Fdtxjr

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Fourier transform of exponential function

0

1

$begingroup$

It is a function $A(f,t)= e^j 2 pi (t/a-af)$
I would like to take the inverse Fourier transform.

So:

$B(tau,t)=int A(f,t) e^j2pi ftaudf=delta(tau-t_0)...???$

How to solve this integral?
Or should I use $e^j 2 pi (t/a)$ as constant?

exponential-function inverse fourier-transform

share|cite|improve this question

edited Mar 20 at 9:14

Nani

asked Mar 20 at 8:03

NaniNani

33

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The delta function expression is right, what is the reason that you doubt that?
  $endgroup$
  – Aminopterin
  Mar 20 at 8:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Aminopterin i am not sure how to solve this integral. If i take that exp(i2 pi (t/a) is constant, shoud I multiply with delta function ( delta(t) or delta (tau)) as fourier transform of constant?
  $endgroup$
  – Nani
  Mar 20 at 9:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, Fourier transform is linear, you should multiply the constant.
  $endgroup$
  – Aminopterin
  Mar 20 at 9:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Aminopterin i have done the calculation, but if i take fourier transform of result I didnt get the same function
  $endgroup$
  – Nani
  Mar 20 at 9:26
  

  
  
  
  

add a comment | 

0

1

$begingroup$

It is a function $A(f,t)= e^j 2 pi (t/a-af)$
I would like to take the inverse Fourier transform.

So:

$B(tau,t)=int A(f,t) e^j2pi ftaudf=delta(tau-t_0)...???$

How to solve this integral?
Or should I use $e^j 2 pi (t/a)$ as constant?

exponential-function inverse fourier-transform

share|cite|improve this question

edited Mar 20 at 9:14

Nani

asked Mar 20 at 8:03

NaniNani

33

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The delta function expression is right, what is the reason that you doubt that?
  $endgroup$
  – Aminopterin
  Mar 20 at 8:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Aminopterin i am not sure how to solve this integral. If i take that exp(i2 pi (t/a) is constant, shoud I multiply with delta function ( delta(t) or delta (tau)) as fourier transform of constant?
  $endgroup$
  – Nani
  Mar 20 at 9:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, Fourier transform is linear, you should multiply the constant.
  $endgroup$
  – Aminopterin
  Mar 20 at 9:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Aminopterin i have done the calculation, but if i take fourier transform of result I didnt get the same function
  $endgroup$
  – Nani
  Mar 20 at 9:26
  

  
  
  
  

add a comment | 

$begingroup$

It is a function $A(f,t)= e^j 2 pi (t/a-af)$
I would like to take the inverse Fourier transform.

So:

$B(tau,t)=int A(f,t) e^j2pi ftaudf=delta(tau-t_0)...???$

How to solve this integral?
Or should I use $e^j 2 pi (t/a)$ as constant?

exponential-function inverse fourier-transform

share|cite|improve this question

edited Mar 20 at 9:14

Nani

asked Mar 20 at 8:03

NaniNani

33

$endgroup$

share|cite|improve this question

edited Mar 20 at 9:14

Nani

asked Mar 20 at 8:03

NaniNani

33

share|cite|improve this question

edited Mar 20 at 9:14

Nani

asked Mar 20 at 8:03

NaniNani

33

asked Mar 20 at 8:03

NaniNani

33

asked Mar 20 at 8:03

NaniNani

33

2 Answers
2

active

oldest

votes

0

$begingroup$

Just recognize that, I think,
$$
mathcalF^-1_f to t' exp left[ 2pi j left( fracta-af right) right]
= e^2pi j t/a delta ( t'-a )
$$
if $t$ does not depend on $f$.
Also, there are several different conventions on FT, you have to specify, but usually they differ up to a constant multiple. (See Wikipedia formula 102, 302)

share|cite|improve this answer

edited Mar 20 at 11:03

answered Mar 20 at 10:55

AminopterinAminopterin

1,089213

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how do you get delta function in the answer? I got exp(j pi f (t´-a).
  $endgroup$
  – Nani
  Mar 20 at 11:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you seen the Wiki table formula 102, 302?
  $endgroup$
  – Aminopterin
  Mar 20 at 12:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i have seen, but it is for fourier transform, in my case is inverse FT. I don`t understand how do you and Mostafa Ayaz get delta function.
  $endgroup$
  – Nani
  Mar 20 at 12:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I got the same answer with you, but instead ofdelta i got exp
  $endgroup$
  – Nani
  Mar 20 at 12:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Try to read the table from right to left, since the Wiki convention is symmetrical. If you still got questions, post your calculation in the question body so we can see.
  $endgroup$
  – Aminopterin
  Mar 20 at 12:07
  

  
  
  
  

 | 
show 3 more comments

0

$begingroup$

Yes that's right and you should consider $t$ as constant and define a new, independent variable $tau$ for time. The model in which the Fourier transform changes with some other variable ($t$ in this case) can be used for example to model a time-variant communication channel. In that case, even the channel is modeled stochastically with its FT dependent to time. In that case, the power spectral density is used as the FT of stochastic channel response and therefore varies with time, i.e. $$h(tau;t)=textchannel response varying over time\S(f;t)=int Eh(tau+tau_1;t)h(tau_1;t)e^-i2pi ftau_1dtau_1$$where $Ecdot$ stands for mathematical expectation and channel is assumed to be a stationary process.

share|cite|improve this answer

answered Mar 20 at 8:47

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  in my case, I got exp(i pi (t/a) as constant, but Should i take delta(t) as fourier tansform of constant?
  $endgroup$
  – Nani
  Mar 20 at 9:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No you shouldn't since the FT is 1D only taken over $tau$. $t$ can only be parameter of change in $A(f;t)$ (no matter what the origination is, even it can be else than time). Also in your question it should be $$B(tau,t)=int A(f,t) e^j2pi ftaudf$$not$$B(tau,t)=int A(f,t) e^j2pi ftdf$$and you should double-check your final answer (an exponential factor is neglected)
  $endgroup$
  – Mostafa Ayaz
  Mar 20 at 9:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $B(tau,t)=int e^j2pi (t/a-fa) e^j2pi ftaudf=e^j2pi (t/a)int e^j2pi f(tau-a) df=(e^j2pi (t/a)/(j2pi (tau-a)))e^j2pi f(tau-a)$...is it right?
  $endgroup$
  – Nani
  Mar 20 at 9:21
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right. But you can write alternatively $$int e^j2pi f(tau-a)df=delta(tau+a)$$
  $endgroup$
  – Mostafa Ayaz
  Mar 20 at 9:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how do you get this? Why does the integration of exponential function equal to dirac?
  $endgroup$
  – Nani
  Mar 20 at 9:36
  

  
  
  
  

 | 
show 12 more comments

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0

$begingroup$

Just recognize that, I think,
$$
mathcalF^-1_f to t' exp left[ 2pi j left( fracta-af right) right]
= e^2pi j t/a delta ( t'-a )
$$
if $t$ does not depend on $f$.
Also, there are several different conventions on FT, you have to specify, but usually they differ up to a constant multiple. (See Wikipedia formula 102, 302)

share|cite|improve this answer

edited Mar 20 at 11:03

answered Mar 20 at 10:55

AminopterinAminopterin

1,089213

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how do you get delta function in the answer? I got exp(j pi f (t´-a).
  $endgroup$
  – Nani
  Mar 20 at 11:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you seen the Wiki table formula 102, 302?
  $endgroup$
  – Aminopterin
  Mar 20 at 12:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i have seen, but it is for fourier transform, in my case is inverse FT. I don`t understand how do you and Mostafa Ayaz get delta function.
  $endgroup$
  – Nani
  Mar 20 at 12:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I got the same answer with you, but instead ofdelta i got exp
  $endgroup$
  – Nani
  Mar 20 at 12:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Try to read the table from right to left, since the Wiki convention is symmetrical. If you still got questions, post your calculation in the question body so we can see.
  $endgroup$
  – Aminopterin
  Mar 20 at 12:07
  

  
  
  
  

 | 
show 3 more comments

0

$begingroup$

Just recognize that, I think,
$$
mathcalF^-1_f to t' exp left[ 2pi j left( fracta-af right) right]
= e^2pi j t/a delta ( t'-a )
$$
if $t$ does not depend on $f$.
Also, there are several different conventions on FT, you have to specify, but usually they differ up to a constant multiple. (See Wikipedia formula 102, 302)

share|cite|improve this answer

edited Mar 20 at 11:03

answered Mar 20 at 10:55

AminopterinAminopterin

1,089213

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how do you get delta function in the answer? I got exp(j pi f (t´-a).
  $endgroup$
  – Nani
  Mar 20 at 11:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you seen the Wiki table formula 102, 302?
  $endgroup$
  – Aminopterin
  Mar 20 at 12:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i have seen, but it is for fourier transform, in my case is inverse FT. I don`t understand how do you and Mostafa Ayaz get delta function.
  $endgroup$
  – Nani
  Mar 20 at 12:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I got the same answer with you, but instead ofdelta i got exp
  $endgroup$
  – Nani
  Mar 20 at 12:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Try to read the table from right to left, since the Wiki convention is symmetrical. If you still got questions, post your calculation in the question body so we can see.
  $endgroup$
  – Aminopterin
  Mar 20 at 12:07
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

Just recognize that, I think,
$$
mathcalF^-1_f to t' exp left[ 2pi j left( fracta-af right) right]
= e^2pi j t/a delta ( t'-a )
$$
if $t$ does not depend on $f$.
Also, there are several different conventions on FT, you have to specify, but usually they differ up to a constant multiple. (See Wikipedia formula 102, 302)

share|cite|improve this answer

edited Mar 20 at 11:03

answered Mar 20 at 10:55

AminopterinAminopterin

1,089213

$endgroup$

share|cite|improve this answer

edited Mar 20 at 11:03

answered Mar 20 at 10:55

AminopterinAminopterin

1,089213

answered Mar 20 at 10:55

AminopterinAminopterin

1,089213

answered Mar 20 at 10:55

AminopterinAminopterin

1,089213

0

$begingroup$

Yes that's right and you should consider $t$ as constant and define a new, independent variable $tau$ for time. The model in which the Fourier transform changes with some other variable ($t$ in this case) can be used for example to model a time-variant communication channel. In that case, even the channel is modeled stochastically with its FT dependent to time. In that case, the power spectral density is used as the FT of stochastic channel response and therefore varies with time, i.e. $$h(tau;t)=textchannel response varying over time\S(f;t)=int Eh(tau+tau_1;t)h(tau_1;t)e^-i2pi ftau_1dtau_1$$where $Ecdot$ stands for mathematical expectation and channel is assumed to be a stationary process.

share|cite|improve this answer

answered Mar 20 at 8:47

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  in my case, I got exp(i pi (t/a) as constant, but Should i take delta(t) as fourier tansform of constant?
  $endgroup$
  – Nani
  Mar 20 at 9:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No you shouldn't since the FT is 1D only taken over $tau$. $t$ can only be parameter of change in $A(f;t)$ (no matter what the origination is, even it can be else than time). Also in your question it should be $$B(tau,t)=int A(f,t) e^j2pi ftaudf$$not$$B(tau,t)=int A(f,t) e^j2pi ftdf$$and you should double-check your final answer (an exponential factor is neglected)
  $endgroup$
  – Mostafa Ayaz
  Mar 20 at 9:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $B(tau,t)=int e^j2pi (t/a-fa) e^j2pi ftaudf=e^j2pi (t/a)int e^j2pi f(tau-a) df=(e^j2pi (t/a)/(j2pi (tau-a)))e^j2pi f(tau-a)$...is it right?
  $endgroup$
  – Nani
  Mar 20 at 9:21
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right. But you can write alternatively $$int e^j2pi f(tau-a)df=delta(tau+a)$$
  $endgroup$
  – Mostafa Ayaz
  Mar 20 at 9:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how do you get this? Why does the integration of exponential function equal to dirac?
  $endgroup$
  – Nani
  Mar 20 at 9:36
  

  
  
  
  

 | 
show 12 more comments

0

$begingroup$

Yes that's right and you should consider $t$ as constant and define a new, independent variable $tau$ for time. The model in which the Fourier transform changes with some other variable ($t$ in this case) can be used for example to model a time-variant communication channel. In that case, even the channel is modeled stochastically with its FT dependent to time. In that case, the power spectral density is used as the FT of stochastic channel response and therefore varies with time, i.e. $$h(tau;t)=textchannel response varying over time\S(f;t)=int Eh(tau+tau_1;t)h(tau_1;t)e^-i2pi ftau_1dtau_1$$where $Ecdot$ stands for mathematical expectation and channel is assumed to be a stationary process.

share|cite|improve this answer

answered Mar 20 at 8:47

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  in my case, I got exp(i pi (t/a) as constant, but Should i take delta(t) as fourier tansform of constant?
  $endgroup$
  – Nani
  Mar 20 at 9:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No you shouldn't since the FT is 1D only taken over $tau$. $t$ can only be parameter of change in $A(f;t)$ (no matter what the origination is, even it can be else than time). Also in your question it should be $$B(tau,t)=int A(f,t) e^j2pi ftaudf$$not$$B(tau,t)=int A(f,t) e^j2pi ftdf$$and you should double-check your final answer (an exponential factor is neglected)
  $endgroup$
  – Mostafa Ayaz
  Mar 20 at 9:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $B(tau,t)=int e^j2pi (t/a-fa) e^j2pi ftaudf=e^j2pi (t/a)int e^j2pi f(tau-a) df=(e^j2pi (t/a)/(j2pi (tau-a)))e^j2pi f(tau-a)$...is it right?
  $endgroup$
  – Nani
  Mar 20 at 9:21
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right. But you can write alternatively $$int e^j2pi f(tau-a)df=delta(tau+a)$$
  $endgroup$
  – Mostafa Ayaz
  Mar 20 at 9:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  how do you get this? Why does the integration of exponential function equal to dirac?
  $endgroup$
  – Nani
  Mar 20 at 9:36
  

  
  
  
  

 | 
show 12 more comments

$begingroup$

Yes that's right and you should consider $t$ as constant and define a new, independent variable $tau$ for time. The model in which the Fourier transform changes with some other variable ($t$ in this case) can be used for example to model a time-variant communication channel. In that case, even the channel is modeled stochastically with its FT dependent to time. In that case, the power spectral density is used as the FT of stochastic channel response and therefore varies with time, i.e. $$h(tau;t)=textchannel response varying over time\S(f;t)=int Eh(tau+tau_1;t)h(tau_1;t)e^-i2pi ftau_1dtau_1$$where $Ecdot$ stands for mathematical expectation and channel is assumed to be a stationary process.

share|cite|improve this answer

answered Mar 20 at 8:47

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

share|cite|improve this answer

answered Mar 20 at 8:47

Mostafa AyazMostafa Ayaz

18.2k31040

answered Mar 20 at 8:47

Mostafa AyazMostafa Ayaz

18.2k31040

answered Mar 20 at 8:47

Mostafa AyazMostafa Ayaz

18.2k31040