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Fourier Space: Fourier Transform

00:00:12.04 This time we're going to talk about
00:00:14.01 Fourier transform and Fourier space,
00:00:16.04 which is a very important concept
00:00:17.16 in understanding microscopy.
00:00:20.27 Well, Fourier transform is
00:00:22.18 all about math, like here.
00:00:24.26 So, I'm going to make it simple.
00:00:27.21 Let's start with a simple sine wave.
00:00:32.20 As simple as possible,
00:00:34.05 you see sine oscillation, very smooth oscillations.
00:00:37.12 To describe this sine wave,
00:00:39.11 well, you have to have some mathematical formula,
00:00:41.10 like here, as simple as we can go:
00:00:43.26 I, which is the intensity at a certain point,
00:00:45.27 equals sine of the coordinate x.
00:00:51.02 And there are more properties of this wave.
00:00:53.15 For example, you can have different spacing
00:00:55.16 or different frequencies,
00:00:57.03 and in this case you can see
00:00:59.14 3 white stripes in the entire view,
00:01:01.16 so we can say this has a frequency of 3.
00:01:04.29 And in the case when you have a frequency of 4,
00:01:07.15 k=4,
00:01:09.01 you have 4 white stripes.
00:01:11.17 And higher frequency, k=6,
00:01:13.06 and even higher frequencies,
00:01:16.02 where you have k=30,
00:01:18.26 then there's a lot of little white stripes showing here...
00:01:21.17 very dense, high frequency waves.
00:01:24.28 So, that's one aspect
00:01:26.26 of a pure sine wave.
00:01:28.09 Now, in addition to frequency,
00:01:30.22 there's also another thing,
00:01:32.13 which is the amplitude.
00:01:33.21 Now you have a coefficient, A,
00:01:35.15 in front of the sine function.
00:01:37.15 And in this case you have amplitude 1,
00:01:40.03 and when you have amplitude 0.5
00:01:42.05 it's dimmer or lower amplitude,
00:01:45.03 smaller oscillations.
00:01:46.20 And an amplitude of 0.25,
00:01:48.20 it's going to be even dimmer in the image,
00:01:51.11 or even lower contrast.
00:01:53.02 That's now two of the properties of the sine wave:
00:01:59.02 the frequency and the amplitude.
00:02:01.06 Getting a little bit more complicated,
00:02:02.27 we have an additional phase factor
00:02:05.27 in this sine function.
00:02:08.05 What we can see here is
00:02:10.22 one additional ɸ in the bracket of the sine function.
00:02:13.07 And in this case,
00:02:15.27 this phase is zero,
00:02:17.09 that means the sine oscillation,
00:02:19.01 here, starts from zero,
00:02:20.15 and continues oscillating.
00:02:22.20 If we have an additional phase shift of 45°,
00:02:26.13 and you can see the entire wave
00:02:28.14 gets translated
00:02:30.28 and it starts somewhere mid-way.
00:02:33.08 And we can have more phase shifts,
00:02:35.13 like 90°, 135, 180, 225, 270, 315,
00:02:44.28 until we get a phase shift of 360°,
00:02:47.00 that's one entire circle,
00:02:48.12 so it's come back to the original sine wave.
00:02:52.12 This is the phase factor.
00:02:54.20 And a fourth one to consider is that
00:02:58.11 we're describing a two-dimensional image.
00:03:00.14 In this case, the sine wave function
00:03:03.11 is basically oscillating along the x direction
00:03:06.26 -- you can oscillate it along an axis
00:03:09.09 that's slightly off from the x direction.
00:03:13.24 And in this case
00:03:15.21 we have to have a little bit more complicated math
00:03:18.14 to describe it.
00:03:20.11 And you can see in the sine function here,
00:03:22.05 we've changed both the k and x
00:03:25.09 into vectors,
00:03:26.29 and the k essentially describes,
00:03:29.03 as a vector...
00:03:31.02 and the k, as a vector, now,
00:03:33.05 essentially describes the propagation function
00:03:35.12 of this sine wave.
00:03:36.24 So, we have our x and y axis,
00:03:39.23 and we have a vector k.
00:03:40.23 In this case, the k has
00:03:43.14 a 30° angle to the x axis
00:03:45.11 and that is the propagation angle.
00:03:48.17 And in a different case,
00:03:50.10 this vector has been rotated
00:03:51.26 to have a 60° angle to the x axis
00:03:53.28 and you can see the entire waveform gets rotated.
00:03:57.11 So those are the four parameters
00:04:00.17 to describe a two-dimensional sine wave.
00:04:03.12 And we have our frequency,
00:04:06.13 and we have our orientation.
00:04:08.11 Both of these are included
00:04:10.29 in this k vector.
00:04:14.04 And then we have our amplitude in a,
00:04:16.17 and we have our phase in ɸ0.
00:04:20.00 We need to find
00:04:22.27 a graphical way of describing these sine functions,
00:04:25.05 because looking at the mathematical formula
00:04:27.12 and looking at pure numbers,
00:04:29.02 that's very boring.
00:04:30.15 Then we can have our coordinate system,
00:04:33.21 and in this case to avoid confusion
00:04:35.04 I called it kx and ky, instead of x and y,
00:04:37.25 but it's basically the same thing.
00:04:39.11 And then we have the k vector.
00:04:42.27 Just now, I said
00:04:44.28 the orientation or the direction of the k vector
00:04:46.26 tells the propagation direction of the sine wave,
00:04:49.09 and in fact the length
00:04:52.24 of this k vector, the wave vector,
00:04:55.04 tells the frequency.
00:04:56.21 And so the k
00:05:00.02 describes the frequency and orientation as a vector.
00:05:02.21 For each point on this two-dimensional kx-ky plane,
00:05:07.06 we can have two values.
00:05:09.20 One value is A
00:05:12.18 as a function of the k vector,
00:05:13.26 that tells the amplitude of the sine wave,
00:05:17.15 and another value is ɸ,
00:05:19.13 that tells the phase of the sine wave.
00:05:21.29 So a total of four numbers
00:05:25.28 describes this sine wave
00:05:28.16 and then, because we're looking at a 2-D plane,
00:05:32.09 we can draw them into a picture,
00:05:34.24 and this is basically
00:05:37.25 a full picture of describing
00:05:40.23 a lot of different sine waves.
00:05:42.01 Each point on this picture
00:05:43.28 describes one sine wave with
00:05:46.20 a certain frequency and orientation
00:05:49.26 and the value of this point
00:05:52.06 describes the amplitude and the phase.
00:05:53.24 And that is how we can graphically represent
00:05:56.09 a collection of different sine waves.
00:05:59.21 So, so far we are talking about simple sine waves,
00:06:02.18 but whatever cell structure we're looking at,
00:06:06.02 they are not sine waves.
00:06:07.08 They have fairly complicated structures,
00:06:09.18 and how can we use sine waves
00:06:11.15 to describe the structure.
00:06:13.19 Well, again, we start with
00:06:15.09 an extremely simple example.
00:06:16.19 Now, instead of sine waves,
00:06:18.23 we have black and white stripes.
00:06:20.25 That's simple enough.
00:06:22.00 But even that differs significantly
00:06:25.16 from a pure sine wave.
00:06:26.25 Just in cross-section,
00:06:28.07 you can see a sine wave
00:06:29.21 is a fairly smooth oscillation,
00:06:31.27 but here, now,
00:06:33.24 we have very sharp steps.
00:06:35.15 We can try to compensate for that, though,
00:06:39.26 for example, here, at the edge,
00:06:41.09 this sine wave has a smooth drop,
00:06:44.00 whereas this black and white pattern
00:06:46.11 has a very sharp transition.
00:06:48.08 So we can imagine if we have
00:06:50.20 yet another peak, here.
00:06:52.09 And that will somewhat compensate for this.
00:06:55.05 And at the same time,
00:06:56.17 if we have a negative peak in the middle,
00:06:58.13 that will also compensate
00:07:00.04 for this peaky behavior of the sine wave
00:07:02.15 to make it more flat.
00:07:04.05 So this is what we do,
00:07:05.20 we have yet another sine wave
00:07:07.03 which is 3 times the frequency
00:07:09.06 and, summed up, you can see,
00:07:12.13 because its 3 times the frequency
00:07:14.01 and it has a negative in the middle
00:07:16.00 and two positive peaks at the edge,
00:07:18.04 and that makes this overall shape
00:07:21.24 more step-like.
00:07:23.28 But this is still not perfect yet.
00:07:25.23 We can consider more,
00:07:27.06 further compensating for it.
00:07:28.19 This time, we have another one,
00:07:29.28 5 times the frequency.
00:07:31.10 And when you have an infinite amount of these adding up,
00:07:34.15 then we will approximate
00:07:37.22 our black and white streak pattern.
00:07:40.16 So this is the process
00:07:43.05 of how we can use a collection of sine waves,
00:07:45.20 or a collection of different spatial frequencies,
00:07:48.01 to describe a real image.
00:07:51.02 And we can write it in mathematical form
00:07:52.20 -- and this is important --
00:07:54.06 as, well, our original image,
00:07:58.09 F(x) equals
00:08:01.21 a sum of different sine functions,
00:08:05.03 and then G(x)
00:08:07.14 actually describes the amplitude and phase
00:08:09.29 of these functions,
00:08:11.06 it's how they are summed.
00:08:12.15 So now we have two functions:
00:08:14.08 one function is F(x),
00:08:15.22 that describes the real image
00:08:17.24 in the coordinate x;
00:08:19.06 the other function is G(k),
00:08:21.10 that describes our collection of sine waves
00:08:25.00 that can sum up into this F(x) function,
00:08:27.07 and that's in the coordinate of k(x).
00:08:28.22 If you look into image,
00:08:30.22 you can see discrete spots.
00:08:33.07 The position of these spots
00:08:34.23 describes the spatial frequency,
00:08:37.25 k equals 1, 3, 5,
00:08:39.27 and, very dim, 7,
00:08:41.15 and there can be an infinite number of them.
00:08:44.05 And the intensity of these spots
00:08:46.14 essentially describes the amplitude.
00:08:48.23 In this case,
00:08:50.04 we're not considering the phase shift yet.
00:08:52.10 So, in this case,
00:08:53.28 you can see that
00:08:56.05 we have some relationship between
00:08:59.00 F(x) and G(k),
00:09:01.02 and to derive G(k) from F(x)
00:09:03.13 is called the Fourier transform.
00:09:06.21 In this case, we established
00:09:08.29 the connection between the original image,
00:09:12.08 which is in so-called real space,
00:09:13.22 and the Fourier image,
00:09:15.13 which is in frequency space.
00:09:19.12 This is a very, very simple example.
00:09:22.26 Now, let's give you a slightly more complicated image
00:09:26.23 to show you how you can
00:09:28.20 read the Fourier space information.
00:09:31.04 We're looking at the Fourier transform
00:09:33.24 of an image of Fourier himself.
00:09:37.09 That's the Fourier transform of Fourier,
00:09:39.02 and in this case
00:09:42.01 the original image has an x and a y coordinate.
00:09:44.18 The Fourier space image has
00:09:46.18 a kx and ky coordinate.
00:09:49.18 And one very simple thing to think about this...
00:09:52.25 the things close to the origin
00:09:55.16 will have very short k vectors
00:09:58.23 and that's describing the low frequency,
00:10:02.09 low spatial frequency information.
00:10:05.03 And things further away from the original
00:10:08.26 will describe the highest spatial frequency information.
00:10:11.20 Just to illustrate that,
00:10:13.11 if we take this image
00:10:15.01 and we mask out
00:10:18.07 the highest spatial frequency information
00:10:20.10 and we inverse transform it back
00:10:23.06 -- so, this is call the inverse Fourier transform --
00:10:25.04 well, we still see the image of Fourier,
00:10:28.08 but at a much lower resolution.
00:10:30.10 For example, here,
00:10:32.14 the details of his coat are completely lost,
00:10:35.14 and this is because is on Fourier space
00:10:37.11 we masked out all
00:10:39.06 the high spatial frequency information.
00:10:41.24 We can do it the other way,
00:10:43.18 we can mask out the
00:10:45.18 low spatial frequency information
00:10:46.27 and do the inverse Fourier transform
00:10:49.09 -- this is what we get.
00:10:51.03 Well, we lose the overall image shape,
00:10:54.03 we lose the overall shape of Fourier,
00:10:56.25 but we do preserve these fine details in the coat,
00:10:59.18 these very sharp edges,
00:11:01.20 so that's what we see if we
00:11:04.13 only keep the high spatial frequency information
00:11:06.26 out of the image.
00:11:08.05 And that's one simple example
00:11:11.13 of how you can read
00:11:14.07 an image in Fourier space.
00:11:16.18 Okay, that's enough math
00:11:19.02 and that's enough very abstract concepts.
00:11:22.27 Why are we talking about Fourier transform?
00:11:25.01 How is that related to microscopy?
00:11:29.01 Okay, let's come to the
00:11:31.24 very basics of image formation of a microscope,
00:11:34.22 and the key element in the microscope
00:11:36.25 is the objective.
00:11:38.10 And you can think of it as a lens,
00:11:40.21 so it has the back focal plane,
00:11:43.01 which is behind the objective,
00:11:44.19 and the sample is placed
00:11:47.18 at the front focal plane of this objective.
00:11:51.08 And what our lens does,
00:11:53.02 for any parallel light from the back,
00:11:57.03 it's going to focus it down
00:12:00.02 to one point in the
00:12:02.13 front focal plane or the sample plane.
00:12:04.10 And if the angle of this parallel light
00:12:08.05 has an angle of α to the optical axis,
00:12:11.06 it's going to focus down to a point
00:12:13.18 at x = f sinα,
00:12:18.04 f is the focal length.
00:12:19.12 This is basically the design of an objective,
00:12:21.09 so this is how the objectives are designed.
00:12:24.07 Now, making this simple,
00:12:27.00 we only consider two light rays.
00:12:29.11 And the question we ask is that,
00:12:32.20 at the back focal plane,
00:12:33.21 because this is where light comes in,
00:12:35.17 we have some light intensity distribution,
00:12:37.22 say, at a distance of k
00:12:40.16 from the center of the objective,
00:12:43.11 we're going to have some light intensity of A
00:12:47.03 as a function of k.
00:12:48.28 Now, what's the light intensity
00:12:50.28 in the sample plane when it's focused down,
00:12:52.20 with all these different light rays?
00:12:54.25 We can calculate that.
00:12:56.14 Now, two light rays are going to
00:12:58.13 propagate from the back focal plane
00:13:00.04 all the way to the sample,
00:13:01.19 and they go through
00:13:04.01 two different paths
00:13:05.14 and there are two different lengths to these paths.
00:13:07.04 To calculate the distance difference
00:13:10.02 of these two light rays
00:13:11.22 to get to the sample,
00:13:13.18 we simply draw a perpendicular line
00:13:16.21 through the center of the back focal plane,
00:13:18.05 and in this case this distance, d,
00:13:21.01 will tell you how much extra length
00:13:23.17 this light ray has traveled
00:13:25.08 to get to the sample.
00:13:27.20 And to... this distance, d,
00:13:30.00 we can calculate it very easily
00:13:32.07 assuming the refractive index is 1,
00:13:34.16 that's in air.
00:13:36.22 Using very simple triangular mathematics,
00:13:39.00 d equals k times the sine
00:13:42.08 of this angle α.
00:13:44.11 And we know, here,
00:13:46.23 x = f sinα,
00:13:48.22 we can plug that back in,
00:13:51.16 we get d=kx/f,
00:13:53.17 which is the focal length of the objective.
00:13:55.04 So, what does this extra distance mean?
00:13:58.18 Light propagates as waves
00:14:00.23 from this plane all the way to the sample
00:14:03.26 and then these two waves,
00:14:08.03 or many waves,
00:14:09.26 can then add together,
00:14:12.04 one wave, here, another wave
00:14:15.00 has traveled an extra distance,
00:14:16.20 so there's some phase delay.
00:14:18.11 And light is also a sine wave,
00:14:20.23 so we can write that in,
00:14:22.22 so the intensity, here,
00:14:24.12 at the sample, I,
00:14:26.05 is going to be
00:14:28.29 the sum of the intensity over the back focal plane
00:14:32.23 times the effect of different phase delays,
00:14:35.03 and that is a sine function.
00:14:36.26 So, what we get, here,
00:14:38.17 is the sum of A(k) times sine of d,
00:14:41.23 that's the delay,
00:14:44.11 [times 2π over λ],
00:14:46.13 λ is the [wavelength] of the light.
00:14:48.19 And then we can plug in what we get from d.
00:14:52.07 Here is... inside the sin function,
00:14:54.26 we we have kx times...
00:14:57.24 kx times 2π over λf.
00:15:01.05 If you look at this,
00:15:03.19 this appears similar.
00:15:04.29 If you remember the mathematical formula
00:15:07.26 I showed before,
00:15:09.09 which I said is very important,
00:15:10.21 we can see this is exactly
00:15:12.12 a Fourier transform.
00:15:13.27 That's what a lens does.
00:15:16.02 It takes the image
00:15:18.23 at its front focal plane
00:15:20.26 and produces a Fourier transform of the image
00:15:24.11 at the back focal plane,
00:15:25.25 and that's why Fourier transform
00:15:28.29 is so tightly associated to microscopy.
00:15:32.24 Indeed, if we take
00:15:35.27 an image of Fourier,
00:15:38.02 put it at the sample plane of the microscope,
00:15:40.27 at the back focal plane we will observe
00:15:43.09 exactly the same thing
00:15:45.02 as we have mathematically calculated here
00:15:47.19 -- the Fourier transform of Fourier.
00:15:49.28 And this means a lot
00:15:53.19 to microscope image formation.
00:15:58.19 For example, as well we have discussed,
00:16:01.04 in the Fourier space,
00:16:03.08 towards the center,
00:16:05.08 that's where the low spatial frequency information is,
00:16:08.03 towards the edge,
00:16:09.15 that's the high spatial frequency.
00:16:11.21 And the high spatial frequency
00:16:13.12 means very fine features.
00:16:15.13 On the other hand,
00:16:17.15 well, a microscope objective
00:16:20.08 cannot be infinitely large.
00:16:21.20 Here, we're talking about some real space
00:16:24.11 in the Fourier... in the back focal plane,
00:16:26.02 and so it has only
00:16:29.22 some finite size of the back aperture.
00:16:31.25 This means that low spatial frequency information,
00:16:34.25 within this back aperture,
00:16:36.21 will be able to go through the objective
00:16:38.27 and reach out camera and get detected.
00:16:41.03 But high spatial frequency information,
00:16:43.14 outside of this back aperture,
00:16:45.11 will get cut off by the objective.
00:16:47.22 And this is essentially
00:16:51.03 the fundamental reason
00:16:53.12 why microscopes have a finite spatial resolution.
00:16:55.11 It cannot resolve anything infinitely sharp.
00:16:59.12 Let's be more quantitative.
00:17:01.06 The size of the back aperture
00:17:05.11 determines some maximum k value
00:17:08.01 that the objective will allow to go through,
00:17:10.29 and this k value
00:17:12.24 is related to the spatial frequency
00:17:14.23 in the sample domain
00:17:16.07 in this formula, k...
00:17:18.09 spatial frequency = k times 2π over λf,
00:17:20.14 where f is the focal length,
00:17:22.09 λ is the wavelength of light.
00:17:24.07 Then, what is the maximum value of k?
00:17:27.00 Well, from the previous slides,
00:17:29.26 we know that given a light ray
00:17:32.12 that has an angle of α
00:17:34.25 to the optical axis,
00:17:36.14 the k will be f times sinα,
00:17:40.28 and the maximum value of k
00:17:43.17 will be determined by the maximum value of α,
00:17:46.12 or how high of an angle
00:17:49.03 this light ray can be from the optical axis.
00:17:51.29 And that is essentially described
00:17:54.00 by the numerical aperture of this objective.
00:17:57.07 So, kmax = f times sin,
00:18:00.18 or maximum value, of α,
00:18:02.10 and sin of maximum value of α
00:18:06.03 is exactly the numerical aperture of the objective,
00:18:10.03 so kmax = f times numerical aperture.
00:18:13.25 And then... and what does this mean to resolution?
00:18:17.02 Okay, now we're talking about resolving two sine waves,
00:18:21.04 and one sine wave like this,
00:18:22.23 another sine wave like this,
00:18:26.13 and if we're not considering this periodicity,
00:18:30.02 we're just singly looking at two peaks,
00:18:31.23 and you can think,
00:18:33.27 or an easy criteria is,
00:18:36.05 that to be able to resolve these two peaks,
00:18:38.14 the sum of these two peaks needs to have a dip in the middle,
00:18:41.04 and in this case the maximum value of the second peak
00:18:45.00 is going to fall at exactly the same point
00:18:46.28 as the minimum value of the first peak,
00:18:49.07 and then the two sine waves
00:18:52.27 will have a phase difference of π.
00:18:55.01 We can plug that in, here.
00:18:57.14 The resolution is related to the maximum spatial frequency
00:19:00.14 as π over maximum spatial frequency.
00:19:04.08 And then, putting all these numbers together,
00:19:06.08 we get that the resolution equals λ/2NA.
00:19:10.22 This is exactly what we have seen
00:19:13.05 a lot of the time
00:19:14.28 in the microscopy literature,
00:19:16.21 how the numerical aperture of the objective
00:19:18.11 is connected to the resolution of the objective.
00:19:20.28 We can also understand it in this way.
00:19:22.22 Now, we think about
00:19:26.00 the microscope objective focusing the light,
00:19:28.13 then we have our circular numerical aperture
00:19:30.14 is completely filled by light,
00:19:31.26 and that is going to
00:19:33.19 converge down to one single point,
00:19:35.12 and, by definition, this point,
00:19:38.09 or this focal point,
00:19:39.29 is the point spread function of the microscope objective.
00:19:43.09 And now we also know that
00:19:46.17 from the left side and the right side,
00:19:48.15 the connection here is Fourier transform,
00:19:50.23 so it means that the Fourier transform
00:19:53.21 of the light intensity,
00:19:55.06 and of course the phase,
00:19:57.00 at the back aperture,
00:19:59.25 is the point spread function of the microscope.
00:20:03.05 So, that's the lecture.
00:20:06.11 We've talked about real space
00:20:08.12 and that's connected to frequency space
00:20:12.13 through the Fourier transform,
00:20:14.07 and in microscopy
00:20:16.00 this Fourier transform is done
00:20:17.20 by the objective itself,
00:20:18.29 connecting the sample plane
00:20:20.25 and the back focal plane.

This Talk
Speaker: Bo Huang
Audience:
  • Researcher
Recorded: April 2012
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Talk Overview

The Fourier transform is intimately associated with microscopy, since the alternating planes occurring in the microscope (focal plane – back-focal plane, etc.) are related to each other by a function very similar to the Fourier transform. This lectures explains the Fourier transform in terms understandable to non-mathematicians, and explains the relations with microscopy.

Speaker Bio

Bo Huang

Bo Huang

Dr. Huang’s research focuses on using super-resolution microscopy and single-molecule imaging to understand how proteins form large complexes and how proteins interact to regulate signaling. Huang is an Assistant Professor in Pharmaceutical Chemistry and in Biochemistry and Biophysics at UC San Francisco. Continue Reading

Playlist: Microscopy Series

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