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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 14 - Digital-Analog Systems"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 14_1 Page No. 420"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"n= 3.00\n",
"L=2**(n)= 8.00\n",
"VFS= 10.00 volts\n",
"Q.E=VFS/L= 1.25\n",
"Q.E= +0.625,-0.625\n",
"Resolution=(100/2**(n))= 12.50 percent\n",
"Resolution= 12.50 percent,-12.50 percent\n"
]
}
],
"source": [
"from __future__ import division \n",
"n=3\n",
"print \"n= %0.2f\"%(n) # Number of bits \n",
"L=2**(n)\n",
"print \"L=2**(n)= %0.2f\"%(L)# Number of quantization levels\n",
"VFS=10\n",
"print \"VFS= %0.2f\"%(VFS),\" volts\" # Maximum value of analog input voltage\n",
"QE=VFS/L\n",
"print \"Q.E=VFS/L= %0.2f\"%(QE)# Quantization error\n",
"print \"Q.E= +0.625,-0.625\"# To make Quantization error symmetrical ittaken as (-Q.E/2) negative and positive value(+Q.E/2)\n",
"Resolution=(100/2**(n))#Formulae\n",
"print \"Resolution=(100/2**(n))= %0.2f\"%(Resolution),\" percent\"#Resolution\n",
"print \"Resolution= %0.2f\"%(+Resolution),\" percent,%0.2f\"%(-Resolution),\" percent\"# Since Resolution is (+)as well as (-)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 14_2 Page No. 420"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"n= 3.00\n",
"L=2**(n)= 8.00\n",
"VFS= 1.02 volts\n",
"part(i)\n",
"LSB=VFS/(2**n)= 0.13 volts\n",
"part(ii)\n",
"vh= 64 to 192 mV with offset\n",
"part(iii)\n",
"Inherent error,I.E= (LSB)/2= -0.06 V,+0.06 V\n",
"part(iv)\n",
"Resolution= 0.00 V\n",
"VFS= 1.00 V\n",
"k=VFS/(Resolution)= 1.00e+03\n",
"number of bits=10\n",
"so 10-bit ADC required\n"
]
}
],
"source": [
"from __future__ import division \n",
"n=3\n",
"print \"n= %0.2f\"%(n) # Number of bits \n",
"L=2**(n)\n",
"print \"L=2**(n)= %0.2f\"%(L) # Number of quantization levels\n",
"VFS=1024*10**(-3)\n",
"print \"VFS= %0.2f\"%(VFS),\" volts\" # Maximum value of analog input voltage\n",
"\n",
"print \"part(i)\"# Part(i)\n",
"LSB=VFS/(2**n)\n",
"print \"LSB=VFS/(2**n)= %0.2f\"%(LSB),\" volts\" # Lowest significant bit of 3-bit ADC\n",
"\n",
"print \"part(ii)\"# Part(ii)\n",
"print \"vh= 64 to 192 mV with offset\" # Analog voltage corresponding to binary word 001\n",
"\n",
"print \"part(iii)\"# Part(iii)\n",
"IE=(LSB)/2\n",
"print \"Inherent error,I.E= (LSB)/2= -%0.2f\"%(IE),\" V,+%0.2f\"%(IE),\" V\"# Inherent error in each binary word\n",
"\n",
"print \"part(iv)\"# Part(iv)\n",
"Resolution=(1*10**(-3))\n",
"print \"Resolution= %0.2f\"%(Resolution),\" V\"#Resolution\n",
"VFS=1\n",
"print \"VFS= %0.2f\"%(VFS),\" V\" # Maximum value of analog input voltage2\n",
"k=VFS/(Resolution)\n",
"print \"k=VFS/(Resolution)= %0.2e\"%(k) # 'k' taken only for calculation purpose\n",
"print \"number of bits=10\"# since k =[VFS/(Resolution)]is approximately equal to 2**10,\n",
"print \"so 10-bit ADC required\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 14_3 Page No. 421"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"VREF= -10.00 V\n",
"RF= 5000.00 ohm\n",
"R= 10000.00 ohm\n",
"vLSB=(-RF*VREF)/(8*R)=0.62 V\n",
"vo = -2*vLSB =-1.25 V\n",
"vo= -15*vLSB =-9.38 V\n"
]
}
],
"source": [
"from __future__ import division \n",
"VREF=-10\n",
"print \"VREF= %0.2f\"%(VREF),\" V\" # Reference voltage\n",
"RF=5*10**(3)\n",
"print \"RF= %0.2f\"%(RF)+ \" ohm\" #Feedback resistance\n",
"R=10*10**(3)\n",
"print \"R= %0.2f\"%(R)+ \" ohm\" # resistance\n",
"vLSB=(-RF*VREF)/(8*R)# Since IF=I/8,so vLSB=(-RF*IF)=(-RF*I/8)=(-RF*VREF/8*R)\n",
"print \"vLSB=(-RF*VREF)/(8*R)=%0.2f\"%(vLSB),\" V\" # Equivalent voltage for binary word 0001\n",
"vo=-2*vLSB# Since current IF=I/4\n",
"print \"vo = -2*vLSB =%0.2f\"%(vo),\" V\" # Equivalent voltage for binary word 0010=2 (in decimal)\n",
"vo=-15*vLSB# Since current IF=I+(I/2)+(I/4)+(I/8)=(15*I/8),so vo=15*VLSB\n",
"print \"vo= -15*vLSB =%0.2f\"%(vo),\" V\" # Equivalent voltage for binary word 0010=2 (in decimal)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 14_4 Page No. 422"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"VREF= -10.00 V\n",
"RF= 5000.00 ohm\n",
"R= 10000.00 ohm\n",
"vMSB=-(RF*VREF)/(2*R)=2.50 V\n",
"vo2 = vMSB/2 =1.25 V\n",
"vo3= (15/8)*vMSB =4.69 V\n"
]
}
],
"source": [
"from __future__ import division \n",
"VREF=-10\n",
"print \"VREF= %0.2f\"%(VREF),\" V\" # Reference voltage\n",
"RF=5*10**(3)\n",
"print \"RF= %0.2f\"%(RF)+ \" ohm\" #Feedback resistance\n",
"R=10*10**(3)\n",
"print \"R= %0.2f\"%(R)+ \" ohm\" # resistance\n",
"vMSB=-(RF*VREF)/(2*R)# Since IF=I/2,so vMSB=(-RF*IF)=(-RF*I/2)=(-RF*VREF/2*R)\n",
"print \"vMSB=-(RF*VREF)/(2*R)=%0.2f\"%(vMSB),\" V\" # Equivalent voltage for binary word 1000=8(in decimal)\n",
"vo2=vMSB/2# Since current IF=I/4\n",
"print \"vo2 = vMSB/2 =%0.2f\"%(vo2),\" V\" # Equivalent voltage for binary word 0100=4 (in decimal)\n",
"vo3=(15/8)*vMSB# Since current IF=I+(I/2)+(I/4)+(I/8)+(I/16)=(15*I/6),so vo=(15/8)*VMSB\n",
"print \"vo3= (15/8)*vMSB =%0.2f\"%(vo3),\" V\" # Equivalent voltage for binary word 1111=15 (in decimal)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 14_5 Page No. 422"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"n= 12.00\n",
"VFS= 50.00 volts\n",
"S=VFS/(2**n)= 0.01 volts\n",
"Resolution=(100/2**(n))= -0.02 percent, +0.02 percent\n"
]
}
],
"source": [
"from __future__ import division \n",
"n=12\n",
"print \"n= %0.2f\"%(n) # Number of bits\n",
"VFS=50\n",
"print \"VFS= %0.2f\"%(VFS),\" volts\" # Maximum value of analog input voltage\n",
"S=VFS/(2**n)\n",
"print \"S=VFS/(2**n)= %0.2f\"%(S),\" volts\" # Maximum quantization error\n",
"Resolution=(100/2**(n))#Formulae\n",
"print \"Resolution=(100/2**(n))= -%0.2f\"%(Resolution),\" percent, +%0.2f\"%(Resolution),\" percent\"# Since Resolution is (+)as well as (-)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 14_7 Page No. 423"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"n= 12.00\n",
"t= 5.00e-06 A\n",
"Vsp= 10.00 volts\n",
"LSB=Vsp/(2**n)= 0.0024 volts\n",
"LSB/2= -0.0012 V, -0.00 V\n",
"SR=(LSB/2)/t= 244.14 V/s\n",
"f = SR/(2*pi*Vsp)=3.89 Hz\n"
]
}
],
"source": [
"from math import pi\n",
"from __future__ import division \n",
"n=12\n",
"print \"n= %0.2f\"%(n) # Number of bits\n",
"t=5*10**(-6)\n",
"print \"t= %0.2e\"%(t),\" A\"\n",
"Vsp=10\n",
"print \"Vsp= %0.2f\"%(Vsp),\" volts\" # value of analog input voltage\n",
"LSB=Vsp/(2**n)\n",
"print \"LSB=Vsp/(2**n)= %0.4f\"%(LSB),\" volts\" # Lowest significant bit of 12-bit ADC\n",
"print \"LSB/2= -%0.4f\"%(LSB/2),\" V, -%0.2f\"%(LSB/2),\" V\" \n",
"SR=(LSB/2)/t\n",
"print \"SR=(LSB/2)/t= %0.2f\"%(SR),\" V/s\"\n",
"fmax=SR/(2*pi*Vsp)\n",
"print \"f = SR/(2*pi*Vsp)=%0.2f\"%(fmax),\" Hz\"# Highest frequency allowed at the input"
]
}
],
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|
