Time Integration property of the Fourier Transform

Given that $ F(\omega)=\mathcal{F}[f(t)] $, then
$$\mathcal{F}\left[\int_{-\infty}^{t} f(t) d t\right]=\frac{F(\omega)}{j \omega}+\pi F(0) \delta(\omega) \tag{1}$$
that is, the transform of the integral of $ f(t) $ is obtained by dividing the transform of $ f(t) $ by $ j \omega $ and adding the result to the impulse term that reflects the dc component $ F(0) $. Someone might ask, "How do we know that when we take the Fourier transform for time integration, we should integrate over the interval $ [-\infty, t] $ and not $ [-\infty, \infty] $ ?" When we integrate over $ [-\infty, \infty] $, the result does not depend on time anymore, and the Fourier transform of a constant is what we will eventually get. But when we integrate over $ [-\infty, t] $, we get the integral of the function from the past to time $ t $, so that the result depends on $ t $ and we can take the Fourier transform of that.
If $ \omega $ is replaced by 0 in Eq. (17.8), $$F(0)=\int_{-\infty}^{\infty} f(t) d t$$ indicating that the dc component is zero when the integral of $ f(t) $ over all time vanishes. The proof of the time integration in Eq. (1) will be given later when we consider the convolution property.
For example, we know that $ \mathcal{F}[\delta(t)]=1 $ and that integrating the impulse function gives the unit step function. By applying the property in Eq. (1), we obtain the Fourier transform of the unit step function as
$$\mathcal{F}[u(t)]=\mathcal{F}\left[\int_{-\infty}^{t} \delta(t) d t\right]=\frac{1}{j \omega}+\pi \delta(\omega)$$

Do you want to say or ask something?

Only 250 characters are allowed. Remaining: 250
Please login to enter your comments. Login or Signup .
Be the first to comment here!
Terms and Condition
Copyright © 2011 - 2024 realnfo.com
Privacy Policy