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The class named Transform

Listing 2 presents the beginning of the class named Transform . Listing 2 also presents the beginning of an instance method of that class named doIt . The doIt method computes and returns the complex transform (via output parameters) of an incoming complex series.

Listing 2. The class named Transform.
class Transform{ void doIt(double[]realIn, double[]imagIn, double scale,double[] realOut,double[] imagOut){

The method parameters

The doIt method receives five incoming parameters. The first two parameters are references to two array objects of type double containing the real and imaginary parts of the input series.

The third parameter is a scale factor that is applied to the transform output in an attempt to keep the values in a range suitable for plotting if desired.

The last two parameters are references to array objects of type double . The results of performing the transform are used to populate these two arrays. This is the mechanism by which the object returns thetransform results to the calling program. It is assumed that all of the elements in these two array objects contain values of zero upon entry to the doIt method.

Performing the transform

The body of the doIt method is presented in Listing 3 . The code in Listing 3 iterates on the input arrays, passing each complex sample contained in those two arrays to a method named correctAndRecombine .

Listing 3. Performing the transform.
for(int cnt = 0;cnt<realIn.length;cnt++){ correctAndRecombine(realIn[cnt], imagIn[cnt], cnt,realIn.length, scale,realOut, imagOut);}//end for loop }//end doIt

The transforms of the complex input samples

Each complex value in the incoming arrays represents both a complex sample and the transform of that complex sample under the assumption that the complexsample appears at the origin of the input series.

Correct for actual position and recombine

The method named correctAndRecombine corrects the transform result for each of the complex samples in the series so as to reflect the actualposition of the complex sample in the original input series.

Then the method named correctAndRecombine adds the corrected transform result into a pair of accumulators, one for the real part and one forthe imaginary part. This accomplishes the recombination of the corrected transforms of the input samples in order to produce the transform of the entireoriginal complex input series.

The correctAndRecombine method

The correctAndRecombine method is shown in Listing 4 . Listing 4 also signals the end of the Transform class.

Listing 4. The correctAndRecombine method.
void correctAndRecombine(double realSample, double imagSample,int position, int length,double scale, double[]realOut, double[]imagOut){ //Calculate the complex transform values for// each sample in the complex output series. for(int cnt = 0; cnt<length; cnt++){ double angle =(2.0*Math.PI*cnt/length)*position; //Calculate output based on real inputrealOut[cnt] +=realSample*Math.cos(angle)/scale; imagOut[cnt]+= realSample*Math.sin(angle)/scale;//Calculate output based on imag input realOut[cnt]-= imagSample*Math.sin(angle)/scale;imagOut[cnt] +=imagSample*Math.cos(angle)/scale; }//end for loop}//end correctAndRecombine }//end class transform

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
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sure. what is your question?
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I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
I'm not sure why it wrote it the other way
I got X =-6
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
is it a question of log
Commplementary angles
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Kevin Reply
a perfect square v²+2v+_
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
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or infinite solutions?
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
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I start with an easy one. carbon nanotubes woven into a long filament like a string
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
anybody can imagine what will be happen after 100 years from now in nano tech world
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
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how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
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I'm interested in Nanotube
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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how did you get the value of 2000N.What calculations are needed to arrive at it
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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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