path: root/Electronic_Communication_Systems_by_Roy_Blake/Chapter2.ipynb

{
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   "source": [
    "# Chapter 2 : Radio Frequency Circuits"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 1 : pg 50"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
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   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "a)The resonant frequency is  4.5 MHz\n",
      "The bandwidth is  300.0 kHz\n"
     ]
    }
   ],
   "source": [
    "#page no 50\n",
    "#calculate the resonant frequency and bandwidth\n",
    "#prob no 2.1\n",
    "from math import pi,sqrt\n",
    "#Given:\n",
    "#Refer the fig 2.6 of page 50. L1=25uH;C1=50pF\n",
    "L1=25.*10**-6;C1=50.*10**-12;Q=15;\n",
    "#calculations\n",
    "#A) The resonent freqency is given as\n",
    "fo=(1/(2*pi*sqrt(L1*C1)));\n",
    "#B) The bandwidth is given as \n",
    "B=fo/Q;\n",
    "#results\n",
    "print 'a)The resonant frequency is ',round(fo/10**6,2),'MHz'\n",
    "print 'The bandwidth is ',round(B/1000.),'kHz'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 2 : pg 62"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
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   },
   "outputs": [
    {
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     "text": [
      "1)The operating frequency is 5.03 MHz\n",
      "2)Operating gain -10.0\n",
      "The -ve sign denotes a phase inversion\n",
      "3)Operating gain 0.2\n"
     ]
    }
   ],
   "source": [
    "#page no 62\n",
    "# prob no. 2.2\n",
    "#calculate the operating frequency, gain\n",
    "from math import pi,sqrt\n",
    "# Given : Hartley oscillators L=10uH; C=100pF\n",
    "L=10.*10**-6; C=100.*10**-12;N1=10;N2=100.\n",
    "#calculations and results\n",
    "# A)The operating frequency is \n",
    "fo=1/(2.*pi*sqrt(L*C));\n",
    "print '1)The operating frequency is',round(fo/10**6,2),'MHz'\n",
    "# The feedback fraction is given by\n",
    "B=-N1/N2;\n",
    "#Operating gain is given as \n",
    "A=1/B;\n",
    "print '2)Operating gain',A\n",
    "print 'The -ve sign denotes a phase inversion'\n",
    "#B) The operating frequency is same as in part A)\n",
    "N1=20;N2=80.;\n",
    "# The feedback fraction is given by\n",
    "B=(N1+N2)/N1;\n",
    "#Operating gain is given as \n",
    "A=1./B;\n",
    "print '3)Operating gain',A"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 3 : pg 66"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
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   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The effective capacitance (pF)= 9.09\n",
      "1)The operating frequency is 52.8 Hz\n",
      "The feedback fraction is -0.1\n",
      "The feedback fraction is 0.0909\n"
     ]
    }
   ],
   "source": [
    "# page no 66\n",
    "#prob no 2.3\n",
    "#calculate the operating frequency and feedback fraction\n",
    "from math import pi,sqrt\n",
    "#Given:\n",
    "C1=10.*10**-12; C2=100.*10**-12; L=1*10**-6;\n",
    "#calculations and results\n",
    "# The effective capacitance is \n",
    "CT=(C1*C2)/(C1+C2);\n",
    "# The operating frequency is \n",
    "f0=1/(2*pi*sqrt(L*CT));\n",
    "print 'The effective capacitance (pF)=',round(CT*10**12,2);\n",
    "print '1)The operating frequency is',round(f0/10**6,1),'Hz'\n",
    "# The feedback fraction is given approximately by\n",
    "B=-C1/C2; \n",
    "print 'The feedback fraction is',B\n",
    "# For the common-base ckt, the op-freq is same but the feedback fraction willbe different.\n",
    "C1=100*10**-12; C2=10*10**-12;\n",
    "# It is given by \n",
    "B=C2/(C1+C2);\n",
    "print 'The feedback fraction is',round(B,4)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4 : pg 68"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The effective total capacitance is 9.01 pF\n",
      "The operating freq is 53.03 MHz\n"
     ]
    }
   ],
   "source": [
    "# page no 68\n",
    "#prob no 2.4\n",
    "#caclculate the effective total capacitance and operating frequency\n",
    "#Refer fig 2.22\n",
    "from math import pi,sqrt\n",
    "#Given:\n",
    "c1=1000.;c2=100.;c3=10.;#all values are in pf\n",
    "#calculations and results\n",
    "#The effective total capacitance\n",
    "Ct=1/((1/c1)+(1/c2)+(1/c3));\n",
    "print 'The effective total capacitance is',round(Ct,2),'pF'\n",
    "CT=Ct*10**-12;L=10**-6;\n",
    "#The operating freq is\n",
    "f0=1/(2*pi*sqrt(L*CT));\n",
    "print 'The operating freq is',round(f0/10**6,2),'MHz'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 5 : pg 70"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
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   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The resonent freq is 1.78 MHz\n",
      "The tuning voltage is  7.5 V\n"
     ]
    }
   ],
   "source": [
    "# page no 70\n",
    "#prob no 2.5\n",
    "#calculate the resonant freq and tuning voltage\n",
    "from math import sqrt,pi\n",
    "#Given:\n",
    "C=80*10**-12; L= 100*10**-6;\n",
    "#calculations\n",
    "#Part a) The resonent frequency is \n",
    "f0=1/(2*pi*sqrt(L*C));\n",
    "# Part b) In this part the circuit is resonate on doubling the frequency,therefore\n",
    "f1=2*f0;\n",
    "# from the equation of resonent frequency \n",
    "C1=1/(4*(pi*f1)**2*L);\n",
    "# Now for tuning voltage  we have to use equation C1=Co/sqrt(1+2V)\n",
    "Co=C;\n",
    "# after solving the expression\n",
    "v=((Co/C1)**2 -1)/2;\n",
    "#results\n",
    "print 'The resonent freq is',round(f0/10**6,2),'MHz'\n",
    "print 'The tuning voltage is ',v,'V'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 7 : pg 76"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The output frequencies at the output of square-law mixer are :  21.0 MHz  1.0 MHz\n"
     ]
    }
   ],
   "source": [
    "\n",
    "#page no 76\n",
    "#calculate the output frequencies\n",
    "#problem 2.7\n",
    "#Given:\n",
    "# all frequencies are in MHz\n",
    "f1=11.;f2=10;\n",
    "#calculations\n",
    "# output frequencies at the output of square-law mixer\n",
    "a=f1+f2;\n",
    "b=f1-f2;\n",
    "#results\n",
    "print 'The output frequencies at the output of square-law mixer are : ',a,'MHz ',b,'MHz'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 8 : pg 85"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
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   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The capture range is  4.0 MHz\n",
      "The lock range is  8.0 MHz\n",
      "The freq at which the lock is acquired, moving downward in freq is  14.0 MHz\n",
      "Lock will be lost on the way down at 8.0 MHz\n"
     ]
    }
   ],
   "source": [
    "#page no 85\n",
    "#calculate the capture range, lock range, frequency\n",
    "#problem no. 2.8\n",
    "# all the frequencies are in MHz\n",
    "#Given:\n",
    "freq_free_run =12.; \n",
    "freq_lock1 =10.;\n",
    "freq_lock2 =16.;\n",
    "#calculations and results\n",
    "# capture range is approximately twice the difference between the free-running freq and the freq at which lock is first achieved\n",
    "capture_range =2*(freq_free_run - freq_lock1 );\n",
    "print 'The capture range is ',capture_range,'MHz'\n",
    "# lock range is approximately twice the the difference between the freq where lock is lost and free-running freq\n",
    "lock_range = 2*(freq_lock2 - freq_free_run);\n",
    "print 'The lock range is ',lock_range,'MHz'\n",
    "# The PLL freq response id approximate symmetrical.\n",
    "#This means the free-running freq is in the center of the lock range and capture range. Therefore\n",
    "freq_lock_acquired = freq_free_run + (capture_range/2);\n",
    "freq_lock_lost = freq_free_run - capture_range\n",
    "print 'The freq at which the lock is acquired, moving downward in freq is ',freq_lock_acquired,'MHz'\n",
    "print 'Lock will be lost on the way down at',freq_lock_lost,'MHz'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 9 : pg 86"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
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   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The values of N at high end is 170.0\n",
      "The values of N at low end is 54.0\n"
     ]
    }
   ],
   "source": [
    "#page no 86\n",
    "#calculate the values of N at high and low ends\n",
    "# prob no 2.9\n",
    "# refer fig 2.38\n",
    "#Given:\n",
    "#Here we are using a 10MHz crystal, it will be necessar to devide it by a factor to get 10kHz\n",
    "f_osc = 10.*10**6; f_ref=10.*10**3;f0_1=540.*10**3;f0_2=1700.*10**3;\n",
    "#calculations\n",
    "Q=f_osc/f_ref;\n",
    "# we have to specify the range of values of N. Find N at each and of the tuning range\n",
    "N1=f0_1/f_ref;\n",
    "N2=f0_2/f_ref;\n",
    "#results\n",
    "print 'The values of N at high end is',N2\n",
    "print 'The values of N at low end is',N1"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 10 : pg 89"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
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   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The step size that would have been obtained without prescaling 10000.0 Hz\n"
     ]
    }
   ],
   "source": [
    "#page no 89\n",
    "# prob no 2.10\n",
    "# refer fig 2.40\n",
    "#Given:\n",
    "P=10.;f_ref=10.*10**3;M=10.;\n",
    "#consider \n",
    "N=1.;\n",
    "#calculations\n",
    "# With a fixed-modulus prescalar, the min freq step is \n",
    "step_size=M*f_ref;\n",
    "# With the two-modulus system, let the main divider modulus N remain constant & \n",
    "#increase the modulus m to (m+1) to find how much the freq changes.\n",
    "# for 1st case, o/p freq \n",
    "fo=(M+N*P)*f_ref;\n",
    "# for 2nd case where leave N alone but changes M to M+1, new o/p freq \n",
    "fo_=(M+1+N*P)*f_ref;\n",
    "# The difference is \n",
    "f= fo_-fo;\n",
    "#results\n",
    "print 'The step size that would have been obtained without prescaling',f,'Hz'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 11 : pg 91"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The output frequencies are 10.2 MHz 12.0 MHz\n",
      "The step size is 20.0 kHz\n"
     ]
    }
   ],
   "source": [
    "#page no 91\n",
    "#prob no 2.11\n",
    "#refer fig 2.42\n",
    "#Given:\n",
    "f_ref= 20.*10**3;\n",
    "f_osc= 10.*10**6;\n",
    "N1=10;N2=100.;\n",
    "#calculations\n",
    "f0=(N1*f_ref) + f_osc;\n",
    "f1=(N2*f_ref) + f_osc;\n",
    "step_size=(f1-f0)/(N2-N1);\n",
    "#results\n",
    "print 'The output frequencies are',f0/10**6,'MHz',f1/10**6,'MHz'\n",
    "print 'The step size is',step_size/1000.,'kHz'"
   ]
  }
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