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+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 8 Analog Multiplier"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 8.1 Pg 267"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "the output voltage of inverting amplifier (V2) is = 3.00 V\n"
+ ]
+ }
+ ],
+ "source": [
+ "from math import sqrt, pi\n",
+ "from __future__ import division\n",
+ "# to determine the output voltage of inverting amplifier (V2)\n",
+ "Vin = 18 # # V\n",
+ "V1 = -6 # # V\n",
+ "\n",
+ "# in the op-amp due to the infinite i/p resiostance the input current is = 0\n",
+ "# i1+i2 = 0\n",
+ "# it gives relation\n",
+ "Vo = -Vin #\n",
+ "\n",
+ "# the output of multiplier is defined as\n",
+ "#Vo = K*V1*V2\n",
+ "\n",
+ "K = 1 # # we assume\n",
+ "\n",
+ "V2 = (Vo/(K*V1))#\n",
+ "print 'the output voltage of inverting amplifier (V2) is = %0.2f'%(V2),'V'"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 8.2 Pg 267"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "the output voltage of multiplier is = 225.00 V\n"
+ ]
+ }
+ ],
+ "source": [
+ "# to determine the output voltage of multiplier\n",
+ "Vin = 15 # # V\n",
+ "\n",
+ "# the output of multiplier is defined as\n",
+ "#Vo = K*V1*V2\n",
+ "# because of i/p terminal the circuit performs mathematical operation squaring\n",
+ "# i.e V1 = V2 = Vin\n",
+ "K = 1 # # we assume\n",
+ "Vo = K*(Vin)**2#\n",
+ "print 'the output voltage of multiplier is = %0.2f'%(Vo),'V' "
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 8.3 Pg 268"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "the output voltage of inverting amplifier is = 4.00 V \n",
+ "the output voltage of multiplier is = 16.00 V \n"
+ ]
+ }
+ ],
+ "source": [
+ "from math import sqrt, pi\n",
+ "# to determine the output voltage of multiplier and inverting amplifier\n",
+ "Vin = 16 #\n",
+ "# the output of the inverting amplifier\n",
+ "K =1 # # we assume\n",
+ "Vos = sqrt(abs(Vin)/K) #\n",
+ "print 'the output voltage of inverting amplifier is = %0.2f'%(Vos),' V '\n",
+ "\n",
+ "# the output of the multiplier\n",
+ "Vo = K*Vos**2 #\n",
+ "print 'the output voltage of multiplier is = %0.2f'%(Vo),' V '"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 8.5 Pg 269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "output voltage of RMS detector is = 10.00 V \n"
+ ]
+ }
+ ],
+ "source": [
+ "# output voltage of of RMS detector\n",
+ "Vin = 10 # \n",
+ "T = 1 # # we assume that the charging and discharging period of capacitor\n",
+ "\n",
+ "# the output voltage of RMS detector\n",
+ "# Vo =sqrt(1/T*)integrate(Vin**2*(t),t,0,1 [,atol [,rtol]]) #\n",
+ "Vo = 10 #\n",
+ "print 'output voltage of RMS detector 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
+}