{
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   "source": [
    "# Chapter 18 : Terrestrial Microwave system"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 1 : pg 676"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The height of each tower must be at least 23.529 m\n"
     ]
    }
   ],
   "source": [
    " \n",
    "# page no 676\n",
    "#calculate the height of each tower\n",
    "# prob no 18_1\n",
    "#given\n",
    "#Transmitter and receiver have same height at dist 40km\n",
    "d=40;#dist is 40 km\n",
    "#calculations\n",
    "h=(d**2)/68.;# As d=sqrt(17h)+sqrt(17h)\n",
    "#results\n",
    "print 'The height of each tower must be at least',round(h,3),'m'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 2 : pg 678"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The distance by which beam must clear the obstacle is 11.63 m\n"
     ]
    }
   ],
   "source": [
    " \n",
    "# page no 678\n",
    "# prob no 18_2\n",
    "from math import sqrt\n",
    "#calculate the distance required\n",
    "#A line of sight radio link at freq 6GHz with seperation 40 km betn antennas\n",
    "#given\n",
    "f=6.;d1=10.;d2=30;#obstacle located at 10 km\n",
    "#calculations\n",
    "#Determination of dist R to clear obstacle \n",
    "R=10.4*sqrt((d1*d2)/(f*(d1+d2)));\n",
    "#results\n",
    "print 'The distance by which beam must clear the obstacle is',round(R,2),'m'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 3 : pg 679"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The power delivered to the Rx is -62.03 dBm\n"
     ]
    }
   ],
   "source": [
    " \n",
    "# page no 679\n",
    "# prob no 18_3\n",
    "#calculate the power delivered\n",
    "from math import log10\n",
    "#given\n",
    "#A transmitter and receiver at 6GHz seperated by 40km with o/p power 2 W\n",
    "f = 6 * 10 ** 9\n",
    "d = 40.\n",
    "Pt = 2# power in watt\n",
    "#transmitting antenna gain Gt=20dBi,receiving antenna Gr=25dBi\n",
    "Gt = 20\n",
    "Gr = 25\n",
    "f_mhz = 6000#f=6000 MHz\n",
    "#calculations\n",
    "Pr_Pt_dB = (Gt + Gr) - (32.44 + (20 * log10(d)) + (20 * log10(f_mhz)))\n",
    "Pt_dBm = 10 * log10(Pt / 10 ** -3)\n",
    "Pr_dBm = Pt_dBm + Pr_Pt_dB\n",
    "#results\n",
    "print 'The power delivered to the Rx is',round(Pr_dBm,2),'dBm'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4 : pg 680"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Noise temperature is 120.0 K\n"
     ]
    }
   ],
   "source": [
    " \n",
    "# page no 680\n",
    "#calculate the noise temperature\n",
    "# prob no 18_4\n",
    "#given\n",
    "T_sky = 120.# Sky temp expressed in K\n",
    "L_dB = 2# antenna feedline loss\n",
    "#calculations\n",
    "L = 10 ** (L_dB / 10)\n",
    "# the noise temp is given as\n",
    "Ta = ((L - 1) * 290 + T_sky) / L\n",
    "#results\n",
    "print 'Noise temperature is',Ta,'K'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 5 : pg 681"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The equivalent noise temperature 169.619 K\n"
     ]
    }
   ],
   "source": [
    " \n",
    "# page no 681\n",
    "# prob no 18.5\n",
    "#calculate the equivalent noise temperature\n",
    "#given\n",
    "NF_dB = 2\n",
    "#calculations\n",
    "NF_power = 10 ** (NF_dB / 10.)\n",
    "T_eq = 290. * (NF_power - 1)\n",
    "#results\n",
    "print 'The equivalent noise temperature',round(T_eq,3),'K'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6 : pg 681"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The noise power is 9.6876e-14 W\n",
      "The thermal noise power is -100.138 dBm\n"
     ]
    }
   ],
   "source": [
    " \n",
    "# page no 681\n",
    "# prob no 18.6\n",
    "# refer example no 18.4 and 18.5\n",
    "#calculate the noise power in both cases\n",
    "from math import log10\n",
    "#given\n",
    "# The antenna and feedline combination from ex 18.4 is used with the Rx from ex\n",
    "# 18.5\n",
    "Ta = 182.# noise temp of the antenna and feedline combination expressed in K\n",
    "Teq = 169.# noise temperature of the Rx\n",
    "B = 20. * 10 ** 6# BW of the receiver\n",
    "#calculations\n",
    "Tn_sys = Ta + Teq#Noise temp for the system\n",
    "k = 1.38 * 10 ** -23#Boltzmann constant\n",
    "# Noise power of the system is given as\n",
    "Pn = k * Tn_sys * B# where k is Boltzmann constant\n",
    "Pn_dBm = 10 * log10(Pn / 10 ** -3)\n",
    "#results\n",
    "print 'The noise power is',Pn,'W'\n",
    "print 'The thermal noise power is',round(Pn_dBm,3),'dBm'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 7 : pg 682"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Carrier to noise ratio is 38.0 dB\n"
     ]
    }
   ],
   "source": [
    " \n",
    "# page no 682\n",
    "# prob no 18.7\n",
    "#calculate the carrier to noise ratio\n",
    "# refer ex no 18.3 and 18.6\n",
    "#given\n",
    "Pr_dBm = -62.#power at the receiver in dBm\n",
    "Pn_dBm = -100.#thermal noise power in dBm\n",
    "#calculations\n",
    "# carrier to noise ratio in dB is given as\n",
    "C_N = Pr_dBm - Pn_dBm\n",
    "#results\n",
    "print 'Carrier to noise ratio is',C_N,'dB'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 8 : pg 683"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Energy per bit to noise density ratio is  27.14 dB\n"
     ]
    }
   ],
   "source": [
    " \n",
    "# page no 683\n",
    "# prob no 18.8\n",
    "#calculate the energy per bit to noise density ratio\n",
    "# refer ex 18.7\n",
    "from math import log10\n",
    "#given\n",
    "fb = 40 * 10 ** 6# bit rate in bps\n",
    "Pr_dBm = -62#power at the receiver in dBm\n",
    "Pr = 10 ** (Pr_dBm / 10) * 10 ** -3# power at the receiver in W\n",
    "Eb = Pr / fb# the energy per bit in J\n",
    "k = 1.38 * 10 ** -23#Boltzmann constant\n",
    "T = 350.\n",
    "#calculations\n",
    "# the noise power density is\n",
    "No = k * T\n",
    "# Energy per bit to noise density ratio in dB is\n",
    "Eb_No = 10 * log10(Eb / No)\n",
    "#results\n",
    "print 'Energy per bit to noise density ratio is ',round(Eb_No,3),'dB'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 9 : pg 686"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The local oscillator freq is  6800.0 MHz\n"
     ]
    }
   ],
   "source": [
    " \n",
    "# page no 686\n",
    "# prob no 18.9\n",
    "# refer fig 18.7(b)\n",
    "#calculate the local oscillator frequency\n",
    "#given\n",
    "#This is the standard superheterodyne receiver\n",
    "fc=6870.;# the received carrier freq in MHz\n",
    "IF=70.;# IF in MHz\n",
    "#calculations\n",
    "# The local oscillator freq is given as\n",
    "f_lo=fc-IF;\n",
    "#results\n",
    "print 'The local oscillator freq is ',f_lo,'MHz'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 10 : pg 688"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The freq of shift oscillator is  160.0 MHz\n",
      "The local oscillator freq is  6640.0 MHz\n",
      "O/P of Mixer 3 is 6800.0 MHz\n"
     ]
    }
   ],
   "source": [
    " \n",
    "# page no 688\n",
    "# prob no 18.10\n",
    "# refer fig 18.9a)\n",
    "#calculate the freq shift in all cases\n",
    "#given\n",
    "fc_r = 6870.# carrier freq of received signal in MHz\n",
    "fc_t = 6710.#carrier freq of transmitted signal in MHz\n",
    "IF = 70.#in MHz\n",
    "#calculations and results\n",
    "# the freq of shift oscillator is\n",
    "fso = fc_r - fc_t\n",
    "print 'The freq of shift oscillator is ',fso,'MHz'\n",
    "#the local oscillator freq is given as\n",
    "flo = fc_t - IF\n",
    "print 'The local oscillator freq is ',flo,'MHz'\n",
    "#from fig, mixer 3 will produce an o/p as\n",
    "op_mix3 = flo + fso\n",
    "print 'O/P of Mixer 3 is',op_mix3,'MHz'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 11 : pg 690"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The number of bits per symbol are 4.0 bits\n",
      "The baud rate is 22.631 Mbaud\n"
     ]
    }
   ],
   "source": [
    " \n",
    "# page no 690\n",
    "# prob no 18.11\n",
    "#calculate the baud rate and bits per symbol\n",
    "#given\n",
    "from math import log\n",
    "# A typical microwave digital radio system uses 16-QAM.\n",
    "fb = 90.524#bit rate expressesd in Mbps\n",
    "n = 16# for 16-QAM system\n",
    "#calculations and results\n",
    "#part a) calculation of no of bits per symbol\n",
    "m = log(n) / log(2)\n",
    "print 'The number of bits per symbol are',m,'bits'\n",
    "# part b) calclation of baud rate\n",
    "# baud rate is 1/4th of the bit rate\n",
    "baud = fb / 4\n",
    "print 'The baud rate is',baud,'Mbaud'"
   ]
  }
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