Time-Frequency Analysis

Time-Frequency Analysis Fourier Transform and Fast Fourier Transform Philippe B. Laval KSU Fall 2015 Philippe B. Laval (KSU) FFT Fall 2015 1/8 ...
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Time-Frequency Analysis Fourier Transform and Fast Fourier Transform Philippe B. Laval KSU

Fall 2015

Philippe B. Laval (KSU)

FFT

Fall 2015

1/8

Introduction One way to look at Fourier series is that it is a transformation from the time domain to the frequency domain. Given a signal f (x), …nding its Fourier series on an interval [ L; L], 2n amounts to …nding how much each frequency of the form L contributes to the signal. The Fourier transform is similar but it is de…ned over the entire real line that is on ( 1; 1).

It also covers all the possible frequencies, not just frequencies of the 2n . form L The Fourier transform has proven to be an incredible tool in mathematics as well as many applied …elds.

Philippe B. Laval (KSU)

FFT

Fall 2015

2/8

De…nitions The Fourier transform is an integral transform, de…ned over the entire real line, that is for x 2 [ 1; 1]. It decomposes a function of time (some signal) into the frequencies that make up that signal. The Fourier transform and its inverse always go in pair. They can be viewed as transformations between the time domain and the frequency domain. There are di¤erent notations used for the Fourier transform. We can think of it as an operator F which acts on functions (we compute the Fourier transform of a function). In this case, the Fourier transform of a function f is denoted F ff (x)g. We can also think of it as a function which lives in the frequency domain; the notation used is b f (w ). The variable !, used to represent frequencies reminds us that the Fourier transform operates in the frequency domain. F ff (x)g or b f (w ) tells us how much the frequency ! contributes to the signal f (x).

Philippe B. Laval (KSU)

FFT

Fall 2015

3/8

De…nitions

De…nition Let f (x) be some 1-dimensional signal ; x 2 ( 1; 1). 1

2

The Fourier transform of f is de…ned by 1 F ff (x)g (!) = b f (w ) = p 2

Z

1

f (x) e

i !x

dx

1

Its inverse is de…ned by f (x) = F

Philippe B. Laval (KSU)

1

Z n o 1 b f (w ) (x) = p 2

FFT

1 1

b f (w ) e i !x dw

Fall 2015

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De…nitions Let us make some remarks:

Remark 1

2

The Fourier transform is a continuous function of !. If we think of ! as the frequency, then b f (w ) gives us the contribution the frequency ! makes to the signal. The analogy with Fourier series is that the Fourier transform corresponds to the coe¢ cients of the Fourier series.

3

Using the inverse Fourier transform, we can recover the signal from its Fourier transform.

4

The Fourier transform allows us to identify the frequencies which play an important role in a signal.

5

The Fourier transform allows us to …lter a signal by removing certain frequencies from it. Philippe B. Laval (KSU)

FFT

Fall 2015

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Example

Though we will not focus on computing Fourier transforms, we give an example of how it is computed.

Example Find the Fourier transform of f (x) = e > 0.

Philippe B. Laval (KSU)

FFT

jx j

where x 2 ( 1; 1) and

Fall 2015

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The Fast Fourier Transform (FFT) In theory a signal is a continuous function of x (or t) and its transform a continuous function of !. In practice, we sample a signal, in other words we measure the signals at a …nite number of time values. Hence, instead of having an integral transform (the Fourier transform), we have a discrete transform, also known as the discrete Fourier transform or DFT . One problem with the DFT is that the number of operations needed to evaluate it are of the order of n2 that is the algorithm id O n2 . For large n, this is signi…cant. In the 1960,s, Cooley and Tukey developed a much faster algorithm to compute the DFT (see the Cooley and Tukey algorithm).It is now commonly known as the Fast Fourier Transform (FFT) algorithm. This algorithm is so good that it is considered one of the top ten algorithms of the twentieth century. It reduced the number of operations needed to compute the DFT drastically. The FFT algorithm is O (n log n).

Philippe B. Laval (KSU)

FFT

Fall 2015

7/8

Exercises

See the problems at the end of the notes on the Fourier transform.

Philippe B. Laval (KSU)

FFT

Fall 2015

8/8