Connexions

You are here: Home » Content » Fourier Series: Eigenfunction Approach
Content Actions
Lenses

What is a lens?

Lenses

A lens is a custom view of Connexions content. You can think of it as a fancy kind of list that will let you see Connexions through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to Connexions materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual Connexions member, a community, or a respected organization.

This content is ...
Affiliated with (?)
This content is either by members of the organizations listed or about topics related to the organizations listed. Click each link to see a list of all content affiliated with the organization.

  • This module is included inLens: Rice University OpenCourseWare
    By: OpenCourseWare ConsortiumAs a part of collection:"Signals and Systems"

    Click the "Rice University OCW" link to see all content affiliated with them.

    Rice University OCW
Also in these lenses

  • This module is included inLens: richb's DSP resources
    By: Richard BaraniukAs a part of collection:"Signals and Systems"

    Comments:

    "My introduction to signal processing course at Rice University."

    Click the "richb's DSP" link to see all content selected in this lens.

    richb's DSP
Tags

(?)

These tags come from the endorsement, affiliation, and other lenses that include this content.

Fourier Series: Eigenfunction Approach

Module by: Justin Romberg

Summary: This module will introduce the Fourier Series and its Fourier coefficients using the concepts of eigenfunctions and basis. We will show several examples of how to decompose a signal and find the Fourier coefficients.

Introduction

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)
simpleLTIsys.png
Figure 1: Simple LTI system.
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:
  1. 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
  2. ∑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

Fourier Series

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.

Examples

For each of the given functions below, break it down into its "simpler" parts and find its fourier coefficients. Click to see the solution.
Problem 1
ft=cos ω 0 t f t ω 0 t
[ Click for Solution 1 ]
Solution 1
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
[ Hide Solution 1 ]
Problem 2
ft=sin2 ω 0 t f t 2 ω 0 t
[ Click for Solution 2 ]
Solution 2
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
[ Hide Solution 2 ]
Problem 3
ft=3+4cos ω 0 t+2cos2 ω 0 t f t 3 4 ω 0 t 2 2 ω 0 t
[ Click for Solution 3 ]
Solution 3
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
[ Hide Solution 3 ]

Fourier Coefficients

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.

Example: Using Fourier Coefficient Equation

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:
Problem 4
ft=1if|t|≤T0otherwise f t 1 t T 0
[ Click for Solution 4 ]
Solution 4
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
[ Hide Solution 4 ]

Summary: Fourier Series Equations

Our first equation is the synthesis equation, which builds our function, ft f t , by combining sinusoids.
Synthesis 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 .
Analysis 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 .
note: 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.
Example 1 
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.

Comments, questions, feedback, criticisms?

Send feedback