# 1.6 The fft algorithm

 Page 1 / 1
The FFT, an efficient way to compute the DFT, is introduced and derived throughout this module.

FFT
(Fast Fourier Transform) An efficient computational algorithm for computing the DFT .

## The fast fourier transform fft

DFT can be expensive to compute directly $\forall k, 0\le k\le N-1\colon X(k)=\sum_{n=0}^{N-1} x(n)e^{-(i\times 2\pi \frac{k}{N}n)}$

For each $k$ , we must execute:

• $N$ complex multiplies
• $N-1$ complex adds
The total cost of direct computation of an $N$ -point DFT is
• $N^{2}$ complex multiplies
• $N(N-1)$ complex adds
How many adds and mults of real numbers are required?

This " $O(N^{2})$ " computation rapidly gets out of hand, as $N$ gets large:

 $N$ 1 10 100 1000 $10^{6}$ $N^{2}$ 1 100 10,000 $10^{6}$ $10^{12}$

The FFT provides us with a much more efficient way of computing the DFT. The FFT requires only " $O(N\lg N)$ " computations to compute the $N$ -point DFT.

 $N$ 10 100 1000 $10^{6}$ $N^{2}$ 100 10,000 $10^{6}$ $10^{12}$ $N\lg N$ 10 200 3000 $6E6$

How long is $10^{12}\mathrm{sec}$ ? More than 10 days! How long is $6E6\mathrm{sec}$ ?

The FFT and digital computers revolutionized DSP (1960 - 1980).

## How does the fft work?

• The FFT exploits the symmetries of the complex exponentials ${W}_{N}^{(kn)}=e^{-(i\frac{2\pi }{N}kn)}$
• ${W}_{N}^{(kn)}$ are called " twiddle factors "

## Complex conjugate symmetry

${W}_{N}^{(k(N-n))}={W}_{N}^{-(kn)}=\overline{{W}_{N}^{(kn)}}$ $e^{-(i\times 2\pi \frac{k}{N}(N-n))}=e^{i\times 2\pi \frac{k}{N}n}=\overline{e^{-(i\times 2\pi \frac{k}{N}n)}}$

## Periodicity in n and k

${W}_{N}^{(kn)}={W}_{N}^{(k(N+n))}={W}_{N}^{((k+N)n)}$ $e^{-(i\frac{2\pi }{N}kn)}=e^{-(i\frac{2\pi }{N}k(N+n))}=e^{-(i\frac{2\pi }{N}(k+N)n)}$ ${W}_{N}=e^{-(i\frac{2\pi }{N})}$

## Decimation in time fft

• Just one of many different FFT algorithms
• The idea is to build a DFT out of smaller and smaller DFTs by decomposing $x(n)$ into smaller and smaller subsequences.
• Assume $N=2^{m}$ (a power of 2)

## Derivation

$N$ is even , so we can complete $X(k)$ by separating $x(n)$ into two subsequences each of length $\frac{N}{2}$ . $x(n)\to \begin{cases}\frac{N}{2} & \text{if n=\mathrm{even}}\\ \frac{N}{2} & \text{if n=\mathrm{odd}}\end{cases}$ $\forall k, 0\le k\le N-1\colon X(k)=\sum_{n=0}^{N-1} x(n){W}_{N}^{(kn)}$ $X(k)=\sum_{n=2r} x(n){W}_{N}^{(kn)}+\sum_{n=2r+1} x(n){W}_{N}^{(kn)}$ where $0\le r\le \frac{N}{2}-1$ . So

$X(k)=\sum_{r=0}^{\frac{N}{2}-1} x(2r){W}_{N}^{(2kr)}+\sum_{r=0}^{\frac{N}{2}-1} x(2r+1){W}_{N}^{((2r+1)k)}=\sum_{r=0}^{\frac{N}{2}-1} x(2r){W}_{N}^{2}^{(kr)}+{W}_{N}^{k}\sum_{r=0}^{\frac{N}{2}-1} x(2r+1){W}_{N}^{2}^{(kr)}$
where ${W}_{N}^{2}=e^{-(i\frac{2\pi }{N}\times 2)}=e^{-(i\frac{2\pi }{\frac{N}{2}})}={W}_{\frac{N}{2}}$ . So $X(k)=\sum_{r=0}^{\frac{N}{2}-1} x(2r){W}_{\frac{N}{2}}^{(kr)}+{W}_{N}^{k}\sum_{r=0}^{\frac{N}{2}-1} x(2r+1){W}_{\frac{N}{2}}^{(kr)}$ where $\sum_{r=0}^{\frac{N}{2}-1} x(2r){W}_{\frac{N}{2}}^{(kr)}$ is $\frac{N}{2}$ -point DFT of even samples ( $G(k)$ ) and $\sum_{r=0}^{\frac{N}{2}-1} x(2r+1){W}_{\frac{N}{2}}^{(kr)}$ is $\frac{N}{2}$ -point DFT of odd samples ( $H(k)$ ). $\forall k, 0\le k\le N-1\colon X(k)=G(k)+{W}_{N}^{k}H(k)$ Decomposition of an $N$ -point DFT as a sum of 2 $\frac{N}{2}$ -point DFTs.

Why would we want to do this? Because it is more efficient!

Cost to compute an $N$ -point DFT is approximately $N^{2}$ complex mults and adds.
But decomposition into 2 $\frac{N}{2}$ -point DFTs + combination requires only $\left(\frac{N}{2}\right)^{2}+\left(\frac{N}{2}\right)^{2}+N=\frac{N^{2}}{2}+N$ where the first part is the number of complex mults and adds for $\frac{N}{2}$ -point DFT, $G(k)$ . The second part is the number of complex mults and adds for $\frac{N}{2}$ -point DFT, $H(k)$ . The third part is the number of complex mults and adds for combination. And the total is $\frac{N^{2}}{2}+N$ complex mults and adds.

## Savings

For $N=1000$ , $N^{2}=10^{6}$ $\frac{N^{2}}{2}+N=\frac{10^{6}}{2}+1000$ Because 1000 is small compared to 500,000, $\frac{N^{2}}{2}+N\approx \frac{10^{6}}{2}$

So why stop here?! Keep decomposing. Break each of the $\frac{N}{2}$ -point DFTs into two $\frac{N}{4}$ -point DFTs, etc. , ....

We can keep decomposing: $\frac{N}{2^{1}}=\{\frac{N}{2}, \frac{N}{4}, \frac{N}{8}, , \frac{N}{2^{(m-1)}}, \frac{N}{2^{m}}\}=1$ where $m=\log_{2}N=\text{times}$

Computational cost: $N$ -pt DFTtwo $\frac{N}{2}$ -pt DFTs. The cost is $N^{2}\to 2\left(\frac{N}{2}\right)^{2}+N$ . So replacing each $\frac{N}{2}$ -pt DFT with two $\frac{N}{4}$ -pt DFTs will reduce cost to $2(2\left(\frac{N}{4}\right)^{2}+\frac{N}{2})+N=4\left(\frac{N}{4}\right)^{2}+2N=\frac{N^{2}}{2^{2}}+2N=\frac{N^{2}}{2^{p}}+pN$ As we keep going $p=\{3, 4, , m\}$ , where $m=\log_{2}N$ . We get the cost $\frac{N^{2}}{2^{\log_{2}N}}+N\log_{2}N=\frac{N^{2}}{N}+N\log_{2}N=N+N\log_{2}N$ $N+N\log_{2}N$ is the total number of complex adds and mults.

For large $N$ , $\mathrm{cost}\approx N\log_{2}N$ or " $O(N\log_{2}N)$ ", since $(N\log_{2}N, N)$ for large $N$ .

Weird order of time samples

can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
im not good at math so would this help me
yes
Asali
I'm not good at math so would you help me
Samantha
what is the problem that i will help you to self with?
Asali
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
what is nanomaterials​ and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!