# Classle

## What is the difference between DTFT and DFT?

Suppose you have discrete signal $x[n], n=0..N-1$ and you need to decompose it in the sum of $N$sine waves (different amplitudes and phase) (again discrete and again having $N$ points) with periods from $N$ points up to 2 points + constant discrete value signal (because our original signal could have a shift in value relative to 0) Think for this only as a mathematical problem. This task is solved by DFT. This is what FFT (algorithm for DFT calculation) solves in the computers.

$X[k]=\Bigsum_{n=0}^{N-1}x[n] e^{\frac{-j 2 \pi n k}{N}}$

Now about DTFT. Suppose you have analog real world signal. All the signals in real world are infinite in time.
Then you sample this signal with analog sampling function $\tilde{x(t)}=\Bigsum_{k=-\infty}^{+\infty}x(t)\delta(t-kT)$ where $T$ is a time sampling step and $\delta(t)$ is the infinity high and infinity short peak at the origin (Don’t confuse it with $\delta[n]$ which is 1 at the origin). Then you take the normal (analog) Fourier
transform of the sampled version as it is shown below. That is all about DTFT.

$\tilde{X(\omega)} =\Bigint_{-\infty}^{+\infty} \tilde{x(t)} e^{-j \omega t} \, dt$

Note that there is relation between DFT and DTFT if we made some assumption about analog signal above.
Which actually makes the DFT (FFT actually) useful for calculation (with computer) of the real world (analog) problems.

Answered by: AnanthKK     On: 04-Apr-2011

Suppose you have discrete signal $x[n], n=0..N-1$ and you need to decompose it in the sum of $N$sine waves (different amplitudes and phase) (again discrete and again having $N$ points) with periods from $N$ points up to 2 points + constant discrete value signal (because our original signal could have a shift in value relative to 0) Think for this only as a mathematical problem. This task is solved by DFT. This is what FFT (algorithm for DFT calculation) solves in the computers.

$X[k]=\Bigsum_{n=0}^{N-1}x[n] e^{\frac{-j 2 \pi n k}{N}}$

Now about DTFT. Suppose you have analog real world signal. All the signals in real world are infinite in time.
Then you sample this signal with analog sampling function $\tilde{x(t)}=\Bigsum_{k=-\infty}^{+\infty}x(t)\delta(t-kT)$ where $T$ is a time sampling step and $\delta(t)$ is the infinity high and infinity short peak at the origin (Don’t confuse it with $\delta[n]$ which is 1 at the origin). Then you take the normal (analog) Fourier
transform of the sampled version as it is shown below.
That is all about DTFT.

$\tilde{X(\omega)} =\Bigint_{-\infty}^{+\infty} \tilde{x(t)} e^{-j \omega t} \, dt$

Note that there is relation between DFT and DTFT if we made some assumption about analog signal above.
Which actually makes the DFT (FFT actually) useful for calculation (with computer) of the real world (analog) problems.

Note that there is relation between DFT and DTFT if we made some assumption about analog signal above.
Which actually makes the DFT (FFT actually) useful for calculation (with computer) of the real world (analog) problems.

Now about DTFT. Suppose you have analog real world signal. All the signals in real world are infinite in time.
Then you sample this signal with analog sampling function $\tilde{x(t)}=\Bigsum_{k=-\infty}^{+\infty}x(t)\delta(t-kT)$ where $T$ is a time sampling step and $\delta(t)$ is the infinity high and infinity short peak at the origin (Don’t confuse it with $\delta[n]$ which is 1 at the origin). Then you take the normal (analog) Fourier
transform of the sampled version as it is shown below.
That is all about DTFT.

$\tilde{X(\omega)} =\Bigint_{-\infty}^{+\infty} \tilde{x(t)} e^{-j \omega t} \, dt$

Note that there is relation between DFT and DTFT if we made some assumption about analog signal above.
Which actually makes the DFT (FFT actually) useful for calculation (with computer) of the real world (analog) problems.

$X[k]=\Bigsum_{n=0}^{N-1}x[n] e^{\frac{-j 2 \pi n k}{N}}$

Now about DTFT. Suppose you have analog real world signal. All the signals in real world are infinite in time.
Then you sample this signal with analog sampling function $\tilde{x(t)}=\Bigsum_{k=-\infty}^{+\infty}x(t)\delta(t-kT)$ where $T$ is a time sampling step and $\delta(t)$ is the infinity high and infinity short peak at the origin (Don’t confuse it with $\delta[n]$ which is 1 at the origin). Then you take the normal (analog) Fourier
transform of the sampled version as it is shown below.
That is all about DTFT.

$\tilde{X(\omega)} =\Bigint_{-\infty}^{+\infty} \tilde{x(t)} e^{-j \omega t} \, dt$

Note that there is relation between DFT and DTFT if we made some assumption about analog signal above.
Which actually makes the DFT (FFT actually) useful for calculation (with computer) of the real world (analog) problems.

Suppose you have discrete signal $x[n], n=0..N-1$ and you need to decompose it in the sum of $N$sine waves (different amplitudes and phase) (again discrete and again having $N$ points)

with periods from $N$ points up to 2 points + constant discrete value signal (because our original signal could have a shift in value relative to 0)

Think for this only as a mathematical problem.This task is solved by DFT. This is what FFT (algorithm for DFT calculation) solves in the computers.

$X[k]=\Bigsum_{n=0}^{N-1}x[n] e^{\frac{-j 2 \pi n k}{N}}$

Now about DTFT. Suppose you have analog real world signal. All the signals in real world are infinite in time.
Then you sample this signal with analog sampling function $\tilde{x(t)}=\Bigsum_{k=-\infty}^{+\infty}x(t)\delta(t-kT)$ where $T$ is a time sampling step and $\delta(t)$ is the infinity high and infinity short peak at the origin (Don’t confuse it with $\delta[n]$ which is 1 at the origin). Then you take the normal (analog) Fourier
transform of the sampled version as it is shown below.
That is all about DTFT.

$\tilde{X(\omega)} =\Bigint_{-\infty}^{+\infty} \tilde{x(t)} e^{-j \omega t} \, dt$

Note that there is relation between DFT and DTFT if we made some assumption about analog signal above.
Which actually makes the DFT (FFT actually) useful for calculation (with computer) of the real world (analog) problems.

Suppose you have discrete signal and you need to decompose it in the sum of sine waves (different amplitudes and phase) (again discrete and again having points)

with periods from points up to 2 points + constant discrete value signal (because our original signal could have a shift in value relative to 0)

Think for this only as a mathematical problem.This task is solved by DFT. This is what FFT (algorithm for DFT calculation) solves in the computers.

Now about DTFT. Suppose you have analog real world signal. All the signals in real world are infinite in time.
Then you sample this signal with analog sampling function where $T$ is a time sampling step and is the infinity high and infinity short peak at the origin (Don’t confuse it with which is 1 at the origin). Then you take the normal (analog) Fourier
transform of the sampled version as it is shown below.
That is all about DTFT.

Note that there is relation between DFT and DTFT if we made some assumption about analog signal above.
Which actually makes the DFT (FFT actually) useful for calculation (with computer) of the real world (analog) problems.

Answered by: Tarapriya     On: 14-Mar-2012

Suppose you have discrete signal and you need to decompose it in the sum of sine waves (different amplitudes and phase) (again discrete and again having points)

with periods from points up to 2 points + constant discrete value signal (because our original signal could have a shift in value relative to 0)

Think for this only as a mathematical problem.This task is solved by DFT. This is what FFT (algorithm for DFT calculation) solves in the computers.

Now about DTFT. Suppose you have analog real world signal. All the signals in real world are infinite in time.
Then you sample this signal with analog sampling function where is a time sampling step and is the infinity high and infinity short peak at the origin (Don’t confuse it with <img alt="\delta[n]" "="" title="\delta[n]" align="absmiddle"> which is 1 at the origin). Then you take the normal (analog) Fourier<br> transform of the sampled version as it is shown below.</span><span class="postbody"> That is all about DTFT.<br> <br> <img alt="\tilde{X(\omega)} =\Bigint_{-\infty}^{+\infty} \tilde{x(t)} e^{-j \omega t} \, dt " title="\tilde{X(\omega)} =\Bigint_{-\infty}^{+\infty} \tilde{x(t)} e^{-j \omega t} \, dt " align="absmiddle"><br> </span></p> <p> <span class="postbody">Note that there is relation between DFT and DTFT if we made some assumption about analog signal above.<br> Which actually makes the DFT (FFT actually) useful for calculation (with computer) of the real world (analog) problems.<br> </span></p> <p></p>

Answered by: Nithy09cs12     On: 27-Jun-2012

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Mon, 04/04/2011 - 03:35