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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 1 - Operational Amplifiers"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Exa 1.1 - page 11"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given data\n",
      "G= -100 \n",
      "R1= 2.2 # in kohm\n",
      "R1=R1*10**3 # in ohm\n",
      "# Formula G=-Rf/R1\n",
      "Rf= -G*R1 \n",
      "print \"The value of Rf = %0.f kohm \" %(Rf*10**-3)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The value of Rf = 220 kohm \n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Exa 1.2 - page 11"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given data\n",
      "Rf= 200 # in kohm\n",
      "R1= 2 # in kohm\n",
      "vin=2.5 # in mV\n",
      "vin=vin*10**-3 # in volt\n",
      "G= -Rf/R1 \n",
      "vo= G*vin # in V\n",
      "print \"The output voltage = %0.2f Volt \" %vo"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The output voltage = -0.25 Volt \n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Exa 1.3 - page 12"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given data\n",
      "G=-10 \n",
      "Ri= 100 # in kohm\n",
      "R1= Ri # in kohm\n",
      "R1=R1*10**3 # in ohm\n",
      "# Formula G=-R2/R1\n",
      "R2= R1*abs(G) # ohm\n",
      "print \"Value of R1 = %0.f kohm \" %(R1*10**-3)\n",
      "print \"and value of R2 = %0.f Mohm \" %(R2*10**-6)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Value of R1 = 100 kohm \n",
        "and value of R2 = 1 Mohm \n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Exa 1.4 - page 37"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given data\n",
      "R1= 100 # in kohm\n",
      "R2= 500 # in kohm\n",
      "V1= 2 # in volt\n",
      "Vo= (1+R2/R1)*V1 # in volt\n",
      "print \"Output voltage for noninverting amplifier = %0.f Volt\" %Vo"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Output voltage for noninverting amplifier = 12 Volt\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Exa 1.5 - page 38"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "# Given data\n",
      "Rf= 1 # in Mohm\n",
      "Rf=Rf*10**6 #in ohm\n",
      "\n",
      "# Part(a)\n",
      "V1=1 #in volt\n",
      "V2=2 #in volt\n",
      "V3=3 #in volt\n",
      "R1= 500 # in kohm\n",
      "R1=R1*10**3 #in ohm\n",
      "R2= 1 # in Mohm\n",
      "R2=R2*10**6 #in ohm\n",
      "R3= 1 # in Mohm\n",
      "R3=R3*10**6 #in ohm\n",
      "Vo= -Rf*(V1/R1+V2/R2+V3/R3) # in volt\n",
      "print \"(a) Output voltage = %0.f Volt \" %Vo\n",
      "\n",
      "# Part(b)\n",
      "V1=-2 #in volt\n",
      "V2=3 #in volt\n",
      "V3=1 #in volt\n",
      "R1= 200 # in kohm\n",
      "R1=R1*10**3 #in ohm\n",
      "R2= 500 # in kohm\n",
      "R2=R2*10**3 #in ohm\n",
      "R3= 1 # in Mohm\n",
      "R3=R3*10**6 #in ohm\n",
      "Vo= -Rf*(V1/R1+V2/R2+V3/R3) # in volt\n",
      "print \"(b) Output voltage = %0.f Volt\" %Vo"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a) Output voltage = -7 Volt \n",
        "(b) Output voltage = 3 Volt\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Exa 1.6 - page 38"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "# Given data\n",
      "print \"Minimum closed loop voltage gain for R2=0 and R1= 2 kohm\"\n",
      "R2=0 \n",
      "R1=2 # in kohm\n",
      "R1=R1*10**3 # in ohm\n",
      "Av_min= (1+R2/R1)\n",
      "print \"Av(min) =\",Av_min\n",
      "\n",
      "print \"Maximum closed loop voltage gain for maximum value of R2=100 kohm and R1= 2 kohm\"\n",
      "R2=100 # in kohm\n",
      "R1=2 # in kohm\n",
      "Av_max= (1+R2/R1)\n",
      "print \"Av(max) =\",Av_max"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Minimum closed loop voltage gain for R2=0 and R1= 2 kohm\n",
        "Av(min) = 1.0\n",
        "Maximum closed loop voltage gain for maximum value of R2=100 kohm and R1= 2 kohm\n",
        "Av(max) = 51.0\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Exa 1.7 - page 39"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "# Given data\n",
      "V1= 745 # in \u00b5V\n",
      "V2= 740 # in \u00b5V\n",
      "V1=V1*10**-6 # in volt\n",
      "V2=V2*10**-6 # in volt\n",
      "CMRR=80 # in dB\n",
      "Av=5*10**5 \n",
      "# (i)\n",
      "# CMRR in dB= 20*log(Ad/Ac)\n",
      "Ad=Av \n",
      "Ac= Ad/10**(CMRR/20) \n",
      "# (ii)\n",
      "Vo= Ad*(V1-V2)+Ac*(V1+V2)/2 \n",
      "print \"Output voltage = %0.2f Volt\" %Vo\n",
      "\n",
      "# Note:- In the book, there is calculation error to evaluate the value of Ac,\n",
      "#so the value of Ac is wrong ans to evaluate the output voltage there is also calculation error "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Output voltage = 2.54 Volt\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Exa 1.8 - page 40"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "# Given data\n",
      "R1= 1 # in Mohm\n",
      "Ri=R1 # in Mohm\n",
      "Rf=1 # in Mohm\n",
      "A_VF= -Rf/R1 \n",
      "print \"Voltage gain = %0.f\" %A_VF"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Voltage gain = -1\n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Exa 1.10 - page 41"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "# Given data\n",
      "V1=2 # in V\n",
      "V2=3 # in V\n",
      "Rf=3 # in kohm\n",
      "R1=1 # in kohm\n",
      "Vo1= (1+Rf/R1)*V1 \n",
      "print \"Output voltage when only 2V voltage source is acting is %0.f Volt\" %Vo1\n",
      "Vo2= (1+Rf/R1)*V2 \n",
      "print \"Output voltage due to 3V voltage source is %0.f Volt\" %Vo2\n",
      "Vo= Vo1+Vo2 # in volts\n",
      "print \"Total output voltage is %0.f Volts\" %Vo"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Output voltage when only 2V voltage source is acting is 8 Volt\n",
        "Output voltage due to 3V voltage source is 12 Volt\n",
        "Total output voltage is 20 Volts\n"
       ]
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Exa 1.11 - page 42"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "# Given data\n",
      "Rf=500 # in kohm\n",
      "min_vvs= 0 # minimum value of variable resistor in ohm\n",
      "max_vvs= 10 # maximum value of variable resistor in ohm\n",
      "Ri_min= 10+min_vvs # in kohm\n",
      "Ri_max= 10+max_vvs #in kohm\n",
      "# Av= Vo/Vi= -Rf/Ri\n",
      "Av=-Rf/Ri_min \n",
      "print \"Closed loop voltage gain corresponding to Ri(min) is\",Av\n",
      "Av=-Rf/Ri_max \n",
      "print \"and closed loop voltage gain corresponding to Ri(max) is\",Av"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Closed loop voltage gain corresponding to Ri(min) is -50.0\n",
        "and closed loop voltage gain corresponding to Ri(max) is -25.0\n"
       ]
      }
     ],
     "prompt_number": 21
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Exa 1.12 - page 43"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given data\n",
      "Rf=200 # in kohm\n",
      "R1= 20 # in kohm\n",
      "# Av= Vo/Vi= -Rf/Ri\n",
      "Av= -Rf/R1 \n",
      "Vi_min= 0.1 # in V\n",
      "Vi_max= 0.5 # in V\n",
      "# Vo= Av*Vi\n",
      "Vo_min= Av*Vi_min # in V\n",
      "Vo_max= Av*Vi_max # in V\n",
      "print \"Output voltage ranges from\",Vo_min,\"V to\",Vo_max,\"V\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Output voltage ranges from -1.0 V to -5.0 V\n"
       ]
      }
     ],
     "prompt_number": 23
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Exa 1.13 - page 43"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "# Given data\n",
      "Rf= 250 # in kohm\n",
      "# Output voltage expression, Vo= -5*Va+3*Vb\n",
      "# and we know that for a difference amplifier circuit, \n",
      "# Vo= -Rf/R1*Va + [R2/(R1+R2)]*[1+Rf/R1]*Vb\n",
      "# Comparing both the expression, we get\n",
      "# -Rf/R1*Va= -5*Va, or\n",
      "R1= Rf/5 # in kohm\n",
      "print \"The value of R1 = %0.2f kohm\" %R1\n",
      "# and \n",
      "R2= 3*R1**2/(R1+Rf-3*R1)\n",
      "print \"The value of R2 = %0.2f kohm\" %R2\n",
      "\n",
      "# Note : Answer in the book is wrong"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The value of R1 = 50.00 kohm\n",
        "The value of R2 = 50.00 kohm\n"
       ]
      }
     ],
     "prompt_number": 24
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Exa 1.14 - page 44"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given data\n",
      "Vi_1= 150 # in \u00b5V\n",
      "Vi_2= 140 # in \u00b5V\n",
      "Vd= Vi_1-Vi_2 # in \u00b5V\n",
      "Vd=Vd*10**-6 # in V\n",
      "Vc= (Vi_1+Vi_2)/2 # in \u00b5V\n",
      "Vc=Vc*10**-6 # in V\n",
      "# Vo= Ad*Vd*(1+Vc/(CMRR*Vd))\n",
      "\n",
      "# (i) For Ad=4000 and CMRR= 100\n",
      "Ad=4000 \n",
      "CMRR= 100 \n",
      "Vo= Ad*Vd*(1+Vc/(CMRR*Vd)) # in volt\n",
      "print \"(a) Output voltage = %.1f mV\" %(Vo*10**3)\n",
      "\n",
      "# (ii) For Ad=4000 and CMRR= 10**5\n",
      "Ad=4000 \n",
      "CMRR= 10**5 \n",
      "Vo= Ad*Vd*(1+Vc/(CMRR*Vd)) # in volt\n",
      "print \"(b) Output voltage = %0.1f mV\" %(Vo*10**3)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a) Output voltage = 45.8 mV\n",
        "(b) Output voltage = 40.0 mV\n"
       ]
      }
     ],
     "prompt_number": 31
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Exa 1.15 - page 45"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "# Given data\n",
      "Rf=470 # in kohm\n",
      "R1=4.3 # in kohm\n",
      "R2=33 # in kohm\n",
      "R3=33 # in kohm\n",
      "Vi= 80 # in \u00b5V\n",
      "Vi=Vi*10**-6 # in volt\n",
      "A1= 1+Rf/R1 \n",
      "A2=-Rf/R2 \n",
      "A3= -Rf/R3 \n",
      "A=A1*A2*A3 \n",
      "Vo= A*Vi # in volt\n",
      "print \"Output voltage = %0.2f Volts\" %Vo"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Output voltage = 1.79 Volts\n"
       ]
      }
     ],
     "prompt_number": 33
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Exa 1.16 - page 46"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from sympy import *\n",
      "t = symbols('t')\n",
      "# Given data\n",
      "R1= 33 # in k\u03a9\n",
      "R2= 10 # in k\u03a9\n",
      "R3= 330 # in k\u03a9\n",
      "V1 = simplify(50*sin(1000*t)) # in mV\n",
      "V2 = simplify(10*sin(3000*t)) # in mV\n",
      "Vo = -(R3/R1*V1+R3/R2*V2)/1000  # in V\n",
      "print \"Output voltage is\",Vo"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Output voltage is -0.5*sin(1000*t) - 0.33*sin(3000*t)\n"
       ]
      }
     ],
     "prompt_number": 37
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Exa 1.17 - page 47"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "# Given data\n",
      "R1=10 # in kohm\n",
      "R2=150 # in kohm\n",
      "R3=10 # in kohm\n",
      "R4=300 # in kohm\n",
      "V1= 1 # in V\n",
      "V2= 2 # in V\n",
      "Vo= ((1+R4/R2)*(R3*V1/(R1+R3))-(R4/R2)*V2) \n",
      "print \"Output voltage = %0.2f Volts\" %Vo"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Output voltage = -2.50 Volts\n"
       ]
      }
     ],
     "prompt_number": 39
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Exa 1.18 - page 47"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "# Given data\n",
      "R1=12 # in kohm\n",
      "Rf=360 # in kohm\n",
      "V1= -0.3 # in V\n",
      "Vo= (1+Rf/R1)*V1 # in V\n",
      "print \"(a) Output voltage result in %0.2f Volts\" %Vo\n",
      "\n",
      "# Part(b)\n",
      "Vo= 2.4 # in V\n",
      "# We know, Vo= (1+Rf/R1)*V1\n",
      "V1= Vo/(1+Rf/R1) \n",
      "print \"(b) To result in an output of 2.4 Volt, Input voltage = %0.2f mV\" %(V1*10**3)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a) Output voltage result in -9.30 Volts\n",
        "(b) To result in an output of 2.4 Volt, Input voltage = 77.42 mV\n"
       ]
      }
     ],
     "prompt_number": 41
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Exa 1.19 - page 48"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "# Given data\n",
      "Rf=68 # in kohm\n",
      "R1=33 # in kohm\n",
      "R2=22 # in kohm\n",
      "R3=12 # in kohm\n",
      "V1= 0.2 # in V\n",
      "V2=-0.5 # in V\n",
      "V3= 0.8 # in V\n",
      "Vo= -Rf/R1*V1 + (-Rf/R2)*V2 + (-Rf/R3)*V3 # in volts\n",
      "print \"Output voltage = %0.3f Volts\" %Vo\n",
      "#Answer in the textbook is not accurate."
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Output voltage = -3.400 Volts\n"
       ]
      }
     ],
     "prompt_number": 44
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Exa 1.20 - page 48"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "# Given data\n",
      "Rf=100 # in kohm\n",
      "R1=20 # in kohm\n",
      "V1= 1.5 # in V\n",
      "Vo1= V1 \n",
      "Vo= -Rf/R1*Vo1 # in volts\n",
      "print \"Output voltage = %0.2f Volts\" %Vo"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Output voltage = -7.50 Volts\n"
       ]
      }
     ],
     "prompt_number": 45
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Exa 1.22 - page 50"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "# Given data\n",
      "vo= -10 # in V\n",
      "i_f= 1 # in mA\n",
      "i_f= i_f*10**-3 #in A\n",
      "# Formula vo= -i_f*Rf\n",
      "Rf= -vo/i_f # in \u03a9\n",
      "# The output voltage, vo= -(v1+5*v2)       (i)\n",
      "# vo= -Rf/R1*v1 - Rf/R2*v2                     (ii)\n",
      "# Comparing equations (i) and (2)\n",
      "R1= Rf/1 # in \u03a9\n",
      "R2= Rf/5 # in \u03a9\n",
      "print \"The value of Rf = %0.2f k\u03a9\" %(Rf*10**-3)\n",
      "print \"The value of R1 = %0.2f k\u03a9\" %(R1*10**-3)\n",
      "print \"The value of R2 = %0.2f k\u03a9\" %(R2*10**-3)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The value of Rf = 10.00 k\u03a9\n",
        "The value of R1 = 10.00 k\u03a9\n",
        "The value of R2 = 2.00 k\u03a9\n"
       ]
      }
     ],
     "prompt_number": 46
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Exa 1.24 - page 52"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "from sympy import *\n",
      "v1,v2 = symbols('v1 v2')\n",
      "# Given data\n",
      "R1= 9 # in k\u03a9\n",
      "R2= 1 # in k\u03a9\n",
      "R3= 2 # in k\u03a9\n",
      "R4= 3 # in k\u03a9\n",
      "# for node 1\n",
      "va = R4/(R4+R3)*v1\n",
      "vo1 = (1+R1/R2)*va\n",
      "# for node 2\n",
      "va=R3/(R3+R4)*v2\n",
      "vo2 = (1+R1/R2)*va\n",
      "vo = vo1+vo2\n",
      "print \"Total voltage is, vo =\",vo"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Total voltage is, vo = 6.0*v1 + 4.0*v2\n"
       ]
      }
     ],
     "prompt_number": 60
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Exa 1.25 - page 54"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "from sympy import *\n",
      "v1,v2,v3 = symbols('v1 v2 v3')\n",
      "# Given data\n",
      "R1= 9 # in k\u03a9\n",
      "R2= 1 # in k\u03a9\n",
      "R3= 2 # in k\u03a9\n",
      "R4= 3 # in k\u03a9\n",
      "# Voltage at node 1\n",
      "va= R4*v1/(R3+R4)\n",
      "vo1= (1+R1/R2)*va\n",
      "# Voltage at node 2\n",
      "va= R3*v2/(R3+R4)\n",
      "# From (i) and (ii)\n",
      "vo2= (1+R1/R2)*va\n",
      "# Voltage at node 3\n",
      "va= R3*v2/(R3+R4)\n",
      "vo3= (-R1/R2)*v3\n",
      "vo = vo1+vo2+vo3\n",
      "print \"Total voltage is\",vo"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Total voltage is 6.0*v1 + 4.0*v2 - 9.0*v3\n"
       ]
      }
     ],
     "prompt_number": 67
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Exa 1.26 - page 55"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given data\n",
      "import numpy as np\n",
      "from __future__ import division\n",
      "# omega_t= Ao*omega_b\n",
      "# 2*pi*f_t = Ao*2*pi*f_b\n",
      "# f_t= Ao*f_b\n",
      "# Part (i)\n",
      "Ao1= 10**5 \n",
      "f_b1= 10**2 # in Hz\n",
      "f_t1= Ao1*f_b1 # in Hz\n",
      "row1 = np.array([Ao1,f_b1,f_t1])\n",
      "# Part (ii)\n",
      "Ao2= 10**6 \n",
      "f_t2= 10**6 # in Hz\n",
      "f_b2= f_t2/Ao2 # in Hz\n",
      "row2 = np.array([Ao2,f_b2,f_t2])\n",
      "# Part (iii)\n",
      "f_b3= 10**3 # in Hz\n",
      "f_t3= 10**8 # in Hz\n",
      "Ao3= f_t3/f_b3 \n",
      "row3 = np.array([Ao3,f_b3,f_t3])\n",
      "# Part (iv)\n",
      "f_b4= 10**-1 # in Hz\n",
      "f_t4= 10**6 # in Hz\n",
      "Ao4= f_t4/f_b4 \n",
      "row4 = np.array([Ao4,f_b4,f_t4])\n",
      "# Part (v)\n",
      "Ao5= 2*10**5 \n",
      "f_b5= 10 # in Hz\n",
      "f_t5= Ao5*f_b5 # in Hz\n",
      "row5 = np.array([Ao5,f_b5,f_t5])\n",
      "print \"-\"*33\n",
      "print \"Ao           fb(Hz)       ft(Hz)\"\n",
      "print \"-\"*33\n",
      "print \"%.e        %.e        %.e\" %(row1[0], row1[1], row1[2])\n",
      "print \"%.e        %.f           %.e\" %(row2[0], row5[1], row2[2])\n",
      "print \"%.e        %.e        %.e\" %(row3[0], row3[1], row3[2])\n",
      "print \"%.e        %.e        %.e\" %(row4[0], row4[1], row4[2])\n",
      "print \"%.e        %.f           %.e\" %(row5[0], row5[1], row5[2])\n",
      "# Answer for f_b2 is wrong in the textbook."
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "---------------------------------\n",
        "Ao           fb(Hz)       ft(Hz)\n",
        "---------------------------------\n",
        "1e+05        1e+02        1e+07\n",
        "1e+06        10           1e+06\n",
        "1e+05        1e+03        1e+08\n",
        "1e+07        1e-01        1e+06\n",
        "2e+05        10           2e+06\n"
       ]
      }
     ],
     "prompt_number": 37
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Exa 1.27 - page 56"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "from numpy import sqrt\n",
      "# Given data\n",
      "Ao= 86 # in dB\n",
      "A= 40 # in dB\n",
      "f=100 # in kHz\n",
      "f=f*10**3 # in Hz\n",
      "# From  20*log(S) = 20*log(Ao/A), where S, stands for sqrt(1+(f/fb)**2)\n",
      "S= 10**((Ao-A)/20) \n",
      "# S= sqrt(1+(f/fb)**2)\n",
      "fb= f/sqrt(S**2-1) # in Hz\n",
      "Ao= 10**(Ao/20) \n",
      "ft= Ao*fb # in Hz\n",
      "print \"The value of Ao = %0.3e\" %Ao\n",
      "print \"The value of fb = %0.f Hz\" %fb\n",
      "print \"The value of ft = %0.f MHz\" %round(ft*10**-6)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The value of Ao = 1.995e+04\n",
        "The value of fb = 501 Hz\n",
        "The value of ft = 10 MHz\n"
       ]
      }
     ],
     "prompt_number": 43
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Exa 1.28 - page 56"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "from numpy import pi, sqrt\n",
      "# Given data\n",
      "Ao= 10**4 # in V/V\n",
      "f_t= 10**6 # in Hz\n",
      "R2byR1= 20 \n",
      "omega_t= 2*pi*f_t \n",
      "omega_3dB= omega_t/(1+R2byR1) \n",
      "f3dB= omega_3dB/(2*pi) # in Hz\n",
      "print \"3-dB frequency of the closed loop amplifier is %0.1f kHz\" %(f3dB*10**-3)\n",
      "f3dB= 0.1*f3dB # in Hz\n",
      "voBYvi= -R2byR1/sqrt(1+(2*pi*f3dB/omega_3dB)**2) \n",
      "voBYvi= abs(voBYvi) # in v/v\n",
      "print \"Gain = %0.1f v/v\" %(voBYvi)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "3-dB frequency of the closed loop amplifier is 47.6 kHz\n",
        "Gain = 19.9 v/v\n"
       ]
      }
     ],
     "prompt_number": 48
    }
   ],
   "metadata": {}
  }
 ]
}