{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 7 : Active Filters"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.1 Page No.269"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import numpy as np\n",
"import math\n",
"# Given Data\n",
"fh = 10**3\n",
"C = 0.1*10**-6\n",
"Rf = 5.86*10**3\n",
"Ri = 10**4\n",
"\n",
"# Solution \n",
"\n",
"R = 1/(2*np.pi*C*fh)\n",
"Ao = (1 + Rf/Ri)\n",
"\n",
"a = np.array([100, 200, 500, 1000, 5000, 10000])\n",
"print \" The value of R is =\",round(R/1000,1),\"Kilo ohms\"\n",
"print \" The value of Ao =\",Ao\n",
"print \" Frequency in Hz Gain magnitude in dB 2o log (Vo/Vi)\"\n",
"print \" ========================================================\" \n",
"for i in a:\n",
"\n",
" val = round(20 * math.log10(Ao/(math.sqrt(1 + (i/fh)**4))),3)\n",
" print i,\" || \",val\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The value of R is = 1.6 Kilo ohms\n",
" The value of Ao = 1.586\n",
" Frequency in Hz Gain magnitude in dB 2o log (Vo/Vi)\n",
" ========================================================\n",
"100 || 4.006\n",
"200 || 4.006\n",
"500 || 4.006\n",
"1000 || 0.996\n",
"5000 || -23.96\n",
"10000 || -35.994\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.2 Page No.270"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Design of a 4th order Butter worth Low pss filter \n",
"\n",
"fh = 10**3\n",
"C = 0.1*10**-6\n",
"a1 = 0.765\n",
"a2 = 1.848\n",
"\n",
"Ao1 = 3 - a1\n",
"Ao2 = 3 - a2\n",
"\n",
"# let Rf1 = 12.35 kilo Ohm and Rf2 = 15.2 kilo ohm finding the Ri1 and Ri2 \n",
"\n",
"Rf1 = 12.35*10**3\n",
"Rf2 = 15.2*10**3\n",
"\n",
"Ri1 = Rf1/(Ao1 - 1)\n",
"Ri2 = Rf2/(Ao2 - 1)\n",
"\n",
"print \" The value of Ao1 = \",Ao1\n",
"print \" The value of Ao2 = \",Ao2\n",
"print \" The value of Rf1 = \",Rf1/1000,\"Kilo Ohm\"\n",
"print \" The value of Rf2 = \",Rf2/1000,\"Kilo Ohm\"\n",
"print \" The value of Ri1 = \",int(Ri1/1000),\"Kilo Ohm\"\n",
"print \" The value of Ri2 = \",int(Ri2/1000),\"Kilo Ohm\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The value of Ao1 = 2.235\n",
" The value of Ao2 = 1.152\n",
" The value of Rf1 = 12.35 Kilo Ohm\n",
" The value of Rf2 = 15.2 Kilo Ohm\n",
" The value of Ri1 = 10 Kilo Ohm\n",
" The value of Ri2 = 100 Kilo Ohm\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.3 Page No.271 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Solution to fing out the value of n or order\n",
"\n",
"# We know the equation is 20 log|(H(jw)| = 20 log (Ao/(math.sqrt(1+(w/wh)**n))\n",
"# putig the values into the equation we will get 0.01**2 = 1/(1 + 2**2*n)\n",
"# solving the equation to get the value of n\n",
"\n",
"# 2**(2*n) = 1/(1 + (0.01)**2)\n",
"# taking log to the base 2 on both sides\n",
"n = math.log(((10**4)-1),2)/2\n",
"\n",
"print \" The order of the filter will be =\",int(math.ceil(n))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The order of the filter will be = 7\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.4 Page No.272"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import numpy as np\n",
"import math\n",
"# Given Data\n",
"fl = 10**3\n",
"C = 0.1*10**-6\n",
"Rf = 5.86*10**3\n",
"Ri = 10**4\n",
"\n",
"# Solution \n",
"\n",
"R = 1/(2*np.pi*C*fl)\n",
"Ao = (1 + Rf/Ri)\n",
"\n",
"a = np.array([100, 200, 500, 1000, 5000, 10000])\n",
"print \" The value of R is =\",round(R/1000,1),\"Kilo ohms\"\n",
"print \" The value of Ao =\",Ao\n",
"print \" Frequency in Hz Gain magnitude in dB 2o log (Vo/Vi)\"\n",
"print \" ========================================================\" \n",
"for i in a:\n",
"\n",
" val = round(Ao/math.sqrt(1 + (fl/i)**4),3)\n",
" print i,\" || \",val\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The value of R is = 1.6 Kilo ohms\n",
" The value of Ao = 1.586\n",
" Frequency in Hz Gain magnitude in dB 2o log (Vo/Vi)\n",
" ========================================================\n",
"100 || 0.016\n",
"200 || 0.063\n",
"500 || 0.385\n",
"1000 || 1.121\n",
"5000 || 1.586\n",
"10000 || 1.586\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.5 Page No.276"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Given Data \n",
"\n",
"fl = 400\n",
"fh = 2*10**3\n",
"Ao = 4\n",
"\n",
"# For design top get Ao = 2 the values of Rf = Ri\n",
"\n",
"Rf = Ri = 10*10**3\n",
"\n",
"# for LPF \n",
"\n",
"C1 = 0.01*10**-6\n",
"R1 = 1/(2*math.pi*fh*C1)\n",
"\n",
"# for HPF \n",
"\n",
"C2 = 0.01*10**-6\n",
"R2 = 1/(2*math.pi*fl*C2)\n",
"\n",
"fo = math.sqrt(fh*fl)\n",
"Q = fo/(fh - fl)\n",
"\n",
"print \"The value of c1 =\",C1*10**6,\"uF\"\n",
"print \"The value of R1 =\",round(R1/1000,1),\"kilo Ohms\"\n",
"print \"The value of c2 =\",C2*10**6,\"uF\"\n",
"print \"The value of R2 =\",round(R2/1000,1),\"Kilo Ohms\"\n",
"print \"The value of Q =\",round(Q,2)\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of c1 = 0.01 uF\n",
"The value of R1 = 8.0 kilo Ohms\n",
"The value of c2 = 0.01 uF\n",
"The value of R2 = 39.8 Kilo Ohms\n",
"The value of Q = 0.56\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.6 Page No.279"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"# Given Data \n",
"\n",
"fo = 50\n",
"C = 0.1*10**-6\n",
"\n",
"# Solution \n",
"\n",
"R = 1/(2*math.pi*fo*C)\n",
"\n",
"print \"The value of Capasitor = \",C*10**6,\"uF\"\n",
"print \"The value of Resistor = \",round(R/1000,1),\"Kilo Ohm\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of Capasitor = 0.1 uF\n",
"The value of Resistor = 31.8 Kilo Ohm\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.7 Page No.280"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# Given Data \n",
"\n",
"fl = 400\n",
"fh = 2*10**3\n",
"Ao = 4\n",
"\n",
"# For design top get Ao = 2 the values of Rf = Ri\n",
"\n",
"Rf = Ri = 10*10**3\n",
"\n",
"# for LPF \n",
"\n",
"C1 = 0.1*10**-6\n",
"R1 = 1/(2*math.pi*fh*C1)\n",
"\n",
"# for HPF \n",
"\n",
"C2 = 0.1*10**-6\n",
"R2 = 1/(2*math.pi*fl*C2)\n",
"\n",
"fo = math.sqrt(fh*fl)\n",
"Q = fo/(fh - fl)\n",
"\n",
"print \"The value of c1 =\",C1*10**6,\"uF\"\n",
"print \"The value of R1 =\",round(R1/1000,1),\"kilo Ohms\"\n",
"print \"The value of c2 =\",C2*10**6,\"uF\"\n",
"print \"The value of R2 =\",round(R2/1000),\"Kilo Ohms\"\n",
"print \"The value of Q =\",round(Q,2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of c1 = 0.1 uF\n",
"The value of R1 = 0.8 kilo Ohms\n",
"The value of c2 = 0.1 uF\n",
"The value of R2 = 4.0 Kilo Ohms\n",
"The value of Q = 0.56\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7.10 Page No.298 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# given data \n",
"\n",
"fc = 400\n",
"Ao = -2\n",
"Vcc = 5\n",
"R1 = 10*10**3\n",
"HoLP = 2\n",
"\n",
"R2 = HoLP * R1 \n",
"\n",
"# for second order butterworth filter Q = 0.707\n",
"Q = 0.707\n",
"R3 = Q * R2\n",
"\n",
"fclock = 50 * fc\n",
"\n",
"# Dispalying the outputs \n",
"\n",
"print \"The value of R1 =\",R1/1000,\"Ohms\"\n",
"print \"The value of R2 =\",R2/1000,\"Kilo Ohms\"\n",
"print \"The value of R3 =\",R3/1000,\"Kilo Ohms\"\n",
"print \"The value of clock frequency =\",fclock/1000,\"Kilo Hertz\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of R1 = 10 Ohms\n",
"The value of R2 = 20 Kilo Ohms\n",
"The value of R3 = 14.14 Kilo Ohms\n",
"The value of clock frequency = 20 Kilo Hertz\n"
]
}
],
"prompt_number": 18
},
{
"cell_type": "code",
"collapsed": false,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}