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It is a significant concept in signal processing that there is more than one way to express the information in a given signal. One way is by defining the signal in terms of the value it has at any given time $n$, thus the notation $x[n]$. But is also possible to express a signal in terms of how it is be a combination of signals from a particular signal set. We could express the information in the signal by noting how much of each signal in the signal set contributes in the combination.

Definition of the dtft

The discrete-time Fourier transform , or DTFT, is a common way to express infinite-length discrete-time signals in another way. As noted above, the signal $x[n]$ can be understood as an entity as having a particular value for every time $n$ in the range $-\infty\lt n\lt \infty$. But it also can be understood in this way:$x[n]=\frac{1}{2\pi}\int\limits_{-\pi}^{\pi}X(\omega)e^{-j\omega n}d\omega$$x[n]$ is expressed in terms of how much of each signal $e^{-j\omega n}$ goes in to the integral which composes it. The weighting of each signal is according to the function $X(\omega)$. The values of that function are found according to this formula:$X(\omega)=\sum_{n=-\infty}^{\infty} x[n]\, e^{-j \omega n}$This formula for $X(\omega)$ is known as the DTFT of $x[n]$, while that first integral formula is known as the inverse DTFT.

Interpretations of the dtft

We have given above the formulas for the DTFT and the inverse DTFT: --$X(\omega)=\sum_{n=-\infty}^{\infty} x[n]\, e^{-j \omega n}$ --$x[n]=\frac{1}{2\pi}\int\limits_{-\pi}^{\pi}X(\omega)e^{-j\omega n}d\omega$ But what do these formulas mean? There are a few ways of understanding them. One is to say that the two formulas are expressing the same entity, but in different domains: $x[n]$ is the time-domain representation of the signal, whereas $X(\omega)$ is the frequency-domain representation of it. In that sense, it is like two different language translations of the same book (this analogy is not perfect, for translations are not exactly one-to-one; there are many English-language translations of the Spanish book Don Quixote, but there is only one frequency-domain representation $X(\omega)$ for a time-domain signal $x[n]$).A second way to understand the DTFT is in terms of the concepts of analysis and synthesis. The DTFT formula gives us a frequency analysis of the signal $x[n]$. The signal $X(\omega)$ analyzes $x[n]$ from a different point of view; it tells us the frequency content of $x[n]$, whether it is composed of only low frequencies, or just high ones, or whatever the case may be. This is because every value of $X(\omega)$ is simply the inner product of $x[n]$ with $e^{j\omega n}$; it tells us the "strength" of $e^{j\omega n}$ in the signal $x[n]$. The inverse DTFT then synthesizes, or composes, $x[n]$ based upon its weighting at each different frequency.

Relationship between the dtft and the dft

The DTFT is not the only Fourier transform for discrete-time signals. While the DTFT operates on infinite-length signals, the DFT (discrete Fourier transform) is a tool for frequency analysis of finite-length signals. A finite-length signal (say, of length $N$) $x[n]$ has an $N$ length frequency representation $X[k]$, according to these formulas: $X[k]=\sum\limits_{n=0}^{N-1}x[n]e^{-j\frac{2\pi}{N}kn}$$x[n]=\frac{1}{N}\sum\limits_{k=0}^{N-1}X[k]e^{j\frac{2\pi}{N}kn}$ As $x[n]$ and $X[k]$ are N-periodic, we can also represent them in this way, shifted about $n$ and $k$ at the origin:$X[k]=\sum\limits_{n=-\frac{N}{2}}^{\frac{N}{2}-1}x[n]e^{-j\frac{2\pi}{N}kn}$ $x[n]=\frac{1}{N}\sum\limits_{k=-\frac{N}{2}}^{\frac{N}{2}-1}X[k]e^{j\frac{2\pi}{N}kn}$We can develop the DTFT by seeing what happens to the DFT as $N$ gets larger and larger. As $N$ approaches $\infty$, so also do $n$ and $k$ run from $\infty$ to $\infty$. Consider also what happens to $e^{-j\frac{2\pi}{N}kn}$. As $N$ gets infinitely large, the exponent term $\frac{2\pi}{N}k$ becomes a continuous value, running from $-\pi$ to $\pi$, and we can call that value $\omega$. We have then that the DFT sum becomes: $X(\omega)=\sum\limits_{n=-\infty}^{\infty}x[n]e^{-j\omega n}$ which is the definition of the DTFT.Similarly, on the synthesis side of the transform, the DFT "sum" will not run over a discrete number of values $k$, but rather become an integral over the continuous variable $\omega$: $x[n]=\int\limits_{-\pi}^{\pi}X(\omega)e^{j\omega n}\frac{d\omega}{2\pi}$ which is our inverse DTFT.Graphically, we can see how increasing the length of a signal by adding more and more zeros on either side of it results in the signal's DFT (which is expressed in terms of $k$) appearing to be more and more like a continuous function (expressed in terms of $\omega$):
Image

Questions & Answers

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Lukman Reply
The denominator of a certain fraction is 9 more than the numerator. If 6 is added to both terms of the fraction, the value of the fraction becomes 2/3. Find the original fraction. 2. The sum of the least and greatest of 3 consecutive integers is 60. What are the valu
SABAL Reply
1. x + 6 2 -------------- = _ x + 9 + 6 3 x + 6 3 ----------- x -- (cross multiply) x + 15 2 3(x + 6) = 2(x + 15) 3x + 18 = 2x + 30 (-2x from both) x + 18 = 30 (-18 from both) x = 12 Test: 12 + 6 18 2 -------------- = --- = --- 12 + 9 + 6 27 3
Pawel
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Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113?
mariel Reply
Mark = x,. Don = 3x + 1 x + 3x + 1 = 113 4x = 112, x = 28 Mark = 28, Don = 85, 28 + 85 = 113
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4
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x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
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Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
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(61/11,41/11,−4/11)
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x=61/11 y=41/11 z=−4/11 x=61/11 y=41/11 z=-4/11
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Need help solving this problem (2/7)^-2
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x+2y-z=7
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-1
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A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place.
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Jeannette has $5 and $10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
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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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Source:  OpenStax, Discrete-time signals and systems. OpenStax CNX. Oct 07, 2015 Download for free at https://legacy.cnx.org/content/col11868/1.2
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