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Fourier transform of delta function

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0

$begingroup$

I would like to take the Fourier transform of that:

$$x(tau ,t)=delta (tau -frac14t)+frac14delta (tau -T)$$

for $0leq tleq T$

My solution: I took the FT (first dirac function)+ FT( the sevond).
I found the first FT, but I Have a problem with second.
What i should to do with T?

$x(tau ,t)$ is given in $0leq tleq T$. Does it mean that T in this case is t, variable/

does it mean that if i would like to take the inverse transform later, T will be integration variable?

fourier-transform dirac-delta

share|cite|improve this question

edited 16 hours ago

Nani

asked 18 hours ago

NaniNani

11

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That's not a function, that's a measure. Please stop confusing those. FT of $delta_a$ is $f(t) = e^iat$
  $endgroup$
  – Jakobian
  18 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you seeking the FT of $x(tau,t)times (H(t)-H(t-T))$ where $H$ is the Heaviside (aka, unit step) function?
  $endgroup$
  – Mark Viola
  17 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sure you dont mean Furrier Trasorm?
  $endgroup$
  – mathreadler
  17 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Dirac $delta$ is a distribution, not a function in normal sense. It does not take values, it only produces values when integrated together with another function.
  $endgroup$
  – mathreadler
  17 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

I would like to take the Fourier transform of that:

$$x(tau ,t)=delta (tau -frac14t)+frac14delta (tau -T)$$

for $0leq tleq T$

My solution: I took the FT (first dirac function)+ FT( the sevond).
I found the first FT, but I Have a problem with second.
What i should to do with T?

$x(tau ,t)$ is given in $0leq tleq T$. Does it mean that T in this case is t, variable/

does it mean that if i would like to take the inverse transform later, T will be integration variable?

fourier-transform dirac-delta

share|cite|improve this question

edited 16 hours ago

Nani

asked 18 hours ago

NaniNani

11

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That's not a function, that's a measure. Please stop confusing those. FT of $delta_a$ is $f(t) = e^iat$
  $endgroup$
  – Jakobian
  18 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you seeking the FT of $x(tau,t)times (H(t)-H(t-T))$ where $H$ is the Heaviside (aka, unit step) function?
  $endgroup$
  – Mark Viola
  17 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sure you dont mean Furrier Trasorm?
  $endgroup$
  – mathreadler
  17 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Dirac $delta$ is a distribution, not a function in normal sense. It does not take values, it only produces values when integrated together with another function.
  $endgroup$
  – mathreadler
  17 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I would like to take the Fourier transform of that:

$$x(tau ,t)=delta (tau -frac14t)+frac14delta (tau -T)$$

for $0leq tleq T$

My solution: I took the FT (first dirac function)+ FT( the sevond).
I found the first FT, but I Have a problem with second.
What i should to do with T?

$x(tau ,t)$ is given in $0leq tleq T$. Does it mean that T in this case is t, variable/

does it mean that if i would like to take the inverse transform later, T will be integration variable?

fourier-transform dirac-delta

share|cite|improve this question

edited 16 hours ago

Nani

asked 18 hours ago

NaniNani

11

$endgroup$

share|cite|improve this question

edited 16 hours ago

Nani

asked 18 hours ago

NaniNani

11

share|cite|improve this question

edited 16 hours ago

Nani

asked 18 hours ago

NaniNani

11

asked 18 hours ago

NaniNani

11

asked 18 hours ago

NaniNani

11