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As the frequency of the sine wave continues to increase beyond that point, without a corresponding change in the sampling frequency, it is impossible todetermine from the samples so obtained whether the frequency is increasing or decreasing.

An ambiguity in the spectrum analysis

As a result, the spectrum analysis process was unable to determine if the peak in the frequency spectrum was below or above the folding frequency. Thus,the bottom trace in Figure 3 shows two peaks which are mirror images of one another with the folding frequency being half way between the two peaks.

(As a practical matter, when doing spectrum analysis, there is no point in computing the values above the folding frequency. I did that here just toillustrate that there is a folding frequency, which is equal to one-half the sampling frequency.)

Let's see some code

The class used to produce the data displayed in Figure 3 is named Dsp002 . A complete listing of this class definition is shown in Listing 38 near the end of the module.

I will break this class down into fragments and briefly discuss it to show how you can define significant classes and easily connect them to the generalizedplotting program named Graph01 .

As before, having compiled the class named Dsp002 , you would exercise it by entering the following at a command prompt:

java Graph01 Dsp002

Different from the previous example class

This class differs from the class named Graph01Demo in one very significant way. In that class, all the values returned by the methods named f1 through f5 were computed on the fly as the methods were called.

In this new class named Dsp002 , all the data is generated and stored in array objects when an object of the class named Dsp002 is instantiated. When the methods named f1 through f5 are called later, they simply retrieve the data from the array objects and return that data to the plotting program.

Basic operation of the program

As mentioned earlier, this program applies a narrow-band convolution filter to white noise, and then computes the amplitude spectrum of the filtered noiseusing a Discrete Fourier Transform (DFT) algorithm. The spectrum of the white noise is also computed. All of the processing occurs when an object of the classis instantiated, and the processed results are saved in arrays.

The input noise, the filter, the filtered output, and the two spectra are deposited in five arrays for later retrieval and display. The data in the fivearrays are returned by the methods named f1 , f2 , f3 , f4 , and f5 respectively.

The values that are returned by the methods are scaled for appropriate display in the plotting areas provided by the program named Graph01 .

Beginning of the class named Dsp002

The code in Listing 9 establishes the data lengths for the white noise, the convolution filter, the filtered output, and the spectrum.

Listing 9. Beginning of the class named Dsp002.
class Dsp002 implements GraphIntfc01{ int operatorLen = 33;int dataLen = 256+operatorLen; int outputLen = dataLen - operatorLen;int spectrumPts = outputLen;

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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