Fourier transform

From Class Wiki
Revision as of 10:51, 9 December 2004 by Guenan (talk | contribs)

Jump to: navigation, search

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