Difference between revisions of "Fourier transform"

From Class Wiki
Jump to: navigation, search
Line 2: Line 2:
 
<math>
 
<math>
   
x(t)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt
+
X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt
   
 
</math>
 
</math>
Line 10: Line 10:
 
This means that <math> \beta(t)=0 </math> for some <math> T_\alpha < \left | t \right | </math>
 
This means that <math> \beta(t)=0 </math> for some <math> T_\alpha < \left | t \right | </math>
 
<br><br>
 
<br><br>
Now let's make a periodic function <math> \gamma(t) </math> by repeating <math> \beta(t) </math> with a fundamental period <math> T_\zeta </math>.
 
  +
Now let's make a periodic function
Note that <math> \lim_{T_\zeta \to \infty}\gamma(t)=\beta(t) </math>
 
  +
<math>
  +
\gamma(t)
  +
</math>
  +
by repeating
  +
<math>
  +
\beta(t)
  +
</math>
  +
with a fundamental period
  +
<math>
  +
T_\zeta
  +
</math>.
  +
Note that
  +
<math>
 
\lim_{T_\zeta \to \infty}\gamma(t)=\beta(t)
  +
</math>
 
<br>
 
<br>
 
The Fourier Series representation of <math> \gamma(t) </math> is
 
The Fourier Series representation of <math> \gamma(t) </math> is
 
<br>
 
<br>
<math> \gamma(t)=\sum_{k=-\infty}^\infty \alpha_k e^{j2\pi fkt} </math> where <math> f={1\over T_\zeta}
 
  +
<math>
</math> <br>and <math> \alpha_k={1\over T_\zeta}\int_{-{T_\zeta\over 2}}^{{T_\zeta\over 2}} \gamma(t) e^{-j2\pi kt}\,dt</math>
 
 
\gamma(t)=\sum_{k=-\infty}^\infty \alpha_k e^{j2\pi fkt}
  +
</math>
  +
where
  +
<math>
  +
f={1\over T_\zeta}
  +
</math>
  +
<br>and
  +
<math>
 
\alpha_k={1\over T_\zeta}\int_{-{T_\zeta\over 2}}^{{T_\zeta\over 2}} \gamma(t) e^{-j2\pi kt}\,dt
  +
</math>
 
<br>
 
<br>
<math> \alpha_k </math> can now be rewritten as <math> \alpha_k={1\over T_\zeta}\int_{-\infty}^{\infty} \beta(t) e^{-j2\pi kt}\,dt </math>
+
<math> \alpha_k </math> can now be rewritten as
  +
<math>
  +
\alpha_k={1\over T_\zeta}\int_{-\infty}^{\infty} \beta(t) e^{-j2\pi kt}\,dt
  +
</math>
 
<br>From our initial identity then, we can write <math> \alpha_k </math> as
 
<br>From our initial identity then, we can write <math> \alpha_k </math> as
 
<math>
 
<math>
\alpha_k={1\over T_\zeta}\Beta(kf)
+
\alpha_k={1\over T_\zeta}\Beta(kf)
 
</math>
 
</math>
 
<br> and
 
<br> and
 
<math>
 
<math>
\gamma(t)
+
\gamma(t)
 
</math>
 
</math>
 
becomes
 
becomes
 
<math>
 
<math>
\gamma(t)=\sum_{k=-\infty}^\infty {1\over T_\zeta}\Beta(kf) e^{j2\pi fkt}
+
\gamma(t)=\sum_{k=-\infty}^\infty {1\over T_\zeta}\Beta(kf) e^{j2\pi fkt}
  +
</math>
  +
<br>
  +
Now remember that
  +
<math>
  +
\beta(t)=\lim_{T_\zeta \to \infty}\gamma(t)
  +
</math>
  +
and
  +
<math>
  +
{1\over {T_\zeta}} = f.
  +
</math>
  +
<br>
  +
Which means that
  +
<math>
  +
\beta(t)=\lim_{f \to 0}\gamma(t)=\lim_{f \to 0}\sum_{k=-\infty}^\infty f \Beta(kf) e^{j2\pi fkt}
  +
</math>
  +
<br>
  +
Which is just to say that
  +
<math>
  +
\beta(t)=\int_{-\infty}^\infty f \Beta(f) e^{j2\pi fkt}\,df
  +
</math>
  +
<br>
  +
<br>
  +
So we have that the Fourier Transform of
  +
<math>
  +
\beta(t)
  +
</math>
  +
is
  +
<math>
  +
X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt
 
</math>
 
</math>

Revision as of 11:21, 9 December 2004

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 f \Beta(f) e^{j2\pi fkt}\,df

So we have that the Fourier Transform of 
\beta(t)
is 
X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt