0:06

Now in order to understand the data collection strategies for

tootie crystallography, and the image processing that follows.

We need to understand what the foray transform of a two dimensional crystal

looks like in 3D.

And to understand that let's begin

by thinking about the Fourier transform of an asymmetric object.

For instance here this image of a duck.

0:42

Obviously this a very corse pixelation in the, and

the image is higher resolution than my pixels.

But for the sake of explanation,

each pixel would have a particular density value for that pixel.

Some are obviously black, some are obviously white.

>> And let's imagine that there was, say,

ten by ten pixels in this real space image.

And so, as explained in the lectures on Fourier transforms,

the Fourier transform of such an image will have approximately five times

ten series of amplitudes associated with phases.

So the same number of values,

100 numbers basically representing it in reciprocal space.

There's one extra column here representing a zero spatial frequency column,

but to a first approximation, it's the same number of values in the transform and

the real space image.

And so we have a series of pixels here.

Notice that the transform has two fold symmetry.

So the unique part of the transform is just one half of it.

Okay, and so it's full of pixels as well and

each pixel has both an amplitude and a face.

And so each of these pixels is going to have a substantial amplitude.

It's an important amplitude in phase that's required

to specify the image of the duck.

So I'll just put a check mark in each of these pixels,

because they're all required.

Now let's think about what would happen, though, if instead of having one duck in

the picture, we now had four copies of the duck in this regular lattice.

And now we're going to need 20 pixels by 20 pixels

to fill out this complete image with the same fidelity as the original.

And so in the transform we're now going to have ten columns of

20 pixels and each one will have an amplitude and phase associated with it.

So if we start to draw, and if you'll remember the pixel at the origin

represents the DC component, which is just a, a plain wave across the image.

2:53

And the next pixel over here represents the wave.

That pixel right there represents a wave that has one maximum within the image.

Just like this.

And what we can see is that there really isn't going to be a significant amplitude

to such a wave, because the pattern in the image is actually a pattern that repeats

twice across the box.

So instead of that wave being significant, the wave that's going to be significant

3:37

In addition to specify all the fine details of the shape of this duck,

all of the harmonics of this wave will be necessary.

So for instance there's going to be a wave with four maxima across the box,

and six, and eight.

And so this pixel is going to be, have a significant amplitude and so

will this one and so will this one.

4:01

Similarly, there's really no power in a wave that goes across the box

vertically in that fashion, because the image here is a duplicate.

There's two of them.

And so the first wave with significant amplitude is going to be the one with

two maxima across the box and all its harmonics.

And so that one appears here, and

then its harmonics will also be present in the image.

4:33

All of these pixels will have significant values.

And so we start to establish a pattern where only every other pixel

has a significant value.

But if you count the number of pixels

that actually have these important amplitudes and phases.

It's the same as the number of pixels that were required to specify just

one duck in the previous slide.

It's that same 100 values, 50 amplitudes and

50 phases that were required to specify a duck.

And so that, all that information is still here, despite the zeros in between them.

5:31

Now this process can be extended.

What if we had four by four ducks?

Now we're going to have an image with 40 pixels by 40 pixels and a transform

with 20 columns of 40 pixels of amplitudes and phases.

And I won't be able to draw all those pixels.

So instead, if I select just the center of the transform,

6:28

The third pixel represents a wave with three oscillations across the box.

And, again, there's not much of that present.

However there's a substantial

presence of a wave that has four maxima across the box.

Because of course there are four ducks across the box now.

And all of its harmonics.

So the next one will be essentially zero.

But then the eighth pixel will have a substantial amplitude and face.

Following the same pattern, there's very little of the wave that crosses once,

so that's going to be a zero.

Or twice, or three times.

But there will be a substantial wave that has

four maxima across the box to represent the pattern of the four ducks.

So this is 0, 0, and then that's going to be substantial.

And then all of its harmonics off to the edge of the transform and

then the cross terms will also be substantial.

Okay, so, you see the pattern emerging.

Now if we could see the entire transform and

just not the center here, we would find that once again,

there are exactly 50 pixels here with significant amplitudes and phases.

The same 100 values that were needed to specify the shape of one duck.

But now there pattern and the zeroes between them,

their pattern reveals the pattern of the spacing of the ducks.

And in one further elaboration of this, imagine that now we have.

Eight ducks across the box.

We now have 80 pixels by 80 pixels, and this is going to be 40

columns of 80 pixels each with their unique amplitude in phase.

8:28

And as you might have guessed, there's going to be a significant amplitude in

phase and the eighth pixel, both in the horizontal and vertical direction, and

their cross peaks, and their harmonics out here where it pass where we can see.

9:18

And so, in a slightly more realistic example.

If we have a crystal here, of unit cells, so

each of these is a unit cell with a regular pattern inside of it,

then it's Fourier transform will be a series of discrete spots.

9:37

And each spot of course has an amplitude in phase, and.

The information about what is the structure of a unit cell

is contained in all of the amplitudes and phases that are present here.

But between these discrete spots there's a lot of pixels of just zero or

very low numbers here between them.

And so the pattern of where the spots are, that there's one here, that there's

a spot there and there's a spot there and a spot there and a spot there.

This lattice of spots contains the information of how these unit

cells are arranged in real space.

10:31

And the Fourier transform of an object convolved with a lattice,

let's represent lattice with the small case l,

according to the convolution theorem is

equal to the Fourier transform of the object

times the Fourier transform of the lattice.

Now, the Fourier transform of the lattice is another lattice itself.

We'll represent it with a capital L.

Meaning this series of discrete spots.

So if you think of the original lattice as a series of delta functions.

Let us just write in delta functions of each of these positions.

11:30

And so if we multiply that series of delta functions with the Fourier

transform of the object, basically what it means,

it's like a screen on a filter where we only see the values of the Fourier

transform of the object in the locations where we have a lattice spot.

So it's as if we had the Fourier transform in the background in this image and

we could only see its value in these particular spots where

the Fourier transform of the lattice has a value of 1.

And so you see it there.

12:05

Now, on first hearing this you might think that there must be something wrong with

this, because it seems like, just the pixels that are showing up

here are not enough to fully specify this complete image.

This image seems to contain more information than this transform.

But, remember that the image itself is just a small,

unique image repeated over, and over.

And so the information content of a single unit cell in fact,

12:36

is not more than just the amplitudes and

phases of these discrete lattice points can reveal.

And so that all applied to these hypothetical perfect crystals

where every unit cell is exactly like the next cell exactly like the next unit cell.

Of course in real crystals they're not exactly identical.

And the information about how they're different,

is actually found in the pixels in between the lattice spots.

So these pixels in between the lattice spots are actually non zero for

a real crystal and

they contain that information about how the unit cells are actually different.

The data on the lattice points

contains the information of the average unit cell here.

The information between the lattice spots gives you the details

of how each unit cell is subtly different than its neighbor.

Okay, now those were examples in two dimensions.

Now it's difficult for me to draw a three dimensional crystal, but

I'll give it a try here.

14:17

So it's discreet spots in this plane.

And then above that there will also be another row of discrete spots.

You know, in the plane above that we'll have the same pattern of discrete spots.

And like I say, it's very difficult for me to draw this in 3D.

14:37

But the principle is that the same effect that we described about having

waves that need to have

multiple maximum as they cross the crystal in all three directions.

Means that in the transform of a 3D crystal you have discrete spots

in all three directions.

The lattice here is a 3D lattice.

The lattice here will be the transform of that lattice which is also a 3D lattice.

But what happens if,

instead you have just a two dimensional crystal.

And by that I mean a crystal that's a single unit cell thick.

15:25

In this case its Fourier transform has

discrete spots in the xy plane, but

there's no discretization of the amplitudes and

phases in the vertical direction.

And so each of these spots is not just a spot in the plane but

it is, it's a column of amplitudes and

phases in Z and that's because a, a,

an object that's just a single unit cell thick.

All of the waves are required to specify the kinds of

structures that might exist in one single unit cell.

For instance, if we have a single protein in this unit cell,

16:19

we're going to need all possible sine waves that cross that

box with one maximum and two maxima and three maxima, et cetera.

In order to reproduce the structure of that protein.

And so, there needs to be a value here and right next to it, and

right next to it, and right next to it, and right next to it.

And so you have a whole column of amplitudes and

phases in z at each of these discrete spots.

17:17

seeing these spots in the xy plane.

And this is not surprising because this image is the image of a crystal.

And that as we went through previously, the Fourier transform of such an image

of a crystal should be a series of discrete spots.

And so here this is what you see.

18:00

But the spacings in between the proteins,

the spacings of this lattice, would be reduced or compressed

with respect to the spacings that we're seeing previously in the untilted view.

If we were to calculate its Fourier transform,

what we would get is the amplitudes in phases on a tilted

plane through this three dimensional transform.

And it would sample the same amplitudes and phases on the xy plane there.

But for other lattice lines is what these are called,

we would sample values that were lower off the xy plane.

Fortunately, we have a better picture of the situation here.

This represents the three dimensional Fourier transform of a 2D crystal.

And so the axes are spatial frequency in one direction, spatial frequency

in another, and then here spatial frequency in the third direction.

And in the xy plane, here labeled as a star and b star for other reasons,

we see that we have a series of discrete spots just as expected,

because the object is periodic in the x and y directions as a crystal.

But as we move away from that xy plane, instead of having another series

of discrete spots in z, as you would have for a three dimensional crystal.

Instead for a two dimensional crystal, you have a continuous line of amplitudes and

phases, and this we call a lattice line.

So that's a lattice line, and here there's another lattice line of continuous

amplitudes of phases that go both above the xy plane and down below the xy plane.

And because of this, if we were to take an image of the crystal at a tilted angle,

20:19

In the transform of an image of the tilted crystal, we would again see a series

of spots as they cross through that plane, here, here.

Okay, at all of these crossing points, which would reveal the amplitude and

phase of that lattice line in the position where it crosses the lattice line.

The lattice line has a continuous progression of amplitudes and

phases that vary as you rise above or below the xy plane.

And each image of a tilted crystal measures the amplitude and

phase in one position along that lattice line, as well as a position on

all the other lattice lines in the 3D Fourier transform of the crystal.

21:09

To further clarify the situation, here's a depiction of the three

dimensional Fourier transform of a two dimensional crystal,

given in Orlova and Saibil's review here.

And here plane A is the xy plane, and you

can see a series of discrete spots in the transform of this two-dimensional crystal.

Shown here in this plane C, which includes

the y axis and this plane B, which includes the x axis.

21:47

You can see that each of these discrete spots has significant amplitudes

as it rises above, and so their lattice lines complete continuous rods.

And any particular image of a tilted crystal

would give a cross-section through these rods.

Now if we turn our attention here to this cross section through the 3D

Fourier transform of a 2D crystal, we see that each image

will give us data about some particular cross section through that transform.

And give us information about the lattice lines, the amplitudes and

phases along the lattice lines that cross through there.

But because, just like in tomography, we can't tilt the sample to 90 degrees.

Instead, we can only go to about 6 or 70 degrees.

There will be a missing cone of information here of amplitudes and

phases that we can't sample.

23:20

And just like in tomography or the random conical tilt reconstruction,

because of this missing cone here of information.

If we have an original object that looks like this, and we produce a three

dimensional reconstruction through electron crystallography.

The result of the missing cone here is that all the objects in

the reconstruction will be smeared somewhat in the vertical direction.

And, so the, the resolution would be isotropic,

just as we've discussed in the other cases.