Lecture 17

Lecture 17 - ECE52 Spring 11 Lecture 17 2/21/11 Lab this...

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1 ECE52 Spring 11 Lecture 17 2/21/11 Lab this week: ADC Lunch: Wednesday, noon-1ish, LSRC
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2 1’s, 2’s complement - examples 63-27: What is minimum number of bits this calculation can be conducted in?
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3 1’s, 2’s complement - examples 63-27: What is minimum number of bits this calculation can be conducted in? 7, since 2 6 =64; 63=0111111; MUST have leading zero to indicate positive number!
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4 1’s, 2’s complement - examples 63-27: 1’s 2’s 0111111 0111111 +1100100 (since 011011=27) + 1100101 10100011 10100100 +1 0100100
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2’s complement overflow rules Convert all add/subtract to addition x-y=x+(-y) in 4 bit world: 2+1 is fine, but 6+5? (-3)+(-2) is fine, but (-4)+(-7)? how about any legal positive number plus any legal negative number? RULE: If signs of operands match and differ from sign of answer, OVERFLOW! if signs of operands differ, overflow not possible! 5
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6 Downside of 1’s, 2’s – loss of simple positional representation • In 4-bit unsigned, 1011 = 11 10 means 1x2 3 +0x2 2 +1x2 1 +1x2 0 In 4-bit 2’s complement, 1011 = -5 no longer has this nice simple interpretation! Your book, and ECE152, discuss some tricks for interpreting values of 2’s complement numbers rapidly – turns out we are ok: 1011= -1x2 3 +0x2 2 +1x2 1 +1x2 0
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7 Multiplication: repeated addition (not the ultimate way to do multiplication though!) circuit to multiply 8-bit number by 3 7 x 0 y 7 y 0 x 7 x 0 y 8 y 0 y 7 x 8 s 0 s 7 c 7 0 s 0 s 8 c 8 P 9 P 8 P 0 P 3 A = : (a) Naive approach a 7 A : a 0 x 3A=A+A+A top adder: A+A=2A 8-bit adder, 9-bit result bottom adder: 2A+A=3A 9-bit adder, 10-bit result
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8 Better solution: drops an important concept in passing – shifting! recognize 2A is just A shifted over one bit position to the left! 11010 26 1101 13 number) binary digit (9 0 law) ive (distribut 0 2 2 ... 2 2 2 2 ... 2 2 10 10 0 1 2 3 4 5 6 7 1 0 2 1 7 6 8 7 0 1 1 6 6 7 7 0 1 2 3 4 5 6 7 = = = + × + × + + × + × = + × + + × + × = = a a a a a a a a a a a a A a a a a a a a a a a a a A
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9 So 3A is 2A+A x 1 x 0 y 8 y 0 y 7 x 8 s 0 s 8 c 8 0 0 a 7 A : P 9 P 8 P 0 P 3 A = : (b) Efficient design a 0 Circuit that multiplies an eight-bit unsigned number by 3 with a single 9-bit adder with 10-bit result.
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10 So 3A is 2A+A x 1 x 0 y 8 y 0 y 7 x 8 s 0 s 8 c 8 0 0 a 7 A : P 9 P 8 P 0 P 3 A = : (b) Efficient design a 0 Circuit that multiplies an eight-bit unsigned number by 3 with a single 9-bit adder with 10-bit result. We note that a circuit that only multiplies by 3 is of rather limited utility in most situations! Much better solutions in ECE152.
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11 VHDL, take 2 We introduced VHDL with just BIT types – legal values 0 and 1 only, but we have seen that Z and d (high impedance and don’t care) are also important states in practical
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Lecture 17 - ECE52 Spring 11 Lecture 17 2/21/11 Lab this...

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