# 4.5 Java1486-fun with java, understanding the fast fourier transform  (Page 7/14)

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The shape of the imaginary part of the transform is an upside down sine curve.

## Number of output samples equals number of input samples

This transform program computes real and imaginary values from zero to an output index that is one output sample interval less than the samplingfrequency. The number of output values is equal to the number of samples in the input series. This is very typical of FFT algorithms.

In this case, I set the applet up to accept sixteen input samples and to produce sixteen output samples.

## Representing time and frequency

For the moment, lets think in terms of time and frequency. Assume that the input series f(x) is a time series and the output series F(k) is a frequencyspectrum.

To make the arithmetic easy, let's assume that the sampling interval for the input time series in the upper left box of Figure 5 is one second. This gives a sampling frequency of one sample per second, and a total elapsed time of sixteenseconds.

The sine and cosine curves in Figure 5 each go through one complete period between a frequency of zero and the sampling frequency, which one sample persecond. Thus, the period of the sine and cosine curves along the frequency axis is one sample per second. This is the reciprocal of the time shift of one sampleinterval at a sampling frequency of one sample per second.

Stated differently, the number of periods of the sine and cosine curves in the real and imaginary parts of the transform between a frequency of zero and afrequency equal to the sampling frequency is equal to the shift in sample intervals. A shift of one sample interval produces sine and cosine curves havingone period in the frequency range from zero to the sampling frequency. A shift of two sample intervals produces sine and cosine curves having two periods inthe frequency range from zero to the sampling frequency, etc. This is illustrated by Figure 6 .

## A shift of two sample intervals

Figure 6 shows the transform of an impulse with a shift equal to two sample intervals and a positive value.

Figure 6. Transform of an impulse with a shift equal to two sample intervals and a positive value.

The real part of the transform has the shape of a cosine curve with two complete periods between zero and an output index equal to the samplingfrequency.

The imaginary part of the transform has the shape of a sine curve with two complete periods within the same output interval. This agrees with theconclusions stated in the previous section.

## A shift of four sample intervals

Finally, Figure 7 shows the transform of an impulse with a shift equal to four sample intervals.

Figure 7. Transform of an impulse with a shift equal to four sample intervals and a positive value.

The cosine and sine curves that represent the real and imaginary parts of the transform each have four complete periods between zero and an output index equalto the sampling frequency.

In this case the cosine and sine curves are very sparsely sampled.

## Equations to describe the real and imaginary parts of the transform

The main point is:

find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
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
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
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?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
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Sherica
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Sherica
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Tamia
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Uday
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salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
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if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
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Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
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Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
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Porter
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Yasmin
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Cesar
I'm interested in nanotube
Uday
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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Stotaw
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Azam
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Prasenjit
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Azam
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Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
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.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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