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analisis de señales y sistemas
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Lesson No. 7
Signal and System Analysis
Centro Universitario de los Valles
lunes 19 de mayo de 2014
Introduction to Fourier transform
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.
lunes 19 de mayo de 2014
Introduction to Fourier transform
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.
lunes 19 de mayo de 2014
Introduction to Fourier transform
lunes 19 de mayo de 2014
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
lunes 19 de mayo de 2014
{w(k)}k=0,1,...,N1
w(k)n = ej 2piN nk
The Fourier Basis for CN
In vector notation
with
is an orthogonal basis
lunes 19 de mayo de 2014
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
lunes 19 de mayo de 2014
< 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:
lunes 19 de mayo de 2014
Remarks
N orthogonal vectors basis for CN
vectors are not orthonormal
lunes 19 de mayo de 2014
The Fourier Basis for CN
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
lunes 19 de mayo de 2014
Xk =< w(k),x >
x =1N
N1k=0
Xkw(k)
Introduction to Fourier transformAnalysis formula
Synthesis formula
lunes 19 de mayo de 2014
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
lunes 19 de mayo de 2014
DFT of x[n]=[n], x[n] CN
X[k] =N1n=0
x[n] ej 2piN nk
= 1
plot the DFT using MATLAB
lunes 19 de mayo de 2014
DFT of x[n]=1, x[n] CN
X[k] =N1n=0
ej2piN nk
= N[k]
plot the DFT using MATLAB
lunes 19 de mayo de 2014
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
plot the DFT using MATLAB
Euler equation
Two Fourier basis Vectors
Periodicity
lunes 19 de mayo de 2014
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
DFT of x[n]=3 cos(2pi16
n), x[n] C64
lunes 19 de mayo de 2014
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]]
DFT of x[n]=3 cos(2pi16
n+pi
3), x[n] C64
lunes 19 de mayo de 2014
X[k] = < wk[n], x[n] >
=
96ej pi3 for k = 496ej pi3 for k = 600 otherwise
DFT of x[n]=3 cos(2pi16
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.
lunes 19 de mayo de 2014
DFT of x[n]=3 cos(2pi10
), x[n] C64
2pi64
6