2 Fourier series implications

Since a sum of even functions is itself an even function it is not unreasonable to suggest that a Fourier series containing only cosine terms (and perhaps a constant term which can also be considered as an even function) can only represent an even periodic function. Similarly a series of sine terms (and no constant) can only represent an odd function. These results can readily be shown more formally using the expressions for the Fourier coefficients a n and b n .

Task!

Recall that for a 2 π -periodic function

b n = 1 π π π f ( t ) sin n t d t

If f ( t ) is even, deduce whether the integrand is even or odd (or neither) and hence evaluate b n . Repeat for the Fourier coefficients a n .

We have, if f ( t ) is even,

f ( t ) sin n t = ( even ) × ( odd ) = odd

hence b n = 1 π π π (odd function) d t = 0

Thus an even function has no sine terms in its Fourier series.

Also f ( t ) cos n t = ( even ) × ( even ) = even

a n = 1 π π π ( even function ) d t = 2 π 0 π f ( t ) cos n t d t .

It should be obvious that, for an odd function f ( t ) ,

a n = 1 π π π f ( t ) cos n t d t = 1 π π π ( odd function ) d t = 0

b n = 2 π 0 π f ( t ) sin n t d t

Analogous results hold for functions of any period, not necessarily 2 π . For a periodic function which is neither even nor odd we can expect at least some of both the a n and b n to be non-zero. For example consider the square wave function:

Figure 17 :

{ Square wave}

This function is neither even nor odd and we have already seen in Section 23.2 that its Fourier series contains a constant 1 2 and sine terms.

This result could be expected because we can write

f ( t ) = 1 2 + g ( t )

where g ( t ) is as shown:

Figure 18

No alt text was set. Please request alt text from the person who provided you with this resource.

Clearly g ( t ) is odd and will contain only sine terms. The Fourier series are in fact

f ( t ) = 1 2 + 2 π sin t + 1 3 sin 3 t + 1 5 sin 5 t +

and

g ( t ) = 2 π sin t + 1 3 sin 3 t + 1 5 sin 5 t +

Task!

For each of the following functions deduce whether the corresponding Fourier series contains

  1. sine terms only or cosine terms only or both
  2. a constant term

    No alt text was set. Please request alt text from the person who provided you with this resource.

  1. cosine terms only (plus constant).
  2. cosine terms only (no constant).
  3. sine terms only (no constant).
  4. cosine terms only (plus constant).
  5. sine terms only (no constant).
  6. sine and cosine terms (plus constant).
  7. cosine terms only (plus constant).
Task!

Confirm the result obtained for the triangular wave, function 7 in the last Task, by finding the Fourier series fully. The function involved is

f ( t ) = t π < t < π f ( t + 2 π ) = f ( t )

Since f ( t ) is even we can say immediately

b n = 0 n = 1 , 2 , 3 ,

Also

a n = 2 π 0 π t cos n t d t = 0 n even 4 n 2 π n odd (after integration by parts)

Also a 0 = 2 π 0 π t d t = π  so the Fourier series is

f ( t ) = π 2 4 π cos t + 1 9 cos 3 t + 1 25 cos 5 t +