1. **Exercise 1: Sign and Decimal Value of Binary Numbers in SAV, CP1, CP2** Given: $X=11010101$, $Y=11001110$, $Z=01110111$, $T=01010101$ (8 bits each). - **SAV (Sign and Absolute Value):** The first bit is the sign (0 positive, 1 negative), the rest is magnitude. - **CP1 (One's Complement):** Negative numbers are represented by inverting all bits of the positive number. - **CP2 (Two's Complement):** Negative numbers are represented by inverting all bits and adding 1. **Step 1:** Determine sign and magnitude for each number in SAV. - $X$: sign bit=1 (negative), magnitude=$1010101_2=85_{10}$, so $X=-85$. - $Y$: sign bit=1 (negative), magnitude=$1001110_2=78_{10}$, so $Y=-78$. - $Z$: sign bit=0 (positive), magnitude=$1110111_2=119_{10}$, so $Z=119$. - $T$: sign bit=0 (positive), magnitude=$1010101_2=85_{10}$, so $T=85$. **Step 2:** Convert each to decimal in CP1. - For negative numbers, invert bits. - $X=11010101$, sign bit=1, so negative. Invert bits: $00101010_2=42_{10}$, so $X=-42$. - $Y=11001110$, negative. Invert bits: $00110001_2=49_{10}$, so $Y=-49$. - $Z=01110111$, sign bit=0, positive, value $119$. - $T=01010101$, positive, value $85$. **Step 3:** Convert each to decimal in CP2. - For negative numbers, invert bits and add 1. - $X=11010101$, invert: $00101010$, add 1: $00101011_2=43_{10}$, so $X=-43$. - $Y=11001110$, invert: $00110001$, add 1: $00110010_2=50_{10}$, so $Y=-50$. - $Z=01110111$, positive, $119$. - $T=01010101$, positive, $85$. --- 2. **Exercise 2: Express integers X=26, Y=38 in 7 bits SAV, CP1, CP2 and operations** **Step 1:** Express $X=26$ and $Y=38$ in 7 bits. - Binary $26=0011010$, $38=0100110$. **SAV:** - $X=0011010$ (positive) - $-X=1011010$ (sign bit 1, magnitude 26) - $Y=0100110$ - $-Y=1100110$ **CP1:** - $X=0011010$ - $-X$: invert bits $1100101$ - $Y=0100110$ - $-Y$: invert bits $1011001$ **CP2:** - $X=0011010$ - $-X$: invert bits $1100101$ add 1 $1100110$ - $Y=0100110$ - $-Y$: invert bits $1011001$ add 1 $1011010$ **Step 2:** Is 6-bit codification possible? - Max positive in 6 bits SAV is $011111=31$, so yes for 26 and 38 no (38>31). - So 6 bits not possible for 38. **Step 3:** Perform operations $X-Y$, $Y-X$, $X-Y$ in 7 bits SAV, CP1, CP2. - $X-Y=26-38=-12$ - $Y-X=38-26=12$ **SAV:** - $X=0011010=26$ - $Y=0100110=38$ - $X-Y$: sign bit 1, magnitude 12 = $1001100$ - $Y-X$: sign bit 0, magnitude 12 = $0001100$ **CP1:** - $X=0011010$ - $Y=0100110$ - $X-Y$: $X + (-Y)$, $-Y$ invert $1011001$ Add: $0011010 + 1011001 = 1110011$ (no overflow), decimal -12 - $Y-X$: $Y + (-X)$, $-X$ invert $1100101$ Add: $0100110 + 1100101 = 1001011$ (overflow ignored), decimal 12 **CP2:** - $X=0011010$ - $Y=0100110$ - $-Y=1011010$ - $X-Y$: $0011010 + 1011010 = 1110100$ decimal -12 - $Y-X$: $0100110 + 1100110 = 1001100$ decimal 12 --- 3. **Exercise 3: Binary expressions and arithmetic for A=109, B=18, C=36** **Step 1:** Express in base 2: - $A=109_{10} = 1101101_2$ - $B=18_{10} = 10010_2$ - $C=36_{10} = 100100_2$ **Step 2:** Perform $A+B$, $A+C$, $B+C$ in SAV, CP1, CP2. - $A+B=109+18=127$ - $A+C=109+36=145$ - $B+C=18+36=54$ Number of bits: - Max sum 145 requires at least 8 bits. **SAV:** - $A=01101101$ - $B=00010010$ - $C=00100100$ - $A+B=01111111$ (127) - $A+C=10010001$ (sign bit 1 invalid for positive, so 8 bits needed) - $B+C=00110110$ (54) **CP1 and CP2:** Same binary addition applies, with sign handling. **Step 3:** Perform $A-B-C$ and $B-A-C$ in SAV, CP1, CP2 on 8 bits if possible. - $A-B-C=109-18-36=55$ - $B-A-C=18-109-36=-127$ In 8 bits: - $55=00110111$ - $-127$ in CP2: invert $01111111$ to $10000000$ add 1 $10000001$ --- 4. **Exercise 4: Fixed-point binary and hexadecimal coding for X=25.125, Y=39.625, Z=123** **Step 1:** Express X and Y in fixed-point binary (assume 3 fractional bits): - $25.125 = 11001.001_2$ - $39.625 = 100111.101_2$ **Step 2:** Express X and Y in fixed-point hexadecimal: - $25.125 = 19.2_{16}$ (since 0.125 = 0.2 in hex) - $39.625 = 27.A_{16}$ **Step 3:** Perform $X+Y$ and $X-Y$ in binary: - $X+Y=11001.001 + 100111.101 = 111000.110_2 = 64.75_{10}$ - $X-Y=11001.001 - 100111.101 = -11110.100_2 = -14.5_{10}$ --- 5. **Exercise 5: IEEE-754 coding and decoding** **Step 1:** Encode $X=70.75$, $Y=-250.5$, $Z=0.375$ in IEEE-754 single precision. - $70.75_{10} = 1.000111011 imes 2^6$ - $Y=-250.5 = -1.111101001 imes 2^7$ - $Z=0.375 = 1.1 imes 2^{-2}$ Encode sign, exponent (bias 127), mantissa accordingly. **Step 2:** Decode $X=(11011000011010110000000000000000)_2$. - Sign=1 (negative) - Exponent bits=10110000 (176 decimal) - Mantissa=11010110000000000000000 - Real value $= -1.M imes 2^{176-127} = -1.4140625 imes 2^{49}$ (approximate) --- 6. **Exercise 6: Hexadecimal and base conversions, roots, IEEE-754** **Step 1a:** $X=4015.9375_{10}$ to base 16: - Integer part: $4015_{10} = FA7_{16}$ - Fractional part: $0.9375 = 0.F_{16}$ - So $X=FA7.F_{16}$ - Base 2: $111110100111.1111_2$ - Base 8: $17517.74_8$ **Step 1b:** $Y=(50)_{16} = 80_{10}$ **Step 1b-1:** $Z=X+Y=4015.9375 + 80 = 4095.9375$ - Base 16: $FFF.F_{16}$ - Base 2: $111111111111.1111_2$ **Step 1b-2:** Calculate cube roots: - $ oot 3 oot 3 Z = oot 3 {4095.9375} hickapprox 16$ (approximate) - Express in base 2: $10000_2$ **Step 1b-3:** Largest $n$ so that $n$-th root of $Z > 1$ is $n=12$ (since $2^{12}=4096$) **Step 2a:** $A=(FAF.F)_{16}$ - Express as $A = ext{sign} imes 1.M imes 2^E$ **Step 2b:** IEEE-754 single precision binary representation of $A$ (32 bits) **Step 2c:** Internal hexadecimal representation of $A$ **Step 2d:** Hex representation of $B=-A$ --- Final answers are detailed above for each exercise.