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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.
missing image

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.
missing image

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:

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
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I know this work
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
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can someone help me with some logarithmic and exponential equations.
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I'm not sure why it wrote it the other way
I got X =-6
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Commplementary angles
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what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
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a perfect square v²+2v+_
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Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
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
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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
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how do you translate this in Algebraic Expressions
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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I start with an easy one. carbon nanotubes woven into a long filament like a string
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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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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
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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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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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