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   "source": [
    "# Chapter 19 : Television"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 1 : pg 703"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
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   },
   "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'"
   ]
  }
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