Digital Signal Processing - Fourier transform (FT) and fast fourier transform (FFT) Algorithms

angular window
Rectangular window width
5DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
6DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
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
Frequency resolution
7DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
Hamming window in the time and frequency domain
The minimum resolvable frequency difference
8
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 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 
9DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
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 
10DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
2. DTFT computation
2.1. DTFT at a single frequency
Rectangular and hamming windows with L=40 and 100l i i i
DTFT of length-L signal
11
DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
Equivalent Nyquist Interval
2.2. DFT over frequency range: Compute DFT over
12
DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
2.3. DFT
The N points DFT of a L-length signal defined the DFT frequency as follows,i l i l i ll
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.
13DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
Evaluation of z-transforml i
Nth roots of unity for N=8i
The periodicity of X() with a period of 2 or 
DFT X(k)=X(k) in the index k with period N
i i i i i
i i i i
N-point DTFTs over [0,2) and over subinterval [a, b), for N=10
14DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
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 
15DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
The DTFT and DFT 
2.5. The matrix form of DFT
Denoted 
Where the matrix components
defined by twiddle factors
i
i i l
(matrix form of DFT)
16DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
The twiddle factor defined by
For example: L=N and N=2, 4, 8
The corresponding 2-point and 4-point DFT matrices are:
17DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
17
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.
18
Twiddle factor look up tables for N=2, 4, 8
5. Modulo N reduction
Example L=4N
DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
19DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
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
20
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:
21DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
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:
22DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
23
DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
N-point DFTs of the full and wrapped signal are equivalent
24DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
The same DFT can be computed by the DFT matrix 
x~ 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
25DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
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 byi i
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:
*~A
A
~
A
~
26DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
Similar for FFT
Example for an inverse 4-point DFT
27DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
Therefore the alternative form of IDFT
DFT and IDFT
In term of the DFT frequencies k , we have Xk = X(k ) and 
28DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
7. Sampling of periodic signals and DFT
Discrete Fourier series (DFS)i i i
X~ is periodic in n with period N
Sampling rate is a multiple of the fundamental frequency of signalli i l i l l i l
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
29DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
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 the 
signal x(t) is sampled, it will give rise to the samples 
30DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
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
31DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
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
32DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
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. 
33DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
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, 
12/0220  NnNn
Similarly, for the odd-indexed terms, we must have 
1120  Nn
12/02201121  NnNnnn
34DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
35DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
36DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
The butterfly merging builds
upper and lower halves of 
length-N DFT 
37DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
and N=8
38
DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
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
39DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
40
DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
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
41DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien
Example: Using FFT algorithm, compute 8-point DFT of the 8 point signal
42
Example: 8-point Inverse FFT
DSP lectured by Assoc. Prof. Dr. Thuong Le-Tien