Fdtxjr

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Discrete-time Fourier transform of $x[n] = 1$

1

$begingroup$

How to calculate the DTFT of $1$? The sequence $x[n] = 1$ is not absolutely summable, so one can not compute the DTFT by using the definition

$$X(Omega) space = sum limits_n=-infty^inftyx[n]e^-jOmega n$$

Can anyone point me to a derivation of DTFT of $1$ from the first principles?

The derivations I came across used the fact that DTFT of $e^jOmega_0n$ is

$$2pisum limits_k=-infty^infty delta(Omega-Omega_0-2kpi)$$

which again brings me to the question: how?

fourier-analysis fourier-transform signal-processing

share|cite|improve this question

edited Mar 24 at 23:30

Navin

asked Mar 24 at 7:47

NavinNavin

266

$endgroup$

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  $begingroup$
  Rodrigo de Azevedo, thanks for the edit! Yes, I'm aware that it should be an impulse train and computing the IDFT will prove it. My quest however is to compute the DTFT and get the result as the impulse train. I want to deduce it mathematically without guessing its DTFT and then taking IDFT to prove it.
  $endgroup$
  – Navin
  Mar 24 at 10:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The DTFT of the complex exponential is not the result of a computation (as the DTFT series does not converge, of course). It is just a convenient definition, which we take for granted. It is intuitively justified because it gives the expected result when applying the IDTFT formula. Other than that, there is nothing more to say.
  $endgroup$
  – Stelios
  Mar 24 at 12:21
  

  
  
  
  

add a comment | 

1

$begingroup$

How to calculate the DTFT of $1$? The sequence $x[n] = 1$ is not absolutely summable, so one can not compute the DTFT by using the definition

$$X(Omega) space = sum limits_n=-infty^inftyx[n]e^-jOmega n$$

Can anyone point me to a derivation of DTFT of $1$ from the first principles?

The derivations I came across used the fact that DTFT of $e^jOmega_0n$ is

$$2pisum limits_k=-infty^infty delta(Omega-Omega_0-2kpi)$$

which again brings me to the question: how?

fourier-analysis fourier-transform signal-processing

share|cite|improve this question

edited Mar 24 at 23:30

Navin

asked Mar 24 at 7:47

NavinNavin

266

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Rodrigo de Azevedo, thanks for the edit! Yes, I'm aware that it should be an impulse train and computing the IDFT will prove it. My quest however is to compute the DTFT and get the result as the impulse train. I want to deduce it mathematically without guessing its DTFT and then taking IDFT to prove it.
  $endgroup$
  – Navin
  Mar 24 at 10:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The DTFT of the complex exponential is not the result of a computation (as the DTFT series does not converge, of course). It is just a convenient definition, which we take for granted. It is intuitively justified because it gives the expected result when applying the IDTFT formula. Other than that, there is nothing more to say.
  $endgroup$
  – Stelios
  Mar 24 at 12:21
  

  
  
  
  

add a comment | 

$begingroup$

How to calculate the DTFT of $1$? The sequence $x[n] = 1$ is not absolutely summable, so one can not compute the DTFT by using the definition

$$X(Omega) space = sum limits_n=-infty^inftyx[n]e^-jOmega n$$

Can anyone point me to a derivation of DTFT of $1$ from the first principles?

The derivations I came across used the fact that DTFT of $e^jOmega_0n$ is

$$2pisum limits_k=-infty^infty delta(Omega-Omega_0-2kpi)$$

which again brings me to the question: how?

fourier-analysis fourier-transform signal-processing

share|cite|improve this question

edited Mar 24 at 23:30

Navin

asked Mar 24 at 7:47

NavinNavin

266

$endgroup$

share|cite|improve this question

edited Mar 24 at 23:30

Navin

asked Mar 24 at 7:47

NavinNavin

266

share|cite|improve this question

edited Mar 24 at 23:30

Navin

asked Mar 24 at 7:47

NavinNavin

266

asked Mar 24 at 7:47

NavinNavin

266

asked Mar 24 at 7:47

NavinNavin

266