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| author | hardythe1 | 2014-07-25 13:33:31 +0530 |
|---|---|---|
| committer | hardythe1 | 2014-07-25 13:33:31 +0530 |
| commit | 1c1ea29e3e213559fef5f928df109b7d17c21f24 (patch) | |
| tree | a1d38eb039e19d9e84b6e769cec04d59e7179e66 /Electrical_Circuit_Theory_And_Technology/chapter_41.ipynb | |
| parent | efb9ead5d9758d5d0bed7a22069320b14f972e40 (diff) | |
| download | Python-Textbook-Companions-1c1ea29e3e213559fef5f928df109b7d17c21f24.tar.gz Python-Textbook-Companions-1c1ea29e3e213559fef5f928df109b7d17c21f24.tar.bz2 Python-Textbook-Companions-1c1ea29e3e213559fef5f928df109b7d17c21f24.zip | |
removing unwanted and adding book
Diffstat (limited to 'Electrical_Circuit_Theory_And_Technology/chapter_41.ipynb')
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diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_41.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_41.ipynb new file mode 100644 index 00000000..4bdaf71c --- /dev/null +++ b/Electrical_Circuit_Theory_And_Technology/chapter_41.ipynb @@ -0,0 +1,1000 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:c5300e3088f971e53c15dd8fc5e0190bde02f3350126276135402bbe82ffef31"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1>Chapter 41: Attenuators</h1>"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 1, page no. 763</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\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": [
+ "<h3>Example 2, page no. 763</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\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": [
+ "<h3>Example 3, page no. 764</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\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": [
+ "<h3>Example 4, page no. 764</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\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": [
+ "<h3>Example 5, page no. 769</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\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": [
+ "<h3>Example 6, page no. 769</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\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": [
+ "<h3>Example 7, page no. 770</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\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": [
+ "<h3>Example 8, page no. 770</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\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": [
+ "<h3>Example 9, page no. 771</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\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": [
+ "<h3>Example 10, page no. 772</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\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": [
+ "<h3>Example 11, page no. 773</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\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": [
+ "<h3>Example 12, page no. 774</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\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": [
+ "<h3>Example 13, page no. 776</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\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": [
+ "<h3>Example 14, page no. 777</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\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": [
+ "<h3>Example 15, page no. 780</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\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": [
+ "<h3>Example 16, page no. 783</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\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": [
+ "<h3>Example 17, page no. 784</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\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": {}
+ }
+ ]
+}
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