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What happened to the peak structure:

In case you are wondering why the peaks in Figure 9 have less structure than the peaks in Figure 7 , this is because the points at which I computed the spectral data in Figure 9 were twice as far apart as the points at which I computed the spectral data in Figure 7 .

(The total frequency range in Figure 9 is twice as wide as in Figure 7 , but I computed the same number of points in both cases.)

Although it's not obvious at this plotting scale, there are zero-valued points between the side lobes on the peaks in Figure 7 .

The points in the spectral display of Figure 9 simply missed the side lobes and hit the zeros between the side lobes. I willhave a lot more to say about this in a future module discussing spectral analysis.

Back to the case with the sampling problem

Now let's take another look at the case with the sampling problem. This is the case where the sampling frequency was reduced from four samples per secondto two samples per second but the frequencies of the sinusoids was not changed. This view of the problem is shown in Figure 10 . It will probably be useful for you to compare Figure 10 with Figure 8 .

Figure 10. Spectral analyses of five sinusoids with sampling problem
missing image

Folding frequency is at the center

As with the previous case, each of the plotted spectra in Figure 10 shows the frequency range from zero frequency to the sampling frequency of two samples persecond. The folding frequency of one cycle per second appears in the center of each plot.

Peaks to left correspond to Figure 8

The peaks to the left of center in Figure 10 correspond to the peaks in Figure 8 . Because the right edge of Figure 8 is the folding frequency, the peaks to the right of center in Figure 10 don't appear in Figure 8 .

A mirror image

As is always the case, everything to the right of the folding frequency in Figure 10 is a mirror image of everything to the left of the folding frequency.

I have identified the artifacts created by the sampling process with a red oval in Figure 10 .

Raw data frequency exceeds folding frequency

The problem, as you will recall, is that the frequency of the sinusoids corresponding to the two bottom plots in Figure 10 is above the folding frequency. Thus the peaks to the right of center in the bottom two plots of Figure 10 actually represent the frequencies of the corresponding sinusoids.

Unfortunately, these two peaks appear to the right of the folding frequency, which the area of the spectra that we normally ignore.

Artifacts to the left of the folding frequency

Furthermore, these two peaks are reflected through the folded mirror image process into the area to the left of the folding frequency. For these twosinusoids, the peaks to the left of the folding frequency are artifacts, and I have identified them as such with ovals.

Normally can't identify artifacts

I am able to identify these two peaks as artifacts only because I know the true frequency makeup of the raw data. In most real-world situations withunknown data, there would be no way for me to identify these particular peaks to the left of the folding frequency as artifacts.

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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