{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Chapter 19: Television" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 19.1: example_1.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc;\n", "// page no 703\n", "// prob no 19.1\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.Therefore the range from min to max luminnance has 100-7.5=92.5 units.\n", "//Therefore the level is\n", "IRE=7.5 + (0.5*92.5);\n", "disp('IRE units',IRE,'Level of video signals in IRE units');" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 19.2: example_2.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc;\n", "// page no 704\n", "// prob no 19.2\n", "// part a) horizontal blanking\n", "// Horizontal blanking occupies approximately 10 us of the 63.5 us duration of each line,\n", "Hztl_blnk=10/63.5 *100;\n", "disp('of the signal','%',Hztl_blnk,'Horizontal blanking occupies');\n", "// part b) vertical blanking\n", "// Vertical blanking occupies approximately 21 lines per field or 42 lines per frame. A frame has 525 lines altogether,so\n", "Vert_blnk=42/525 *100;\n", "disp('of the signal','%',Vert_blnk,'vertical blanking occupies');\n", "// part c) active signal\n", "// since 8% of the time is lost in vertical blanking, 92% of the time is involved in the tansmission of the active lines.\n", "act_vid = (100-Hztl_blnk)*(100-Vert_blnk)/100;\n", "disp('%',act_vid,'The acive video is');" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 19.3: example_3.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc;\n", "// page no 707\n", "// prob no 19.3\n", "// A typical low-cost monochrome receiver has a video bandwidth of 3MHz\n", "B=3;// bandwidth in MHz\n", "// The horizontal resolution in lines is given as\n", "L_h=B*80;\n", "disp('lines',L_h,'The horizontal resolution in lines is');" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 19.4: example_4.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc;\n", "// page no 709\n", "// prob no 19.4\n", "// A RGB video signal has normalized values as\n", "R=0.2;G=0.4;B=0.8;\n", "//The luminance signal is given as\n", "Y=0.30*R+0.59*G+0.11*B;\n", "disp(Y,'The luminance signal is');\n", "//The in-phase component of the color signal is given as\n", "I=0.60*R-0.28*G-0.32*B;\n", "disp(I,'The in-phase component of the color signal is');\n", "//The quadrature component of the color signal is given as\n", "Q=0.21*R-0.52*G+0.31*B;\n", "disp(Q,'The quadrature component of the color signal is');" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 19.5: example_5.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc;\n", "// page no 712\n", "// prob no 19.5\n", "......///// refer table 19.1///////.......\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", "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", "disp('of the maximum transmitter power is used to transmit a black setup','%',P_black_setup,)" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 19.6: example_6.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc;\n", "// page no 728\n", "// prob no 19.6\n", "// refer fig 19.27 of the page no 729\n", "// from fig, we can write down the values directly as given\n", "In1=100*10^-3;//expressed in mV\n", "In1_dBmV=20*log10(In1/1);\n", "disp('dBmV',In1_dBmV,'The input of Amp 1 is');\n", "// this above calculated signal is applied to the input of the first 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", "disp('dBmV',Out1,'The output of Amp 1 is');\n", "L=15;//expressed in dB\n", "// The input level of Amp 2 is\n", "In2_dBmV=Out1-L;\n", "disp('dBmV',In2_dBmV,'The input of Amp 2 is');\n", "G2=25;//gain of Amp2 expressed in dB\n", "// o/p level of Amp 2 is\n", "Out2=In2_dBmV+G2;\n", "disp('dBmV',Out2,'The output of Amp 2 is');\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", "disp('dBmV',In3_dBmV,'The input of Amp 3 is');\n", "G3=20;//gain of Amp3 expressed in dB\n", "Out3=In3_dBmV+G3;\n", "disp('dBmV',Out3,'The output of Amp 3 is');\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", "disp('mV',V_drop,'Signal strength at subscriber tap is');\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", "disp('dBm',Pdrop_dBm,'The power at the end is');" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 19.7_a: example_7.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc;\n", "// page no 731\n", "// prob no 19.7\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", "f_lo=Fc+IF;\n", "a=f_lo+ch/2; b=f_lo-ch/2;\n", "disp('band','MHz',a,'to','MHz',b,'The interference would in');" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 19.7_b: example_8.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc;\n", "// page no 740\n", "// prob no 19.8\n", "Nh=640; 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", "// By using the product of Horizontal & vertical resolution, no of luminance 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", "disp('bps',fb,'The bit rate for the signal is');" ] } ], "metadata": { "kernelspec": { "display_name": "Scilab", "language": "scilab", "name": "scilab" }, "language_info": { "file_extension": ".sce", "help_links": [ { "text": "MetaKernel Magics", "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" } ], "mimetype": "text/x-octave", "name": "scilab", "version": "0.7.1" } }, "nbformat": 4, "nbformat_minor": 0 }