18.03 Class 26, Apr 12
Fourier Series
A function f(t) is "periodic of period 2L" if f(t+2L) = f(t).
Waves, vibrations, ... many things phenomena are periodic.
A basic example of a periodic function is given by cos(t), which
is periodic of minimal period 2pi. The is such a basic example
that for today we'll consider only period 2pi: L = pi .
Other examples are cos(nt), n = 2, 3, ...; sin(nt), n = 1, 2, ...,
and the constant function 1 = cos(0t).
We know a lot about these functions. We know for example how to solve
LTI equations having these as signals. So by superposition we know
how to deal with linear combinations of them as well. The most general
linear combination is
(a0/2) + a1 cos(t) + a2 cos(2) + ... + b1 sin(t) + b2 sin(2t) + ...
(We'll see in a minute why it was smart to write the constant term this way.)
This is a "Fourier series."
Theorem: Essentially any periodic function with period 2pi can be written
as a Fourier series: f(t) = this sum.
This will allow us to use very general periodic functions as signals.
To be effective though we have to figure out how to compute an and bn
from f(t).
a0 is a good place to start: All the terms beyond the first one have
zero average (over any interval of length 2pi). Thus
a0/2 = average of f(t) = (1/2pi) integral_{-pi}^pi f(t) dt so
a0 = (1/pi) integral_{-pi}^pi f(t) dt .
Recalling that cos(0t) = 1, it might be sensible to try to compute an
by looking at
integral_{-pi}^pi f(t) cos(nt) dt
Putting the various terms of the proposed Fourier series for f(t) into
this integral one at a time, we see that we need the following trigonometric
integrals: for n, k > 0:
integral_{-pi}^pi cos(kt) cos(nt) dt = pi if n = k
otherwise
integral_{-pi}^pi sin(kt) cos(nt) dt = 0
integral_{-pi}^pi sin(kt) sin(nt) dt = pi if n = k
0 otherwise
We find that only one term is nonzero in the sum:
integral_{-pi}^pi f(t) cos(nt) = an pi .
(Notice that this is correct for n = 0 as well as n > 0. This is
the reason for writing the constant term as a0/2.) Similarly,
integral_{-pi}^pi f(t) sin(nt) = bn pi .
That's it: each coefficient turns out to be, necessarily, the integral
of a certain trig function times f(t).
Example: sq(t) = 1 for 0 < t < pi,
-1 for -pi < t < 0 .
The average is 0 : a0 = 0.
sq(t) is an odd function:
A function f(t) is "odd" if f(-t) = -f(t),
"even" if f(-t) = f(t).
Any odd function has average (over -pi < t < pi) zero.
Even times odd is odd
Even times even is even
Odd times odd is even
and
integral_{-pi}^pi odd dt = 0
integral_{-pi}^pi even dt = 2 integral_0^pi even dt
If f(t) is any odd function then f(t) cos(nt) is again odd and has
zero integral: an = 0. (Similarly, if f(t) is an even function then
bn = 0.)
Thus an = 0, and
bn = (2/pi) integral_0^pi sin(nt) dt
= - (2/(pi n)) [cos(nt)]^pi_0.
Since cos(n pi) = +1 if n is even
-1 if n is odd
bn = - (2/(pi n)) [1 - 1] = 0 if n is even
= - (2/(pi n)) [-1 -1] = 4/(n pi) if n is odd.
We have computed the Fourier series for sq(t):
sq(t) = (4/pi)( sin(t) + (1/3) sin(3t) + (1/5) sin(5t) + ... )
I showed some slides of how this builds up.
Once you have computed one Fourier series you can find lots of others.
For example suppose f(t) is periodic of period 2pi and f(t) = 0 for
0 < t < pi, f(t) = -1 for pi < t < 2pi.
Then also f(t) = -1 for -pi < t < 0. We could compute the Fourier
coefficients just as we just did; or we can realize that
f(t) = -(1/2) + (1/2) sq(t) and so
f(t) = -(1/2) + (2/pi)( sin(t) + (1/3) sin(3t) + ...) .
Fourier series constitute a new kind of "transform." Notice the analogy
with Laplace transform:
Laplace: (functions of t > 0) ----> (Functions of s)
Fourier: (functions periodic of period 2pi) ---> (a0 , a1, ..., b1, b2, ...)
Each coefficient an , bn , is like a value of the Laplace transform;
it is given by integrating f(t) multiplied, not by e^{-st} but rather
by a sinusiodal function. We are in a better situation here, in that
we can be explicit about the inverse transform; this is given exactly by
the Fourier series.