Fourier Analysis - Redes de Computadores - UPV

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In expression (1), f is the fundamental frequency of signal s(t). At the other hand, a0 is a constant, we call DC component of signal s(t), that is calculated with expression (2). The rest of terms in (1) correspond to the infinite components of signal s(t), we call them harmonics or frequencial components. So, the n-th harmonic of signal s(t) is given by the following expression sn n n ( t) = a cos( 2π nft) + b sen( 2πnft) (5) Related to (5), if nf=1, then cos( 2π t) + sen( 2πt) corresponds with the sum of one sine and cosine of period 1. So, nf components define the sine and cosine frequency. As a consequence, the frequency of n-th harmonic will be f n = nf Finally, the amplitude of a sine (or cosine) of n harmonic is given by coefficients an and bn. These coefficients weight the contribution of the corresponding harmonic in the original signal s(t), and are calculated by mean of expressions (3) and (4). The power or amplitude of harmonic n may be calculated as: P = a + b (6) n 2 n 2 n The graphical representation of all harmonics of original signal in the frequency domain is known as spectral domain of signal s(t). Latter, we will show some examples. Impact of physical medium in signal transmission When using a wire for transmitting a signal, the resulting signal at the other end is received as a attenuated version of the original signal. However, in the frequency domain, not all harmonics are uniform attenuated. The resulting behaviour is similar to a low-pass filter, where low frequency harmonics cross the filter without attenuation, but as harmonic frequency increase they are more attenuated and even filtered. So, we will consider that a physical transmission medium transport all signal harmonics with frequency inferior to fc (cut frequency), and the rest of signal harmonics are totally filtered. This assumption is not realistic because there is a progressive attenuation of harmonics up to the cut frequency. Anyway, under the above assumption we can say that a physical transmission medium has a bandwidth BW corresponding to the maximum frequency it is able to transport (fc). A digital signal is made up of an infinite number of harmonics. This implies that we would need an infinite bandwidth channel for transmitting the signal without distortion. Unfortunately, there is no physical medium with infinite bandwidth, so in practice the received signal will be a distorted version from the original one (not all harmonics will be received). It is obvious that as more harmonics be received the reconstructed signal will have better quality. However, not all harmonics contribute in the same manner to the reconstruction of the signal. IN particular, the first harmonics carry the most part of the 2
total signal power. So, we have to determine the minimum number of harmonics required by the received to recognize and reconstruct the signal. How to compute the Fourier series of an ASCII character. Suppose that we are going to continuously transmit one character (one byte). If T is the repetition period, we can formally define the transmitted signal s(t) as: ⎧ ⎡ T ⎡ ⎪ bit( 0) when t ∈ ⎢0 + Tn, + Tn⎢ ⎪ ⎣ 8 ⎣ ⎪ ⎡T 2T ⎡ s( t) = bit( 1) when t ∈ ⎨ ⎢ + Tn, + Tn⎢ ⎣8 8 ⎣ ⎪ M ⎪ ⎪ ⎡7T ⎡ bit( 7) whent ∈ ⎪ ⎢ + Tn, T + Tn⎢ ⎩ ⎣ 8 ⎣ ∀n∈ ? (7) Where bit(i) are constants that define the value of bit i (of transmitted character). If the signal is not periodic, we can not use Fourier Series, instead we are obliged to use the Fourier transform which is out of the course scope. Now, we are ready to calculate the terms a0, an and bn de (1) for the signal defined in (7). So, we are going to use the equations (2), (3) and (4). Let start with the DC component a0, a 0 1 = T ∫ T 0 s( t) dt As function s(t) is defined as (7), we can perform the integral calculation by intervals: a 0 T 2T 1 ⎛ ⎜ 8 8 = bit ( 0) dt + bit ( 1) dt + K + ⎜∫ ∫T T 0 ∫ ⎝ 8 3 T 7T 8 ⎞ bit ( 7) dt⎟ ⎟ ⎠ Remember that bit(i) are constants, so we can consider the following a 0 T 2T 1 ⎛ ⎜ 8 8 = bit ( 0) dt + bit ( 1) dt + K + bit ( 7) ⎜ ∫ ∫T T 0 ∫ ⎝ 8 Solving the simple integral and a 0 1 ⎛ = ⎜bit ( 0) T ⎜ ⎝ T 2T [ t] 8 + bit ( 1) [ t] 8 + K + bit ( 7) [ t] 0 1 ⎛ T T T a0 = ⎜ bit ( 0) + bit ( 1) + K + bit ( 7) T ⎝ 8 8 8 T 8 ⎞ ⎟ ⎠ T 7T 8 T 7T 8 ⎞ ⎟ ⎟ ⎠ ⎞ dt⎟ ⎟ ⎠
Page 1: Laboratory 2 Fourier Analysis Study
Page 5 and 6: Execution example Before start prog
Page 7 and 8: In the right, the required informat
Page 9 and 10: To verify this effect we will use o

Fourier Analysis - Redes de Computadores - UPV

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