fft2 algorithm

7/28/2019 FFT2 Algorithm

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DSP lectured by Assoc. Prof. Dr.Thuong Le-Tien

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DISCRETE FOURIER TRANSFORM (DFT)

AND FAST FOURIER TRANSFORM (FFT)

ALGORITHM

Lectured by Assoc Prof. Dr. Thuong Le-Tien

DIGITAL SIGNAL PROCESSING


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1. Frequency resolution and windowing

Spectrum of sampled analog signal

But if the replicas overlap they will contribute to the right hand side of spectrum

In terms of the time samples x(nT), the original sampled spectrum

and its time-windowed version are given by:

)( fX

)( fXL


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Rectangular window of length L

Then define the windowed signal

The DTFT of windowed signal is

Where W( ) is the DTFT of the rectangular window w(n)


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Setting W(n) = 1

Magnitude spectrum

of rectangular window

Rectangular window width


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These equation can be rewritten to give the minimum number of samples required

To achieve a desired freq resolution f. The smaller the desired separation, theLonger the data record

The Hamming window

At its center, n=(L-1)/2, the value of w(n) is 0.54+0.46 = 1, and at its endpoint,

n=0 and n=L-1, its value is 0.54-0.46 = 0.08

For any type of window, the effective of the

mainlobe is inversely proportional to L

c is a constant and always c=>1

Frequency resolution


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Hamming window in the time and frequency domain

The minimum resolvable frequency difference


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Example:

A signal consisting of four sinusoids of frequencies of 1, 1.5, 2.5, and 2.75kHz, is sampled at a rate of 10 kHz. What is the minimum number of

samples that should be collected for the frequency spectrum to exhibit four

distinct peaks at these frequencies? How many samples should be collected

if they are going to be preprocessed by a Hamming window and then Fourier

transformed?

Solution:

The smallest frequency separation that must be resolved by the DFT is

f = 2.75-2.5=0.25 kHz, for rectangular window:

Because the mainlobe width of the Hamming window is twice as wide as

that of the rectangular window, it follows that twice as many samples must

be collected, that is L=80 then c can be calculated to be c=2


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Example:

A 10ms portion of a signal is sampled at a rate of 10kHz. It is known

that the signal consists of two sinusoids of frequencies f1=1kHz and

f2=2khz. It is also known that the signal contains a third component

of frequency f3 that lies somewhere between f1 and f2.a. How close to f1 could f3 be in order for the spectrum of the collected

samples to exhibit three distinct peak? How close to f2 could f3 be?

b. What are the answers if the collected samples are windowed by a

Hamming window?

Solution:

The total number of samples collected is L= fsTL =10x10=100.

The frequency resolution of the rectangular window is

f = fs/L = 10/100 = 0.1kHz

Thus the closest f3

to f1

and f2

will be

f3 = f1 + f = 1.1kHz and f3 = f2 - f = 1.9kHz

In the hamming case, the minimum resolvable frequency separation doubles,

That is,

f = cfs/L = 2.10/100 = 0.2kHz which give f3 = 1.2kHz or f3 = 1.8kHz


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2. DTFT computation

2.1. DTFT at a single frequency

Rectangular and hamming windows with L=40 and 100

DTFT of length-L signal


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Equivalent Nyquist Interval

2.2. DFT over frequency range: Compute DFT over


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2.3. DFT

The N points DFT of a L-length signal defined the DFT frequency as follows,

The only difference between DFT and DTFT is that the former has its N

frequencies distributed evenly over the full Nyquist interval [0, 2 ) whereas

the later has them distributed over any desired subinterval.


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Evaluation of z-transform

Nth roots of unity for N=8

The periodicity of X( ) with a period of 2 or

DFT X(k)=X( k) in the index k with period N

N-point DTFTs over [0,2 ) and over subinterval [ a, b), for N=10


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2.4. Zeros padding

Note that evaluation at the N frequencies DFT are the same for the cases of padding

D zeros at front or delay D samples


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The DTFT and DFT

2.5. The matrix form of DFT

Denoted

Where the matrix components

defined by twiddle factors

(matrix form of DFT)


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The twiddle factor defined by

For example: L=N and N=2, 4, 8

The corresponding 2-point and 4-point DFT matrices are:


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And the 2-point and 4ponit DFT of a length 2 and length 4 signals will be

Thus, the 2-point DFT is formed by taking the sum and difference of the two time

Samples. It will be a convenience starting point for the merging in FFT by hand.


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Twiddle factor look up tables for N=2, 4, 8

5. Modulo N reduction

Example L=4N


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Example: Determine the mod-4 and mod-3 reduction of the length-8 signal vector

For N=4 and N=3

For n=0, 1, 2, , N-1


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Periodic extension interpretation of mod-N reduction of a signal

The connection of the mod-N reduction to the DFT is the theorem that the

Length-N wrapped signal x~ has the same N-point DFT as the originalUnwrapped signal x, that is:


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The DFT matrices A and A~ have the same definition, except they differ in

their dimensions, which are NxL and NxN, respectively. We can write the

DFT of x~ in the compact matrix form:

In general A is partitioned in the form:


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DSP lectured by Assoc. Prof. Dr.Thuong Le-Tien 23

N-point DFTs of the full and wrapped signal are equivalent


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Example: Compute the 4-point DFT of the length-8 signal in two way: (a) working

With the full unwrapped vector x and (b) computing the DFT of its mod-4 reduction

Solution: The corresponding DFT is

The same DFT can be computed by the DFT matrix x~ acted on the wrapped signal x~

The two methods are the same


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6. Inverse DFTThe problem for inverse DFt is the length L of signal greater than N-point DFT,

i.e. the matrix A is not invertible

The inverse DFT defined by

Where IN is the N-dimensional identity matrix and is the complexconjugate of , obtained by conjugating every matrix element of .

For example, for N=4, we can verify easily:

*~A

A~A~


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Similar for FFT

Example for an inverse 4-point DFT


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Therefore the alternative form of IDFT

DFT and IDFT

In term of the DFT frequencies k , we have Xk = X( k ) and


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7. Sampling of periodic signals and DFT

Discrete Fourier series (DFS)

X~ is periodic in n with period N

Sampling rate is a multiple of the fundamental frequency of signal

Taking the Nyquist interval to be the right-sided one [0, fs], we note that

harmonics within that interval are none other than the N DFT frequencies


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Given an integer m, we determine its quotient and reminder of the division

And therefore the corresponding harmonic will be

Defining the aliased Fourier series amplitudes


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If the sampled signal x(nT) be reconstructed by an ideal reconstructor,

the aliased analog waveform is

Example: determine the aliased signal xal(t) resulting by sampling a square

Wave of frequency f1=1 Hz. For a sampling rate of fs = 4Hz, consider one period

Consisting of N=4 samples and perform its 4-point DFT

The Fourier coefficients:

Corresponding to the harmonic

Where f3 = 3 was replaced by its negative version f3-fs = 3-4 = -1. It follows thatthe aliased signal will be


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Similarly, for N=8 corresponding to fs=8 Hz, we perform the 8-point DFT of one

period of the square wave, and divide by 8 to get the aliased amplitudes

These amplitudes corresponding to the frequencies fk = k f1


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8. Fast Fourier Transform FFT

Is a fast implementation of DFT. It is based on a divide and conquer

approach in which the DFT computation is divided into smaller, simpler,

problems and the final DFT is rebuilt from the simpler DFTs.

It is required the initial dimension of N to be power of two

The problem of computing the N-point DFT is replaced by the simpler problems

Of computing two (N/2)-point DFT. Each of these is replaced by two (N/4)-point

DFTs , and so on.


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The summation index n ranges over both even and odd values in the range

[0,N-1]. By grouping the even-indexed and odd-indexed terms, we get

To determine the proper range of summation over n, we consider the twoTerms seperately. For even-indexed terms, the index 2n must be within the

range [0,N-1]. But, because N is even (a power of two), the upper limit

N-1 will be odd. Therefore, the highest even index will be N-2,

12/0220 NnNn

Similarly, for the odd-indexed terms, we must have

1120 Nn

12/02201121 NnNnnn


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The butterfly merging builds

upper and lower halves of length-N DFT


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and N=8


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The typical algorithm consists of three conceptual parts:

1. Shuffling the N-dimensional input into N one-dimensional signals

2. Performing N one-point DFTs

3. Merging the N one-point DFTs into one N-point DFT


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Example: Using FFT algorithm, compute the 4-point wrapped signal

(5, 0, -3, 4)

Solution:

The DFT merging stage merges the two 2-DFTs into the final 4-DFT


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DSP l t d b A P f D 41

Example: Using FFT algorithm, compute 8-point DFT of the 8 point signal

fft2 algorithm

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