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New Module Object Example (Derived copy of Fourier Series in a Nutshell)

Module by: Chinwei Hu. E-mail the author

Based on: Fourier Series in a Nutshell by Michael Haag, Justin Romberg

Summary: [abstract] This module will give a brief over of the key concepts involving the Fourier series and the tools used to decompose and approximate a given signal.

[Introduction]

The convolution integral is the fundamental expression relating the input and output of an LTI system. However, it has three shortcomings:

  1. It can be tedious to calculate.
  2. It offers only limited physical interpretation of what the system is actually doing.
  3. It gives little insight on how to design systems to accomplish certain tasks.
The Fourier Series, along with the Fourier Transform and Laplace Transofrm, provides a way to address these three points. Central to all of these methods is the concept of an eigenfunction (or eigenvector). We will look at how we can rewrite any given signal, f⁢t f t , in terms of complex exponentials.

In fact, by making our notions of signals and linear systems more mathematically abstract, we will be able to draw enlightening parallels between signals and systems and linear algebra.

[Theory] Eigenfunctions and LTI Systems

The action of a LTI system ℋ⁢… ℋ … on one of its eigenfunctions es⁢t s t is

  1. extremely easy (and fast) to calculate
    ℋ⁢es⁢t=H⁢s⁢es⁢t ℋ s t H s s t
    (1)
  2. easy to interpret: ℋ⁢… ℋ … just scales es⁢t s t , keeping its frequency constant.
If only every function were an eigenfunction of ℋ⁢… ℋ … ...

LTI System

... of course, not every function can be, but for LTI systems, their eigenfunctions span the space of periodic functions, meaning that for (almost) any periodic function f⁢t f t we can find c n c n where n∈Z n and c i ∈C c i such that:

f⁢t=∑ n =−∞∞ c n ⁢ei⁢ ω 0 ⁢n⁢t f t n c n ω 0 n t
(2)
Given Equation 2, we can rewrite ℋ⁢f⁢t=y⁢t ℋ f t y t as the following system

Figure 1: Transfer Functions modeled as LTI System.
Figure 1 (Transferfunc.png)

where we have: f⁢t==∑n c n ⁢ei⁢ ω 0 ⁢n⁢t f t n c n ω 0 n t y⁢t=ℋ⁢f⁢t=∑n c n ⁢H⁢i⁢ ω 0 nt⁢ei⁢ ω 0 ⁢n⁢t y t ℋ f t n c n H ω 0 n t ω 0 n t This transformation from f⁢t f t to y⁢t y t can also be illustrated through the process below. Note that each arrow indicates an operation on our signal or coefficients.

f⁢t→ c n → c n ⁢H⁢i⁢ ω 0 ⁢n→y⁢t f t c n c n H ω 0 n y t
(3)
where the three steps (arrows) in the above illustration represent the following three operations:
  1. Transform with analysis ( Fourier Coefficient equation): c n =1T⁢∫0Tf⁢t⁢e−(i⁢ ω 0 ⁢n⁢t)d t c n 1 T t T 0 f t ω 0 n t
  2. Action of ℋℋ on the Fourier series - equals a multiplication by H⁢i⁢ ω 0 ⁢n H ω 0 n
  3. Translate back to old basis - inverse transform using our synthesis equation from the Fourier series: y⁢t=∑ n =−∞∞ c n ⁢ei⁢ ω 0 ⁢n⁢t y t n c n ω 0 n t

[Application] Physical Interpretation of Fourier Series

The Fourier series c n c n of a signal f⁢t f t , defined in Equation 2, also has a very important physical interpretation. Coefficient c n c n tells us "how much" of frequency ω 0 ⁢n ω 0 n is in the signal.

Signals that vary slowly over time - smooth signals - have large c n c n for small nn.

Figure 2: We begin with our smooth signal f⁢t f t on the left, and then use the Fourier series to find our Fourier coefficients - shown in the figure on the right.
(a) (b)
Figure 2(a) (fsnut_1.png)Figure 2(b) (fsnut_2.png)

Signals that vary quickly with time - edgy or noisy signals - will have large c n c n for large nn.

Figure 3: We begin with our noisy signal f⁢t f t on the left, and then use the Fourier series to find our Fourier coefficients - shown in the figure on the right.
(a) (b)
Figure 3(a) (fsnut_3.png)Figure 3(b) (fsnut_4.png)

[Examples]

Example 1: Periodic Pulse

We have the following pulse function, f⁢t f t , over the interval −T2 T2 T 2 T 2 :

Figure 4: Periodic Signal f⁢t f t
Figure 4 (fsnut_e1.png)

Using our formula for the Fourier coefficients,

c n =1T⁢∫0Tf⁢t⁢e−(i⁢ ω 0 ⁢n⁢t)d t c n 1 T t T 0 f t ω 0 n t
(4)
we can easily calculate our c n c n . We will leave the calculation as an exercise for you! After solving the the equation for our f⁢t f t , you will get the following results:
c n ={2⁢ T 1 T  if  n=02⁢sin ω 0 ⁢n⁢ T 1 n⁢π  if  n≠0 c n 2 T 1 T n 0 2 ω 0 n T 1 n n 0
(5)
For T 1 =T8 T 1 T 8 , see the figure below for our results:

Figure 5: Our Fourier coefficients when T 1 =T8 T 1 T 8
Figure 5 (fsnut_e2.png)

Our signal f⁢t f t is flat except for two edges (discontinuities). Because of this, c n c n around n=0 n 0 are large and c n c n gets smaller as nn approaches infinity.

Question:

Why does c n =0 c n 0 for n=…-44816… n … -4 4 8 16 … ? (What part of e−(i⁢ ω 0 ⁢n⁢t) ω 0 n t lies over the pulse for these values of nn?)

[Lab]

Numerical Evaluation of Fourier Series http://www.rose-hulman.edu/DSPFirst/visible3/chapters/10confrq/labsLV/lab13/lab13.pdf

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[Summary]

Fourier analysis is very useful!

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