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Fourier Series

Module by: Paul Padley

Summary: A brief introduction to Fourier Series starting from the normal modes of an oscillating string. The concept is then extended to Fourier's integral theorem.

Fourier Analysis

Fourier Series

Lets go back to the case of a string fixed at 0 0 and L L , its n t h n t h harmonic is y n ( x , t ) = A n sin ( n π x L ) cos ( ω n t − δ n ) y n ( x , t ) = A n sin ( n π x L ) cos ( ω n t − δ n ) In fact all the modes could be permitted, and so any possible motion of the string can be completely specified by: y ( x , t ) = ∑ n = 1 ∞ A n sin ( n π x L ) cos ( ω n t − δ n ) . y ( x , t ) = ∑ n = 1 ∞ A n sin ( n π x L ) cos ( ω n t − δ n ) . This has been rigorously shown by mathematicians but the complete proof is beyond our scope in this course. Lets accept the mathematicians word on this. We could take a snapshot of this function at a time t = t 0 t = t 0 . Then we could write y ( x ) = ∑ n = 1 ∞ B n sin ( n π x L ) y ( x ) = ∑ n = 1 ∞ B n sin ( n π x L ) where B n = A n cos ( ω n t 0 − δ n ) . B n = A n cos ( ω n t 0 − δ n ) . Likewise we could look at one point at space and look at the oscillations as a function of time. In that case we would get. y ( t ) = ∑ n = 1 ∞ C n cos ( ω n t − δ n ) y ( t ) = ∑ n = 1 ∞ C n cos ( ω n t − δ n ) Lets work with the time snapshot, y ( x ) = ∑ n = 1 ∞ B n sin ( n π x L ) y ( x ) = ∑ n = 1 ∞ B n sin ( n π x L ) We need to figure out what the B n B n factors are and this is what Fourier figured out. We can multiply both sides by the sin sin of a particular harmonic y ( x ) s i n ( n i π x L ) = ∑ n = 1 ∞ B n sin ( n π x L ) s i n ( n i π x L ) y ( x ) s i n ( n i π x L ) = ∑ n = 1 ∞ B n sin ( n π x L ) s i n ( n i π x L ) and now we can integrate both sides Recall cos ( θ − φ ) = cos θ cos φ + sin θ sin φ cos ( θ − φ ) = cos θ cos φ + sin θ sin φ cos ( θ + φ ) = cos θ cos φ − sin θ sin φ cos ( θ + φ ) = cos θ cos φ − sin θ sin φ So sin θ sin φ = 1 2 [ c o s ( θ − φ ) − cos ( θ + φ ) ] sin θ sin φ = 1 2 [ c o s ( θ − φ ) − cos ( θ + φ ) ] Thus This is equal to zero at the limits 0 , L 0 , L except for the particular case when n = n i n = n i . In that case ∫ sin ( n π x L ) sin ( n i π x L ) ⅆ x = ∫ sin 2 ( n π x L ) ⅆ x ∫ sin ( n π x L ) sin ( n i π x L ) ⅆ x = ∫ sin 2 ( n π x L ) ⅆ x So you get After all that we should see that for each term in the sum is zero, except the case where n i = n n i = n . Thus we can simplify the equation: ∫ 0 L y ( x ) sin ( n π x L ) ⅆ x = L 2 B n . ∫ 0 L y ( x ) sin ( n π x L ) ⅆ x = L 2 B n . or B n = 2 L ∫ 0 L y ( x ) sin ( n π x L ) ⅆ x B n = 2 L ∫ 0 L y ( x ) sin ( n π x L ) ⅆ x The above is a very specific form of the Fourier Series for a function spanning an interval from 0 0 to L L and passing through zero at x = 0 x = 0 .

More General Case

One could write a more general case for the Fourier Series which applies to an interval spanning − L − L to L L and not constrained to pass through zero. In that case one can write y ( x ) = a 0 2 + ∑ n = 1 ∞ [ a n cos ( n π x L ) + b n sin ( n π x L ) ] y ( x ) = a 0 2 + ∑ n = 1 ∞ [ a n cos ( n π x L ) + b n sin ( n π x L ) ] where A n = 1 L ∫ − L L y ( x ) cos ( n π x L ) ⅆ x    n = 0 , 1 , 2 , 3 , … A n = 1 L ∫ − L L y ( x ) cos ( n π x L ) ⅆ x    n = 0 , 1 , 2 , 3 , … and B n = 1 L ∫ − L L y ( x ) sin ( n π x L ) ⅆ x    n = 1 , 2 , 3 , … B n = 1 L ∫ − L L y ( x ) sin ( n π x L ) ⅆ x    n = 1 , 2 , 3 , … You can then look at the symmetry of the problem and see if just sin sin or cos cos can be used. For example if y ( − x ) = y ( x ) y ( − x ) = y ( x ) then use cosines. If y ( − x ) = − y ( x ) y ( − x ) = − y ( x ) use the sines.

Fourier Integral Theorem

In fact Fourier's theorem can be taken to a next step. This is Fourier's integral theorem. That is any function (even if it is not periodic) can be represented by f ( x ) = 1 π ∫ 0 ∞ [ A ( k ) cos ( k x ) + B ( k ) s i n ( k x ) ] d k f ( x ) = 1 π ∫ 0 ∞ [ A ( k ) cos ( k x ) + B ( k ) s i n ( k x ) ] d k where A ( k ) = ∫ − ∞ ∞ f ( x ) cos ( k x ) ⅆ x A ( k ) = ∫ − ∞ ∞ f ( x ) cos ( k x ) ⅆ x B ( k ) = ∫ − ∞ ∞ f ( x ) sin ( k x ) ⅆ x B ( k ) = ∫ − ∞ ∞ f ( x ) sin ( k x ) ⅆ x A A and B B are called the Fourier transforms of f ( x ) f ( x ) Lets look at an example.

Figure 1
Figure 1 (fourier-box.png)
f ( x ) = E o | x | < L / 2 f ( x ) = 0 | x | > L / 2 f ( x ) = E o | x | < L / 2 f ( x ) = 0 | x | > L / 2 right away you can set B ( x ) = 0 B ( x ) = 0 from symmetry arguments A ( k ) = ∫ − ∞ ∞ f ( x ) cos ( k x ) ⅆ x = ∫ − L / 2 L / 2 E 0 cos ( k x ) ⅆ x = E o k sin ( k x ) | − L / 2 L / 2 = E o k [ sin ( k L 2 ) − sin ( − k L 2 ) ] = 2 E o k sin ( k L 2 ) = E 0 L sin ( k L 2 ) k L 2 A ( k ) = ∫ − ∞ ∞ f ( x ) cos ( k x ) ⅆ x = ∫ − L / 2 L / 2 E 0 cos ( k x ) ⅆ x = E o k sin ( k x ) | − L / 2 L / 2