5 Fourier series for functions of general period
This is a straightforward extension of the period case that we have already discussed.
Using (instead of ) temporarily as the variable. We have seen that a periodic function has a Fourier series
with
Suppose we now change the variable to where
Thus corresponds to and corresponds to
Hence regarded as a function of , we have a function with period .
Making the substitution , and hence , in the expressions for and we obtain
These integrals give the Fourier coefficients for a function of period whose Fourier series is
Various other notations are commonly used in this case e.g. it is sometimes convenient to write the period . (This is particularly useful when Fourier series arise in the solution of partial differential equations.) Another alternative is to use the angular frequency and put .
Task!
Write down the form of the Fourier series and expressions for the coefficients if
- .
-
with
and similarly for . -
with
and similarly for .
You should note that, as usual, any convenient integration range of length (or or ) can be used in evaluating and .
Example 1
Find the Fourier series of the function shown in Figure 11 which is a saw-tooth wave with alternate portions removed.
Figure 11
Solution
Here the period so . The Fourier series will have the form
The coefficients are given by
where
Hence
The integration is readily performed using integration by parts:
Hence, since
The constant term is where
Similarly
where
The second integral gives zero. Hence
Hence, using all these results for the Fourier coefficients, the required Fourier series is
Notice that because the Fourier coefficients depend on (rather than as was the case for the square wave) the sinusoidal components in the Fourier series have quite rapidly decreasing amplitudes. We would therefore expect to be able to approximate the original saw-tooth function using only a quite small number of terms in the series.
Task!
Obtain the Fourier series of the function
First write out the form of the Fourier series in this case:
Since and since the function has a non-zero average value, the form of the Fourier series is
Now write out integral expressions for and . Will there be a constant term in the Fourier series?
Because the function is non-negative there will be a constant term. Since then and we have
The constant term will be where . Now evaluate the integrals. Try to spot the value of the integral for so as to avoid integration. Note that the integrand is an even functions for and an odd functon for .
The integral for is zero for all because the integrand is an odd function of . Since the integrand is even in the integrals for we can write
The constant term will be where
For we must integrate by parts (twice)
The integral in the second term gives zero so
Now writing out the final form of the Fourier series we have
Exercises
For each of the following periodic signals
- sketch the given function over a few periods
- find the trigonometric Fourier coefficients
- write out the first few terms of the Fourier series.
- half-wave rectifier
- where .