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FOURIER TRANSFORM (FT) AND

FAST FOURIER TRANSFORM (FFT)ALGORITHMS

Lectured by Assoc Prof. Dr. Thuong Le-TienSeptember 2011

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

Magnitude spectrum

of rectangular window

Rectangular window width

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To achieve a desired frequency resolution f.The smaller the desired separation, the

Longer the data record

The Hamming window

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

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

Frequency resolution

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Hamming window in the time and frequency domain

The minimum resolvable frequency differenceThe minimum resolvable frequency difference

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

Example:

A signal consisting of four sinusoids of frequencies of 1, 1.5, 2.5,and 2.75 kHz, 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 aswide 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 bef3 = 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 computation2.1. DTFT at a single frequency

Rectangular and hamming windows with L=40 and 100Rectangular 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 over2.2. DFT over frequency range: Compute DFT over

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2.3. DFTThe N points DFT of a L-length signal defined the DFT frequency as follows,

2.3. DFTThe 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.

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-transformEvaluation of z-transform

Nth roots of unity for N=8Nth 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

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

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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 DFT2.5. The matrix form of DFT

Denoted

Where the matrix components

defined by twiddle factors

Where the matrix components

defined by twiddle factors

(matrix form of DFT)(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 4-point 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.

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

5. Modulo N reduction

Example L=4N

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

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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 theLength-N wrapped signal x~ has the same N-point DFT as the original

Unwrapped signal x, that is:

The connection of the mod-N reduction to the DFT is the theorem that theLength-N wrapped signal x~ has the same N-point DFT as the original

Unwrapped 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:

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:In general A is partitioned in the form:

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

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

The two methods are the same

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

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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 byThe inverse DFT defined by

Where IN is the N-dimensional identity matrix and is the complex

conjugate of , obtained by conjugating every matrix element of .

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

*~ ~~

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Similar for FFT

Example for an inverse 4-point DFT

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Therefore the alternative form of IDFTTherefore the alternative form of IDFT

DFT and IDFTDFT 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)Discrete Fourier series (DFS)

X~ is periodic in n with period N

Sampling rate is a multiple of the fundamental frequency of signalSampling 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

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

* Which shows that fm will be aliased with fk. Therefore, if thesignal x(t) is sampled, it will give rise to the samples

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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 that

the 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 FFTIs 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.

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The summation index n ranges over both even and odd valuesin 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 separately. 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 buildsupper and lower halves of

length-N DFT

The butterfly merging buildsupper and lower halves of

length-N DFT

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and N=8and N=8

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The typical algorithm consists of three conceptual parts:1. Shuffling the N-dimensional input into N of 1-D signals2. Performing N one-point DFTs3. 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-DFTsinto the final 4-DFT

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

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Example: 8-point Inverse FFT

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