Fourier Analysis - Redes de Computadores - UPV

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With –o option we specify the name of the resulting binary file. The –lm option is necessary to link with math library (sin and cos trigonometric functions) Impact of physical medium on the transmitted signal. Here we are going to use our program to analyze how the transmitted signal is distorted when crossing an specific transmission channel. For that purpose, we are going to produce four different scenarios corresponding with the transmission of one signal with the same transmission speed under different channel bandwidths. In particular, we will use channels with bandwidths of 400, 1000, 2000 and 9000 Hz. Run the program with the following parameters: ./MiFourier –c a –v 1200 –b 400 –a 10000 > bw400.txt Option –c specifies the character we are going to transmit (also we can specify the ASCII code: –c 97). Option –v fixes the transmission speed (in bits per second). With option --b we can define the desired channel bandwidth. Finally, option –a is aesthetic option used to determine width of the frequency domain plot (in Hz). As the character and transmission speed are just the same than the default ones, we do not need to specify them. Run the program to get results for the rest of proposed bandwidths. ./MiFourier –b 1000 –a 10000 > bw1000.txt ./MiFourier –b 2000 –a 10000 > bw2000.txt ./MiFourier –b 9000 –a 10000 > bw9000.txt Compare the four graphics, using the GNUPlot program. You can observe that as bandwidth increases, more harmonics cross the channel and as a consequence the reconstructed signal is better (closer to the original one). Do you think that a bandwidth of 400 Hz is enough for transmitting this character with a speed of 1.200 bps. ? Should it be necessary a channel bandwidth of 9 KHz or it would be enough with 2 KHz? Impact of transmission speed on the transmitted signal En general, todos tenemos la idea de que al transmitir por un medio físico una señal digital, la velocidad máxima de transmisión está acotada. Hay muchos factores que introducen esta limitación, pero uno de los principales es que el ancho de banda del medio es limitado. Como ya hemos visto, esto provoca que a partir de una determinada frecuencia, el resto de armónicos se atenúan. We can notice that transmission speed is very related with signal period T. So that, as transmission speed increases the signal period becomes shorter. The fundamental frequency that determines the distance between harmonics also increases, so a lower number of harmonics will cross the channel and, as a consequence, the reconstructed signal will be poorer . 8
To verify this effect we will use our program with the following transmission speeds 1200, 2400, 4800 and 9600 bps, and a fixed channel bandwidth of 6 KHz. ./MiFourier –b 6000 –a 10000 –v 1200 > v1200.txt ./MiFourier –b 6000 –a 10000 –v 2400 > v2400.txt ./MiFourier –b 6000 –a 10000 –v 4800 > v4800.txt ./MiFourier –b 6000 –a 10000 –v 9600 > v9600.txt Analyze the results shown in the four resulting graphics. The difference between them stay on the distance between two consecutive harmonics. So, when the transmission speed increases the number of harmonics that arrive at the other end decreases. Is it acceptable a transmission speed of 9600 bps for this physical channel ? In general, is not obvious to determine the maximum transmission speed for a particular physical channel. In theory, we can say that the maximum transmission speed will be the one that allows crossing the channel the enough harmonics to properly reconstruct the signal at the other end. This would be true in this example when the first six harmonics cross the channel. Next, use the program to transmit a character using a channel with a 6 KHz. bandwidth and a transmission speed of 7600 bps. Then, do the same with a channel of 45 KHz. bandwidth and a transmission speed of 56000 bps. Is there any difference between both reconstructed signals? What do you think is more important channel bandwidth, transmission speed o the relation between both ? Justify the answers. Problem We have a physical channel that behaves as an ideal low-pass filter with a 10 KHz. bandwidth (0..10 KHz.). We continuously transmit a signal composed of an ASCII character (whatever char) of 8 bits with a transmission speed of 1000 bps. How many harmonics cross the channel? Idem if transmission speed is 2000 bps ?. Solve the problem analytically, and then run program to verify the results. Analytic solution: The signal period covers the eight character bits. If transmission speed is 1000 bps, then one bit lasts 1 ms, and T = 8 ms. The fundamental frequency (the inverse of period) will be f = 1/0,008 = 125 Hz. If the channel bandwidth is 10000 Hz, the number of passing harmonics will be (10000 / 125) = 80. If transmission speed doubles, the fundamental frequency also doubles (250 Hz) and the number of harmonics will be just the half (40). Additional questions One interesting improvement to the program will be the following: Allow a physical channel with a band-pass behaviour like the local-loop phone lines (with a passing band between 300 and 3300 Hz.) Another proposed addition is modelling the channel taking in a more realistic approach. So harmonics will be attenuated from one frequency fa up to the cut frequency fc. To perform this change, find the attenuation tables (as a function of frequency) for different wire categories, and apply the attenuation over a specified number of harmonics (i.e. the 9
Page 1 and 2: Laboratory 2 Fourier Analysis Study
Page 3 and 4: total signal power. So, we have to
Page 5 and 6: Execution example Before start prog
Page 7: In the right, the required informat

Fourier Analysis - Redes de Computadores - UPV

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