{
"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": {}
}
]
}