Fourier transform

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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)

A Second Approach to Fourier Transforms