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{
"metadata": {
"name": "",
"signature": "sha256:67ee8e7b784b607cbab75ad3758612a5810d648f2de452b74761d1700d1555d3"
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"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 10 - Digital Multiplexers"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1 - pg 469"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the minimum value of permissible sampling rate\n",
"\n",
"#given\n",
"X_1 = 4.*10**3#first analog signal in Hz\n",
"X_2 = 4.5*10**3#second analog signal in Hz\n",
"\n",
"#calculation\n",
"#the highest frequency cmponent of the composite signal consisting among two signal is X_2\n",
"f_sMIN = 2*X_2;\n",
"\n",
"print \"The minimum value of permissible sampling rate (kHz) = \",f_sMIN/1000.\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The minimum value of permissible sampling rate (kHz) = 9.0\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2 - pg 469"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the signalling rate and minimum channel bandwidth\n",
"\n",
"#given\n",
"X_1 = 6.*10**3#Nyquist rate in Hz obtained the table\n",
"X_2 = 2.*10**3#Nyquist rate in Hz obtained the table\n",
"X_3 = 2.*10**3#Nyquist rate in Hz obtained the table\n",
"X_4 = 2.*10**3#Nyquist rate in Hz obtained the table\n",
"\n",
"#calculations\n",
"s = 2000#speed of rotation\n",
"X1 = 3*s#number of samples produced per second for first signal\n",
"X2 = 1*s#number of samples produced per second for second signal\n",
"X3 = X2#number of samples produced per second for third signal\n",
"X3 = X2#number of samples produced per second for fourth signal\n",
"SR = X1 + 3*X2#signalling rate\n",
"BW = .5*SR#minimum channel bandwidth \n",
"\n",
"#results\n",
"print \"If the sampling commutator rotates at the rate of 2000 rotations per second the the signals X_1,X_2,X_3,X_4 will be sampled at their Nyquist rate\"\n",
"print \"Signalling rate (samples per second) = \",SR\n",
"print \"Minimum channel bandwidth (Hz) = \",BW\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"If the sampling commutator rotates at the rate of 2000 rotations per second the the signals X_1,X_2,X_3,X_4 will be sampled at their Nyquist rate\n",
"Signalling rate (samples per second) = 12000\n",
"Minimum channel bandwidth (Hz) = 6000.0\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3 - pg 470"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the sampling rate and Nyquist rate\n",
"\n",
"#given\n",
"SR = 8000.#sampling rate in samples per second\n",
"T = 1.*10**-6#pulse duration\n",
"f = 3.4*10**3#highest frequency component\n",
"\n",
"#calculations\n",
"#second case\n",
"NR = 2.*f#Nyquist rate of sampling\n",
"T2 = 1./NR#time taken for one rotation of commutator\n",
"\n",
"\n",
"#results\n",
"print \"sampling rate for first condition = \",SR\n",
"print \"There are 24 voice signals + 1 synchronizing pulse\"\n",
"print \"Pulse width of each voice channel and synchronizing pulseis 1 microseconds \"\n",
"print \"Now, time taken by the commutator for 1 rotation =1/8000 = 125*10**-6 seconds\"\n",
"print \"Number of pulses produced in one rotation = 24 + 1 = 25\"\n",
"print \"Therefore, the leading edges of the pulses are at 125/25 = 5*10^-6 seconds distance\"\n",
"print \"Nyquist rate for second condition (kHz) = \",NR/1000.\n",
"print \"Time taken for one rotation of commutator (museconds) = \",round(T2*10**6,2)\n",
"print \"Therefore, 147*10**-6 seconds corresponds to 25 pulses\"\n",
"print \"therefore, 1 pulse corresponds to 5.88*10^-6 seconds\"\n",
"print \"As the pulse width of each pulse is 1*10^-6 seconds, the spacing between adjacent pulses will be 4.88*10**-6 seconds\\n and if we assume tou = 0 then the spacing between the adjacent pulses will be 5.88*10**-6 seconds \"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"sampling rate for first condition = 8000.0\n",
"There are 24 voice signals + 1 synchronizing pulse\n",
"Pulse width of each voice channel and synchronizing pulseis 1 microseconds \n",
"Now, time taken by the commutator for 1 rotation =1/8000 = 125*10**-6 seconds\n",
"Number of pulses produced in one rotation = 24 + 1 = 25\n",
"Therefore, the leading edges of the pulses are at 125/25 = 5*10^-6 seconds distance\n",
"Nyquist rate for second condition (kHz) = 6.8\n",
"Time taken for one rotation of commutator (museconds) = 147.06\n",
"Therefore, 147*10**-6 seconds corresponds to 25 pulses\n",
"therefore, 1 pulse corresponds to 5.88*10^-6 seconds\n",
"As the pulse width of each pulse is 1*10^-6 seconds, the spacing between adjacent pulses will be 4.88*10**-6 seconds\n",
" and if we assume tou = 0 then the spacing between the adjacent pulses will be 5.88*10**-6 seconds \n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4 - pg 471"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Signaling rate and minimum channel bandwidth\n",
"\n",
"#given\n",
"N = 6.#number of channels\n",
"f_m = 5.*10**3#bandwidth of each channel\n",
"\n",
"#calculations\n",
"SR1= 2*f_m#minimum sampling rate\n",
"SR = N*SR1#sampling rate\n",
"BW =N*f_m#minimum channel bandwidth\n",
"\n",
"#results\n",
"print \"Signaling rate (Kbits per second) = \",SR/1000.\n",
"print \"Minimum channel bandwidth (kHz) = \",BW/1000.\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Signaling rate (Kbits per second) = 60.0\n",
"Minimum channel bandwidth (kHz) = 30.0\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6 - pg 476"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the number of framing plus signaling bits per frame\n",
"\n",
"#given\n",
"channel=64*10**3 #kb/s\n",
"bitrate=2.048*10**6 #bits/sec\n",
"\n",
"#calculations\n",
"fs_min=channel*2\n",
"x=bitrate/fs_min\n",
"\n",
"#results\n",
"print 'No. of bits per frame = ',x"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"No. of bits per frame = 16.0\n"
]
}
],
"prompt_number": 5
}
],
"metadata": {}
}
]
}
|
