1.7 KiB
1.7 KiB
Chapter 5
The Fourier Transform:
X(\omega) = F[x(t)] = \int\limits_{-\infty}^\infty x(t) e^{-j\omega t} dt
The Inverse Fourier Transform:
x(t) = F^{-1}[X(\omega)] = {1 \over 2\pi} \int\limits_{-\infty}^{\infty} X(\omega)e^{j\omega t} d\omega
The Fourier transform exists if:
x(t)is abolutely integrablex(t)has only a finite number of discontinuities and a finite number of minima and maxima in any finite interval.
If x(t) is even, X(\omega) is real.
If x(t) is odd, X(\omega) is imaginary.
Otherwise, X(t) has a real and imaginary part.
Example
Consider x(t) = \delta(t):
X(\omega) = \int\limits_{-\infty}^\infty \delta(t)e^{-j\omega t} dt = 1
Example
Consider x(t) = \delta(t - \alpha):
X(\omega) = \int\limits_{-\infty}^\infty \delta(t - \alpha)e^{-j\omega t}dt = e^{-j\omega \alpha}
Example
Consider x(t) = u(t+T) - u(t-T):
This is a pulse whose value is 1 on the interval [-T,T].
X(t) = \int\limits_{-T}^T e^{-j\omega t}dt = -{1 \over j \omega} \left[e^{-j\omega T} - e^{-j\omega (-T)}\right] = {2 \over \omega}\left[{e^{j\omega T} - e^{-j\omega T} \over 2j}\right] = 2T\operatorname{sinc}(\omega T)
Example
Consider x(t) = e^{-|t|}
$$X(\omega) = \int\limits_
Method 2 - Laplace Transfor
x(t) \to X(s) \to X(\omega)
s = j\omega
Using a Laplace transform requires that x(t) is a causal signal and the region of convergence of X(s) includes the imaginary axis.
Example - Finite Support Signals
The region of convergence is the entire s-plane (must check s=0). It definitely includes s=j\omeaga.
X(s) = {0.5 \over s^2} - {0.5e^{-2s} \over s^2} - {e^{-2s} \over s}
X(j\omega) = {0.5 \over j\omega^2} - {0.5e^{-2j\omega} \over j\omega^2} - {e^{-2j\omega} \over j\omega}