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Figure 10. The numeric output for Case A.
Case A Real:1.0 0.923 0.707 0.382 0.0 -0.382 -0.707 -0.923 -1.0 -0.923 -0.707 -0.382 0.0 0.382 0.707 0.923imag: 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.00.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

If you plot the real and imaginary values in Figure 10 , you will see that they match the transform output shown in graphic form in Figure 9 .

Case B code

The code from the main method for Case B is shown in Listing 6 . Note that the input complex series contains non-zero values in both the real and imaginaryparts.

Listing 6. Case B code.
System.out.println("\nCase B"); double[]realInB = {0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1};double[] imagInB ={0,-1,0,0,0,0,0,0,0,0,0,0,0,0,0,-1}; double[]realOutB = new double[16];double[] imagOutB = new double[16]; transform.doIt(realInB,imagInB,2.0,realOutB,imagOutB); display(realOutB,imagOutB);

Case B in graphical form

Case B is shown in graphical form in Figure 11 .

Figure 11. Case B in graphical form.
missing image

Case B output in numeric form

The output from the code in Listing 6 is shown in Figure 12 .

Figure 12. Case B output in numeric form.
Case B Real:1.0 0.923 0.707 0.382 0.0 -0.382 -0.707 -0.923 -0.999 -0.923 -0.707 -0.382 0.0 0.382 0.707 0.923imag: -1.0 -0.923 -0.707 -0.382 0.0 0.382 0.707 0.9231.0 0.923 0.707 0.382 0.0 -0.382 -0.707 -0.923

If you plot the values for the real and imaginary parts from Figure 12 , you will see that they match the real and imaginary output shown in Figure 11 .

Case C code

The code extracted from the main method for Case C is shown in Listing 7 .

Listing 7. Case C code.
System.out.println("\nCase C"); double[]realInC = {1.0,0.923,0.707,0.382,0.0,-0.382,-0.707,-0.923,-1.0,-0.923,-0.707,-0.382,0.0, 0.382,0.707,0.923};double[] imagInC ={0.0,-0.382,-0.707,-0.923,-1.0,-0.923, -0.707,-0.382,0.0,0.382,0.707,0.923,1.0,0.923,0.707,0.382}; double[]realOutC = new double[16];double[] imagOutC = new double[16]; transform.doIt(realInC,imagInC,16.0,realOutC,imagOutC); display(realOutC,imagOutC);

The complex input series for Case C is a little more complicated than that for either of the previous two cases. Note in particular that the input complexseries contains non-zero values in both the real and imaginary parts. In addition, very few of the values in the complex series have a value of zero.

(The values of the complex samples actually describe a cosine curve and a negative sine curve as shown in Figure 13 .)

The graphic form of Case C

Case C is shown in graphic form in Figure 13 .

Figure 13. The graphic form of Case C.
missing image

The Fourier transform is reversible

One of the interesting things to note about Figure 13 is the similarity of Figure 13 and Figure 5 . These two figures illustrate the reversible nature of the Fourier transform.

If I had used a positive input real value instead of a negative input real value in Figure 5 , the input of Figure 5 would look exactly like the output in Figure 13 , and the output of Figure 5 would look exactly like the input of Figure 13 .

With that as a hint, you should now be able to figure out how I used a mouse and drew the perfect sine and cosine curves in Figure 13 . In fact, I didn't draw them at all. Rather, I used my mouse and drew the output, andthe applet gave me the corresponding input automatically.

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
Lisa Reply
how to calculate static friction
Lisa
How to calculate a current
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
How is energy being used in bonding?
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.
Justice
t =r×f
Khuthadzo
how to calculate tension by substitution
Precious Reply
hi
Shongi
hi
Leago
use fnet method. how many obects are being calculated ?
Khuthadzo
khuthadzo hii
Hulisani
how to calculate acceleration and tension force
Lungile Reply
you use Fnet equals ma , newtoms second law formula
Masego
please help me with vectors in two dimensions
Mulaudzi Reply
how to calculate normal force
Mulaudzi
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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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