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# Fourier Analysis - PowerPoint PPT Presentation

Fourier Analysis. PSCI 702 November 2, 2005. Even and Odd Functions. Even Functions. Odd Functions. Even and Odd Functions. Kronecker’s Rule. Periodic Functions. Trigonometric System. Trigonometric System of period 2a. Fourier Series.

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### Fourier Analysis

PSCI 702

November 2, 2005

Fourier Series
• The basic Idea behind Fourier series is to express a periodic function in terms of trigonometric system using the orthogonality relations.

Example for Fourier series

source: f(x) =|x|, -8 <= x <= 8

n= 5

n= 1

n= 3

y

Successive approximations of f(x)

f(x)

x

Even & Odd Extensions
• Let the function f be defined on (0,a). The even extension feand odd extension fo of f are the following functions
Square Wave Example

1 + 2 + 3

1 + 2

1

1

0

-

2

3

1 + 2 + 3 + 4 + 5

1 + 2 + 3 + 4

Square wave: Y = 0 for - < x < 0 and Y=1 for 0 < x < 

Y = 1/2 + 2/pi( sinx + sin3x/3 + sin5x/5 + sin7x/7 … + sin(2m+1)x/(2m+1) + …)

1 2 3 4 5

May do with sum of cosines too.

The Euler identity:

The inverse equations:

,

Using the formulas above and some properties of exponential function, the Fourier series can also be written as an expansion in terms of complex exponentials as:

,

Complex version of the Fourier expansion
Fourier Transform
• Let f is a piecewise smooth function defined over R. Since f may not be periodic, we define fL as the peridic extension of f over (-L,L). fL can be expressed as:
Fourier Transform: Properties #1

Linearity

Nth Derivative

Fourier Transform: Properties #2

Convolution

Translation: x-shift & -shift

F(w)

f(t)

t

F

1

w

t

-t/2

0

t/2

-6p

-4p

-2p

2p

4p

6p

0

t

t

t

t

t

t

Fourier Transform Pairs

F

Fourier Transform Pairs

F(w)

f(t)

(p)

(p)

t

w

0

-w0

w0

0

t

F(w)

w

-6p

-4p

-2p

2p

4p

6p

t

t

t

t

t

t

0

Duality
• Forward/inverse transforms are similar
• Example: rect(t/t)  t sinc(wt / 2)
• Apply duality t sinc(t t/2)  2 p rect(-w/t)
• rect(·) is even t sinc(t t /2)  2 p rect(w/t)

f(t)

1

t

-t/2

0

t/2

Parseval’s Equality
• Suppose that the f is piecewise continous function then:

Famous Fourier Transforms

Sine wave

Delta function

Famous Fourier Transforms

Sinc function

Square wave

Famous Fourier Transforms

Exponential

Lorentzian

Example problem

Find the Fourier transform of

Example problem: Answer.

Find the Fourier transform of

f(x) = Π(x /4) – Λ(x /2) + .5Λ(x)

Using the Fourier transforms of Π and Λ

and the linearity and scaling properties,

F(u) = 4sinc(4u) - 2sinc2(2u) + .5sinc2(u)

Example problem: Alternative Answer.

*

–1 -.5 0 .5 1

–2 1 0 1 2

Find the Fourier transform of

f(x) = Π(x /4) – 0.5((Π(x /3) * Π(x))

-

Using the Fourier transforms of Π and Λ

and the linearity and scaling and convolution properties ,

F(u) = 4sinc(4u) – 1.5sinc(3u)sinc(u)

The discrete Fourier transform

Motivation: computer applications of the Fourier transform require that all of the definitions and properties of Fourier transforms be translated into analogous statements appropriate to functions represented by a discrete set of sampling points rather than by continuous functions.

Let f(x) be a function.Let {fk = f(xk)} be a set of N function values, k = 0, 1, …, N-1.Let be the separation of the equidistant sampling points.Assumption: N is even.

The discrete Fourier transform is:

The inverse discrete transform is:

The discrete Fourier transform(2)

Let’s examine more closely the formula of the discrete Fourier transform:

We know that (it’s called n-th root of unity), so the formula above can be rewritten as:

DFT Example -1

Say we want to perform a 8 point DFT on a discretized version of a continuous input signal

having frequency components 1KHz and 2KHz

Calculation of Ts : Suppose

Period of x(t) = 1/1Khz = 1/1000

8 samples per period => sample time (Ts) = 1/8000 sec

Or sample rate = 8000 samples/s

t = nTs so

n = 0,1,…,7

DFT Example -1 (Contd…)

Therefore :

DC Component

And so on...

Where X(k) = (k*8Khz)/8

Actually evaluating we get the values:

X(0) = 0 + i 0 (dc)

X(2) = 1.414 + i1.414 (2Khz)

X(4) = 0 + i0 (4Khz)

X(6) = 1.414 – i 1.414 (6Khz)

X(1) = 0 – i 4 (1KHz)

X(3) = 0 + i 0 (3Khz)

X(5) = 0 + i 0 (5Khz)

X(7) = 0 + i 4 (7KHz)

Fast Fourier Transform
• A direct calculation of N-point DFT requires (N-1)2 multiplications and N(N-1) addition.
• The FFT is begun by noting that W0.5N=-1 and splitting the DFT in to two sums
Fast Fourier Transform
• Now we halve {fk} into two subsequences, according to whether k is even or odd
Some useful links
• http://www.falstad.com/fourier/
• Fourier series java applet