Difference between revisions of "User:Tohline/Appendix/Ramblings/FourierSeries"

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===Alternate===
===Alternate===
Equivalently, we may write the Fourier series expression in the form,
Alternatively, if we set <math>~a_n = c_n \cos\phi_n</math> and <math>~b_n = - c_n \sin\phi_n</math>, then,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~a_n\cos \biggl(\frac{n\pi x}{L}\biggr) + b_n\sin  \biggl(\frac{n\pi x}{L}\biggr) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
c_n \cos\phi_n \cos \biggl(\frac{n\pi x}{L}\biggr) - c_n \sin\phi_n \sin  \biggl(\frac{n\pi x}{L}\biggr)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
c_n \cos \biggl(\frac{n\pi x}{L} + \phi_n \biggr) \, ,
</math>
  </td>
</tr>
</table>
</div>
 
in which case we may rewrite the Fourier series expression in the form,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 100: Line 134:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\phi_n = \tan^{-1}\biggl(\frac{-b_n}{a_n}\biggr)</math>
<math>~\phi_n = \tan^{-1}\biggl(\frac{-b_n}{a_n}\biggr) \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
and in which case, <math>~a_n = c_n\sin\phi_n</math> and <math>~b_n = c_n\cos\phi_n</math> .


===Complex===
===Complex===
Line 143: Line 176:
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{1}{2}\sum_{n = -\infty}^{n = + \infty} c_n e^{i\omega_n x} \, ,
\frac{1}{2}\sum_{n = -\infty}^{n = + \infty} d_n e^{i\omega_n x} \, ,
</math>
</math>
   </td>
   </td>
Line 172: Line 205:
</div>
</div>


and the, now ''complex'', coefficients,
and the ''complex'' coefficients,


<div align="center">
<div align="center">
Line 179: Line 212:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~c_n</math>
<math>~d_n = a_n - ib_n</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 191: Line 224:
   <td align="center">&nbsp; &nbsp;</td>
   <td align="center">&nbsp; &nbsp;</td>
   <td align="right">for</td>
   <td align="right">for</td>
   <td align="left"><math>~n = 0, \pm 1, \pm 2, \pm 3, \dots \, .</math>
   <td align="left"><math>~n = 0, \pm 1, \pm 2, \pm 3, \dots</math>
</tr>
</table>
</div>
 
Let's demonstrate that this rewritten (complex) expression matches the standard Fourier series expression.  First, we will refer to the above ''standard'' definitions of <math>~a_n</math> and <math>~b_n</math> as, respectively, <math>~a_{|n|}</math> and <math>~b_{|n|}</math>, and recognize that, as the summation is extended to negative numbers, the following mapping is appropriate:
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~a_n ~ \rightarrow ~ a_{|n|}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~b_n ~ \rightarrow ~ b_{|n|} \, ,</math>
  </td>
  <td align="right">&nbsp; &nbsp; for </td>
  <td align="left"><math>~n > 0 \, ;</math>
</tr>
<tr>
  <td align="right">
<math>~a_n ~ \rightarrow ~ a_{|n|}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~b_n ~ \rightarrow ~ - b_{|n|} \, ,</math>
  </td>
  <td align="right">&nbsp; &nbsp; for </td>
  <td align="left"><math>~n < 0 \, .</math>
</tr>
</tr>
</table>
</table>
</div>
</div>


Hence, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~2f(x)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sum_{n = -\infty}^{n = + \infty} (a_n - ib_n)e^{i\omega_n x}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sum_{n = 1}^{n = + \infty} (a_{|n|} - ib_{|n|})e^{i\omega_{|n|} x}
+ a_0
+ \sum_{n = \infty}^{n = 1} (a_{|n|} + ib_{|n|})e^{- i\omega_{|n|} x}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sum_{n = 1}^{n = + \infty} (a_{|n|} - ib_{|n|}) [ \cos (\omega_{|n|} x) + i\sin (\omega_{|n|} x)]
+ a_0
+ \sum_{n = 1}^{n = \infty} (a_{|n|} + ib_{|n|}) [ \cos (\omega_{|n|} x) - i\sin (\omega_{|n|} x)]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
a_0 +
\sum_{n = 1}^{n = + \infty}\biggl\{  (a_{|n|} - ib_{|n|}) [ \cos (\omega_{|n|} x) + i\sin (\omega_{|n|} x)]
+ (a_{|n|} + ib_{|n|}) [ \cos (\omega_{|n|} x) - i\sin (\omega_{|n|} x)] \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
a_0 +
\sum_{n = 1}^{n = + \infty}\biggl\{2a_{|n|} \cos (\omega_{|n|} x)  + 2b_{|n|} \sin (\omega_{|n|} x)] \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ f(x)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{a_0}{2} +
\sum_{n = 1}^{n = + \infty}\biggl\{a_{|n|} \cos (\omega_{|n|} x)  + b_{|n|} \sin (\omega_{|n|} x)] \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
Q.E.D.
'''TO BE DONE''':
We should also check that the expression for the complex coefficient, <math>~d_n</math>, also makes sense for all (negative as well as positive) values of the index, <math>~n</math>.  First, for positive numbers, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~d_n = a_{|n|} -i b_{|n|} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{L} \int_{-L}^{L} f(x) \cos\biggl( \frac{n\pi x}{L} \biggr) dx
- i\frac{1}{L} \int_{-L}^{L} f(x) \sin\biggl( \frac{n\pi x}{L} \biggr) dx
</math>
  </td>
</tr>
</table>
</div>
while, for negative numbers, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~d_n = a_{|n|} + i b_{|n|} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{L} \int_{-L}^{L} f(x) \cos\biggl( \frac{-|n|\pi x}{L} \biggr) dx
+ i\frac{1}{L} \int_{-L}^{L} f(x) \sin\biggl( \frac{- |n|\pi x}{L} \biggr) dx
</math>
  </td>
</tr>
</table>
</div>


==One-Dimensional Aperture==
==One-Dimensional Aperture==

Revision as of 05:41, 11 November 2017

Fourier Series

Whitworth's (1981) Isothermal Free-Energy Surface
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Representations

The following Fourier series representations have been drawn primarily from pp. 458 - 460 of the 1971 (19th) edition of the CRC's Standard Mathematical Tables, published by the Chemical Rubber Co., Cleveland, Ohio, U.S.A.

Standard

If <math>~f(x)</math> is a bounded periodic function of period <math>~2L</math>, it may be represented by the Fourier series,

<math>~f(x)</math>

<math>~=</math>

<math>~ \frac{a_0}{2} + \sum_{n=1}^{\infty} \biggl[ a_n\cos \biggl(\frac{n\pi x}{L}\biggr) + b_n\sin \biggl(\frac{n\pi x}{L}\biggr) \biggr] \, , </math>

where,

<math>~a_n</math>

<math>~=</math>

<math>~ \frac{1}{L} \int_{-L}^{L} f(x) \cos\biggl( \frac{n\pi x}{L} \biggr) dx </math>

    for <math>~n = 0, 1, 2, 3, \dots \, ;</math>

<math>~b_n</math>

<math>~=</math>

<math>~ \frac{1}{L} \int_{-L}^{L} f(x) \sin\biggl( \frac{n\pi x}{L} \biggr) dx </math>

    for <math>~n = 1, 2, 3, \dots </math>

Alternate

Alternatively, if we set <math>~a_n = c_n \cos\phi_n</math> and <math>~b_n = - c_n \sin\phi_n</math>, then,

<math>~a_n\cos \biggl(\frac{n\pi x}{L}\biggr) + b_n\sin \biggl(\frac{n\pi x}{L}\biggr) </math>

<math>~=</math>

<math>~ c_n \cos\phi_n \cos \biggl(\frac{n\pi x}{L}\biggr) - c_n \sin\phi_n \sin \biggl(\frac{n\pi x}{L}\biggr) </math>

 

<math>~=</math>

<math>~ c_n \cos \biggl(\frac{n\pi x}{L} + \phi_n \biggr) \, , </math>

in which case we may rewrite the Fourier series expression in the form,

<math>~f(x)</math>

<math>~=</math>

<math>~ \frac{a_0}{2} + \sum_{n=1}^{\infty} c_n\cos \biggl(\frac{n\pi x}{L} + \phi_n\biggr) \, , </math>

where,

<math>~c_n = \sqrt{a_n^2 + b_n^2}</math>

      and      

<math>~\phi_n = \tan^{-1}\biggl(\frac{-b_n}{a_n}\biggr) \, .</math>

Complex

Here we make use of the exponential/complex relation — also referred to as Euler's equation,

<math>~e^{i\alpha} = \cos\alpha + i \sin\alpha \, ,</math>

in which case we may write,

<math>~\cos\alpha = \frac{1}{2} \biggl[ e^{i\alpha} + e^{-i\alpha}\biggr] \, ,</math>

      and      

<math>~\sin\alpha = \frac{1}{2i} \biggl[ e^{i\alpha} - e^{-i\alpha}\biggr]\, .</math>

Employing these definitions of the trigonometric relations <math>~\cos\alpha</math> and <math>~\sin\alpha</math>, the standard representation of the Fourier series may be rewritten as,

<math>~f(x)</math>

<math>~=</math>

<math>~ \frac{1}{2}\sum_{n = -\infty}^{n = + \infty} d_n e^{i\omega_n x} \, , </math>

where,

<math>~\omega_n</math>

<math>~=</math>

<math>~ \frac{n\pi }{L} </math>

    for <math>~n = 0, \pm 1, \pm 2, \dots \, ;</math>

and the complex coefficients,

<math>~d_n = a_n - ib_n</math>

<math>~=</math>

<math>~ \frac{1}{L} \int_{-L}^{L} f(x) e^{-i\omega_n x} dx </math>

    for <math>~n = 0, \pm 1, \pm 2, \pm 3, \dots</math>

Let's demonstrate that this rewritten (complex) expression matches the standard Fourier series expression. First, we will refer to the above standard definitions of <math>~a_n</math> and <math>~b_n</math> as, respectively, <math>~a_{|n|}</math> and <math>~b_{|n|}</math>, and recognize that, as the summation is extended to negative numbers, the following mapping is appropriate:

<math>~a_n ~ \rightarrow ~ a_{|n|}</math>

      and      

<math>~b_n ~ \rightarrow ~ b_{|n|} \, ,</math>

    for <math>~n > 0 \, ;</math>

<math>~a_n ~ \rightarrow ~ a_{|n|}</math>

      and      

<math>~b_n ~ \rightarrow ~ - b_{|n|} \, ,</math>

    for <math>~n < 0 \, .</math>

Hence, we have,

<math>~2f(x)</math>

<math>~=</math>

<math>~ \sum_{n = -\infty}^{n = + \infty} (a_n - ib_n)e^{i\omega_n x} </math>

 

<math>~=</math>

<math>~ \sum_{n = 1}^{n = + \infty} (a_{|n|} - ib_{|n|})e^{i\omega_{|n|} x} + a_0 + \sum_{n = \infty}^{n = 1} (a_{|n|} + ib_{|n|})e^{- i\omega_{|n|} x} </math>

 

<math>~=</math>

<math>~ \sum_{n = 1}^{n = + \infty} (a_{|n|} - ib_{|n|}) [ \cos (\omega_{|n|} x) + i\sin (\omega_{|n|} x)] + a_0 + \sum_{n = 1}^{n = \infty} (a_{|n|} + ib_{|n|}) [ \cos (\omega_{|n|} x) - i\sin (\omega_{|n|} x)] </math>

 

<math>~=</math>

<math>~ a_0 + \sum_{n = 1}^{n = + \infty}\biggl\{ (a_{|n|} - ib_{|n|}) [ \cos (\omega_{|n|} x) + i\sin (\omega_{|n|} x)] + (a_{|n|} + ib_{|n|}) [ \cos (\omega_{|n|} x) - i\sin (\omega_{|n|} x)] \biggr\} </math>

 

<math>~=</math>

<math>~ a_0 + \sum_{n = 1}^{n = + \infty}\biggl\{2a_{|n|} \cos (\omega_{|n|} x) + 2b_{|n|} \sin (\omega_{|n|} x)] \biggr\} </math>

<math>~\Rightarrow ~~~ f(x)</math>

<math>~=</math>

<math>~ \frac{a_0}{2} + \sum_{n = 1}^{n = + \infty}\biggl\{a_{|n|} \cos (\omega_{|n|} x) + b_{|n|} \sin (\omega_{|n|} x)] \biggr\} \, . </math>

Q.E.D.

TO BE DONE:

We should also check that the expression for the complex coefficient, <math>~d_n</math>, also makes sense for all (negative as well as positive) values of the index, <math>~n</math>. First, for positive numbers, we have,

<math>~d_n = a_{|n|} -i b_{|n|} </math>

<math>~=</math>

<math>~ \frac{1}{L} \int_{-L}^{L} f(x) \cos\biggl( \frac{n\pi x}{L} \biggr) dx - i\frac{1}{L} \int_{-L}^{L} f(x) \sin\biggl( \frac{n\pi x}{L} \biggr) dx </math>

while, for negative numbers, we have,

<math>~d_n = a_{|n|} + i b_{|n|} </math>

<math>~=</math>

<math>~ \frac{1}{L} \int_{-L}^{L} f(x) \cos\biggl( \frac{-|n|\pi x}{L} \biggr) dx + i\frac{1}{L} \int_{-L}^{L} f(x) \sin\biggl( \frac{- |n|\pi x}{L} \biggr) dx </math>

One-Dimensional Aperture

General Concept

Hence, we have,

<math>~A(y_1)</math>

<math>~=</math>

<math>~A_0 \sum_j a_j e^{-i[2\pi y_1 Y_j/(\lambda L)]} \, , </math>

 

<math>~=</math>

<math>~A_0 \sum_j a_j \biggl[ \cos\biggl(\frac{2\pi y_1 Y_j}{\lambda L} \biggr) - i \sin\biggl(\frac{2\pi y_1 Y_j}{\lambda L} \biggr) \biggr] \, , </math>

where, now, <math>~A_0 = e^{i2\pi L/\lambda}</math>. When written in this form, it should immediately be apparent why discrete Fourier transform techniques (specifically FFT techniques) are useful tools for evaluation of the complex amplitude, <math>~A</math>.

See Also

  • Tohline, J. E., (2008) Computing in Science & Engineering, vol. 10, no. 4, pp. 84-85 — Where is My Digital Holographic Display? [ PDF ]


Whitworth's (1981) Isothermal Free-Energy Surface

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