Laboratorio 4 práctica PDF

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Laboratorio 4 práctica...

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Noms:

Anna Garcia Martinez

Lloc treball: Grup:

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1. Listen to the music again and relate the musical events with the content of the STFT (Fig. 3.1). Describe the event that produces the STFT change at approximately 5.5s? When hearing to the audio track we can clearly distinguish the sound of bells appearing at around 5,5s. 2. In Figure 3.2, identify the strongest frequency components of the music at t=1s. Give their frequencies. Relate these components with the information given in Figure 3.1. As we can see on the figure 3.2 the eight strongest frequency components appear in 86Hz, 602Hz, 4737Hz, 5426 Hz, 6546Hz, 8872Hz, 13180Hz and 14640 Hz. Although we can see a strong frequency component close to 0 (the 86Hz one), it corresponds to the red line in the lower part of the spectrogram, which is a continuous component the file had. Moreover, if we look at the spectrogram on the frequencies we mentioned earlier, we can also see the have a little bit more amplitude though it is difficult to perceive, thus it is better to look at the figure 3.2 to see those components.

3. Which is the sampling frequency of the decimated signal? As we are decimating the signal by 4, the new sampling frequency should be Fy=Fx/4=44100/100=11025Hz. 4. Which should the cut-off frequency of the anti-aliasing filter be? Our cut-off frequency should be Fx/2M=44100/8=5512Hz.

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5. Relate the spectral content of this new signal to the original signal spectral composition at t=1s (compare Figure 3.2 and Figure 3.3). Identify those components of the original signal that are preserved in the decimated version and those that are filtered out. By comparing the decimated signal to the original signal we can see that now we only have the frequency component at 602Hz, the one at 5426Hz and the one at 4780Hz, and the other we identified on question 2 have now disappeared. We still can see the one close to 0 (at 86Hz), but as we said it should not be taken into account and that. Also, we see that we have only picked up the segment from 0 to 5513 (due to the anti-aliasing filter) so it is normal that the other components have been filtered out.

6. The length of the window used to compute the Fourier transform of the decimated signal is the length used with the original signal divided by M. Could you give a reason for this change? Why and how much the magnitude of the Fourier Transform has decreased? If we look at the same valuer in frequency for the graphics in 3.2 and 3.2 that for the same value (602Hz) on the 3.2 we see an amplitude of 12,45dBs and on the 3.3 we see an amplitude of -0,63dBs, giving us a difference of -13,08dBs, which transformed from dBs is of 0,22 (approximately 1/M=1/4). As we explained, in order to avoid aliasing, we must change our window to adapt it to our decimated signal, being the new window equal to the old but divided by the decimation factor. 7. Do you perceive a significant difference between both versions of the decimated music? Could you describe this difference? There is not much difference between both sounds, but without the antialiasing filter we can see that the repetitive background sound which was high-pitched is now lower-pitched.

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8. As it can be seen in Figure 3.4, the Fourier transform of the new signal shows two spectral components at t=1s that are not present at the Fourier transform of the pre-filtered decimated signal (Figure 3.3). Give their frequencies.

On this figure we can see two new frequency components that didn’t appear on figure 3.3 (marked on 2110Hz and 4457Hz). Those new components have appeared due to aliasing.

9. Identify the frequency components of the original signal at t=1s in Figure 3.2 that more likely can produce aliasing when the signal is decimated by M=4 with no antialiasing filter. The components that can produce aliasing are those strong frequency components higher than 5,5kHz (the one that should be the cut-off frequency of the anti-aliasing filter). The most important ones are in 6546Hz, 8872Hz, 13180Hz and 14640Hz. 10. Give the frequency of the spectral components of the original music that produce the two aliasing components present in Figure 3.4 (Hint: take into account your answer to the question 2 of the previous study). For each of the frequencies mentioned earlier and taking fy=11025Hz f1=2110Hz  fy-f1=8915Hz  2fy-f1=19940Hz  fy+f1=13135Hz For this frequency we can deduce that the spectral component of the original music that produced this aliasing component was 8872Hz or 13180Hz identified in question 2. f2=4457Hz  fy-f2=6568Hz  2fy-f2=17593Hz  fy+f2=15482Hz For this frequency component, and in relation with question 2 again, we can say it was probably produced by the 6546Hz component.

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11. Compare the Fourier transforms in Figure 3.3, Figure 3.6 and Figure 3.7. Point out the main differences between the Fourier transforms in Figure 3.3 and Figure 3.6; repeat this analysis between Figure 3.6 and Figure 3.7. In both cases, explain the causes of the differences you have found out. If we look at figure 3.3 and 3.6 we can see that the first half of 3.6 is equal to 3.3 even though some new components appear in the 3.6 such as in 2885Hz due to the interpolation process. When we interpolate using a filter (in figure 3.7) we can see that it results in a more similar signal to the original one than 3.3, where some samples have been removed. The figures 3.6 and 3.7 are really similar in low frequencies whereas for higher frequencies the interpolated signal with filter, 3.7, attenuates more the frequency components than the interpolation without the filter in 3.6. If we look at the filter in figure 3.5 we can see that it this fact is related to the form of this filter.

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