22.058 - lecture 5, Sampling and the Nyquist Condition Sampling Outline FT of comb function ...
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Transcript of 22.058 - lecture 5, Sampling and the Nyquist Condition Sampling Outline FT of comb function ...
22.058 - lecture 5, Sampling and the Nyquist Condition
Sampling
Outline FT of comb function Sampling• Nyquist Condition sinc interpolation• Truncation• Aliasing
22.058 - lecture 5, Sampling and the Nyquist Condition
Sampling
As we saw with the CCD camera and the pinhole imager, the detector plane is not a continuous mapping, but a discrete set of sampled points. This of course limits the resolution that can be observed.
22.058 - lecture 5, Sampling and the Nyquist Condition
Limits of Sampling
• Finite # of data points
• Finite field of view
• High spatial frequency features can be missed or recorded incorrectly
22.058 - lecture 5, Sampling and the Nyquist Condition
Formalism of Sampling
€
fn{ }= TopHat2xfov
⎛
⎝ ⎜
⎞
⎠ ⎟
−∞
∞∫ Comb
xΔx
⎛ ⎝ ⎜
⎞ ⎠ ⎟ f x( )dx
recall that
22.058 - lecture 5, Sampling and the Nyquist Condition
Formalism of Sampling
€
fn{ }= TopHat2xfov
⎛
⎝ ⎜
⎞
⎠ ⎟ δ x−nΔx( ) f x( )dxn=−∞
∞∑
−∞
∞∫
= f x( )δ x−nΔx( )−fov
2
+fov2∫
n=−∞
∞∑
fn= f x( )δ x−nΔx( )dx−fov
2
+fov2∫
22.058 - lecture 5, Sampling and the Nyquist Condition
Formalism of Sampling
f(x)
Comb(x/x)
TopHat(2x/fov)
22.058 - lecture 5, Sampling and the Nyquist Condition
Formalism of Sampling
f(x)
Comb(x/x)
TopHat(2x/fov)
22.058 - lecture 5, Sampling and the Nyquist Condition
Formalism of Sampling
{fn}
22.058 - lecture 5, Sampling and the Nyquist Condition
Frequency of Sampling
data =Table @If @Mod@x, 7D<2, 1, - 1D* Exp@-Hx - 128L^2ê1000D, 8x, 1, 256<D;
ListPlot @data, 8PlotJoined -> True , PlotRange -> All , Axes -> False , AspectRatio -> 1ê4,PlotStyle -> Thickness @0.005D<D
FT
x
k
22.058 - lecture 5, Sampling and the Nyquist Condition
Frequency of Sampling
FT
x
k
every point
every second point
22.058 - lecture 5, Sampling and the Nyquist Condition
Frequency of Sampling
FT
x
k
every point
every third point
22.058 - lecture 5, Sampling and the Nyquist Condition
Frequency of Sampling
FT
x
k
every point
every fourth point
22.058 - lecture 5, Sampling and the Nyquist Condition
Sampling
In[19]:= p2 =Plot @Cos@2Pi 0.2 Hx - 1LD, 8x, 1, 6<,8PlotPoints -> 512,PlotStyle -> 8Thickness @0.01D, RGBColor@1, 0, 0D<<D
In[11]:= p4 =ListPlot @Table @Cos@2Pi 1.2 xD, 8x, 0, 5, 1<D,Prolog -> AbsolutePointSize @10DD
2 3 4 5 6
-1
-0.5
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22.058 - lecture 5, Sampling and the Nyquist Condition
Sampling
In[19]:= p2 =Plot @Cos@2Pi 0.2 Hx - 1LD, 8x, 1, 6<,8PlotPoints -> 512,PlotStyle -> 8Thickness @0.01D, RGBColor@1, 0, 0D<<D
In[11]:= p4 =ListPlot @Table @Cos@2Pi 1.2 xD, 8x, 0, 5, 1<D,Prolog -> AbsolutePointSize @10DD
2 3 4 5 6
-1
-0.5
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In[18]:= p1 =Plot @Cos@2Pi 1.2 Hx - 1LD, 8x, 1, 6<,8PlotPoints -> 512,PlotStyle -> 8Thickness @0.01D, RGBColor@0, 0, 1D<<D
2 3 4 5 6
-1
-0.5
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22.058 - lecture 5, Sampling and the Nyquist Condition
Sampling
In[19]:= p2 =Plot @Cos@2Pi 0.2 Hx - 1LD, 8x, 1, 6<,8PlotPoints -> 512,PlotStyle -> 8Thickness @0.01D, RGBColor@1, 0, 0D<<D
In[11]:= p4 =ListPlot @Table @Cos@2Pi 1.2 xD, 8x, 0, 5, 1<D,Prolog -> AbsolutePointSize @10DD
In[20]:= p3 =Plot @Cos@2Pi 2.2 Hx - 1LD, 8x, 1, 6<,8PlotPoints -> 512,PlotStyle -> 8Thickness @0.01D, RGBColor@0, 1, 0D<<D
2 3 4 5 6
-1
-0.5
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2 3 4 5 6
-1
-0.5
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22.058 - lecture 5, Sampling and the Nyquist Condition
Sampling
2 3 4 5 6
-1
-0.5
0.5
1
2 3 4 5 6
-1
-0.5
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1
Nyquist theorem: to correctly identify a frequency you must sample twice a period.
So, if x is the sampling, then π/x is the maximum spatial frequency.
22.058 - lecture 5, Sampling and the Nyquist Condition
Sampling A Simple Cosine Function
Consider a simple cosine function,
€
s1 t( ) = cos 2πfot( )
We know the Fourier Transform of this
€
cos 2πfot( ) ⇔ π δ f − fo( ) + δ f + fo( )( )
What happens when we sample this at a rate of
€
1Δt
⎛ ⎝ ⎜
⎞ ⎠ ⎟
€
where 1
Δt has the units of Hz
and Δt = the dwell of the sampled signal.
22.058 - lecture 5, Sampling and the Nyquist Condition
Sampling A Simple Cosine Function (cont …)
So that,
€
s1 n( ) = cos 2πfonΔt( )
= s1 t( ) δ t − nΔt( )∑F s1 n( ){ } = F s1 t( ) δ t − nΔt( )∑{ }
= F s1 t( ){ }⊗ F δ t − nΔt( )∑{ }
= π δ f − fo( ) + δ f + fo( )[ ]⊗2π
Δtδ f −
2πn
Δt
⎛ ⎝ ⎜
⎞ ⎠ ⎟∑
⎡
⎣ ⎢ ⎤
⎦ ⎥
Note, I’ve taken a short cut and left off the integrals that are needed to sample with the delta function.
22.058 - lecture 5, Sampling and the Nyquist Condition
Define A New Frequency: Case 1
Define a new frequency
€
fs =2π
Δt; "the sampling frequency"
Case 1:
€
fo < f s2 = fn = Nyquist frequency
In this case, there is no overlap and regardless of the complexity of this spectrum (think of having a number or continuum of cosine functions), the frequency spectrum correctly portrays the time evolution of the signal.
fs-fs
2f0
0
22.058 - lecture 5, Sampling and the Nyquist Condition
Case 2
€
fo > f s2 = fn = Nyquist frequency
Now the frequency spectrum overlaps and look what happens to our picture,
So it appears as though we are looking at a frequency of
€
fs − fo < fo - this is an aliased signal.
fs-fs
2f0
0
2f’0
fs0-fs
€
f '0= fs−f0
22.058 - lecture 5, Sampling and the Nyquist Condition
Nyquist Theorem
Consider what happens when there is a complex spectrum. Then the entire spectrum overlaps. Either way, the frequency spectrum does not correspond to a correct picture of the dynamics of the original time domain signal.
Nyquist Theorem: In order to correctly determine the frequency spectrum of a signal, the signal must be measured at least twice per period.
22.058 - lecture 5, Sampling and the Nyquist Condition
Return to the sampled cosine function
€
cos 2πfonΔt( ) ; Δt = 1f s
€
cos2πfon
fs
⎛
⎝ ⎜
⎞
⎠ ⎟
So
Now let
€
fo > f s ∴ fo = f s − f s − fo( )
Δf1 2 4 3 4
22.058 - lecture 5, Sampling and the Nyquist Condition
(cont…)
€
cos2πn
f sf s − Δf[ ]
⎛
⎝ ⎜
⎞
⎠ ⎟
cos 2πn −2πn
f sΔf
⎛
⎝ ⎜
⎞
⎠ ⎟
cos −2πnΔf
f s
⎛
⎝ ⎜
⎞
⎠ ⎟
€
cos2πnΔf
f s
⎛
⎝ ⎜
⎞
⎠ ⎟
Therefore we can see that it is not the Fourier Transform that fails to correctly portray the signal, but by our own sampling process we mis-represented the signal.
€
or use :
cos A + B( ) =
cos A( )cos B( ) − sin A( )sin B( )
A = 2πn ∴ cos A( ) =1;sin A( ) = 0
22.058 - lecture 5, Sampling and the Nyquist Condition
A Fourier Picture of Sampling
€
fn{ }= TopHat2xfov
⎛
⎝ ⎜
⎞
⎠ ⎟
c1 2 4 4 3 4 4
δ x−nΔx( )n=−∞
∞∑
c1 2 4 4 3 4 4
f x( )dx
c1 2 4 3 4
−∞
∞∫
Sinckfov( )⊗2πΔx δ k−n2πΔx
⎛ ⎝ ⎜
⎞ ⎠ ⎟
n=−∞
∞∑ ⊗F(k)
Look at this first1 2 4 4 4 4 4 3 4 4 4 4 4
22.058 - lecture 5, Sampling and the Nyquist Condition
A Fourier Picture of Sampling
22.058 - lecture 5, Sampling and the Nyquist Condition
Bandwidth Limited
time domain
22.058 - lecture 5, Sampling and the Nyquist Condition
Limitations of Sampling
{fn}
22.058 - lecture 5, Sampling and the Nyquist Condition
Limitations of Sampling
perfect sampling
average 3 data points
22.058 - lecture 5, Sampling and the Nyquist Condition
Limitations of Sampling
perfect sampling
average 5 data points
22.058 - lecture 5, Sampling and the Nyquist Condition
Limitations of Sampling
perfect sampling
average 7 data points
22.058 - lecture 5, Sampling and the Nyquist Condition
Reciprocal Space
real space reciprocal space
22.058 - lecture 5, Sampling and the Nyquist Condition
Sampling
real space under sampled
0 10 20 30 40 50 600
10
20
30
40
50
60
22.058 - lecture 5, Sampling and the Nyquist Condition
Sampling
real space zero filled0 20 40 60 80 100 120
0
20
40
60
80
100
120
22.058 - lecture 5, Sampling and the Nyquist Condition
Sinc interpolation
22.058 - lecture 5, Sampling and the Nyquist Condition
Filtering
We can change the information content in the image by manipulating the information in reciprocal space.
Weighting function in k-space.
22.058 - lecture 5, Sampling and the Nyquist Condition
Filtering
We can also emphasis the high frequency components.
Weighting function in k-space.
22.058 - lecture 5, Sampling and the Nyquist Condition
Fourier Convolution
22.058 - lecture 5, Sampling and the Nyquist Condition
Deconvolution
€
I x( ) =O x( )⊗ PSF x( ) + N x( )noise
{
Recall in an ideal world
€
i k( ) =O(k) − PSF(k)
€
∴ Try deconvolution with
€
1PSF(k)
€
iperfect k( ) = O k( ) • PSF(k)[ ] •1
PSF k( )
an "inverse" filter
I perfect x( ) = O x( )⊗ PSF x( ) + N x( )[ ]⊗ PSF −1 x( )means "inverse" not 1
PSF x( )
1 2 4 3 4
22.058 - lecture 5, Sampling and the Nyquist Condition
Deconvolution (cont…)
This appears to be a well-balanced function, but look what happens in a Fourier space however:
€
i k( ) =O(k) • PSF(k) •1
PSF(k)+
n(k)
PSF(k)
Recall that the Fourier Transform is linear where PSF(k) << n(k), then noise is blown up. The inverse filter is ill-conditioned and greatly increases the noise particularly the high-frequency noise since
€
PSF k( )⇒ 0 at high k
normally1 2 4 4 4 4 3 4 4 4 4
This is avoided by employing a “Wiener” filter.
€
PSFw k( ) =PSF * k( )
PSF k( )2
+WN k( ) where * is the complex conjugate
€
SN( )
−2
€
noise power spectral density
= SN k( )( )
−2
22.058 - lecture 5, Sampling and the Nyquist Condition
Original Nyquist Problem
A black & white TV has 500 lines with 650 elements per line. These are scanned through
Electron beam is scanned over the screen by deflection magnets and the intensity of the beam is modulated to give the intensity at each point. The refresh rate is 30 frames/second.
By the sampling theorem, if the frequency is , independent information is available once every seconds.
€
fo
€
1
2 f
€
30frames
sec×500
lines
frame×650
pixels
line= 9.75 ×106 impulse
sec
∴ f ≥ 4.875 MHz
Deflection magnets
Electron beam
Need information
22.058 - lecture 5, Sampling and the Nyquist Condition
Fourier Transform of a Comb Function
€
Comb x( ) = δ x − nX( )−∞
∞
∑
where
δ x − nX( ) =1, x = nX
0, x ≠ nX
⎧ ⎨ ⎩
⎫ ⎬ ⎭
The Fourier Transform we wish to evaluate is,
€
F Comb t( ){ } = Comb x( )e−ikxdx−∞
∞
∫
One trick to this is to express the comb function as a Fourier series expansion, not the transform.
€
Comb x( ) = anei2πnx
X
n=−∞
∞
∑
where
an =1
XComb x( )e
−i2πnxX dx
−x2
x2
∫Normalization to keep the series unitary Limits chosen since this
is a periodic function
22.058 - lecture 5, Sampling and the Nyquist Condition
Fourier Transform of a Comb Function
Over the interval to the comb function contains only a single delta function.
€
−x2
€
x2
€
an =1
Xδ x( )e
− i2πnxX dx
−x2
x2
∫
select the x= 0 po int,note this is only trueof a Dirac delta function.
1 2 4 4 3 4 4
an =1
X
€
∴ A series representation of the comb function is
€
Comb x( ) =1X
ei 2πnx
X
n=−∞
∞
∑
22.058 - lecture 5, Sampling and the Nyquist Condition
Fourier Transform of a Comb Function
Now,
€
F Comb{ } =1X
ei 2πnx
Xe− ikxdxn=−∞
∞
∑−∞
∞
∫
=1X
e− ix k−
2πn
X
⎛ ⎝ ⎜
⎞ ⎠ ⎟
dx−∞
∞
∫
2πδ k−2πn
X
⎛ ⎝ ⎜
⎞ ⎠ ⎟
1 2 4 3 4 n=−∞
∞
∑
δ x− nX( )−∞
∞
∑ ⇔2πX
δ k −2πnX
⎛ ⎝ ⎜
⎞ ⎠ ⎟
−∞
∞
∑
0 0 kx
…1 …
€
2πX
€
2πX
€
X
Note: Scaling laws hold