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{
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{
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"metadata": {},
"source": [
"# Chapter 12 : Digital Modulation and Modems"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 1 : pg 407"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"a)The theoretical max data rate is 50.278 kb/s\n",
"b)The data rate for 4 states is 40.0 kb/s\n"
]
}
],
"source": [
" \n",
"# page no 407\n",
"# prob no 12_1\n",
"#calculate the data rate and theoretical max\n",
"#A radio channel with BW=10KHz and SNR=15 dB\n",
"from math import log\n",
"#given\n",
"B=10.*10**3;\n",
"snr=15.;\n",
"#calculations and results\n",
"#converting dB in power ratio\n",
"SNR=10**(snr/10);\n",
"#a)Determination of theoretical max data rate\n",
"C1=B*log(1+SNR) /log(2);\n",
"print 'a)The theoretical max data rate is',round(C1/1000,3),'kb/s'\n",
"#b)Determination of data rate with 4 states i.e M=4\n",
"M=4.;\n",
"C2=2*B*log(M) / log(2);\n",
"print 'b)The data rate for 4 states is',C2/1000,'kb/s'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 2 : pg 408"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The baud rate is 10.0 kbaud\n",
"The bit rate is 60.0 kb/s\n"
]
}
],
"source": [
" \n",
"# page no 408\n",
"# prob no 12_2\n",
"#calculate the baud and bit rates\n",
"from math import log\n",
"#A modulator transmit symbol with symbol rate=10k/sec with 64 states\n",
"#given\n",
"M=64.;\n",
"S=10000.;\n",
"#calculations and results\n",
"#Baud rate is simply symbol rate\n",
"print 'The baud rate is ',S/1000,'kbaud'\n",
"#Determination of bit rate\n",
"C=S*log(64) / log(2);\n",
"print 'The bit rate is ',C/1000,'kb/s'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 3 : pg 411"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"a)The frequency shift is 135416.5 Hz\n",
"b)The maximum frequency is 880067708.25 Hz\n",
"The minimum frequency is 879932291.75 Hz\n",
"The bandwidth efficiency in b/s/Hz is 1.354 b/s/Hz\n"
]
}
],
"source": [
" \n",
"# page no 411\n",
"# prob no 12_3\n",
"#calculate the frequency shift, max, min frequency\n",
"#given\n",
"f=200.*10**3;\n",
"fb=270.833 *10**3;\n",
"data_rate=270.833 *10**3\n",
"fc=880*10**6;\n",
"bandwidth=200*10**3;\n",
"#calculations and results\n",
"freq_shift=0.5*fb;\n",
"print 'a)The frequency shift is',freq_shift,'Hz'\n",
"# The shift each way from the carrier frequency is half the freq_shift\n",
"f_max=fc+0.25*fb;\n",
"print 'b)The maximum frequency is',round(f_max,3),'Hz'\n",
"f_min=fc-0.25*fb;\n",
"print 'The minimum frequency is',round(f_min,3),'Hz'\n",
"bandwidth_efficiency=data_rate/bandwidth;\n",
"print 'The bandwidth efficiency in b/s/Hz is',round(bandwidth_efficiency,3),'b/s/Hz'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 4 : pg 412"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The channel data rate is 48.6 kb/s\n"
]
}
],
"source": [
" \n",
"# page no 412\n",
"# prob no 12_4\n",
"#calculate the channel data rate\n",
"#given\n",
"baud_rate=24.3;# in kilobaud\n",
"#calculations\n",
"# In this problem dibit system is used.\n",
"#Therefore symbol_rate=baud_rate=0.5*bit_rate\n",
"bit_rate=2*baud_rate;\n",
"#results\n",
"print 'The channel data rate is',bit_rate,'kb/s'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 5 : pg 413"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The no. of bits per symbol is 6.0\n"
]
}
],
"source": [
" \n",
"# page no 413\n",
"# prob no 12_5\n",
"#calculate the no. of bits per symbol\n",
"from math import log\n",
"#given\n",
"no_of_phase_angles=16;\n",
"no_of_amplitudes=4;\n",
"#calculations\n",
"no_of_states_per_symbol=no_of_phase_angles*no_of_amplitudes;\n",
"bit_per_symbol=log(no_of_states_per_symbol)/ log(2);\n",
"#results\n",
"print 'The no. of bits per symbol is',bit_per_symbol"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6 : pg 415"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Shannon limit 29901.68 b/s\n"
]
}
],
"source": [
" \n",
"# page no 415\n",
"# prob no 12_6\n",
"#calculate the shannon limit\n",
"from math import log\n",
"#given\n",
"B=3.*10**3;SNR_dB=30.;\n",
"#calculations\n",
"SNR_power=10**(30./10);\n",
"C=B*log(1+SNR_power)/ log(2);\n",
"print 'Shannon limit',round(C,2),'b/s'"
]
}
],
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