{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 10 : D-A and A-D Converters"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.1 Page No.357"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Given Data \n",
"\n",
"n = 9\n",
"s = 10.3*10**-3\n",
"a = '0b101101111'\n",
"\n",
"# Solution \n",
"# Converting the given value into its equivalent decimal value\n",
"x = int(a,2)\n",
"#multiplying with the resolution to get the output\n",
"\n",
"V = round(x*s,2)\n",
"\n",
"print \"The output value will be =\",V,\"V\" \n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The output value will be = 3.78 V\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.2 Page No.357"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from fractions import Fraction \n",
"# Given data \n",
"\n",
"n = 8\n",
"v = 10.0\n",
"\n",
"# Solution \n",
"LSB = 1.0/2**n\n",
"Vlsb =v/2**n\n",
"MSB = v/2\n",
"Vo = v - Vlsb\n",
"\n",
"# Dispalying the output\n",
"\n",
"print \"The value of LSB = \",Fraction(LSB).limit_denominator(256)\n",
"print \"The voltage of LSB = \",int(Vlsb*10**3),\"mV\"\n",
"print \"The MSB = \",int(MSB)\n",
"print \"The value of full scale output is = \",round(Vo,3),\"V\"\n",
" \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of LSB = 1/256\n",
"The voltage of LSB = 39 mV\n",
"The MSB = 5\n",
"The value of full scale output is = 9.961 V\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.3 Page No.357"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import numpy as np\n",
"V = 10.0 # Range of the DAC is 0 -10 voltage \n",
"\n",
"# define a function for performing the DAC action\n",
"\n",
"def DAC(Vo):\n",
" j = 1\n",
" sum = 0.0\n",
" x = len(Vo)\n",
" for i in range(x):\n",
" sum = sum + ((Vo[i])*(0.5**j))\n",
" j += 1\n",
" \n",
" return (V*sum)\n",
"# part 1 \n",
"Vo1 = np.array([1, 0])\n",
"# part 2\n",
"Vo2 = np.array([0, 1, 1, 0])\n",
"#part 3\n",
"Vo3 = np.array([1, 0, 1, 1, 1, 1, 0, 0])\n",
"\n",
"# Finding the solution for all 3 parts and printing the outputs \n",
"\n",
"print \" The output for part 1\",int(DAC(Vo1)),\"V\"\n",
"print \" The output for part 2\",(DAC(Vo2)),\"V\"\n",
"print \" The output for part 3\",round(DAC(Vo3),2),\"V\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The output for part 1 5 V\n",
" The output for part 2 3.75 V\n",
" The output for part 3 7.34 V\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.4 Page No.365"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Given data \n",
"n = 16 \n",
"Clockrate = 4*10**6\n",
"V = 10.0\n",
"c = 0.1*10**-6\n",
"va = -8.0\n",
"\n",
"# Solution \n",
"\n",
"t21 = (2.0**n)/(Clockrate)\n",
"R = ((-V/va)*t21)/c\n",
"\n",
"print \"The value of (t2-t1) =\",round(t21*10**3,2),\"ms\"\n",
"print \"The value of Resistor R =\",int(round(R*10**-3)),\"kilo Ohms\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of (t2-t1) = 16.38 ms\n",
"The value of Resistor R = 205 kilo Ohms\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.5 Page No.365"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Given data \n",
"\n",
"Va = 4.129\n",
"n = 16\n",
"Vr = 8\n",
"\n",
"# Solution \n",
"\n",
"N = int(round((2**n)*(Va/Vr)))\n",
"out = bin(N) # Converting the voltage value into its binary equivalent\n",
"\n",
"print \"The binary equivalent is = \",out\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The binary equivalent is = 0b1000010000100001\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.6 Page No.368"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Given data \n",
"\n",
"d = 3.5 \n",
"print \"The last 3 digit can be\",000,\"To\",(10**int(d))-1\n",
"print \"Hense the 3 1/2 digit DVM reading varies from\",0000,\"to\",\"1\"+str((10**int(d))-1)\n",
"# Reference voltage is 2V\n",
"Vref = 2.0\n",
"# Resolution R \n",
"\n",
"R = Vref/2000\n",
"print \"The resolution of a 3 1/2 digit DVM is =\",int(R*10**3),\"mV\"\n",
"\n",
"# Similarly for 4 1/2 digit DVM\n",
"\n",
"R1 = Vref/20000\n",
"\n",
"print \"Thus resolution of a 4 1/2 digit DVM is =\",R1*10**3,\"mV\"\n",
"\n",
"print \"So the resolution of 4 1/2 digit DVM is better than 3 1/2 digit DVM\"\n",
" \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The last 3 digit can be 0 To 999\n",
"Hense the 3 1/2 digit DVM reading varies from 0 to 1999\n",
"The resolution of a 3 1/2 digit DVM is = 1 mV\n",
"Thus resolution of a 4 1/2 digit DVM is = 0.1 mV\n",
"So the resolution of 4 1/2 digit DVM is better than 3 1/2 digit DVM\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}