ecse221_as1 - ECSE221Assignment#1 Question1 a.)...

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ECSE-221 Assignment #1 Question 1 a.) i.) 153 10  => N 2 Q 1  = 153/2 = 76 R = 1 Q 2  = 76/2 = 38 R = 0 Q 3  = 38/2 = 19 R = 0 Q 4  = 19/2 = 9 R = 1 Q = 9/2 = 4 R = 1 Q 6  = 4/2= 2 R = 0 Q 7  = 2/2= 1 R = 0 Q 8  = 1/2= 0 R = 1 Significant figures: d b  = d a  * (log a / log b) = 3 * (log 10 / log 2) = 9.97 ~ 10 Thus, 153 10  => 10011001 2 ii.) 153 10  => N 16 Using the binary form found above, separating into parts of 4: 10011001 => |1001|1001| - the binary parts correspond to 2 hex numbers (9 and 9) Significant figures: d b  = d a  * (log a / log b) = 3 * (log 10 / log 16) = 2.49 ~ 3 (2 is sufficient) Thus, 153 10  => 10011001 2  = 99 16 b.) i.) 100011101001 2   => N 16
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As was done above, the binary is separated into groups of 4: 100011101001 2  => |1000|1110|1001| which corresponds to 3 hex numbers (8, 14 = E, 6) Significant figures: d b  = d a  * (log a / log b) = 12 * (log 2 / log 16) ~ 3  Thus, 100011101001 2  => 8E6 16 ii.) 100011101001 2  => N 10 Computing: 100011101001 2  = 1x2^0 + 1x2^3 + 1x2^5 + 1x2^6 + 1x2^7 + 1x2^11 = 2281 10 Significant figures: d b  = d a  * (log a / log b) = 12 * (log 2 / log 10) = 3.6 ~ 4  Thus, 100011101001 2  => 2281 10 iii.) 100011101001 2  => N 3 Using the decimal form found above,  Q 1  = 2281/3 = 760 R = 1 Q 2  = 760/3 = 253 R = 1 Q 3  = 253/3 = 84 R = 1 Q 4  = 84/3 = 28 R = 0 Q = 28/3 = 9 R = 1 Q 6  = 9/3= 3 R = 0 Q 7  = 3/3= 1 R = 0 Q 8  = 1/3= 0 R = 1
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Significant figures: d b  = d a  * (log a / log b) = 12 * (log 2 / log 3) = 7.57 ~ 8 Thus, 100011101001 2  => 10010111 3 c.) i.) A42.159 16  => N 2 Translating into binary groups of 4: A = 1010, 4 = 0100, 2 = 0010, 1 = 0001, 5 = 0101, 9 = 1001 So, |1010|0100|0010|.|0001|0101|1001|  Significant figures: d b  = d a  * (log a / log b) = 6 * (log 16 / log 2) ~ 24 Thus, A42.159 16  => 101001000010.000101011001 2 ii.) A42.159 16  => N 10 Computing: A42.159 16  => 10x16^2 + 4x16^1 + 2x16^0 + 1x16^(-1) + 5x16^(-2) + 9x16^(-3) =  2626.08423. .. Significant figures: d b  = d a  * (log a / log b) = 6 * (log 16 / log 10) = 7.22 ~ 8 Thus, A42.159 16  => 2626.0842 10   iii.) A42.159 16  => N 5 Converting the found decimal form: Q 1  = 2626/5 = 525 R = 1 Q 2  = 525/5 = 105 R = 0 Q 3  = 105/5 = 21 R = 0
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Q 4  = 21/5 = 4 R = 1 Q = 4/5 = 0 R = 4 Hence, A42 16  = 2626 10  = 41001 5 . Converting the fractional part 0.159 16  =   0.0842 10 : Significant figures: d b  = d a  * (log a / log b) = 6 * (log 16 / log 5) = 10.3 ~ 11 0.0842 = 0.0842 x (5/5) = 0.421x5^(-1)x(5/5) = 2.105x5^(-2) = 
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This note was uploaded on 12/01/2009 for the course ELEC ENG ECSE 221 taught by Professor Ferrie during the Fall '08 term at McGill.

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ecse221_as1 - ECSE221Assignment#1 Question1 a.)...

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