Changes between Version 4 and Version 5 of u/afrank

Timestamp:

12/22/14 20:55:51 (11 years ago)

Author:

Adam Frank

Comment:

—

u/afrank

@@ -6 +6 @@
 The basic equation for the fourier transform.
 
- [[latex($ F(k) = \int^{+\infty}_{-\infty} f(x) e^{(-i 2 \pi k x)} dx   $)]] 
+ [[latex($ \Xi[f] = F(k) = \int^{+\infty}_{-\infty} f(x) e^{(-i 2 \pi k x)} dx   $)]] 
+
+with the inverse
+
+ [[latex($ \Xi[F]^{-1} = f(x) = \int^{+\infty}_{-\infty} F(k)) e^{(i 2 \pi k x)} dk   $)]] 
 
 Here we want to solve the signup function which is runs from -1 to 1 with a step at x=a
@@ -12 +16 @@
  [[latex($ f(x) = sgn(x-a)   $)]]
 
-[[Image(http://www.pas.rochester.edu/~afrank/Fig1.jpg)]]
+[[Image(http://www.pas.rochester.edu/~afrank/Fig1.jpg, 250px)]]
+
+Use the derivative property of  [[latex($ \Xi[f]   $)]]
+
+ [[latex($ \Xi[f ] = \frac{1}{2\pi i k} \Xi[ \frac{df}{dx} ]  $)]] 
+
+Since derivative of f is the direct delta function
+
+ [[latex($ \frac{df}{dx} = \delta(x-a)   $)]] 
+
+and
+
+ [[latex($ \Xi[ \delta(x-a) ] =  e^{-i2\pi k a} $)]] 
+
+Thus we have 
+
+ [[latex($ \Xi[f ] = F(k) = \frac{1}{2\pi i k} e^{-i2\pi k a}  $)]] 
+
+The next step is to define the Energy Spectral Density (ESD) 
+
+We use Parcivals Thm which tell us "energy" under the curve f(x) 
+
+ [[latex($ \int^{+\infty}_{-\infty} | f(x)|^2 dx  = \int^{+\infty}_{-\infty} | F(k)|^2 dk $)]] 
+
+Thus the ESD which we write as E(k) is
+
+[[latex($  E(k) =  | F(k)|^2  $)]] 
+
+which for the sgn function is
+
+ [[latex($ E(k) = (\frac{1}{2\pi i k} e^{-i2\pi k a})^*  (\frac{1}{2\pi i k} e^{-i2\pi k a}) $)]] 
+
+ [[latex($ E(k) = \frac{1}{4\pi^2 k^2}   $)]]
+
+So for a step function
+
+ [[latex($ E(k) \propto k^{-2}   $)]]
+
+
+
+
+
 
 == Description of Problem ==