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diff --git a/Electronic_Communication_Systems_by_Roy_Blake/Chapter1.ipynb b/Electronic_Communication_Systems_by_Roy_Blake/Chapter1.ipynb new file mode 100644 index 00000000..3a9aac5c --- /dev/null +++ b/Electronic_Communication_Systems_by_Roy_Blake/Chapter1.ipynb @@ -0,0 +1,513 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 1: Introduction to Communication System" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1: pg 7" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "WAVELENGTH IN FREE SPACE IS 300.0 m\n", + "WAVELENGTH IN FREE SPACE IS 11.1 m\n", + "WAVELENGTH IN FREE SPACE IS 7.5 cm\n" + ] + } + ], + "source": [ + "# page no 7\n", + "#calculate the wavelength in all cases\n", + "# prob no 1.1\n", + "#part a) freq= 1MHz(AM radio broadcast band)\n", + "# We have the equation c=freq*wavelength\n", + "c=3.*10**8;\n", + "f=1.*10**6;\n", + "#calculations\n", + "wl=c/f;\n", + "print 'WAVELENGTH IN FREE SPACE IS ',wl,'m'\n", + "#part B) freq= 27MHz(CB radio band)\n", + "f=27.*10**6;\n", + "wl=c/f;\n", + "print 'WAVELENGTH IN FREE SPACE IS ',round(wl,1),'m'\n", + "#part C) freq= 4GHz(used for satellite television)\n", + "f=4*10**9;\n", + "wl=c/f;\n", + "print 'WAVELENGTH IN FREE SPACE IS ',wl*100,'cm'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4: pg 18" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "noise power contributed by resistor 4.14e-17 W\n" + ] + } + ], + "source": [ + "\n", + "# page no 18\n", + "#calculate the noise power\n", + "# prob no. 1.4\n", + "# In given problem noise power bandwidth is 10kHz; resistor temp T(0c)=27\n", + "# First we have to convert temperature to kelvins:\n", + "T0c=27.;\n", + "Tk=T0c+273.;\n", + "# noise power contributed by resistor , Pn= k*T*B\n", + "k=1.38*10**(-23);\n", + "B=10*10**3;\n", + "#calculations\n", + "Pn= k*Tk*B;\n", + "#results\n", + "print 'noise power contributed by resistor',Pn,'W'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5: pg 20" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The noise power is 24.3 pW\n", + "The noise voltage is 5.4 muvolts\n" + ] + } + ], + "source": [ + " \n", + "# page no 20\n", + "#calculate the noise power and voltage\n", + "# prob no 1.5\n", + "from math import sqrt\n", + "# In the given problem B=6MHz, Tk=293, k=1.38*10**-23\n", + "B=6*10**6; Tk=293; k=1.38*10**-23;R=300;\n", + "#calculations\n", + "Pn=k*Tk*B;\n", + "# Th noise voltage is given by Vn=sqrt(4*k*Tk*B*R)\n", + "Vn=sqrt(4*k*Tk*B*R);\n", + "#results\n", + "print 'The noise power is',round(Pn*10**15,1),'pW'\n", + "print 'The noise voltage is',round(Vn*10**6,1),'muvolts'\n", + "# only one-half of this voltage is appears across the antenna terminals, the other appears across the source resistance. Therefore the actual noise voltag at the input is 2.7 uV" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6: pg 21" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Noise current is 0.133 muA\n", + "current through the diode is 278.0 mA\n" + ] + } + ], + "source": [ + " \n", + "# page no 21\n", + "#calculate the current\n", + "# prob no 1.6\n", + "# given: FM broadcast receiver :- Vn=10uV, R=75V, B=200 kHz \n", + "Vn=10.;#in uV\n", + "R=75.; B=200.*10**3;\n", + "#calculations\n", + "#By Ohm's law\n", + "In=Vn/R;\n", + "print 'Noise current is',round(In,3),'muA'\n", + "# Noise votlage is also given as In=sqrt(2*q*Io*B)\n", + "q=1.6*10**-19;\n", + "# solving this for Io=In**2/2*q*B;\n", + "Io=(In*10**-6)**2/(2*q*B);\n", + "print 'current through the diode is',round(Io*1000.,0),'mA'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7: pg 23" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Open-ckt noise voltage is 779.0 nanovolts\n", + "The load power is 5.06e-16 W\n" + ] + } + ], + "source": [ + " \n", + "#page no 23\n", + "#calculate the load power and open-ckt noise voltage\n", + "#pro no 1.7\n", + "from math import sqrt\n", + "#Given:refer fig.1.12 of page no.23;R1=100ohm,300K;R2=200ohm,400k;B=100kHz;Rl=300ohm\n", + "R1=100.;T1=300.;R2=200.;T2=400.;B=100.*10**3;Rl=300.;k=1.38*10**-23;\n", + "#calculations\n", + "#open-ckt noise voltage is given by\n", + "#Vn1 =sqrt(Vr1**2 + Vr2**2)\n", + "# =sqrt[sqrt(4kTBR1)**2 + sqrt(4kTBR2)**2]\n", + "#by solving this we get Vn1=sqrt[4kB(T1R1 + T2R2)]\n", + "Vn1=sqrt(4*k*B*(T1*R1 + T2*R2));\n", + "# since in this case the load is equal in value to the sum of the resistors,\n", + "# one-half of this voltage is appear across the load.\n", + "# Now the load power is P= Vn1**2/Rl\n", + "P= (Vn1/2)**2/Rl;\n", + "#results\n", + "print 'Open-ckt noise voltage is ',round(Vn1*1e9),'nanovolts'\n", + "print 'The load power is',P,'W'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8: pg 24" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The power ratio in dB 14.0\n" + ] + } + ], + "source": [ + " \n", + "# page no 24\n", + "#Calculate the power ratio\n", + "# prob no 1.8\n", + "import math\n", + "# Given: N=0.2W; S+N=5W; :. S=4.8W\n", + "N=0.2; S=4.8;\n", + "#calculations\n", + "p=(S+N)/N;\n", + "pdB=10*math.log10(p);\n", + "#results\n", + "print 'The power ratio in dB',round(pdB)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9: pg 25" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The noise figure is 3.0\n" + ] + } + ], + "source": [ + " \n", + "#page no 25\n", + "#calculate the noise figure\n", + "#prob no 1.9\n", + "#Given: Si=100uW; Ni=1uW; So=1uW; No=0.03W\n", + "Si=100.; Ni=1.; So=1; No= 0.03# all powers are in uW\n", + "#calculations\n", + "r1=Si/Ni;# input SNR\n", + "r2=So/No;# output SNR\n", + "NF=r1/r2;# Amplifier noise figure \n", + "#results\n", + "print 'The noise figure is',NF" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10: pg 25" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "SNR at the output is 36 dB\n" + ] + } + ], + "source": [ + " \n", + "#page no 25\n", + "#calculate the SNR at output\n", + "#prob no 1.10\n", + "#giiven: SNRin=42 dB, NF=6dB\n", + "# NF in dB is given as SNRin(dB)-SNRop(dB)\n", + "SNRin=42 ; NF=6;\n", + "#calculations\n", + "SNRop=SNRin-NF;\n", + "#results\n", + "print 'SNR at the output is ',SNRop,'dB'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11: pg 27" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The noise temperature is 169.65 K\n" + ] + } + ], + "source": [ + " \n", + "#page no 27\n", + "#calculate the noise temperature\n", + "# prob no 1.11\n", + "#Given NFdB=2dB,:.NF=antilog(NFdB)/10=1.585\n", + "NF=1.585;\n", + "#calculations\n", + "Teq=290*(NF-1);\n", + "#results\n", + "print 'The noise temperature is ',Teq,'K'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12: pg 29" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The power gain is 7500.0\n", + "The noise figure is 2.316\n", + "The noise temperature is 382.0 K\n" + ] + } + ], + "source": [ + " \n", + "#page no 29\n", + "#calculate the power gain, noise figure, temperature\n", + "#prob no 1.12\n", + "#Given: \n", + "A1=10.;A2=25.;A3=30.;NF1=2.;NF2=4.;NF3=5.;\n", + "#calculations\n", + "At=A1*A2*A3;\n", + "print 'The power gain is',At\n", + "# The noise figure is given as \n", + "NFt=NF1+((NF2-1)/A1) + ((NF3-1)/(A1*A2));\n", + "print 'The noise figure is',NFt\n", + "# Noise temp can be found as \n", + "Teq=290*(NFt-1);\n", + "print 'The noise temperature is',round(Teq),'K'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13: pg 34" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "A)The freq is 110.0 MHz\n", + "The power in dBm -30 dBm\n", + "The power is 1e-06 W\n", + "The voltage is 0.00707 volts\n", + "B)The freq is 7.4 MHz\n", + "The power is 79.4 W\n", + "The power in dBm 9 dBm\n", + "The voltage is 630.0 volts\n", + "C)The freq is 546 MHz\n", + "The voltage is 16.8 mvolts\n", + "The power is 5.64 uW\n", + "The power in dBm -22.5 dBm\n" + ] + } + ], + "source": [ + " \n", + "# page no 34\n", + "#calculate the power and voltage in all cases\n", + "# prob no1.13 refer fig 1.20 of page no 34\n", + "# part A) The signal frequency is f1=110MHz.\n", + "from math import log10,sqrt\n", + "f=110.;# in MHz\n", + "#calculations and results\n", + "print 'A)The freq is',f,'MHz'\n", + "#The signal peak is two divisions below the reference level of -10dBm, with 10dB/division ,so its -30dBm.\n", + "PdBm=-30;\n", + "print 'The power in dBm',PdBm,'dBm'\n", + "# The equivalent power can be found from P(dBm)=10logP/1 mW\n", + "#P(mW)=antilog dBm/10= antilog -30/10=1*10**-3mW=1uW\n", + "#the voltage can be found from the graph but it is more accurately from P=V**2/R\n", + "P=10**-6; R=50;\n", + "print 'The power is',P,'W'\n", + "V=sqrt(P*R);\n", + "print 'The voltage is',round(V*10e3,2),'mvolts'\n", + "\n", + "# part B)The signal is 1 division to theleft of center, with 100kHz/div. The freq is 100kHz less than the ref freq of 7.5MHz\n", + "f=7.5-0.1;# in MHz\n", + "print 'B)The freq is',f,'MHz'\n", + "# With regards to the amplitude, the scale is 1dB/div & the signal is 1 div below the reference level. Therefore the signal has a power level given as\n", + "PdBm=10-1;# in dBm\n", + "# This can be converted to watts & volts as same in part A\n", + "#P(mW)=antilog dBm/10= antilog 9/10=7.94mW\n", + "P=7.94*10**-3; R=50;\n", + "print 'The power is',round(P*10e3,2),'mW'\n", + "print 'The power in dBm',PdBm,'dBm'\n", + "V=sqrt(P*R);\n", + "print 'The voltage is',round(V*1000.),'mvolts'\n", + "\n", + "#part C) The signal is 3 divisions to the right of the center ref freq of 543MHz, with 1MHz/div. Therefore the freq is \n", + "f=543+3*1;# in MHz\n", + "print 'C)The freq is',f,'MHz'\n", + "# from the spectrum, signal level is\n", + "V=22.4*6/8;\n", + "print 'The voltage is',V,'mvolts'\n", + "# power is given as\n", + "P=V**2/R;\n", + "print 'The power is',round(P,2),'uW'\n", + "PdBm=10*log10(P*10**-6/10**-3);\n", + "print 'The power in dBm',round(PdBm,1),'dBm'" + ] + } + ], + "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 +} |