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diff --git a/Electronic_Communication_Systems_by_Roy_Blake/Chapter3.ipynb b/Electronic_Communication_Systems_by_Roy_Blake/Chapter3.ipynb new file mode 100644 index 00000000..c0bd9d20 --- /dev/null +++ b/Electronic_Communication_Systems_by_Roy_Blake/Chapter3.ipynb @@ -0,0 +1,414 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 3 : The Amplitude Modulation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1 : pg 105" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "V(t) = ( 2.83 + 1.41 *sin( 3142.0 *t))*sin( 9424778.0 *t) V\n" + ] + } + ], + "source": [ + "# page no 105\n", + "# prob no 3.1\n", + "#calculate the Voltage equation\n", + "from math import pi, sqrt\n", + "# given\n", + "Erms_car=2; f_car=1.5*10**6;f_mod=500;Erms_mod=1;\n", + "# Equation requires peak voltages & radian frequencies\n", + "#calculations\n", + "Ec=sqrt(2)*Erms_car; Em=sqrt(2)*Erms_mod;\n", + "wc=2*pi*f_car; \n", + "wm=2*pi*f_mod;t=1;\n", + "#results\n", + "# Therefore the equation is \n", + "print 'V(t) = (',round(Ec,2),'+ ',round(Em,2),'*sin(',round(wm),'*t))*sin(',round(wc),'*t) V'\n", + "#print 'v(t) = (2.83+1.41*sin(3.14*10**3*t))*sin(9.42*10**6*t) V'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2 : pg 106" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "m = 0.5\n", + "The equation can be obtained as v(t) = 2.83(1+ 0.5 *sin(3.14*10**3*t))*sin(9.42*10**6*t) V\n" + ] + } + ], + "source": [ + "#page no 106\n", + "#prob no 3.2\n", + "#calculate the voltage equation\n", + "# To avoid the round-off errors we should use the original voltage values\n", + "#given\n", + "Em=1.;Ec=2.;\n", + "#Calculations\n", + "m=Em/Ec;\n", + "#results\n", + "print 'm =',m\n", + "print 'The equation can be obtained as','v(t) = 2.83(1+ ',m,'*sin(3.14*10**3*t))*sin(9.42*10**6*t) V'," + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3 : pg 109" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The modulation index is 0.374\n" + ] + } + ], + "source": [ + "#page no 109\n", + "#prob no 3.3\n", + "#calculate the modulation index\n", + "from math import sqrt\n", + "#given\n", + "E_car=10.;E_m1=1.;E_m2=2.;E_m3=3.;\n", + "#calculations\n", + "m1=E_m1/E_car;\n", + "m2=E_m2/E_car;\n", + "m3=E_m3/E_car;\n", + "mT=sqrt(m1**2+m2**2+m3**2);\n", + "#results\n", + "print 'The modulation index is',round(mT,3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4 : pg 110" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The modulation index is 0.364\n" + ] + } + ], + "source": [ + "#page no 110\n", + "#prob no 3.4\n", + "#calculate the modulation index\n", + "#refer fig 3.2\n", + "#given\n", + "E_max=150.; E_min=70;# voltages are in mV\n", + "#calculations\n", + "m=(E_max-E_min)/(E_max+E_min);\n", + "#results\n", + "print 'The modulation index is',round(m,3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6 : pg 114" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum modulation freq is 5000.0 Hz\n" + ] + } + ], + "source": [ + "#page no 114\n", + "#prob no 3.6\n", + "#calculate the max modulation frequency\n", + "#given\n", + "B=10.*10**3;\n", + "#calculations\n", + "# maximum modulation freq is given as \n", + "fm=B/2;\n", + "#results\n", + "print 'The maximum modulation freq is',fm,'Hz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7 : pg 116" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total power is 66.0 kW\n" + ] + } + ], + "source": [ + "#page no 116\n", + "#prob no 3.7\n", + "# AM broadcast transmitter\n", + "#calculate the total power\n", + "#given\n", + "Pc=50.;m=0.8;#power is in kW\n", + "#calculations\n", + "Pt=Pc*(1+m**2 /2);\n", + "#results\n", + "print 'The total power is',Pt,'kW'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8 : pg 118" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Signal Frequency in Hz = 426000\n" + ] + } + ], + "source": [ + "# page no 118\n", + "# prob no 8.6\n", + "#calculate the signal frequency\n", + "#2 kHz tone is present on channel 5 of group 3 of supergroup\n", + "#signal is lower sided so\n", + "#given\n", + "fc_channel_5=92*10**3;\n", + "#calculations\n", + "fg=fc_channel_5 - (2*10**3);# 2MHz baseband signal\n", + "# we know group 3 in the supergroup is moved to the range 408-456 kHz with a suppressed carrier frequency of 516kHz\n", + "f_s_carr=516*10**3;\n", + "fsg=f_s_carr - fg;\n", + "#results\n", + "print'The Signal Frequency in Hz =',fsg;" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9 : pg 122" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total power is 200.0 uW\n", + "The modulating freq is 2.0 kHz\n", + "The carrier freq 10.0 MHz\n" + ] + } + ], + "source": [ + "#page no 122\n", + "#prob no. 3.9\n", + "#calculate the total power, modulating frequency and carrier frequency\n", + "# refer fig 3.14\n", + "#given\n", + "# from spectrum we can see that each of the two sidebands is 20dB below the ref level of 10dBm. \n", + "#Therefore each sideband has a power of -10dBm i.e. 100uW.\n", + "power_of_each_sideband = 100.;\n", + "#calculations and results\n", + "Total_power = 2.* power_of_each_sideband;\n", + "print 'The total power is',Total_power,'uW'\n", + "div=4; freq_per_div=1.;\n", + "sideband_separation = div * freq_per_div;\n", + "f_mod= sideband_separation/2;\n", + "print 'The modulating freq is ',f_mod,'kHz'\n", + "# Even if this siganl has no carrier, it still has a carrier freq which is midway between the two sidebands. Therefore\n", + "carrier_freq = 10.;\n", + "print 'The carrier freq',carrier_freq,'MHz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10 : pg 126" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The o/p freq f_out1 is 7 MHz\n", + "The o/p freq f_out2 is 7.9965 MHz\n" + ] + } + ], + "source": [ + "# page no 126\n", + "# prob no 3.10\n", + "#calculate the o/p frequency in both cases\n", + "#given\n", + "f_car=8*10**6;f_mod1=2*10**3;f_mod2=3.5*10**3;\n", + "#calculations\n", + "#Signal is LSB hence o/p freq is obtained by subtracting f_mod from f_car\n", + "f_out1=f_car-f_mod1; \n", + "f_out2=f_car-f_mod2; \n", + "#results\n", + "print 'The o/p freq f_out1 is ',f_out1/(10**6),'MHz'\n", + "print 'The o/p freq f_out2 is ',f_out2/(10**6),'MHz'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11 : pg 127" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of average power of signal is 1.5625 W\n" + ] + } + ], + "source": [ + "# page no 127\n", + "# prob no 3.11\n", + "#calculate the value of average power of signal\n", + "from math import sqrt\n", + "#Refering the fig. 3.17\n", + "#From fig it is clear that thee waveform is made from two sine waves \n", + "#given\n", + "Vp=12.5;#Since Vp-p is 25V from fig hence individual Vp is half of Vp-p\n", + "Rl=50.;#Load resistance is 50 ohm\n", + "#Determination of average power\n", + "#calculations\n", + "Vrms=Vp/sqrt(2);\n", + "P=((Vrms)**2)/Rl;\n", + "#results\n", + "print 'The value of average power of signal is ',P,'W'" + ] + } + ], + "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": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |