Added(A)/Deleted(D) following books

 A  College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch15.ipynb
A  College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch16.ipynb
A  College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch17.ipynb
A  College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch18.ipynb
A  College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch19.ipynb
A  College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch20.ipynb
A  College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch21.ipynb
A  College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch22.ipynb
A  College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch23.ipynb
A  College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch24.ipynb
A  College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch25.ipynb
A  College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch26.ipynb
A  College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch27.ipynb
A  College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch28.ipynb
A  College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch29.ipynb
A  College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/Ch30.ipynb
A  College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/screenshots/1RefractionOfLaserLight.png
A  College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/screenshots/2PropertiesOfImage.png
A  College_Physics_(volume_2)_by_R._A._Serway_and_J._S._Faughn/screenshots/3PositionOf1DarkFringe.png
A  Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/Chapter9_7.ipynb
A  Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter1_7.ipynb
A  Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter2_7.ipynb
A  Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter3_7.ipynb
A  Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter4_7.ipynb
A  Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter6_7.ipynb
A  Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter7_7.ipynb
A  Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter8_7.ipynb
A  Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter_5_7.ipynb
A  Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/1.2_2.png
A  Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/3.7_2.png
A  Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/6.7_2.png
A  Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch11.ipynb
A  Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch12.ipynb
A  Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch13.ipynb
A  Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch14.ipynb
A  Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch3.ipynb
A  Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch4.ipynb
A  Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch5.ipynb
A  Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch6.ipynb
A  Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch7.ipynb
A  Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch8.ipynb
A  Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch9.ipynb
A  Linear_Integrated_Circuit_by_M._S._Sivakumar/screenshots/BiasingVoltage.png
A  Linear_Integrated_Circuit_by_M._S._Sivakumar/screenshots/InOutCrrnt.png
A  Linear_Integrated_Circuit_by_M._S._Sivakumar/screenshots/inAndOutCurrnt.png
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter10.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter28.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/Chapter29.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter1.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter11.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter12.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter13.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter14.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter15.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter16.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter17.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter18.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter19.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter2.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter20.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter21.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter22.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter23.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter24.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter25.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter26.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter27.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter3.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter4.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter5.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter6.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter7.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter8.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/chapter9.ipynb
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/screenshots/image_1.png
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/screenshots/image_2.png
A  principle_of_physics_by_V.K.MEHTA_,_ROHIT_MEHTA_/screenshots/image_3.png
A  sample_notebooks/KhushbuPattani/chapter1_1.ipynb

1 files changed, 722 insertions, 0 deletions

diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch7.ipynb b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch7.ipynb
new file mode 100644
index 00000000..b665ce02
--- /dev/null
+++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch7.ipynb
@@ -0,0 +1,722 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 7 Filters and Rectifiers"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 7.1 Pg 232"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The resistor value is = 1.59  k ohm \n"
+     ]
+    }
+   ],
+   "source": [
+    "from __future__ import division\n",
+    "from math import pi, sqrt\n",
+    "# Design active low filter with cut-off frequency 10 kHz\n",
+    "fc = 10 # # kHz\n",
+    "C = 0.01 # #uF # we assume\n",
+    "\n",
+    "# the cut-off frequency of active low pass filter is defined as\n",
+    "# fc = (1/2*pi*R3*C)#\n",
+    "\n",
+    "# R3 can be calculated as\n",
+    "R3 = (1/(2*pi*fc*C))#\n",
+    "print 'The resistor value is = %0.2f'%R3,' k ohm '"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 7.2 Pg 233"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The resistor value is = 106  ohm \n",
+      "The pass band gain is = 1.50  \n"
+     ]
+    }
+   ],
+   "source": [
+    " # Design active low filter with cut-off frequency 15 kHz\n",
+    "fc = 15*10**3 # # Hz \n",
+    "C = 0.1*10**-6 # #F # we assume\n",
+    "\n",
+    "# the cut-off frequency of active low pass filter is defined as\n",
+    "# fc = (1/2*pi*R3*C)#\n",
+    "\n",
+    "# R3 can be calculated as\n",
+    "R3 = (1/(2*pi*fc*C))#\n",
+    "print 'The resistor value is = %0.f'%R3,' ohm '\n",
+    "\n",
+    "# the pass band gain of filter is given by\n",
+    "# Af = 1+(R2/R1)#\n",
+    "# assume that the inverting terminal resistor R2=0.5*R1#\n",
+    "# in Af equation if we put R2=0.5R1 in R1 R1 cancellout each other \n",
+    "Af = 1+(0.5)\n",
+    "print 'The pass band gain is = %0.2f'%Af,' '"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 7.3 Pg 234"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The resistor value is = 159  Kohm \n",
+      "The resistor R2 value is = 900.00  k ohm \n",
+      "The magnitude of an active low pass filter is = 1.96  \n",
+      "The phase angle of the filter is = -78.69  \n"
+     ]
+    }
+   ],
+   "source": [
+    "from __future__ import division\n",
+    "# Design active low filter with cut-off frequency 20 kHz\n",
+    "fc = 20 # # kHz \n",
+    "f = 100 # # frequency of filter\n",
+    "Af = 10 #  # desired pass band gain\n",
+    "C = 0.05 # #nF # we assume\n",
+    "\n",
+    "# the cut-off frequency of active low pass filter is defined as\n",
+    "# fc = (1/2*pi*R3*C)#\n",
+    "\n",
+    "# R3 can be calculated as\n",
+    "R3 = (1/(2*pi*fc*1e3*C*1e-9))/1e3 # Kohm\n",
+    "print 'The resistor value is = %0.f'%R3,' Kohm '\n",
+    "\n",
+    "# the pass band gain of filter is given by\n",
+    "# Af = 1+(R2/R1)#\n",
+    "# assume that the inverting terminal resistor R1= 100 k ohm#\n",
+    "R1 = 100 # # k ohm\n",
+    "R2 = (Af*R1)-R1#\n",
+    "print 'The resistor R2 value is = %0.2f'%R2,' k ohm '\n",
+    "\n",
+    "# the magnitude of an active low pass filter is given as\n",
+    "A = Af/(sqrt(1+(f/fc)**2))#\n",
+    "print 'The magnitude of an active low pass filter is = %0.2f'%A,' '\n",
+    "\n",
+    "#the phase angle of the filter\n",
+    "from math import atan , degrees\n",
+    "Angle = -degrees(atan(f/fc))#\n",
+    "print 'The phase angle of the filter is = %0.2f'%Angle,' '"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 7.4 Pg 236"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The frequency of the first order low pass filter is = 2.65  kHz \n",
+      "The pass band gain of filter is = 13.00 \n"
+     ]
+    }
+   ],
+   "source": [
+    " # to determine the cut-off frequency and pass band gain Af\n",
+    "R1 = 1 # # k ohm\n",
+    "R2 = 12 # # k ohm\n",
+    "R3 = 1.2 #  # k ohm\n",
+    "C = 0.05 # #uF # we assume\n",
+    "\n",
+    "# the frequency of the first order low pass filter is defined as\n",
+    "fc = (1/(2*pi*R3*C))#\n",
+    "print 'The frequency of the first order low pass filter is = %0.2f'%fc,' kHz '\n",
+    "\n",
+    "# the pass band gain of filter is given by\n",
+    "Af =(1+R2/R1)#\n",
+    "print 'The pass band gain of filter is = %0.2f'%Af,''"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 7.5 Pg 236"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The capacitor of high pass filter is = 25.13  uF \n",
+      "The second resistor value is = 90.00  K ohm \n"
+     ]
+    }
+   ],
+   "source": [
+    " # to design a first order high pass filter with cut-off frequency 2kHz\n",
+    "Af = 10 #\n",
+    "fc = 2 # # kHz \n",
+    "R3 = 2 # #K ohm # we assume\n",
+    "R1 = 10 # # k ohm\n",
+    "# the capacitor of high pass filter is given by\n",
+    "C = 2*pi*R3*fc#\n",
+    "print 'The capacitor of high pass filter is = %0.2f'%C,' uF '\n",
+    "\n",
+    "# the voltage gain of the high pass filter is\n",
+    "# Af = 1+(R2/R1)#\n",
+    "R2 = R1*(Af-1)#\n",
+    "print 'The second resistor value is = %0.2f'%R2,' K ohm '"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 7.6 Pg 237"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The resistance R3 is = 1.59  K ohm \n"
+     ]
+    }
+   ],
+   "source": [
+    " # to design an active high pass filter with cut-off frequency 10kHz\n",
+    "fc = 10 # # kHz \n",
+    "C = 0.01 # #uF # we assume\n",
+    "# the cut-off frequency of active high pass filter is given by\n",
+    "# fc = 2*pi*R3*C#\n",
+    "# R3 can be calculated as\n",
+    "R3 = (1/(2*pi*fc*C))#\n",
+    "print 'The resistance R3 is = %0.2f'%R3,' K ohm '"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 7.7 Pg 238"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The resistance R3 is = 64  Kohm \n",
+      "The pass band gain is = 1.20  \n"
+     ]
+    }
+   ],
+   "source": [
+    "# to design an active high pass filter with cut-off frequency 25kHz\n",
+    "fc = 25 # # kHz \n",
+    "C = 0.1 # #nF # we assume\n",
+    "# the cut-off frequency of active high pass filter is given by\n",
+    "# fc = 2*pi*R3*C#\n",
+    "# R3 can be calculated as\n",
+    "R3 = (1/(2*pi*fc*1e3*C*1e-9)) / 1e3 # Kohm\n",
+    "print 'The resistance R3 is = %0.f'%R3,' Kohm '\n",
+    "\n",
+    "# the desire pass band gain of filter is given by \n",
+    "#Af = 1+(R2/R1)#\n",
+    "# assume that the inverting terminal resistor R2=0.2*R1#\n",
+    "# in Af equation if we put R2=0.2R1 in R1 R1 cancellout each other \n",
+    "Af = 1+(0.2)\n",
+    "print 'The pass band gain is = %0.2f'%Af,' '"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 7.8 Pg 239"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The resistance R3 is = 159  K ohm \n",
+      "The resistance R2 is = 700.00  K ohm \n",
+      "The magnitude of an active high pass filter is = 14.55  \n",
+      "The phase angle of the filter is = 14.04  degree\n"
+     ]
+    }
+   ],
+   "source": [
+    "## # to design an active high pass filter with cut-off frequency 20kHz \n",
+    "Af = 15 #\n",
+    "fc = 20 # #kHz\n",
+    "f = 80 # # kHz  the frequency of filter \n",
+    "C = 0.05 # #nF # we assume\n",
+    "# the cut-off frequency of active high pass filter is given by\n",
+    "# fc = 2*pi*R3*C#\n",
+    "# R3 can be calculated as\n",
+    "R3 = (1/(2*pi*fc*C))#\n",
+    "print 'The resistance R3 is = %0.f'%(R3*1000),' K ohm '  # Round Off Error\n",
+    "\n",
+    "# the desire pass band gain of filter is given by \n",
+    "#Af = 1+(R2/R1)#\n",
+    "# assume that the inverting terminal resistor R1=50 K ohm#\n",
+    "R1 = 50 # # K ohm\n",
+    "R2 = (R1*Af)-(R1)\n",
+    "print 'The resistance R2 is = %0.2f'%R2,' K ohm '\n",
+    "\n",
+    "# the magnitude of an active high pass filter is given as\n",
+    "A = Af*(f/fc)/(sqrt(1+(f/fc)**2))#\n",
+    "print 'The magnitude of an active high pass filter is = %0.2f'%A,' '\n",
+    "\n",
+    "#the phase angle of the filter\n",
+    "from numpy import inf\n",
+    "Angle = degrees(-atan(f/fc)+atan(inf))\n",
+    "print 'The phase angle of the filter is = %0.2f'%Angle,' degree'  # Round Off Error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 7.9 Pg 241"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The lower cut-off frequency FLC of band pass filter is = 159.2  Hz \n",
+      "The upper cut-off frequency FUC of band pass filter is = 15.92  kHz \n"
+     ]
+    }
+   ],
+   "source": [
+    "# to calculate upper and lower cut-off frequency of the band pass filter\n",
+    "R1 = 10*10**3 # #K ohm\n",
+    "R2 = 10 # #K ohm\n",
+    "C1 = 0.1*10**-6 # # uF\n",
+    "C2 = 0.001 #  #uF\n",
+    "\n",
+    "# the lower cut-off frequency of band pass filter is\n",
+    "fLC = 1/(2*pi*R1*C1)#\n",
+    "print 'The lower cut-off frequency FLC of band pass filter is = %0.1f'%fLC,' Hz '\n",
+    "\n",
+    "# The upper cut-off frequency of band pass filter is\n",
+    "fUC = 1/(2*pi*R2*C2)#\n",
+    "print 'The upper cut-off frequency FUC of band pass filter is = %0.2f'%fUC,' kHz '"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 7.10 Pg 242"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The resistance R3 Value is = 7.96  M ohm \n",
+      "The resistance R6 value is = 1.59  M ohm \n"
+     ]
+    }
+   ],
+   "source": [
+    "# to design an active band pass filter with lower cut-off frequency 10 kHz an upper 50 kHz\n",
+    "fL = 10 # # kHz\n",
+    "fH = 50 # # kHz\n",
+    "C1 = 0.002 # # nF\n",
+    "C2 = 0.002 #  # nF\n",
+    "\n",
+    "# the lower cut-off frequency of band pass filter is\n",
+    "# fL = 1/(2*pi*R3*C1)#\n",
+    "R3 = 1/(2*pi*fL*C1)#\n",
+    "print 'The resistance R3 Value is = %0.2f'%R3,' M ohm '\n",
+    "\n",
+    "# The upper cut-off frequency of band pass filter is\n",
+    "# fH = 1/(2*pi*R6*C2)#\n",
+    "R6 = 1/(2*pi*fH*C2)#\n",
+    "print 'The resistance R6 value is = %0.2f'%R6,' M ohm '"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 7.11 Pg 243"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 20,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The resistance R3 Value is = 7.96  M ohm \n",
+      "The resistance R6 value is = 3.98  M ohm \n",
+      "The desire pass band gain of filter is = 15  \n"
+     ]
+    }
+   ],
+   "source": [
+    "# to design an active band pass filter with lower cut-off frequency 20 kHz an upper 40 kHz\n",
+    "fL = 20 # # kHz\n",
+    "fH = 40 # # kHz\n",
+    "# the inverting terminal resistance 2R1=R2 and 4R4=R5\n",
+    "C1 = 0.001 # # nF\n",
+    "C2 = 0.001 #  # nF\n",
+    "\n",
+    "# the lower cut-off frequency of band pass filter is\n",
+    "# fL = 1/(2*pi*R3*C1)#\n",
+    "R3 = 1/(2*pi*fL*C1)#\n",
+    "print 'The resistance R3 Value is = %0.2f'%R3,' M ohm '\n",
+    "\n",
+    "# The upper cut-off frequency of band pass filter is\n",
+    "# fH = 1/(2*pi*R6*C2)#\n",
+    "R6 = 1/(2*pi*fH*C2)#\n",
+    "print 'The resistance R6 value is = %0.2f'%R6,' M ohm '\n",
+    "\n",
+    "# the desire pass band gain of filter is defined as\n",
+    "R1 = 1 # # M ohm we assume\n",
+    "#we define inverting terminal resistance 2R1=R2\n",
+    "R2 = 2 # # M ohm\n",
+    "# then\n",
+    "R4 = 1 # #M ohm\n",
+    "R5 = 4 # # M ohm\n",
+    "Af = (1+(R2/R1))*(1+(R5/R4))#\n",
+    "print 'The desire pass band gain of filter is = %d'%Af,' '"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 7.12 Pg 244"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 23,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The resistance R3 Value is = 7.96  M ohm \n",
+      "The resistance R6 value is = 1.99  M ohm \n",
+      "The desire pass band gain of filter is = 15.00  \n",
+      "The magnitude of gain of band pass filter  is = 11.49  \n",
+      "The phase angle of gain of band pass filter  is = 50 degree\n"
+     ]
+    }
+   ],
+   "source": [
+    "# to design an active band pass filter with lower cut-off frequency 20 kHz an upper 80 kHz\n",
+    "f = 100 # # kHz  the frequency of band pass filter\n",
+    "fL = 20 # # kHz\n",
+    "fH = 80 # # kHz\n",
+    "# the inverting terminal resistance R1=0.5*R2 and R4=0.25*R5\n",
+    "C1 = 0.001 # # nF\n",
+    "C2 = 0.001 #  # nF\n",
+    "\n",
+    "# the lower cut-off frequency of band pass filter is\n",
+    "# fL = 1/(2*pi*R3*C1)#\n",
+    "R3 = 1/(2*pi*fL*C1)#\n",
+    "print 'The resistance R3 Value is = %0.2f'%R3,' M ohm '\n",
+    "\n",
+    "# The upper cut-off frequency of band pass filter is\n",
+    "# fH = 1/(2*pi*R6*C2)#\n",
+    "R6 = 1/(2*pi*fH*C2)#\n",
+    "print 'The resistance R6 value is = %0.2f'%R6,' M ohm '  # Round Off Error\n",
+    "\n",
+    "# the desire pass band gain of filter is defined as\n",
+    "R1 = 1 # # M ohm we assume\n",
+    "#we define inverting terminal resistance R1=0.5*R2\n",
+    "R2 = 2 # # M ohm\n",
+    "# then\n",
+    "R4 = 1 # #M ohm\n",
+    "R5 = 4 # # M ohm\n",
+    "Af = (1+(R2/R1))*(1+(R5/R4))#\n",
+    "print 'The desire pass band gain of filter is = %0.2f'%Af,' '\n",
+    "\n",
+    "# the magnitude of gain of band pass filter is given as\n",
+    "A = Af*(f**2/(fL*fH))/((sqrt(1+(f/fL)**2))*(sqrt(1+(f/fH)**2)))#\n",
+    "print 'The magnitude of gain of band pass filter  is = %0.2f'%A,' '  # Round Off Error\n",
+    "\n",
+    "#the phase angle of the filter\n",
+    "from numpy import inf\n",
+    "Angle = degrees(2*atan(inf)-atan(f/fL)-atan(f/fH))\n",
+    "print 'The phase angle of gain of band pass filter  is = %0.f'%Angle,'degree'  # Round Off Error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 7.13 Pg 247"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 24,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The output of the half wave precision rectifier Vo is = -20.00  V \n"
+     ]
+    }
+   ],
+   "source": [
+    " # to determine the output voltage of the precision rectifier circuit\n",
+    "Vi = 10 # #V i/p volt\n",
+    "R1 = 20 # # K ohm\n",
+    "R2 = 40 # # K ohm\n",
+    "Vd = 0.7 # # V   the diode voltage drop\n",
+    "\n",
+    "# the output of the half wave precision rectifier is defined as\n",
+    "# Vo = -(R2/R1)*Vi # for Vi < 0\n",
+    "#    = 0 otherwise\n",
+    "# i.e for Vi > 0\n",
+    "#              Vo = 0\n",
+    "# for Vi < 0\n",
+    "Vo = -(R2/R1)*Vi\n",
+    "print 'The output of the half wave precision rectifier Vo is = %0.2f'%Vo,' V '"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 7.14 Pg 247"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 25,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The output of the half wave precision rectifier Vo is = -15.00  V \n",
+      "The output of the half wave precision rectifier Vo is = 15.00  V \n"
+     ]
+    }
+   ],
+   "source": [
+    "# to determine the output voltage of the precision rectifier circuit for i/p voltage a) Vi = 5 b) Vi = -5\n",
+    "Vi = 5 # #V i/p volt\n",
+    "R1 = 5 # # K ohm\n",
+    "R2 = 15 # # K ohm\n",
+    "Vd = 0.7 # # V   the diode voltage drop\n",
+    "\n",
+    "# the output of the half wave precision rectifier is defined as\n",
+    "# Vo = -(R2/R1)*Vi # for Vi < 0\n",
+    "#    = 0 otherwise\n",
+    "\n",
+    "# for Vi = 5 V\n",
+    "# i.e for Vi > 0\n",
+    "#              Vo = 0\n",
+    "# for Vi < 0\n",
+    "Vo = -(R2/R1)*Vi#\n",
+    "print 'The output of the half wave precision rectifier Vo is = %0.2f'%Vo,' V '\n",
+    "\n",
+    "# for Vi = -5 V\n",
+    "# i.e for Vi > 0\n",
+    "#              Vo = 0\n",
+    "# for Vi < 0\n",
+    "Vi =-5 # # V\n",
+    "Vo = -(R2/R1)*Vi#\n",
+    "print 'The output of the half wave precision rectifier Vo is = %0.2f'%Vo,' V '"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 7.15 Pg 248"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 26,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The gain of precision full wave rectifier A is = 6.00  \n",
+      "The output voltage Vo is = 42.00  V \n",
+      "The output voltage Vo is = 42.00  V \n"
+     ]
+    }
+   ],
+   "source": [
+    " # to determine the output voltage of the precision rectifier circuit for i/p voltage a) Vi = 7 b) Vi = -7\n",
+    "Vi = 7 # #V i/p volt\n",
+    "R1 = 5 # # K ohm\n",
+    "R3 = 5 # # K ohm\n",
+    "R4 = 5 # # K ohm\n",
+    "R2 = 15 # # K ohm\n",
+    "R5 = 15 # # K ohm\n",
+    "Vd = 0.7 # # V   the diode voltage drop\n",
+    "\n",
+    "# the output of the full wave precision rectifier is defined as\n",
+    "# Vo = -A*Vi #      for Vi < 0                      equation 1\n",
+    "#    = A*Vi  #      otherwise                       equation 2\n",
+    "\n",
+    "# or  Vo = abs(A*Vi) #\n",
+    "\n",
+    "# The gain of precision full wave rectifier\n",
+    "A = (((R2*R5)/(R1*R3))-(R5/R4)) #\n",
+    "print 'The gain of precision full wave rectifier A is = %0.2f'%A,' '\n",
+    "\n",
+    "\n",
+    "# for Vi = 7 V           the output voltage is\n",
+    "Vi = 7 #\n",
+    "Vo = -A*Vi #      # from equation 1\n",
+    "Vo =  A*Vi #      # from equation 2\n",
+    "Vo = abs(A*Vi) #\n",
+    "print 'The output voltage Vo is = %0.2f'%Vo,' V '\n",
+    "\n",
+    "# for Vi = -7 V           the output voltage is\n",
+    "Vi = -7 #\n",
+    "Vo = -A*Vi #      # from equation 1\n",
+    "Vo =  A*Vi #      # from equation 2\n",
+    "Vo = abs(A*Vi) #\n",
+    "print 'The output voltage Vo is = %0.2f'%Vo,' V '"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}