C6-Discrete Time Fourier (DFT) and Fast Fourier Transform (FFT)

DIGITAL SIGNAL PROCESSING

FOURIER TRANSFORM (FT) AND FAST FOURIER TRANSFORM (FFT) ALGORITHMS

Lectured Lectured by Assoc Prof. Dr. Dr. Thuong Thuong Le-Tien Le-Tien September 2011

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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

ˆ ( f ) In terms of the time samples x(nT), the original sampled spectrum X ˆ ( f ) are given by: and its time-windowed version X L

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

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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

ˆ ( f ) In terms of the time samples x(nT), the original sampled spectrum X ˆ ( f ) are given by: and its time-windowed version X L


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

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Frequency resolution To achieve a desired frequency resolution f. The smaller the desired separation, the Longer 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 DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien

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

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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 f 1=1kHz and f 2=2khz. It is also known that the signal contains a third component of frequency f 3 that lies somewhere between f 1 and f 2. a. How close to f 1 could f 3 be in order for the spectrum of the collected samples to exhibit three distinct peak? How close to f 2 could f 3 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= f sTL =10x10=100. The frequency resolution of the rectangular window is f = f s /L = 10/100 = 0.1kHz Thus the closest f 3 to f 1 and f 2 will be f 3 = f 1 + f = 1.1kHz and f 3 = f 2 - f = 1.9kHz In the hamming case, the minimum resolvable frequency separation doubles, that is, f = cf s /L = 2.10/100 = 0.2kHz which give f 3 = 1.2kHz or f 3 = 1.8kHz DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien

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2. DTFT computation 2.1. DTFT at a single frequency DTFT of length-L signal

Rectangular and hamming windows with L=40 and 100 DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien

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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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N-point DTFTs over [0,2 ) and over subinterval [

a,

b),

for N=10

Evaluation of z-transform

The periodicity of X() with a period of 2 or DFT X(k)=X(k) in the index k with period N Nth roots of unity for N=8 DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien

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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 (matrix form of DFT)

Where the matrix components defined by twiddle factors DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien

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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.


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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 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:

In general A is partitioned in the form:


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N-point DFTs of the full and wrapped signal are equivalent DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien

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

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6. Inverse DFT The 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 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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In term of the DFT frequencies

k,

we have Xk = X(

k

) and

Therefore the alternative form of IDFT

DFT and IDFT


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7. Sampling of periodic signals and DFT X~ is periodic in n with period N

Discrete Fourier series (DFS)

Sampling rate is a multiple of the fundamental frequency of signal

Taking the Nyquist interval to be the right-sided one [0, f s], 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

* Which shows that f m will be aliased with f k . Therefore, if the signal x(t) is sampled, it will give rise to the samples

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 f 1=1 Hz. For a sampling rate of f s = 4Hz, consider one period Consisting of N=4 samples and perform its 4-point DFT

The Fourier coefficients: Corresponding to the harmonic Where f 3 = 3 was replaced by its negative version f 3-f s = 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 f k = k f 1


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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.


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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 two Terms 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,

0  2n  N  2



0  n  N / 2  1

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

0  2n  1  N  1

1  2n  1  n  1



0  2n  N  2



0  n  N / 2  1


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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 of 1-D 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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Example: Using FFT algorithm, compute 8-point DFT of the 8 point signal


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C6-Discrete Time Fourier (DFT) and Fast Fourier Transform (FFT)

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