22.058 - lecture 5, Sampling and the Nyquist Condition Sampling Outline FT of comb function Sampling Nyquist Condition sinc interpolation Truncation

22.058 - lecture 5, Sampling and the Nyquist Condition

Sampling

Outline FT of comb function Sampling• Nyquist Condition sinc interpolation• Truncation• Aliasing


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.


Limits of Sampling

• Finite # of data points

• Finite field of view

• High spatial frequency features can be missed or recorded incorrectly


Formalism of Sampling

€

fn{ }= TopHat2xfov

⎛

⎝ ⎜

⎞

⎠ ⎟

−∞

∞∫ Comb

xΔx

⎛ ⎝ ⎜

⎞ ⎠ ⎟ f x( )dx

recall that



€

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∫



f(x)

Comb(x/x)

TopHat(2x/fov)



f(x)

Comb(x/x)

TopHat(2x/fov)



{fn}


Frequency of Sampling

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ListPlot @data, 8PlotJoined -> True , PlotRange -> All , Axes -> False , AspectRatio -> 1ê4,PlotStyle -> Thickness @0.005D<D

FT

x

k



FT

x

k

every point

every second point



FT

x

k

every point

every third point



FT

x

k

every point

every fourth point


Sampling

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2 3 4 5 6

-1

-0.5

0.5

1


Sampling



2 3 4 5 6

-1

-0.5

0.5

1


2 3 4 5 6

-1

-0.5

0.5

1


Sampling




2 3 4 5 6

-1

-0.5

0.5

1

2 3 4 5 6

-1

-0.5

0.5

1


Sampling

2 3 4 5 6

-1

-0.5

0.5

1

2 3 4 5 6

-1

-0.5

0.5

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.


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.


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.


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


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


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.


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


(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


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


A Fourier Picture of Sampling


Bandwidth Limited

time domain


Limitations of Sampling

{fn}



perfect sampling

average 3 data points



perfect sampling




perfect sampling



Reciprocal Space

real space reciprocal space


Sampling

real space under sampled

0 10 20 30 40 50 600

10

20

30

40

50

60


Sampling

real space zero filled0 20 40 60 80 100 120

0

20

40

60

80

100

120


Sinc interpolation


Filtering

We can change the information content in the image by manipulating the information in reciprocal space.

Weighting function in k-space.


Filtering

We can also emphasis the high frequency components.

Weighting function in k-space.


Fourier Convolution


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


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


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


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



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=−∞

∞

∑



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

Documents

22.058 - lecture 5, Sampling and the Nyquist Condition Sampling Outline FT of comb function Sampling Nyquist Condition sinc interpolation Truncation