊕C+C⋅ A
[Note: with XOR there exist multiple representations for the logic]
Method 1. Start with PUN.
F 㐄 B · D · E ᘘ C ㎗ C · A 㐄 B · D · 㐵E · C ㎗ C · E㐹 ㎗ C · A
㐄 B · D · 㐵E · C ㎗ C · E㐹 · C · A
㐄 䚀B · D ㎗ 㐵E · C ㎗ C · E㐹䚁 · 㐵C ㎗ A㐹
㐄 䙦B · D ㎗ E · C ㎗ C · E䙧 · 㐵C ㎗ A㐹 [Version 1] 㐄 䙦B · D ㎗ 䙦E ㎗ C䙧 · 㐵C ㎗ E㐹䙧 · 㐵C ㎗ A㐹 [Version 2]
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Version 1
Version 2
Any combination of the above pmos and nmos designs will work. One possible solution:
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㐄 Y · Z · 㐵X ㎗ 1 ㎗ X㐹 ㎗ Y · X ㎗ X · Z
㐄 Y · Z ㎗ Y · X ㎗ X · Z
RHS 㐄 䙦X ㎗ Y䙧 · 㐵X ㎗ Z㐹
㐄 X · X ㎗ X · Z ㎗ Y · X ㎗ Y · Z
㐄 0 ㎗ X · Z ㎗ Y · X ㎗ Y · Z
㐄 X · Z ㎗ Y · X ㎗ Y · Z
Therefore, LHS = RHS
- (10%) Convert the following equation to its canonical SOP form and canonical POS form. Represent your canonical form in the format of ΣΣΣΣ and ∏∏∏∏. (Derive your answer use Boolean algebra. Do not use truth table.)
F(A, B,C)=A⋅C+A⋅B+B+ C
㐄 㐵A ㎗ C㐹 ㎗ A · B ㎗ B · C
㐄 A · 䙦1 ㎗ B䙧 ㎗ C ㎗ B · C
㐄 A ㎗ C ㎗ B · C
SOP: Expand each term to encompass all literals: A 㐄 A · 㐵B ㎗ B㐹 · 㐵C ㎗ C㐹 㐄 A · B · C ㎗ A · B · C ㎗ A · B · C ㎗ A · B · C 㐄 m䙦0䙧 ㎗ m䙦1䙧 ㎗ m䙦2䙧 ㎗ m䙦3䙧 C 㐄 C · 㐵B ㎗ B㐹 · 㐵A ㎗ A㐹 㐄 A · B · C ㎗ A · B · C ㎗ A · B · C ㎗ A · B · C 㐄 m䙦0䙧 ㎗ m䙦2䙧 ㎗ m䙦4䙧 ㎗ m䙦6䙧 B · C 㐄 B · C · 㐵A ㎗ A㐹 㐄 A · B · C ㎗ A · B · C 㐄 m䙦5䙧 ㎗ m䙦1䙧
Therefore F䙦A, B, C䙧 㐄 㔳 m䙦0, 1,2,3,4,5,6䙧
POS: Use the property [X+Y·Z = (X+Y) ·(X+Z)] F䙦A, B, C䙧 㐄 A ㎗ C ㎗ B · C 㐄 A ㎗ 㐵C ㎗ B㐹 · 㐵C ㎗ C㐹 㐄 A ㎗ 㐵C ㎗ B㐹 · 䙦1䙧 㐄 A ㎗ B ㎗ C 㐄 M䙦7䙧
Therefore F䙦A, B, C䙧 㐄 㔷 M䙦7䙧
[Note: the terms of the POS plus the terms of the SOP accounts for all the 8 arrangement of literals]
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- (20%) Given the following canonical POS function in B^4 , please (1) Use Karnaugh map method to find a minimal SOP expression. (2) Draw your derived SOP function using CMOS.
F(A,B,C,D)= ∏ M(3,6,7,11,12,13,15)+d(2,8)
Note the cells are indexed in the following manner. This is arranged such that only one bit is changed from one adjacent cell to another.
CD AB^00 01 11 00 0 1 3 2 01 4 5 7 6 11 12 13 15 14 10 8 9 11 10
Step 1. Fill in Maxterms and don’t cares CD (^00 01 11 ) AB 00 0 X 01 0 0 11 0 0 0 10 X 0
Step 2. Fill in remainder with 1s CD (^00 01 11 ) AB 00 1 1 0 X 01 1 1 0 0 11 0 0 0 1 10 X 1 0 1
Step 3. Find Largest Prime Implicants CD AB^00 01 11 00 1 1 0 X 01 1 1 0 0 11 0 0 0 1 10 X 1 0 1
㐄 A · C ㎗ B · C ㎗ A · C · D
The minimal SOP is: SOP 㐄 A · C ㎗ B · C ㎗ A · C · D
