From c8733e4b6b4bffcddf7eb45ff1c72ccc837aa3af Mon Sep 17 00:00:00 2001 From: Jovina Dsouza Date: Tue, 22 Jul 2014 00:00:04 +0530 Subject: adding book --- .../chapter_41-checkpoint.ipynb | 1005 ++++++++++++++++++++ 1 file changed, 1005 insertions(+) create mode 100755 Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint.ipynb (limited to 'Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint.ipynb') diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint.ipynb new file mode 100755 index 00000000..788f6f5e --- /dev/null +++ b/Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint.ipynb @@ -0,0 +1,1005 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Chapter 41: Attenuators

" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 1, page no. 763

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine the power ratio in each case (i) in decibels and (ii) in nepers.\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + " #ratio of output power to input power\n", + "rp1 = 2;\n", + "rp2 = 25; \n", + "rp3 = 1000;\n", + "rp4 = 0.01;\n", + "\n", + "#calculation:\n", + " #power ratio in decibels\n", + "rpd1 = 10*(1/2.303)*math.log(rp1)\n", + "rpd2 = 10*(1/2.303)*math.log(rp2)\n", + "rpd3 = 10*(1/2.303)*math.log(rp3)\n", + "rpd4 = 10*(1/2.303)*math.log(rp4)\n", + " #power ratio in nepers\n", + "rpn1 = (math.log(rp1))/2\n", + "rpn2 = (math.log(rp2))/2\n", + "rpn3 = (math.log(rp3))/2\n", + "rpn4 = (math.log(rp4))/2\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n power ratio in decibels are (a)\",round(rpd1,2),\" dB (b)\",round(rpd2,2),\" dB \"\n", + "print \"(c) \",round(rpd3,2),\" dB and (d) \",round(rpd4,2),\" dB\"\n", + "print \"\\n power ratio in nepers are (a)\",round(rpn1,2),\" Np (b)\",round(rpn2,2),\" Np\"\n", + "print \"(c) \",round(rpn3,2),\" Np and (d) \",round(rpn4,2),\" Np\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " power ratio in decibels are (a) 3.01 dB (b) 13.98 dB \n", + "(c) 29.99 dB and (d) -20.0 dB\n", + "\n", + " power ratio in nepers are (a) 0.35 Np (b) 1.61 Np\n", + "(c) 3.45 Np and (d) -2.3 Np\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 2, page no. 763

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine the attenuation in decibels\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "rp = 0.05;# power ratio P2/P1\n", + "\n", + " #calculation:\n", + " #power ratio in decibels\n", + "rpd = 10*(1/2.303)*math.log(rp)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\nthe attenuation is \",round(abs(rpd),2),\" dB\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + "the attenuation is 13.01 dB" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 3, page no. 764

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#determine the output power.\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "gain = 1.5;# in dB\n", + "Pi = 0.012;# in Watt\n", + "\n", + "#calculation:\n", + " #output power\n", + "Po = Pi*10**gain\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\noutput power is \",round(Po,2),\" W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + "output power is 0.38 W" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 4, page no. 764

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#determine (a) the current input and (b) the current ratio expressed in decibels.\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "I2 = 0.05;# in Amperes\n", + "rin = 1.32;# in Np\n", + "\n", + " #calculation:\n", + " #current input, I1\n", + "I1 = I2*math.e**(rin)\n", + " #current ratio in decibels\n", + "rid = 20*(1/2.303)*math.log(I2/I1)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\ncurrent input, I1 is \",round(I1,2),\" A\"\n", + "print \"\\ncurrent ratio in decibels is \",round(rid,2),\" dB\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + "current input, I1 is 0.19 A\n", + "\n", + "current ratio in decibels is -11.46 dB" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 5, page no. 769

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine the characteristic impedance.\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "Ra1 = 8; # in ohms\n", + "Ra2 = 8; # in ohms\n", + "Ra3 = 21; # in ohms\n", + "Rb1 = 10; # in ohms\n", + "Rb2 = 10; # in ohms\n", + "Rb3 = 15; # in ohms\n", + "Rc1 = 200; # in ohms\n", + "Rc2 = 200; # in ohms\n", + "Rc3 = 56.25; # in ohms\n", + "\n", + "#calculation:\n", + "R01 = (Ra1**2 + 2*Ra2*Ra3)**0.5\n", + "R02 = (Rb1**2 + 2*Rb2*Rb3)**0.5\n", + "R03 = (Rc1**2 + 2*Rc2*Rc3)**0.5\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n(a) the characteristic impedance, R0 is \",R01,\" ohm\"\n", + "print \"\\n(b) the characteristic impedance, R0 is \",R02,\" ohm\"\n", + "print \"\\n(c) the characteristic impedance, R0 is \",R03,\" ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + "(a) the characteristic impedance, R0 is 20.0 ohm\n", + "\n", + "(b) the characteristic impedance, R0 is 20.0 ohm\n", + "\n", + "(c) the characteristic impedance, R0 is 250.0 ohm" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 6, page no. 769

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine (a) the characteristic impedance, and (b) the attenuation (in dB) produced by the pad\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "R1 = 500;# in ohm\n", + "R2 = 1000;# in ohm\n", + "I1 = 1;# in ampere (lets say)\n", + "\n", + "#calculation:\n", + " # for symmetrical pi-attenuator section\n", + " #characteristic impedance, R0\n", + "R0 = (R1*(R2**2)/(R1 + 2*R2))**0.5\n", + " #current Ix\n", + "Ix = (R2/(R2 + R1 + (R2*R0/(R2 + R0))))*I1\n", + " #current I2\n", + "I2 = (R2/(R2 + R0))*Ix\n", + "ri = I1/I2;# retio of currents\n", + " #attenuation\n", + "attn = 20*(1/2.303)*math.log(ri)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n the characteristic impedance is \",round(R0,2),\" ohm\"\n", + "print \"\\n attenuation is \",round(attn,2),\" dB\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " the characteristic impedance is 447.21 ohm\n", + "\n", + " attenuation is 8.36 dB" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 7, page no. 770

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#determine (a) the input resistance when the output port is open-circuited, \n", + "#(b) the input resistance when the output port is short-circuited, and (c) the characteristic impedance.\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "Ra1 = 15;# in ohm\n", + "Ra2 = 15;# in ohm\n", + "Ra3 = 10;# in ohm\n", + "Rb1 = 15;# in ohm\n", + "Rb2 = 5;# in ohm\n", + "Rb3 = 5;# in ohm\n", + "\n", + "#calculation:\n", + "Roc1 = Ra1 + Ra3\n", + "Rsc1 = Ra1 + Ra2*Ra3/(Ra2+Ra3)\n", + "R01 = (Roc1*Rsc1)**0.5\n", + "\n", + "Roc2 = Rb2* (Rb1 + Rb3)/(Rb1 + Rb2 + Rb3)\n", + "Rsc2 = Rb2*Rb1/(Rb2+Rb1)\n", + "R02 = (Roc2*Rsc2)**0.5\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n (a) the input resistance when the output port is open-circuited is\", Roc1,\" ohm for T-Network\"\n", + "print \"and \",Roc2,\" ohm for pi-Network \"\n", + "print \"\\n (b) the input resistance when the output port is short-circuited is,\", Rsc1,\" ohm for T-Network\"\n", + "print \"and \",Rsc2,\" ohm for pi-Network \"\n", + "print \"\\n (c) the characteristic impedance. is,\",round(R01,1),\" ohm for T-Network and ,\",round(R02,2),\" ohm for pi-Network \"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " (a) the input resistance when the output port is open-circuited is 25 ohm for T-Network\n", + "and 4.0 ohm for pi-Network \n", + "\n", + " (b) the input resistance when the output port is short-circuited is, 21.0 ohm for T-Network\n", + "and 3.75 ohm for pi-Network \n", + "\n", + " (c) the characteristic impedance. is, 22.9 ohm for T-Network and , 3.87 ohm for pi-Network \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 8, page no. 770

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Design a T-section symmetrical attenuator pad to provide a voltage attenuation of 20 dB and \n", + "#having a characteristic impedance of 600 ohm.\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "Vat = 20; # in db\n", + "R0 = 600;# in ohm\n", + "\n", + "#calculation:\n", + "N = math.e**(Vat*2.3/20)\n", + "R1 = R0*(N-1)/(N+1)\n", + "R2 = R0*2*N/(N**2 - 1)\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n For a T-section symmetrical attenuator pad, Resistance R1 is\",round(R1,0),\" ohm and Resistance R2 is\",round(R2,1),\" ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " For a T-section symmetrical attenuator pad, Resistance R1 is 491.0 ohm and Resistance R2 is 121.5 ohm\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 9, page no. 771

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Design a pi-section symmetrical attenuator pad to provide a voltage attenuation of 20 dB and \n", + "#having a characteristic impedance of 600 ohm.\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "Vat = 20; # in db\n", + "R0 = 600;# in ohm\n", + "\n", + "#calculation:\n", + "N = math.e**(Vat*2.303/20)\n", + "R1 = R0*(N**2 - 1)/(2*N)\n", + "R2 = R0*(N+1)/(N-1)\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n For a pi-section symmetrical attenuator pad, Resistance R1 is\",round(R1/1000,2),\" Kohm\"\n", + "print \"and Resistance R2 is\",round(R2,0),\" ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " For a pi-section symmetrical attenuator pad, Resistance R1 is 2.97 Kohm\n", + "and Resistance R2 is 733.0 ohm\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 10, page no. 772

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine (a) the characteristic impedance R0, and (b) the insertion loss in decibels\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "R1 = 300;# in ohm\n", + "R2 = 450;# in ohm\n", + "I1 = 1;# in ampere (lets say)\n", + "\n", + "#calculation:\n", + " #the characteristic impedance of a symmetric T-pad attenuator is given by\n", + "R0 = (R1**2 + 2*R1*R2)**0.5\n", + " #By current division\n", + " #current I2\n", + "I2 = (R2/(R2 + R1+ R0))*I1\n", + "ri = I1/I2;# ratio of currents\n", + " #insertion loss\n", + "il = 20*(1/2.303)*math.log(ri)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n the characteristic impedance is \",R0,\" ohm\"\n", + "print \"\\n insertion loss is \",round(il,2),\" dB\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " the characteristic impedance is 600.0 ohm\n", + "\n", + " insertion loss is 9.54 dB" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 11, page no. 773

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#determine the insertion loss at a tapping of (a) 2 kohm, (b) 1 kohm.\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "r = 500;# in ohm\n", + "Rhm = 3000;# in ohm\n", + "RL = 2000;# in ohm\n", + "r1 = 2000;# in ohm\n", + "r2 = 1000;# in ohm\n", + "E = 1;# in volts (lets say)\n", + "\n", + " #calculation:\n", + " #Without the rheostat in the circuit the voltage across the 2 kohm\u0018 load, VL\n", + "VL = (RL/(RL + r))*E\n", + " #voltage V2 with 2kohm tapping\n", + "V2 = ((RL*r1/(r1 + RL))/((RL*r1/(r1 + RL)) + Rhm - r1 + r))*E\n", + "rv1 = VL/V2;# ratio of currents\n", + " #insertion loss \n", + "il1 = 20*(1/2.303)*math.log(rv1)\n", + " #voltage V1 with 1kohm tapping\n", + "V1 = ((RL*r2/(r2 + RL))/((RL*r2/(r2 + RL)) + Rhm - r2 + r))*E\n", + "rv2 = VL/V1;# ratio of currents\n", + " #insertion loss \n", + "il2 = 20*(1/2.303)*math.log(rv2)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n insertion loss for 2kohm tap is \",round(il1,2),\" dB\"\n", + "print \"\\n insertion loss for 1kohm tap is \",round(il2,2),\" dB\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " insertion loss for 2kohm tap is 6.02 dB\n", + "\n", + " insertion loss for 1kohm tap is 11.59 dB" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 12, page no. 774

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine (a) its characteristic impedance, and (b) the insertion loss (in decibels) when feeding a matched load.\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "R1 = 1000;# in ohm\n", + "R2 = 500;# in ohm\n", + "I1 = 1;# in amperes (lets say)\n", + "\n", + "#calculation:\n", + " #characteristic impedance of a symmetrical attenuator\n", + "R0 = (R1*(R2**2)/(R1 + 2*R2))**0.5\n", + " #current Ix\n", + "Ix = (R2/(R2 + R1 + (R2*R0/(R2 + R0))))*I1\n", + " #current I2\n", + "I2 = (R2/(R2 + R0))*Ix\n", + "ri = I1/I2;# retio of currents\n", + " #insertion loss \n", + "il = 20*(1/2.303)*math.log(ri)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n characteristic impedance is \",round(R0,2),\" ohm\"\n", + "print \"\\n insertion loss is \",round(il,2),\" dB\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " characteristic impedance is 353.55 ohm\n", + "\n", + " insertion loss is 15.31 dB" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 13, page no. 776

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine for the section (a) the image impedances, and (b) the iterative impedances\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "R1 = 100;# in ohm\n", + "R2 = 200;# in ohm\n", + "R3 = 300;# in ohm\n", + "I1 = 1;# in amperes (lets say)\n", + "\n", + "#calculation:\n", + " #image impedance Roa\n", + "Roa = ((R1 + R2)*(R2 + (R1*R3/(R1 + R3))))**0.5\n", + " #image impedance Rob\n", + "Rob = ((R1 + R3)*(R3 + (R1*R2/(R1 + R2))))**0.5\n", + " #The iterative impedance at port 1\n", + "Ri1 = (-1*R1 + (R1**2 - (-1*4*((R2*(R1 + R3)) + (R3*R1))))**0.5)/2\n", + " #The iterative impedance at port 2\n", + "Ri2 = (R1 + (R1**2 - (-1*4*((R3*(R1 + R2)) + (R2*R1))))**0.5)/2\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n image impedance are \",round(Roa,2),\" ohm and \",round(Rob,2),\" ohm \"\n", + "print \"\\n iterative impedances are \",round(Ri1,2),\" ohm and \",round(Ri2,2),\" ohm \"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " image impedance are 287.23 ohm and 382.97 ohm \n", + "\n", + " iterative impedances are 285.41 ohm and 385.41 ohm " + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 14, page no. 777

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine for the section (a) the image impedances, and (b) the iterative impedances\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "R1 = 1000;# in ohm\n", + "R2 = 2000;# in ohm\n", + "R3 = 3000;# in ohm\n", + "I1 = 1;# in amperes (lets say)\n", + "\n", + "#calculation:\n", + " #image impedance Roa\n", + "Roa = (((R3 + R2)*R1/(R1 + R2 + R3))*(R1*R3/(R1 + R3)))**0.5\n", + " #image impedance Rob\n", + "Rob = (((R3 + R1)*R2/(R1 + R2 + R3))*(R2*R3/(R2 + R3)))**0.5\n", + " #The iterative impedance at port 1\n", + "Ri1 = (-1*R1 + ((R1**2) - (-1*4*2*R2*R1))**0.5)/(2*2)\n", + " #The iterative impedance at port 2\n", + "Ri2 = (R1 + ((-1*R1)**2 - (-1*4*2*R2*R1))**0.5)/(2*2)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n image impedance are \",round(Roa,2),\" ohm and \",round(Rob,2),\" ohm \"\n", + "print \"\\n iterative impedances are \",round(Ri1,2),\" ohm and \",round(Ri2,2),\" ohm \"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " image impedance are 790.57 ohm and 1264.91 ohm \n", + "\n", + " iterative impedances are 780.78 ohm and 1280.78 ohm " + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 15, page no. 780

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine (a) the values of resistance R1 and R2, (b) the attenuation of the pad in decibels, and (c) its insertion loss.\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "r = 500;# in ohm\n", + "RL = 100;# in ohm\n", + "E = 1;# in volts (lets say)\n", + "\n", + "#calculation:\n", + " #res.\n", + "R1 = (r*(r - RL))**0.5\n", + "R2 = (r*RL**2/(r - RL))**0.5\n", + " #current I1\n", + "I1 = E/(r + R1 + R2*RL/(RL + R2))\n", + " #current I2\n", + "I2 = (R2/(R2 + RL))*I1\n", + " #input power\n", + "P1 = r*I1**2\n", + " #output power\n", + "P2 = RL*I2**2\n", + " #attenuation\n", + "attn = 10*(1/2.303)*math.log(P1/P2)\n", + " #Load current, IL\n", + "IL = E/(r + RL)\n", + " #voltage, VL\n", + "VL = IL*RL\n", + " #voltage, V1\n", + "V1 = E - I1*r\n", + " #voltage, V2\n", + "V2 = V1 - I1*R1\n", + " #insertion loss\n", + "il = 20*(1/2.303)*math.log(VL/V2)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n R1 = \",round(R1,2),\" ohm and R2 = \",round(R2,2),\" ohm \"\n", + "print \"\\n attenuation is \",round(attn,2),\" dB \"\n", + "print \"\\n In decibels, the insertion loss is \",round(il,2),\" dB \"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " R1 = 447.21 ohm and R2 = 111.8 ohm \n", + "\n", + " attenuation is 12.54 dB \n", + "\n", + " In decibels, the insertion loss is 9.98 dB " + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 16, page no. 783

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine (a) the attenuation of each individual attenuation section, \n", + "#(b) the voltage output of the final stage, and (c) the voltage output of the third stage.\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "attnO = 70;# in dB\n", + "n = 5;# numbers of identical atteneurs\n", + "V1 = 0.02;# in Volts\n", + "\n", + "#calculation:\n", + " #attenuation of each section\n", + "attn = attnO/n\n", + " #output of the final stage\n", + "Vo = V1/(10**(attnO/20))\n", + " #voltage output of the third stage\n", + "V3 = V1/(10**(3*attn/20))\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n attenuation of each section = \",round(attn,2),\" dB \"\n", + "print \"\\n output of the final stage is \",round(Vo*1E6,2),\"uV \"\n", + "print \"\\n voltage output of the third stage is \",round(V3*1E3,3),\"mV \"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " attenuation of each section = 14.0 dB \n", + "\n", + " output of the final stage is 6.32 uV \n", + "\n", + " voltage output of the third stage is 0.159 mV " + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 17, page no. 784

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#(a)Design a T-network attenuator pad\n", + "#(b)determine the fraction of the initial current \n", + "#(c) Determine the attenuation in decibels given by four such sections\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "r = 450; # in ohm\n", + "R0 = 450; # in ohms\n", + "x = 1/8\n", + "\n", + "#calculation:\n", + "N = 1/x\n", + "R1 = R0*(N-1)/(N+1)\n", + "R2 = R0*2*N/(N**2 - 1)\n", + "\n", + "Io = x*x\n", + "\n", + "attn = 20*math.log10(N)\n", + "attnO = 4*attn\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n (a)For a T-section symmetrical attenuator pad, Resistance R1 is\",round(R1,2),\" ohm \"\n", + "print \"and Resistance R2 is\",round(R2,0),\" ohm\"\n", + "print \"\\n (b)current flows in the load = \",round(Io,2),\"of the original current.\"\n", + "print \"\\n (c)overall attenuation is \",round(attnO,2),\"dB \"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " (a)For a T-section symmetrical attenuator pad, Resistance R1 is 350.0 ohm \n", + "and Resistance R2 is 114.0 ohm\n", + "\n", + " (b)current flows in the load = 0.02 of the original current.\n", + "\n", + " (c)overall attenuation is 72.25 dB \n" + ] + } + ], + "prompt_number": 5 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit