{
"metadata": {
"name": ""
},
"nbformat": 3,
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 15: Digital Electronics"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.1, Page 771"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"n = 25\n",
"\n",
"#Calculations\n",
"b = bin(n)\n",
"\n",
"#Result\n",
"print \"The binary equivalent of 25 is\",b"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The binary equivalent of 25 is 0b11001\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.2, Page 771"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"def binary_decimal(ni): # Function to convert binary to decimal\n",
" deci = 0;\n",
" i = 0;\n",
" while (ni != 0):\n",
" rem = ni-int(ni/10.)*10\n",
" ni = int(ni/10.);\n",
" deci = deci + rem*2**i;\n",
" i = i + 1;\n",
" return deci\n",
" \n",
"\n",
"def binfrac_decifrac(nf): # Function to convert binary fraction to decimal fraction\n",
" decf = 0;\n",
" i = -1;\n",
" while (i >= -3):\n",
" nf = nf*10;\n",
" rem = round(nf); \n",
" nf = nf-rem;\n",
" decf = decf + rem*2**i;\n",
" i = i - 1;\n",
" return decf\n",
" \n",
"\n",
"n = 101.101; # Initialize the binary number\n",
"n_int = int(n); # Extract the integral part\n",
"n_frac = n-n_int; # Extract the fractional part\n",
"\n",
"#Result\n",
"print \"Decimal equivalent of %7.3f = %5.3f\"%(n, binary_decimal(n_int)+binfrac_decifrac(n_frac))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Decimal equivalent of 101.101 = 5.625\n"
]
}
],
"prompt_number": 33
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.3, Page 772"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"def octal_decimal(n): # Function to convert binary to decimal\n",
" dec = 0;\n",
" i = 0;\n",
" while (n != 0):\n",
" rem = n-int(n/10)*10; \n",
" n = int(n/10);\n",
" dec = dec + rem*8**i;\n",
" i = i + 1;\n",
" return dec\n",
"\n",
"n = 173; # Initialize the octal number\n",
"\n",
"#Result\n",
"print \"Decimal equivalent of %d = %d\"%(n, octal_decimal(n)); \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Decimal equivalent of 173 = 123\n"
]
}
],
"prompt_number": 37
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.4, Page 772"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"n = 278\n",
"\n",
"#Calculations\n",
"o = oct(n)\n",
"\n",
"#Result\n",
"print \"The octal equivalent of 278 is\",o"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The octal equivalent of 278 is 0426\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.5, Page 772"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"n1 = '10001100'\n",
"n2 = '1011010111'\n",
"\n",
"#Calculations\n",
"x1 = hex(int(n1,2))\n",
"x2 = hex(int(n2,2))\n",
"\n",
"#Results\n",
"print \"The hexadecimal equivalent of 10001100 is\",x1\n",
"print \"The hexadecimal equivalent of 1011010111 is\",x2\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The hexadecimal equivalent of 10001100 is 0x8c\n",
"The hexadecimal equivalent of 1011010111 is 0x2d7\n"
]
}
],
"prompt_number": 21
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.6, Page 772"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"n = 72905\n",
"\n",
"#Calculations\n",
"h = hex(n)\n",
"\n",
"#Result\n",
"print \"The hexadecimal equivalent of 278 is\",h"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The hexadecimal equivalent of 278 is 0x11cc9\n"
]
}
],
"prompt_number": 23
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.7, Page 773"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"x1 = 0b11101\n",
"x2 = 0b10111\n",
"\n",
"#Calculations\n",
"x = bin(x1+x2)\n",
"\n",
"#Result\n",
"print \"The required result is\",x"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The required result is 0b110100\n"
]
}
],
"prompt_number": 27
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.8, Page 773"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"def decimal_binary(ni): # Function to convert decimal to binary\n",
" bini = 0;\n",
" i = 1;\n",
" while (ni != 0):\n",
" rem = ni-int(ni/2)*2; \n",
" ni = int(ni/2);\n",
" bini = bini + rem*i;\n",
" i = i * 10;\n",
" return bini\n",
"\n",
"def decifrac_binfrac(nf): # Function to convert binary fraction to decimal fraction\n",
" binf = 0; i = 0.1;\n",
" while (nf != 0):\n",
" nf = nf*2;\n",
" rem = int(nf); \n",
" nf = nf-rem;\n",
" binf = binf + rem*i;\n",
" i = i/10;\n",
" return binf\n",
"\n",
"def binary_decimal(ni): # Function to convert binary to decimal\n",
" deci = 0;\n",
" i = 0;\n",
" while (ni != 0):\n",
" rem = ni-int(ni/10)*10; \n",
" ni = int(ni/10);\n",
" deci = deci + rem*2.**i;\n",
" i = i + 1;\n",
" return deci\n",
"\n",
"def binfrac_decifrac(nf): # Function to convert binary fraction to decimal fraction\n",
" decf = 0;\n",
" i = -1;\n",
" while (i >= -3):\n",
" nf = nf*10;\n",
" rem = round(nf); \n",
" nf = nf-rem;\n",
" decf = decf + rem*2.**i;\n",
" i = i - 1;\n",
" return decf \n",
"\n",
"bin1 = 1011.11; # Initialize the first binary binber\n",
"bin2 = 1011.01; # Initialize the second binary binber\n",
"bin1_int = int(bin1); # Extract the integral part for first\n",
"bin1_frac = bin1-bin1_int; # Extract the fractional part for second\n",
"bin2_int = int(bin2); # Extract the integral part for first\n",
"bin2_frac = bin2-bin2_int; # Extract the fractional part for second\n",
"dec1 = binary_decimal(bin1_int)+binfrac_decifrac(bin1_frac);\n",
"dec2 = binary_decimal(bin2_int)+binfrac_decifrac(bin2_frac);\n",
"dec = dec1+dec2;\n",
"dec_int = int(dec);\n",
"dec_frac = dec-dec_int;\n",
"\n",
"#Result\n",
"print \"%7.2f + %7.2f = %8.2f\"%(bin1, bin2, decimal_binary(dec_int)+decifrac_binfrac(dec_frac))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"1011.11 + 1011.01 = 10111.00\n"
]
}
],
"prompt_number": 49
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.9, Page 773"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"x1 = 0b0111\n",
"x2 = 0b1001\n",
"\n",
"#Calculations\n",
"x = bin(x1-x2)\n",
"\n",
"#Result\n",
"print \"The required result is\",x"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The required result is -0b10\n"
]
}
],
"prompt_number": 30
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.10, Page 773"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"x1 = 0b1101\n",
"x2 = 0b1100\n",
"\n",
"#Calculations\n",
"x = bin(x1*x2)\n",
"\n",
"#Result\n",
"print \"The required result is\",x\n",
"#Incorrect solution in textbook"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The required result is 0b10011100\n"
]
}
],
"prompt_number": 31
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.11, Page 774"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"x1 = 0b11001\n",
"x2 = 0b101\n",
"\n",
"#Calculations\n",
"x = bin(x1/x2)\n",
"\n",
"#Result\n",
"print \"The required result is\",x"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The required result is 0b101\n"
]
}
],
"prompt_number": 32
}
],
"metadata": {}
}
]
}