Time Shifting property of the Fourier Transform

Electrical Circuit Analysis > Fourier Transform > Properties of the Fourier Transform

Likes (0)

Fav (0)

Views (4.6K)

1 year

Facebook

Twitter

If $ F(\omega)=\mathcal{F}[f(t)] $, then

$$\mathcal{F}\left[f\left(t-t_{0}\right)\right]=e^{-j \omega t_{0}} F(\omega)$$

that is, a delay in the time domain corresponds to a phase shift in the frequency domain. To derive the time shifting property, we note that

$$\mathcal{F}\left[f\left(t-t_{0}\right)\right]=\int_{-\infty}^{\infty} f\left(t-t_{0}\right) e^{-j \omega t} d t$$

If we let $ x=t-t_{0} $ so that $ d x=d t $ and $ t=x+t_{0} $, then

$$\begin{aligned}\mathcal{F}\left[f\left(t-t_{0}\right)\right] &=\int_{-\infty}^{\infty} f(x) e^{-j \omega\left(x+t_{0}\right)} d x \\&=e^{-j \omega t_{0}} \int_{-\infty}^{\infty} f(x) e^{-j \omega x} d x=e^{-j \omega t_{0}} F(\omega)\end{aligned}$$

Similarly, $ \mathcal{F}\left[f\left(t+t_{0}\right)\right]=e^{j \omega t_{0}} F(\omega) $. For example, $$\mathcal{F}\left[e^{-a t} u(t)\right]=\frac{1}{a+j \omega}$$ The transform of $ f(t)=e^{-(t-2)} u(t-2) $ is

$$F(\omega)=\mathcal{F}\left[e^{-(t-2)} u(t-2)\right]=\frac{e^{-j 2 \omega}}{1+j \omega}$$

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!