{
"metadata": {
"name": ""
},
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 5 Frequency Modulation : Transmission"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.1 Page no 209"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#given\n",
"v=25*10**-3\n",
"f=750 #deviation constant\n",
"vg=10.0*10**-3 #deviation constant\n",
"\n",
"#calculation\n",
"pfd=v*(f/vg) #positive frequency deviation\n",
"nfd=-v*(f/vg) #negative frequency deviation\n",
"\n",
"#result\n",
"print\"(a) positive frequency deviation = \",pfd,\"Hz\"\n",
"print\"negative frequency deviation = \",nfd,\"Hz\"\n",
"print\"The total deviation is written as +-2.25kHz for the given input signal level\"\n",
"print\"(b) The carrier wil deviate \",pfd,\"Hz\",\"&\",nfd,\"Hz\",\"at 400 Hz\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a) positive frequency deviation = 1875.0 Hz\n",
"negative frequency deviation = -1875.0 Hz\n",
"The total deviation is written as +-2.25kHz for the given input signal level\n",
"(b) The carrier wil deviate 1875.0 Hz & -1875.0 Hz at 400 Hz\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.3 Page no 214"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#given\n",
"d=20*10**3 #maximum deviation\n",
"fi=10.0*10**3 #input frequency\n",
"\n",
"#calculation\n",
"mf=d/fi\n",
"a=mf*40\n",
"\n",
"#result\n",
"print\"total required bandwidth is \",a,\"KHz\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"total required bandwidth is 80.0 KHz\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.4 Page no 214"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#given\n",
"d=20*10**3 #maximum deviation\n",
"fi=5.0*10**3 #input frequency\n",
"\n",
"#calculation\n",
"mf=d/fi\n",
"a=2*35\n",
"\n",
"#print\n",
"print\"the required bandwidth is \",mf\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the required bandwidth is 4.0\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.5 Page no 215"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#given\n",
"Vm=2000\n",
"R=50.0 #resistance, ohm\n",
"\n",
"#calcultion\n",
"import math\n",
"fc=(2*math.pi*(10**8))/2.0*math.pi\n",
"P=(2000/math.sqrt(2))**2/R\n",
"mf=2 #by inspection of FM equation\n",
"fi=(math.pi*10**4)/(2.0*math.pi)\n",
"d=(mf*fi)\n",
"BW=mf*40\n",
"bw=2*(d+fi)\n",
"P1=((0.58*2000/math.sqrt(2))**2)/R\n",
"P2=((0.03*2000/math.sqrt(2))**2)/R\n",
"\n",
"#result\n",
"print\"(a) carrier frequency = \",round(fc,-9),\"Hz\" #by inspection of FM equation\n",
"print\"(b) the peak voltage is 2000V P thus= \",P,\"W\"\n",
"print\"(c) mf = 2\" #by inspection of FM equation\n",
"print\"(d) the intelligence frequency fi = \",fi,\"Hz\"\n",
"print\"(e) BW = \",bw ,\"Hz\" #using carson's rule \n",
"print\"(f) The smallest sideband J4 is 0.03 times the carrier = \",P2,\"W\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a) carrier frequency = 1000000000.0 Hz\n",
"(b) the peak voltage is 2000V P thus= 40000.0 W\n",
"(c) mf = 2\n",
"(d) the intelligence frequency fi = 5000.0 Hz\n",
"(e) BW = 30000.0 Hz\n",
"(f) The smallest sideband J4 is 0.03 times the carrier = 36.0 W\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.6 Page no 218"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#given\n",
"d=75*10**3 #maximum deviation\n",
"fi=30.0 #modulating frequency, Hz\n",
"fi1=15.0*10**3 \n",
"d1=1*10**3\n",
"fi2=100 #Hz\n",
"fi3=2.0*10**3\n",
"\n",
"#calculation\n",
"mf1=d/fi\n",
"mf2=d/fi1\n",
"mf3=d1/fi2\n",
"mf4=d1/fi3\n",
"DR=d1/fi3\n",
"\n",
"#result\n",
"print\"(a)maximum deviation at 30 Hz = \",mf1 \n",
"print\"maximum deviation at 15kHz= \",mf2\n",
"print\"(b) maximum deviation at 100Hz = \",mf3\n",
"print\"maximum deviation at 2KHz= \",mf4\n",
"print \"(c)Deviation Ratio \",DR #deviation ratio"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)maximum deviation at 30 Hz = 2500.0\n",
"maximum deviation at 15kHz= 5.0\n",
"(b) maximum deviation at 100Hz = 10\n",
"maximum deviation at 2KHz= 0.5\n",
"(c)Deviation Ratio 0.5\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.7 Page no 218"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#given\n",
"mf=0.25\n",
"a=0.98\n",
"b=0.12\n",
"x=10*10**3 #power, watt\n",
"\n",
"#calculation\n",
"P=(a**2)*x\n",
"P1=(b**2)*x\n",
"t=P+2*P1\n",
"\n",
"#result\n",
"print\"power of each sideband = \",P1,\"W\"\n",
"print\"total power = \",round(t,-4),\"W\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"power of each sideband = 144.0 W\n",
"total power = 10000.0 W\n"
]
}
],
"prompt_number": 31
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.8 Page no 222"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#given\n",
"phi=0.5 #maximum intelligence frequency\n",
"fi=5.0*10**3\n",
"x=75*10**3\n",
"\n",
"#calculation\n",
"d=phi*fi\n",
"y=x/d\n",
"\n",
"#result\n",
"print\"o/p S/N = \",y\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"o/p S/N = 30.0\n"
]
}
],
"prompt_number": 34
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.9 Page no 222"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#given\n",
"dm=10*10**3\n",
"x=(1/3.0) #N/S input ratio\n",
"\n",
"#calculation\n",
"import math\n",
"phi=math.asin(x)\n",
"phi1=math.asin(x)\n",
"fi=3*10**3\n",
"d=phi1*fi\n",
"a=dm/d\n",
"\n",
"#result\n",
"print\"The S/N output will be \",round(a,0)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The S/N output will be 10.0\n"
]
}
],
"prompt_number": 36
}
],
"metadata": {}
}
]
}