path: root/Electronic_Communication_Systems_by_Roy_Blake/Chapter3.ipynb

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   "source": [
    "# Chapter 3 : The Amplitude Modulation"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 1 : pg 105"
   ]
  },
  {
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   "execution_count": 1,
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   "outputs": [
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     "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": {
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   },
   "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": {
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   },
   "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,
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   "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": {
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   },
   "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,
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   },
   "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'"
   ]
  }
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