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### 02_fouriertransform

Course: ESS 522, Spring 2008
School: Washington
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ESS 522 2. Fourier Transform Spring 2007 I have drawn the content for this lecture mostly from the book Mathematical Methods for the Physical Sciences by K. F. Riley In the last lecture we showed that we could represent a periodic function by a sum of sine and cosine terms of alternately by complex exponentials. ( ( 1 &amp;quot; 2! nt % ( &amp;quot; 2! nt % &amp;quot; i2! nt % g(t) = A0 + ) An cos \$ +...

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ESS 522 2. Fourier Transform Spring 2007 I have drawn the content for this lecture mostly from the book Mathematical Methods for the Physical Sciences by K. F. Riley In the last lecture we showed that we could represent a periodic function by a sum of sine and cosine terms of alternately by complex exponentials. ( ( 1 &quot; 2! nt % ( &quot; 2! nt % &quot; i2! nt % g(t) = A0 + ) An cos \$ + ) Bn sin \$ = ) Cn exp \$ (2-1) ' ' # T &amp; n =1 # T &amp; n = *( # T ' &amp; 2 n =1 Furthermore we showed that we could evaluate the coefficients in the first representation T /2 2 &quot; 2! mt % Am = )/2 g(t)cos \$ T ' dt # &amp; T (T (2-2) T /2 2 &quot; 2! mt % Bm = ) g(t)sin \$ T ' dt # &amp; T (T /2 For the second representation the coefficients are obtained by T /2 1 # i2&quot; mt &amp; Cm = )/2 g(t)exp % ! T ( dt \$ ' T !T (Note the negative sign in the exponential) We can extend the Fourier series to non-periodic functions that are defined over an infinite interval using Fourier transforms. To obtain these we formally let the interval T of the Fourier series become infinite. If we substitute =2m/T into equation (2-3) we get T /2 1 (2-4) C (! ) = # g(t)exp ( &quot;i! t ) dt T &quot;T /2 For any function which is given by a series, we can write a rather trivial result &quot; T &quot; 2\$ g ( t ) = # Gm = (2-5) # Gm 2\$ m = !&quot; T m = !&quot; If we plot Gm against , we can see right hand sum is just the area under the function, which if G is treated as a function of , is just g (t ) = # (2-3) \$ # &quot;# G (! ) d! . We can write &quot;# \$ G (! ) d! = # T 2% T /2 &quot;T /2 \$ C(! )exp ( i! t ) d! (2-6) Now for equations (2-4) and (2-6) we set G() = TC() and let T. This yields G (! ) = &quot;# \$ g(t)exp ( &quot;i! t ) dt # 1 g (t ) = \$ G(! )exp (i! t ) d! 2% &quot;# Alternatively, we can set G() = TC()/(2) and let T. This yields (2-7) 2-1 ESS 522 Spring 2007 G (! ) = g (t ) = 1 1 \$ ( 2&quot; ) 2 #\$ 1 1 \$ % g(t)exp ( #i! t ) dt (2-8) ( 2&quot; ) 2 #\$ # % G(! )exp (i! t ) d! Since = 2f, we can also write (2-7) as G( f ) = (2-9) # # 1 g (t ) = \$ G( f )exp (i2&quot; ft ) 2&quot; df = !# G( f )exp (i2&quot; ft ) df \$ 2&quot; !# Equations (2-7) to (2-9) are three alternate but equivalent representations of the Fourier transform pair (i.e., the Fourier transform and the <a href="/keyword/inverse-fourier-transform/" >inverse fourier transform</a> ) and you need to be careful to make sure which version is being used. In this class, we will follow the convection adopted in Bob Crosson's class notes and use (2-9). Properties of the Fourier Transform Differentiation is very easy in the frequency domain you just multiply by i \$ 1 g ' (t ) = % i&quot; G(&quot; )exp (i&quot; t ) d&quot; 2! #\$ !# \$ g(t)exp ( !i2&quot; ft ) dt (2-10) FT &amp; g ' ( t ) ( = i&quot; G(&quot; ) ' ) Similarly integration involves dividing by i % 1 G(# ) !t g ( s )ds = 2&quot; \$% i# exp (i# t ) d# + constant ! G(# ) FT &amp; ! g ( s )ds ( = + constant ' t ) i# Translation by a involves multiplying by exp(ia) ( 1 g (t + a ) = \$ &amp; ) G(&quot; )exp #i&quot; (t + a )% d&quot; 2! '( (2-11) (2-12) FT # g ( t + a ) % = exp ( i&quot; a ) G(&quot; ) \$ &amp; The -function The -function is a very useful function that we will come across repeatedly when consider filters and the discrete (digital) version of the Fourier transform. It is defined mathematically by 2-2 ESS 522 Spring 2007 ! ( x ) = 0, x &quot; 0 b ! ( x ) dx = 1, a &lt; 0 &lt; b (2-13) # f ( y )! ( y \$ x ) dy = f (x), a a&lt;y&lt;b It can be thought of an infinitely sharp narrow pulse it not physically realizable but for practical purposes it just needs to be narrower than a system can resolve (i.e., shorter in duration than the time between samples in digital system). The Fourier transform of a -function is D (! ) = \$ #\$ % &quot; (t)exp ( #i! t ) dt = 1 0 (2-14) All frequencies are equally represented. For a -function at time t0, the Fourier transform is D (! ) = \$ #\$ % &quot; (t # t )exp ( #i! t ) dt = exp ( #i! t 0 ) (2-15) We get a similar result when we consider the <a href="/keyword/inverse-fourier-transform/" >inverse fourier transform</a> of a -function in the frequency domain. % 1 (2-16) d (t ) = &amp; &quot; (# \$ # 0 )exp (i# t ) d# = exp (i# 0t ) 2! \$% In the exercise you will at a form of &quot;Uncertainty Principle&quot; - it is impossible to produce a signal that is narrow in both the time and frequency domain. Parseval's Theorem Consider a signal of infinite duration. For the Fourier transform to be convergent its energy must be bounded &quot; !&quot; # x(t) dt &lt; &quot...

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