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Unformatted text preview: 18.03 Lecture #23 Oct. 30, 2009: notes Topic for today is (zero) more formulas for Fourier coefficients, then (first) solving differential equations with Fourier series. Stuck in the middle is a bit not covered in lecture about computing Fourier coefficients of a translated function; at the end is a little more about how fast Fourier coefficients go to zero. Part zero begins with something I have mentioned but never really touched: functions having a period other than 2 . Ill just write down some formulas, using notation that seems to be pretty common in engineering, and which is certainly used in EP. Remember that a function of a real variable t is periodic of period p if f ( t + p ) = f ( t ) , (all t R ) . What is standard is to write the period as 2 L (partly by analogy with the period 2 of sin and cos), so that the condition is f ( t + 2 L ) = f ( t ) , (all t R ) . (Periodic of period 2 L ) If f is such a function, its Fourier series looks like a 2 + summationdisplay n =1 ( a n cos( nt/L ) + b n sin( nt/L )) . (Fourier series of f ) The even terms are a 2 + summationdisplay n =1 a n cos( nt/L ) , (Cosine series of f ) and the odd terms summationdisplay n =1 b n sin( nt/L ) . (Sine series of f ) There is also a complex exponential version c + summationdisplay n =1 ( c n e int + c n e int ) . (Complex Fourier series of f ) The Fourier coefficients are given by a n = 1 L integraldisplay L L f ( t )cos( nt/L ) dt ( n = 0 , 1 , 2 , . . . ) , (Fourier cosine coefficients) b n = 1 L integraldisplay L L f ( t )sin( nt/L ) dt ( n = 1 , 2 , 3 , . . . ) . (Fourier sine coefficients) c n = 1 2 L integraldisplay L L f ( t ) e int/L dt ( n = 0 , 1 , 2 . . . ) . (Complex Fourier coefficients) Here are formulas (from the problem set) relating complex to sine and cosine coefficients: c n = 1 2 ( a n + ib n ) , c n = 1 2 ( a n- ib n ) ( n = 0 , 1 , 2 , . . . ) . (Complex from cosine and sine) 1 a n = c n + c n , b n = 1 i ( c n- c n ) . (Cosine and sine from complex) In each case the interval of integration can be replaced by any interval of length 2 L : c n = 1 2 L integraldisplay L + a L + a f ( t ) e int/L dt (Shifted formula for complex coefficients) A useful example is a = L : c n = 1 2 L integraldisplay 2 L f ( t ) e int/L dt. ( L-shifted formula for complex coefficients) I wont write the similar formulas for a n and b n . The reason this is true is that for any c and d , and any 2 L-periodic g , integraldisplay d c g ( t ) dt = integraldisplay d +2 L c +2 L g ( s- 2 L ) ds = integraldisplay d...
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This note was uploaded on 05/06/2010 for the course 18 18.03 taught by Professor Unknown during the Fall '09 term at MIT.
- Fall '09