Take the 2-minute tour ×
Signal Processing Stack Exchange is a question and answer site for practitioners of the art and science of signal, image and video processing. It's 100% free, no registration required.

By definition of fourier transform

$$X(w)=\int_{-\infty}^\infty x(t) e^{-jwt} dt $$

Now what will happen to the answer of transform for example in case of $x(t)= cos(w_ot)$ if limit is 0 to A instead of $-\infty$ to $\infty$?

For $x(t)=cos(w_ot)$ its fourier transform is given by $ X(w)= \pi[\delta(w-w_0) + \delta(w+w_o)]$

so if the limit is changed will it effect the answer?

share|improve this question
add comment

1 Answer 1

up vote 9 down vote accepted

Yes, it will affect the answer. What you're suggesting is known as the short-time Fourier transform. In the sinusoidal case that you proposed, you will observe spectral leakage, as the truncation of the integral limits is equivalent to multiplication of the sinusoid by a rectangular window function. This multiplication in the time domain maps to convolution in the frequency domain. The Fourier transform of a rectangular window is a sinc function, so the convolution will yield two sinc functions centered at the locations of the impulses in your original answer.

share|improve this answer
3  
Moreover, with limits $0$ and $A$, the Fourier transform is definitely going to be a complex-valued function of $\omega$ while if the limits were $-A$ and $A$, the Fourier transform will continue to be purely real-valued. –  Dilip Sarwate Dec 15 '12 at 21:39
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.