Since complex
exponentials are
eigenfunctions of linear time-invariant (LTI)
systems, calculating the output of an LTI system
ℋℋ given
ⅇst
s
t
as an input amounts to simple multiplcation, where
Hs∈ℂ
H
s
is a constant (that depends on s). In the figure
below we have a simple exponential input that yields the
following output:
yt=Hsⅇst
y
t
H
s
s
t
(1)
Using this and the fact that ℋℋ
is linear, calculating
yt
y
t
for combinations of complex exponentials is also
straightforward. This linearity property is depicted in the
two equations below - showing the input to the linear system
HH on the left side and the
output,
yt
y
t
, on the right:
-
c
1
ⅇ
s
1
t+
c
2
ⅇ
s
2
t→
c
1
H
s
1
ⅇ
s
1
t+
c
2
H
s
2
ⅇ
s
2
t
c
1
s
1
t
c
2
s
2
t
c
1
H
s
1
s
1
t
c
2
H
s
2
s
2
t
-
∑n
c
n
ⅇ
s
n
t→∑n
c
n
H
s
n
ⅇ
s
n
t
n
c
n
s
n
t
n
c
n
H
s
n
s
n
t
The action of HH on an input such
as those in the two equations above is easy to explain:
ℋℋ independently
scales each exponential component
ⅇ
s
n
t
s
n
t
by a different complex number
H
s
n
∈ℂ
H
s
n
. As such, if we can write a function
ft
f
t
as a combination of complex exponentials it allows us to:
-
easily calculate the output of
ℋℋ given
ft
f
t
as an input (provided we know the eigenvalues
Hs
H
s
)
-
interpret how ℋℋ manipulates
ft
f
t
Joseph
Fourier demonstrated that an arbitrary T-periodic function
ft
f
t
can be written as a linear combination of harmonic
complex sinusoids
ft=∑n=-∞∞
c
n
ⅇⅈ
ω
0
nt
f
t
n
c
n
ω
0
n
t
(2)
where
ω
0
=2πT
ω
0
2
T
is the fundamental frequency. For almost all
ft
f
t
of practical interest, there exists
c
n
c
n
to make
Equation 2 true. If
ft
f
t
is finite energy (
ft∈L20T
f
t
L
0
T
2
), then the equality in
Equation 2
holds in the sense of energy convergence; if
ft
f
t
is continuous, then
Equation 2 holds
pointwise. Also, if
ft
f
t
meets some mild conditions (the Dirichlet
conditions), then
Equation 2 holds
pointwise everywhere except at points of discontinuity.
The
c
n
c
n
- called the Fourier coefficients -
tell us "how much" of the sinusoid
ⅇⅈ
ω
0
nt
ω
0
n
t
is in
ft
f
t
.
Equation 2 essentially breaks down
ft
f
t
into pieces, each of which is easily processed by an
LTI system (since it is an eigenfunction of
every LTI system). Mathematically,
Equation 2 tells us that the set of
harmonic complex exponentials
∀n,n∈ℤ:ⅇⅈ
ω
0
nt
n
n
ω
0
n
t
form a basis for the space of T-periodic continuous
time functions. Below are a few examples that are intended to
help you think about a given signal or function,
ft
f
t
, in terms of its exponential basis functions.
For each of the given functions below, break it down into
its "simpler" parts and find its fourier coefficients.
Click to see the solution.
The tricky part of the problem is finding a way to
represent the above function in terms of its basis,
ⅇⅈ
ω
0
nt
ω
0
n
t
. To do this, we will use our knowledge of
Euler's Relation to represent our
cosine function in terms of the exponential.
ft=12ⅇⅈ
ω
0
t+ⅇ-ⅈ
ω
0
t
f
t
1
2
ω
0
t
ω
0
t
Now from this form of our function and from Equation 2, by inspection we can see
that our fourier coefficients will be:
c
n
=12if|n|=10otherwise
c
n
1
2
n
1
0
ft=sin2
ω
0
t
f
t
2
ω
0
t
As done in the previous example, we will again use Euler's Relation to represent our sine function in terms
of exponential functions.
ft=12ⅈⅇⅈ
ω
0
t-ⅇ-ⅈ
ω
0
t
f
t
1
2
ω
0
t
ω
0
t
And so our fourier coefficients are
c
n
=-ⅈ2ifn=-1ⅈ2ifn=10otherwise
c
n
2
n
-1
2
n
1
0
ft=3+4cos
ω
0
t+2cos2
ω
0
t
f
t
3
4
ω
0
t
2
2
ω
0
t
Once again we will use the same technique as was used
in the previous two problems. The break down of our
function yields
ft=3+412ⅇⅈ
ω
0
t+ⅇ-ⅈ
ω
0
t+212ⅇⅈ2
ω
0
t+ⅇ-ⅈ2
ω
0
t
f
t
3
4
1
2
ω
0
t
ω
0
t
2
1
2
2
ω
0
t
2
ω
0
t
And from this we can find our fourier coefficients to
be:
c
n
=3ifn=02if|n|=11if|n|=20otherwise
c
n
3
n
0
2
n
1
1
n
2
0
In general
ft
f
t
, the Fourier coefficients can be calculated from
Equation 2 by solving for
c
n
c
n
, which requires a little algebraic manipulation (for
the complete derivation see the
Fourier coefficients derivation). The end
results will yield the following general equation for the
fourier coefficients:
c
n
=1T∫0Tftⅇ-ⅈ
ω
0
ntdt
c
n
1
T
t
T
0
f
t
ω
0
n
t
(3)
The sequence of complex numbers
∀n,n∈ℤ:
c
n
n
n
c
n
is just an alternate representation of the function
ft
f
t
. Knowing the Fourier coefficients
c
n
c
n
is the same as knowing
ft
f
t
explicitly and vice versa. Given a periodic
function, we can
transform it into it Fourier
series representation using
Equation 3. Likewise, we can
inverse
transform a given sequence of complex numbers,
c
n
c
n
, using
Equation 2 to
reconstruct the function
ft
f
t
.
Along with being a natural representation for signals being
manipulated by LTI systems, the Fourier series provides
a description of periodic signals that is convenient in many
ways. By looking at the Fourier series of a signal
ft
f
t
, we can infer mathematical properties of
ft
f
t
such as smoothness, existence of certain symmetries,
as well as the physically meaningful frequency content.
Here we will look at a rather simple example that almost
requires the use of Equation 3 to
solve for the fourier coefficients. Once you understand the
formula, the solution becomes a straightforward calculus
problem. Find the fourier coefficients for the following
equation:
ft=1if|t|≤T0otherwise
f
t
1
t
T
0
We will begin by plugging our above function,
ft
f
t
, into Equation 3.
Our interval of integration will now change to match the
interval specified by the function.
c
n
=1T∫-
T
1
T
1
1
ⅇ-ⅈ
ω
0
ntdt
c
n
1
T
t
T
1
T
1
1
ω
0
n
t
Notice that we must consider two cases:
n=0
n
0
and
n≠0
n
0
. For
n=0
n
0
we can tell by inspection that we will get
∀n,n=0:
c
n
=2
T
1
T
n
n
0
c
n
2
T
1
T
For
n≠0
n
0
, we will need to take a few more steps to
solve. We can begin by looking at the basic integral of
the exponential we have. Remembering our calculus, we
are ready to integrate:
c
n
=1T1ⅈ
ω
0
nⅇ-ⅈ
ω
0
nt|t=-
T
1
T
1
c
n
1
T
1
ω
0
n
t
T
1
T
1
ω
0
n
t
Let us now evaluate the exponential functions for the
given limits and expand our equation to:
c
n
=1T1-ⅈ
ω
0
nⅇ-ⅈ
ω
0
n
T
1
-ⅇⅈ
ω
0
n
T
1
c
n
1
T
1
ω
0
n
ω
0
n
T
1
ω
0
n
T
1
Now if we multiple the right side of our equation by
2ⅈ2ⅈ
2
2
and distribute our negative sign into the
parenthesis, we can utilize Euler's Relation to
greatly simplify our expression into:
c
n
=1T2ⅈⅈ
ω
0
nsin
ω
0
n
T
1
c
n
1
T
2
ω
0
n
ω
0
n
T
1
Now, recall earlier that we defined
ω
0
=2πT
ω
0
2
T
. We can solve this equation for T
T and substitute in.
c
n
=2ⅈ
ω
0
ⅈ
ω
0
n2πsin
ω
0
n
T
1
c
n
2
ω
0
ω
0
n
2
ω
0
n
T
1
And finally, if we make a few simple cancellations we
will arrive at our final answer for the Fourier
coefficients of
ft
f
t
:
∀n,n≠0:
c
n
=sin
ω
0
n
T
1
nπ
n
n
0
c
n
ω
0
n
T
1
n
Our first equation is
the synthesis equation, which builds our
function,
ft
f
t
, by combining sinusoids.
ft=∑n=-∞∞
c
n
ⅇⅈ
ω
0
nt
f
t
n
c
n
ω
0
n
t
(4)
And our
second
equation, termed the
analysis equation,
reveals how much of each sinusoid is in
ft
f
t
.
c
n
=1T∫0Tftⅇ-ⅈ
ω
0
ntdt
c
n
1
T
t
T
0
f
t
ω
0
n
t
(5)
where we have stated that
ω
0
=2πT
ω
0
2
T
.
Understand that our interval of integration does not have to
be
0T
0
T
in our Analysis Equation. We could use any
interval
aa+T
a
a
T
of length TT.
This demonstration lets you synthesize a signal by combining
sinusoids, similar to the synthesis equation for the Fourier
series. See here for
instructions on how to use the demo.
Eigenfunctions of LTI Systems
Fourier Series: Another Approach
"My introduction to signal processing course at Rice University."