# Difference between revisions of "Fourier transform"

## From the Fourier Transform to the Inverse Fourier Transform

An initially identity that is useful: $X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt$

Suppose that we have some function, say $\beta (t)$, that is nonperiodic and finite in duration.
This means that $\beta(t)=0$ for some $T_\alpha < \left | t \right |$

Now let's make a periodic function $\gamma(t)$ by repeating $\beta(t)$ with a fundamental period $T_\zeta$. Note that $\lim_{T_\zeta \to \infty}\gamma(t)=\beta(t)$
The Fourier Series representation of $\gamma(t)$ is $\gamma(t)=\sum_{k=-\infty}^\infty \alpha_k e^{j2\pi fkt}$ where $f={1\over T_\zeta}$
and $\alpha_k={1\over T_\zeta}\int_{-{T_\zeta\over 2}}^{{T_\zeta\over 2}} \gamma(t) e^{-j2\pi kt}\,dt$ $\alpha_k$ can now be rewritten as $\alpha_k={1\over T_\zeta}\int_{-\infty}^{\infty} \beta(t) e^{-j2\pi kt}\,dt$
From our initial identity then, we can write $\alpha_k$ as $\alpha_k={1\over T_\zeta}\Beta(kf)$
and $\gamma(t)$ becomes $\gamma(t)=\sum_{k=-\infty}^\infty {1\over T_\zeta}\Beta(kf) e^{j2\pi fkt}$
Now remember that $\beta(t)=\lim_{T_\zeta \to \infty}\gamma(t)$ and ${1\over {T_\zeta}} = f.$
Which means that $\beta(t)=\lim_{f \to 0}\gamma(t)=\lim_{f \to 0}\sum_{k=-\infty}^\infty f \Beta(kf) e^{j2\pi fkt}$
Which is just to say that $\beta(t)=\int_{-\infty}^\infty \Beta(f) e^{j2\pi fkt}\,df$

So we have that $\mathcal{F}[\beta(t)]=\Beta(f)=\int_{-\infty}^{\infty} \beta(t) e^{-j2\pi ft}\, dt$
Further $\mathcal{F}^{-1}[\Beta(f)]=\beta(t)=\int_{-\infty}^\infty \Beta(f) e^{j2\pi fkt}\,df$

## Some Useful Fourier Transform Pairs $\mathcal{F}[\alpha(t)]=\frac{1}{\mid \alpha \mid}f(\frac{\omega}{\alpha})$ $\mathcal{F}[c_1\alpha(t)+c_2\beta(t)]$ $=\int_{-\infty}^{\infty} (c_1\alpha(t)+c_2\beta(t)) e^{-j2\pi ft}\, dt$ $=\int_{-\infty}^{\infty}c_1\alpha(t)e^{-j2\pi ft}\, dt+\int_{-\infty}^{\infty}c_2\beta(t)e^{-j2\pi ft}\, dt$ $=c_1\int_{-\infty}^{\infty}\alpha(t)e^{-j2\pi ft}\, dt+c_2\int_{-\infty}^{\infty}\beta(t)e^{-j2\pi ft}\, dt=c_1\Alpha(f)+c_2\Beta(f)$ $\mathcal{F}[\alpha(t-\gamma)]=e^{-j2\pi f\gamma}\Alpha(f)$ $\mathcal{F}[\alpha(t)*\beta(t)]=\Alpha(f)\Beta(f)$ $\mathcal{F}[\alpha(t)\beta(t)]=\Alpha(f)*\Beta(f)$