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The locations of the spectral peaks

The spectral peaks in Figure 7 appear where you would expect to see them. For example, the location of the peak in the first plot corresponds to a frequencyof 0.25 cycles per second within a total frequency range extending from zero to two cycles per second. This matches the information given in the above table forthe first sinusoid.

The location of the spectral peak in the fifth plot corresponds to a frequency of 1.75 cycles per second within a total frequency range extendingfrom zero to two cycles per second. This matches the information given in the above table for the fifth sinusoid.

The location of the peak in each of the three plots between the first and the last are correct for the frequency of the sinusoid involved.

Introduce a sampling problem

Now I will introduce a sampling problem by keeping the frequencies of the sinusoids the same and reducing the sampling frequency from four samples persecond to two samples per second.

The result of this change is shown in Figure 8 .

Figure 8. Spectral analyses of five sinusoids with sampling problem.
missing image

As before, each of the plots in Figure 8 shows the frequency spectrum of an individual sinusoid. The spectrum is plotted from zero frequency on the left tothe folding frequency on the right.

Sampling frequency was two samples per second

In this case, the sampling frequency was two samples per second, giving a folding frequency of one cycle per second. Therefore, the horizontal scale oneach plot represents the frequencies from zero on the left to one cycle per second on the right.

The heights of the spectral peaks

Once again, the height of each spectral peak is consistent with the amplitude of the sinusoid.

The locations of the spectral peaks

As before, the spectral peaks in the first three plots appear where you would expect to see them. The peak in the first plot is about twenty-five percent ofthe way across the total spectrum, corresponding to 0.25 cycles per second.

The spectral peak in the second plot is at the center, corresponding to 0.5 cycles per second. The third peak is in the correct location for 0.75 cycles persecond.

A problem with the location of two spectral peaks

However, a problem exists with the spectral peaks in the last two plots.

(I marked the two problem peaks with a red oval to make it obvious which ones I am talking about. You may find it helpful to compare Figure 8 side-by-side with Figure 7 .)

The spectral peak in the fourth plot also appears about midway between zero and one cycle per second. This indicates that the corresponding sinusoid had afrequency of 0.5 cycles per second.

However, the frequency of the sinusoid for the fourth plot was 1.50 cycles per second, not 0.5 cycles per second as indicated. Thus, that spectral peakshould have been off the scale on the right-hand side of the plot.

The folding frequency

Recall, however, that the right edge of the plot is the folding frequency. Therefore, any spectral components that should appear to the right of thefolding frequency fold around and appear to the left of the folding frequency. Therefore, the spectral peak in the fourth plot, which should appear at 0.50cycles per second above the folding frequency, appears instead at 0.50 cycles per second below the folding frequency.

Questions & Answers

if three forces F1.f2 .f3 act at a point on a Cartesian plane in the daigram .....so if the question says write down the x and y components ..... I really don't understand
Syamthanda Reply
hey , can you please explain oxidation reaction & redox ?
Boitumelo Reply
hey , can you please explain oxidation reaction and redox ?
Boitumelo
for grade 12 or grade 11?
Sibulele
the value of V1 and V2
Tumelo Reply
advantages of electrons in a circuit
Rethabile Reply
we're do you find electromagnetism past papers
Ntombifuthi
what a normal force
Tholulwazi Reply
it is the force or component of the force that the surface exert on an object incontact with it and which acts perpendicular to the surface
Sihle
what is physics?
Petrus Reply
what is the half reaction of Potassium and chlorine
Anna Reply
how to calculate coefficient of static friction
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how to calculate static friction
Lisa
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Tumelo
how to calculate the magnitude of horizontal component of the applied force
Mogano
How to calculate force
Monambi
a structure of a thermocouple used to measure inner temperature
Anna Reply
a fixed gas of a mass is held at standard pressure temperature of 15 degrees Celsius .Calculate the temperature of the gas in Celsius if the pressure is changed to 2×10 to the power 4
Amahle Reply
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Raymond Reply
what is acceleration
Syamthanda Reply
a rate of change in velocity of an object whith respect to time
Khuthadzo
how can we find the moment of torque of a circular object
Kidist
Acceleration is a rate of change in velocity.
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t =r×f
Khuthadzo
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Precious Reply
hi
Shongi
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Leago
use fnet method. how many obects are being calculated ?
Khuthadzo
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Hulisani
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Lungile Reply
you use Fnet equals ma , newtoms second law formula
Masego
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Mulaudzi Reply
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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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