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+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 12 : Digital Modulation and Modems"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1 : pg 407"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "a)The theoretical max data rate is 50.278 kb/s\n",
+ "b)The data rate for 4 states is 40.0 kb/s\n"
+ ]
+ }
+ ],
+ "source": [
+ " \n",
+ "# page no 407\n",
+ "# prob no 12_1\n",
+ "#calculate the data rate and theoretical max\n",
+ "#A radio channel with BW=10KHz and SNR=15 dB\n",
+ "from math import log\n",
+ "#given\n",
+ "B=10.*10**3;\n",
+ "snr=15.;\n",
+ "#calculations and results\n",
+ "#converting dB in power ratio\n",
+ "SNR=10**(snr/10);\n",
+ "#a)Determination of theoretical max data rate\n",
+ "C1=B*log(1+SNR) /log(2);\n",
+ "print 'a)The theoretical max data rate is',round(C1/1000,3),'kb/s'\n",
+ "#b)Determination of data rate with 4 states i.e M=4\n",
+ "M=4.;\n",
+ "C2=2*B*log(M) / log(2);\n",
+ "print 'b)The data rate for 4 states is',C2/1000,'kb/s'"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2 : pg 408"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The baud rate is 10.0 kbaud\n",
+ "The bit rate is 60.0 kb/s\n"
+ ]
+ }
+ ],
+ "source": [
+ " \n",
+ "# page no 408\n",
+ "# prob no 12_2\n",
+ "#calculate the baud and bit rates\n",
+ "from math import log\n",
+ "#A modulator transmit symbol with symbol rate=10k/sec with 64 states\n",
+ "#given\n",
+ "M=64.;\n",
+ "S=10000.;\n",
+ "#calculations and results\n",
+ "#Baud rate is simply symbol rate\n",
+ "print 'The baud rate is ',S/1000,'kbaud'\n",
+ "#Determination of bit rate\n",
+ "C=S*log(64) / log(2);\n",
+ "print 'The bit rate is ',C/1000,'kb/s'"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3 : pg 411"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "a)The frequency shift is 135416.5 Hz\n",
+ "b)The maximum frequency is 880067708.25 Hz\n",
+ "The minimum frequency is 879932291.75 Hz\n",
+ "The bandwidth efficiency in b/s/Hz is 1.354 b/s/Hz\n"
+ ]
+ }
+ ],
+ "source": [
+ " \n",
+ "# page no 411\n",
+ "# prob no 12_3\n",
+ "#calculate the frequency shift, max, min frequency\n",
+ "#given\n",
+ "f=200.*10**3;\n",
+ "fb=270.833 *10**3;\n",
+ "data_rate=270.833 *10**3\n",
+ "fc=880*10**6;\n",
+ "bandwidth=200*10**3;\n",
+ "#calculations and results\n",
+ "freq_shift=0.5*fb;\n",
+ "print 'a)The frequency shift is',freq_shift,'Hz'\n",
+ "# The shift each way from the carrier frequency is half the freq_shift\n",
+ "f_max=fc+0.25*fb;\n",
+ "print 'b)The maximum frequency is',round(f_max,3),'Hz'\n",
+ "f_min=fc-0.25*fb;\n",
+ "print 'The minimum frequency is',round(f_min,3),'Hz'\n",
+ "bandwidth_efficiency=data_rate/bandwidth;\n",
+ "print 'The bandwidth efficiency in b/s/Hz is',round(bandwidth_efficiency,3),'b/s/Hz'"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4 : pg 412"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The channel data rate is 48.6 kb/s\n"
+ ]
+ }
+ ],
+ "source": [
+ " \n",
+ "# page no 412\n",
+ "# prob no 12_4\n",
+ "#calculate the channel data rate\n",
+ "#given\n",
+ "baud_rate=24.3;# in kilobaud\n",
+ "#calculations\n",
+ "# In this problem dibit system is used.\n",
+ "#Therefore symbol_rate=baud_rate=0.5*bit_rate\n",
+ "bit_rate=2*baud_rate;\n",
+ "#results\n",
+ "print 'The channel data rate is',bit_rate,'kb/s'"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5 : pg 413"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The no. of bits per symbol is 6.0\n"
+ ]
+ }
+ ],
+ "source": [
+ " \n",
+ "# page no 413\n",
+ "# prob no 12_5\n",
+ "#calculate the no. of bits per symbol\n",
+ "from math import log\n",
+ "#given\n",
+ "no_of_phase_angles=16;\n",
+ "no_of_amplitudes=4;\n",
+ "#calculations\n",
+ "no_of_states_per_symbol=no_of_phase_angles*no_of_amplitudes;\n",
+ "bit_per_symbol=log(no_of_states_per_symbol)/ log(2);\n",
+ "#results\n",
+ "print 'The no. of bits per symbol is',bit_per_symbol"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6 : pg 415"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Shannon limit 29901.68 b/s\n"
+ ]
+ }
+ ],
+ "source": [
+ " \n",
+ "# page no 415\n",
+ "# prob no 12_6\n",
+ "#calculate the shannon limit\n",
+ "from math import log\n",
+ "#given\n",
+ "B=3.*10**3;SNR_dB=30.;\n",
+ "#calculations\n",
+ "SNR_power=10**(30./10);\n",
+ "C=B*log(1+SNR_power)/ log(2);\n",
+ "print 'Shannon limit',round(C,2),'b/s'"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.11"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
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