ClaseNo8_2014

Lesson No. 7

Signal and System Analysis

Centro Universitario de los Valles

lunes 19 de mayo de 2014

Introduction to Fourier transform



From time domain to frequency domain.Find the contribution of different frequencies.

Discover hidden signal properties.



We will consider finite-length in CN.Fourier analysis is a simple change of basis.A change of basis is a change of perspective.A change of perspective can reveal new things.


wk[n] = ej2piN nk, n, k = 0, 1, ..., N 1

=2piNk

The Fourier Basis for CN

is an orthogonal basisk is the index of the vector in the Vector space , i.e, k is the signaln is the index of the each component of the vector k , i. e n indicates each element of the signal.

The fundamental frequency


{w(k)}k=0,1,...,N1

w(k)n = ej 2piN nk


In vector notation

with

is an orthogonal basis


Remember the exponential generating machine

Img

Rew1[0]

w1[1]

2pi/N

w1[2]

w(k)n = ej 2piN nk

k = 1

See the example in the course: Pintando los vectores en C-64


< w(k),w(h) > =N1n=0

(ej2piN nk)ej

2piN nh

=N1n=0

ej2piN (hk)n

=

{N for h = k,1ej2pi(hk)1ej 2piN (hk)

= 0 otherwise

a+ ar + ar2 + ar3 + ...+ arN1 =N1k=0

ark = a1 rN1 r

Proof of the orthogonality

Remeber !!!: The sum of the first n terms of geometric series is:


Remarks

N orthogonal vectors basis for CN

vectors are not orthonormal



in signal notation:

in vector notation:

wk[n] = ej2piN nk, n, k = 0, 1, ..., N 1

w(k)n = ej 2piN nk

{w(k)}k=0,1,...,N1


Xk =< w(k),x >

x =1N

N1k=0

Xkw(k)

Introduction to Fourier transformAnalysis formula

Synthesis formula


X[k] =N1n=0

x[n]ej2piN nk, k = 0, 1, ..., N 1

x[n] =1N

N1k=0

X[k]ej2piN nk, n = 0, 1, ..., N 1

Introduction to Fourier transformAnalysis formula

Synthesis formula

N- point signal in the frequency domain

N- point signal in the time domain


DFT of x[n]=[n], x[n] CN

X[k] =N1n=0

x[n] ej 2piN nk

= 1

plot the DFT using MATLAB


DFT of x[n]=1, x[n] CN

X[k] =N1n=0

ej2piN nk

= N[k]



x[n] = 3 cos(2pi16n)

= 3 cos(2pi64

4n)

=32[ej

2pi64 4n + ej

2pi64 4n]

=32[ej

2pi64 4n + ej

2pi64 60n]

=32[w4[n] + w60[n]]

cos() =ej + ej

2

DFT of x[n]=3 cos(2pi16

n), x[n] C64

j 2pi644n+ j2pin = j

2pi6460n


Euler equation

Two Fourier basis Vectors

Periodicity


X[k] = < wk[n], x[n] >

= < wk[n],32(w4[n] + w60[n]) >

=32< wk[n], w4[n] > +

32< wk[n], w60[n] > because of the linearity

=

{96 for k = 4, 600 otherwise


n), x[n] C64


x[n] = 3 cos(2pi16n+

pi

3)

= 3 cos(2pi64

4n+pi

3)

=32[ej

2pi64 4nej

pi3 + ej

2pi64 4nej

pi3 ]

=32[ej

pi3 w4[n] + ej

pi3 w60[n]]


n+pi

3), x[n] C64


X[k] = < wk[n], x[n] >

=

96ej pi3 for k = 496ej pi3 for k = 600 otherwise


n+pi

3), x[n] C64

Plot the real and imaginary part of the DFT.Plot the magnitude and plot the phase of the DFT.



), x[n] C64

2pi64

6

ClaseNo8_2014

Documents

Transcript of ClaseNo8_2014