Need command for Continuous time fourier transform

Hai, I need command for Continuous time fourier transform.I know the command for Discrete time fourier transform.
One more Question, does the both results of Continuous time fourier transform and Discrete time fourier transform the same, or different.

 Risposta accettata

Continuous and Discrete Fourier Transform are the same in the limit case of the steps being infinitesimals.
Other than that, they cannot be compared as they work on two different kinds of information.

Più risposte (5)

Hi, If you have the Symbolic Toolbox, you can use fourier() to obtain the Fourier transform.
syms x;
f = exp(-x^2);
fourier(f)
Wayne

3 Commenti

Hi Wayne,
Thanks for your quick response.
fourier() is for fourier series but I need for fourier transform.
Hi Ratna, With all due respect, that is not correct.
Please see
>>doc symbolic/fourier
fourier
Fourier integral transform
The examples are not periodic functions of the independent variable.
Wayne
Hi Wayne,
I am wrong, you are correct. But I have a function to find fourier transform over the limits.
How can I do this? As the above function fourier is for [-infinity to infinity]
Thanks,
Ratna.

Accedi per commentare.

Hi Ratna, You can use assume() to place limits on your variable of integration.
For example
syms x
% create Dirac distribution shifted to -1
f = dirac(x+1)
fourier(f)
% gives exp(w*i)
assume(x>0)
fourier(f)
% gives 0
Wayne

1 Commento

hi, how can we find continous time fourier and transform by using for loop. don't use built in func. plz help.

Accedi per commentare.

I would not recommend the approach of using assumptions. Fourier transforms are defined from -infinity to +infinity and attempts to cheat that are likely to go wrong.
Instead, multiply the function of interest by dirac(x-lowerbound) * dirac(upperbound-x) and fourier() the transformed function.
continuous-time Fourier series and transforms:
p(t) = A 0 ≤ t ≤ Tp < T
0 otherwise
how can we write the code for this?

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by