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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

how did you get 1640
Noor Reply
If auger is pair are the roots of equation x2+5x-3=0
Peter Reply
Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne 2 hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.
MATTHEW Reply
420
Sharon
from theory: distance [miles] = speed [mph] × time [hours] info #1 speed_Dennis × 1.5 = speed_Wayne × 2 => speed_Wayne = 0.75 × speed_Dennis (i) info #2 speed_Dennis = speed_Wayne + 7 [mph] (ii) use (i) in (ii) => [...] speed_Dennis = 28 mph speed_Wayne = 21 mph
George
Let W be Wayne's speed in miles per hour and D be Dennis's speed in miles per hour. We know that W + 7 = D and W * 2 = D * 1.5. Substituting the first equation into the second: W * 2 = (W + 7) * 1.5 W * 2 = W * 1.5 + 7 * 1.5 0.5 * W = 7 * 1.5 W = 7 * 3 or 21 W is 21 D = W + 7 D = 21 + 7 D = 28
Salma
Devon is 32 32​​ years older than his son, Milan. The sum of both their ages is 54 54​. Using the variables d d​ and m m​ to represent the ages of Devon and Milan, respectively, write a system of equations to describe this situation. Enter the equations below, separated by a comma.
Aaron Reply
find product (-6m+6) ( 3m²+4m-3)
SIMRAN Reply
-42m²+60m-18
Salma
what is the solution
bill
how did you arrive at this answer?
bill
-24m+3+3mÁ^2
Susan
i really want to learn
Amira
I only got 42 the rest i don't know how to solve it. Please i need help from anyone to help me improve my solving mathematics please
Amira
Hw did u arrive to this answer.
Aphelele
hi
Bajemah
-6m(3mA²+4m-3)+6(3mA²+4m-3) =-18m²A²-24m²+18m+18mA²+24m-18 Rearrange like items -18m²A²-24m²+42m+18A²-18
Salma
complete the table of valuesfor each given equatio then graph. 1.x+2y=3
Jovelyn Reply
x=3-2y
Salma
y=x+3/2
Salma
Hi
Enock
given that (7x-5):(2+4x)=8:7find the value of x
Nandala
3x-12y=18
Kelvin
please why isn't that the 0is in ten thousand place
Grace Reply
please why is it that the 0is in the place of ten thousand
Grace
Send the example to me here and let me see
Stephen
A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of one of the other legs. Find the lengths of the hypotenuse and the other leg
Marry Reply
how far
Abubakar
cool u
Enock
state in which quadrant or on which axis each of the following angles given measure. in standard position would lie 89°
Abegail Reply
hello
BenJay
hi
Method
I am eliacin, I need your help in maths
Rood
how can I help
Sir
hmm can we speak here?
Amoon
however, may I ask you some questions about Algarba?
Amoon
hi
Enock
what the last part of the problem mean?
Roger
The Jones family took a 15 mile canoe ride down the Indian River in three hours. After lunch, the return trip back up the river took five hours. Find the rate, in mph, of the canoe in still water and the rate of the current.
cameron Reply
Shakir works at a computer store. His weekly pay will be either a fixed amount, $925, or $500 plus 12% of his total sales. How much should his total sales be for his variable pay option to exceed the fixed amount of $925.
mahnoor Reply
I'm guessing, but it's somewhere around $4335.00 I think
Lewis
12% of sales will need to exceed 925 - 500, or 425 to exceed fixed amount option. What amount of sales does that equal? 425 ÷ (12÷100) = 3541.67. So the answer is sales greater than 3541.67. Check: Sales = 3542 Commission 12%=425.04 Pay = 500 + 425.04 = 925.04. 925.04 > 925.00
Munster
difference between rational and irrational numbers
Arundhati Reply
When traveling to Great Britain, Bethany exchanged $602 US dollars into £515 British pounds. How many pounds did she receive for each US dollar?
Jakoiya Reply
how to reduced echelon form
Solomon Reply
Jazmine trained for 3 hours on Saturday. She ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. What is her running speed?
Zack Reply
d=r×t the equation would be 8/r+24/r+4=3 worked out
Sheirtina
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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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