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The peak in the fifth plot

Similarly, the frequency of the sinusoid for the fifth plot was 1.75 cycles per second. The peak for this sinusoid should have appeared 0.75 cycles persecond above the folding frequency, but appeared instead 0.75 cycles per second below the folding frequency. In other words, the spectrum folded around thefolding frequency so that this peak appeared below the folding frequency.

I am going to show you two more views of the spectra of these sinusoids to help you better understand the folding phenomena.

Back to the case with no problems

Let's go back and examine another view of the case that has no sampling problems. This view is shown in Figure 9 .

Figure 9. Spectral analyses of five sinusoids with no sampling problems.
missing image

Sampled at four samples per second

This is the case where all five sinusoids are sampled at a sampling frequency of four samples per second, resulting in a folding frequency of two cycles persecond. If you compare Figure 9 with Figure 7 , you will see that the left half of Figure 9 is very similar to Figure 7 .

Figure 9 Shows twice the frequency range

In Figure 7 , the spectral data was computed and displayed from zero frequency on the left to the folding frequency (two cycles per second) on the right. In Figure 9 , the spectral data was computed and displayed from zero frequency on the left to the sampling frequency (four cycles per second) on the right.

Thus, the total frequency range for Figure 9 is twice the frequency range for Figure 7 .

Folding frequency at the center

In Figure 9 , the folding frequency is exactly in the center of each plot. In other words, the center of the plots in Figure 9 corresponds to the right edge of the plots in Figure 7 . Everything to the left of center in Figure 9 corresponds to the plots in Figure 7 . The material to the right of center in Figure 9 was not shown in Figure 7 .

Why is it called the folding frequency?

Hopefully the display in Figure 9 will explain why the frequency that is half the sampling frequency is called the folding frequency. The computed spectrumfolds around that frequency. Everything to the right of the folding frequency is a mirror image of everything to the left of the folding frequency.

Peaks below folding frequency are valid

All the peaks to the left of center in Figure 9 are valid spectral peaks associated with the corresponding sinusoids. However, all the peaks to the rightof center, which I marked with red ovals, are artifacts of the sampling process. Those peaks do not exist in the true spectrum of the original raw data. Theywere created by the sampling process.

Normally don't compute the mirror image

Normally we don't worry about this mirror image above the folding frequency when doing spectral analyses. We know it is there and we simply ignore it.

In fact, for reasons of economy, when doing spectral analyses using discrete Fourier transforms, we usually don't even compute the spectrum at frequenciesabove the folding frequency. Since it is always a mirror image of the spectrum below the folding frequency, we know what it looks like without even computingit.

Questions & Answers

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The denominator of a certain fraction is 9 more than the numerator. If 6 is added to both terms of the fraction, the value of the fraction becomes 2/3. Find the original fraction. 2. The sum of the least and greatest of 3 consecutive integers is 60. What are the valu
1. x + 6 2 -------------- = _ x + 9 + 6 3 x + 6 3 ----------- x -- (cross multiply) x + 15 2 3(x + 6) = 2(x + 15) 3x + 18 = 2x + 30 (-2x from both) x + 18 = 30 (-18 from both) x = 12 Test: 12 + 6 18 2 -------------- = --- = --- 12 + 9 + 6 27 3
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Q2 x+(x+2)+(x+4)=60 3x+6=60 3x+6-6=60-6 3x=54 3x/3=54/3 x=18 :. The numbers are 18,20 and 22
Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113?
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Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
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Source:  OpenStax, Digital signal processing - dsp. OpenStax CNX. Jan 06, 2016 Download for free at https://legacy.cnx.org/content/col11642/1.38
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