ECE438 - Digital Signal Processing

# CTFT

$X(f) = \int_{-\infty}^{\infty} x(t) e^{-2 \pi j f t} dt$

$x(t) = \int_{-\infty}^{\infty} X(f) e^{2 \pi j f t} df$

As a convention, we require $$x(t)$$ has the form $$u(t)f(t)$$ so that infinite negative tail of $$e^{-2 \pi j f t}$$ is eliminated.

An alternative notation:

$X(\omega) = \int_{-\infty}^{\infty} x(t) e^{j f t} dt$

$x(t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} X(f) e^{j f t} df$

## Useful CTFT Pairs

$\text{sinc}(t) = \frac{\sin (\pi t)}{\pi t} \xrightarrow[]{CTFT} \text{rect}(f)$ $\text{comb}_T(x(t)) \xrightarrow[]{CTFT} \frac{1}{T} \text{rep}_{1/T}(X(f))$

## Duality of fourier transform

$x(t) \xrightarrow[]{CTFT} X(f)$

$X(t) \xrightarrow[]{CTFT} x(-f)$

## Time Shifting

$x(t - t_0) \xrightarrow[]{CTFT} X(f)e^{-j 2 \pi ft_0}$

## Time Scaling

$x\left(\frac{t}{a}\right) \xrightarrow[]{CTFT} |a| X(af)$

# DTFT

$X_d (\omega) = \sum_{n = - \infty}^\infty x[n] e^{-j \omega n}$

$x[n] = \frac{1}{2 \pi} \int_{-\infty}^{\infty} X_d(\omega) e^{j \omega n}$