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diff --git a/Electronic_Communication_Systems_by_Roy_Blake/Chapter19.ipynb b/Electronic_Communication_Systems_by_Roy_Blake/Chapter19.ipynb new file mode 100644 index 00000000..38f17535 --- /dev/null +++ b/Electronic_Communication_Systems_by_Roy_Blake/Chapter19.ipynb @@ -0,0 +1,370 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 19 : Television" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1 : pg 703" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Level of video signals in IRE units 53.75 IRE units\n" + ] + } + ], + "source": [ + " \n", + "# page no 703\n", + "# prob no 19.1\n", + "#calculate the level of video signals\n", + "#given\n", + "# In the given problem,a video signal has 50% of the maximum luminance level\n", + "#A black setup level of 7.5 IRE represents zero luminance,and 100 IRE is max brightness.\n", + "#Therefore the range from min to max luminnance has 100-7.5=92.5 units.\n", + "#Therefore the level is\n", + "#calculations\n", + "IRE = 7.5 + (0.5 * 92.5)\n", + "#results\n", + "print 'Level of video signals in IRE units',IRE,'IRE units'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2 : pg 704" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Horizontal blanking occupies 15.75 % of the signal\n", + "vertical blanking occupies 8.0 % of the signal\n", + "The active video is 77.51 %\n" + ] + } + ], + "source": [ + " \n", + "# page no 704\n", + "# prob no 19.2\n", + "#calculate the horizontal, vertical blanking occupies\n", + "# part a) horizontal blanking\n", + "# Horizontal blanking occupies approximately 10 us of the 63.5 us duration of each line,\n", + "#given\n", + "Hztl_blnk = 10 / 63.5 * 100\n", + "#calculations and results\n", + "print 'Horizontal blanking occupies',round(Hztl_blnk,2),'%','of the signal'\n", + "# part b) vertical blanking\n", + "# Vertical blanking occupies approximately 21 lines per field or 42 lines per\n", + "# frame. A frame has 525 lines altogether,so\n", + "Vert_blnk = 42. / 525 * 100\n", + "print 'vertical blanking occupies',Vert_blnk,'%','of the signal'\n", + "# part c) active signal\n", + "# since 8% of the time is lost in vertical blanking, 92% of the time is\n", + "# involved in the tansmission of the active lines.\n", + "act_vid = (100 - Hztl_blnk) * (100 - Vert_blnk) / 100\n", + "print 'The active video is',round(act_vid,2),'%'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3 : pg 707" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The horizontal resolution in lines is 240.0 lines\n" + ] + } + ], + "source": [ + " \n", + "# page no 707\n", + "# prob no 19.3\n", + "#calculate the horizontal resolution\n", + "# A typical low-cost monochrome receiver has a video bandwidth of 3MHz\n", + "#given\n", + "B = 3.# bandwidth in MHz\n", + "#calculations\n", + "# The horizontal resolution in lines is given as\n", + "L_h = B * 80\n", + "#results\n", + "print 'The horizontal resolution in lines is',L_h,'lines'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4 : pg 709" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The luminance signal is 0.384\n", + "The in-phase component of the color signal is -0.248\n", + "The quadrature component of the color signal is 0.082\n" + ] + } + ], + "source": [ + " \n", + "# page no 709\n", + "# prob no 19.4\n", + "#given\n", + "#calculate the components of signal\n", + "# A RGB video signal has normalized values as\n", + "R=0.2;G=0.4;B=0.8;\n", + "#calculations and results\n", + "#The luminance signal is given as\n", + "Y=0.30*R+0.59*G+0.11*B;\n", + "print 'The luminance signal is',Y\n", + "#The in-phase component of the color signal is given as\n", + "I=0.60*R-0.28*G-0.32*B;\n", + "print 'The in-phase component of the color signal is',I\n", + "#The quadrature component of the color signal is given as\n", + "Q=0.21*R-0.52*G+0.31*B;\n", + "print 'The quadrature component of the color signal is',Q" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5 : pg 712" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "45.5625 % of the maximum transmitter power is used to transmit a black setup\n" + ] + } + ], + "source": [ + " \n", + "# page no 712\n", + "# prob no 19.5\n", + "#refer table 19.1\n", + "#calculate the max transmitter power\n", + "#given\n", + "# The proportion in the table are voltage levels and have to be squared to get power.\n", + "# for black setup the voltage level is given as\n", + "#calculations\n", + "v = 0.675\n", + "#Therefore the power level as a fraction of the maximum transmitter power is\n", + "P_black_setup = v ** 2 * 100\n", + "#results\n", + "print P_black_setup,'%','of the maximum transmitter power is used to transmit a black setup'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6 : pg 728" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The input of Amp 1 is -20.0 dBmV\n", + "The output of Amp 1 is 20.0 dBmV\n", + "The input of Amp 2 is 5.0 dBmV\n", + "The output of Amp 2 is 30.0 dBmV\n", + "The input of Amp 3 is 8.0 dBmV\n", + "The output of Amp 3 is 28.0 dBmV\n", + "Signal strength at subscriber tap is 1.778 mV\n", + "The power at the end is -43.751 dBm\n" + ] + } + ], + "source": [ + " \n", + "# page no 728\n", + "# prob no 19.6\n", + "# refer fig 19.27 of the page no 729\n", + "#calculate the output and input in all cases\n", + "#given\n", + "from math import log10\n", + "# from fig, we can write down the values directly as given\n", + "In1 = 100 * 10 ** -3#expressed in mV\n", + " #calculations and results\n", + "In1_dBmV = 20 * log10(In1 / 1)\n", + "print 'The input of Amp 1 is',In1_dBmV,'dBmV'\n", + "# this above calculated signal is applied to the input of the first\n", + "# amplifier,i.e. head-end signal processing\n", + "G1 = 40# gain of Amp 1 expressed in dB\n", + " # o/p level of Amp 1 is\n", + "Out1 = In1_dBmV + G1\n", + "print 'The output of Amp 1 is',Out1,'dBmV'\n", + "L = 15#expressed in dB\n", + "# The input level of Amp 2 is\n", + "In2_dBmV = Out1 - L\n", + "print 'The input of Amp 2 is',In2_dBmV,'dBmV'\n", + "G2 = 25#gain of Amp2 expressed in dB\n", + "# o/p level of Amp 2 is\n", + "Out2 = In2_dBmV + G2\n", + "print 'The output of Amp 2 is',Out2,'dBmV'\n", + "L1 = 10# loss in cable\n", + "L2 = 12#loss in directional coupler\n", + "# The input level of Amp 3 is\n", + "In3_dBmV = Out2 - L1 - L2\n", + "print 'The input of Amp 3 is',In3_dBmV,'dBmV'\n", + "G3 = 20#gain of Amp3 expressed in dB\n", + "Out3 = In3_dBmV + G3\n", + "print 'The output of Amp 3 is',Out3,'dBmV'\n", + "# There is further 3dB cable loss and 20dB loss in the tap\n", + "L3 = 3.#loss in cable\n", + "L4 = 20.# loss in tap\n", + "#signal strength at the tap is\n", + "Vdrop_dBmV = Out3 - L3 - L4\n", + "V_drop = 10 ** (Vdrop_dBmV / 20)# expressed in mV\n", + "print 'Signal strength at subscriber tap is',round(V_drop,3),'mV'\n", + "# Calculation of power into 75 ohm\n", + "R = 75.#expressed in ohm\n", + "Pdrop = (V_drop * 10 ** -3) ** 2 / R\n", + "Pdrop_dBm = 10 * log10(Pdrop / 10 ** -3)\n", + "print 'The power at the end is',round(Pdrop_dBm,3),'dBm'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7 : pg 731" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The interference would in 132.0 MHz to 126.0 MHz band\n", + "The bit rate for the signal is 110592000.0 bps\n" + ] + } + ], + "source": [ + " \n", + "# page no 731\n", + "# prob no 19.7\n", + "#calculate bit rate and interference\n", + "#given\n", + "# In given problem a TV receiver is tuned to channel 6.\n", + "#All modern Rx uses a picture IF of 45.75 MHz with high-side injection of the signal into the cable.\n", + "# The picture carrier of channel 6 is at a frequency of 83.25MHz,so\n", + "ch = 6\n", + "Fc = 83.25# expressed in MHz\n", + "IF = 45.75#expressed in MHz\n", + "Nh = 640.\n", + "Nv = 480# resolution of digital video signal as 640*480 pixels\n", + "Rf = 30.#frame rate expressed in Hz\n", + "m = 8.# bits per sample\n", + "#calculations\n", + "f_lo = Fc + IF\n", + "a = f_lo + ch / 2\n", + "b = f_lo - ch / 2\n", + "# By using the product of Horizontal & vertical resolution, no of luminance\n", + "# pixels per frame are\n", + "Npl = Nh * Nv\n", + "# since each of the color signals has one-fourth the total no of luma pixels\n", + "Npt = 1.5 * Npl\n", + "#therefore bit rate is given as\n", + "fb = Npt * m * Rf\n", + "#results\n", + "print 'The interference would in',a,'MHz','to',b,'MHz','band'\n", + "print 'The bit rate for the signal is',fb,'bps'" + ] + } + ], + "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 +} |