FAQ Course book: Autonomous Mobile Robots

Chapter 2

Chapter 3

Chapter 4

Q: Explain equation (4.10)

Lamba is the wavelength of the modulated transmitted wave. That is, it is not the wavelength of the light, which would be very shorth typically around 800 nm, but the created by modulating (varying) the amplitude. By measuring the difference in phase between the light that has gone directly from the transmitter to the detector and the one that is reflected on the target. The maximum detectable phase difference, theta, is one period of the modulated wave. One period corresponds to 360 degrees or 2pi. So theta / (2pi) * lambda is the distance but it is we need to rememeber that the light has to go back and forth to the obstacle, ie double distance. The final distance is thus

 (theta / (2pi) * lambda) / 2 = lambda * theta / (4pi)

Q: Explain 4.11

Using a pinhole camera model we can use equal triangles and get that x/f = L / D. f is the focal length, i.e. the distance from the lens to the "image plane" where the distance x is measured. If we rearrange the equations we get

 D = f * L / x

Q: How do you get 4.12?

First we need to remind ourselves that cot(a) = cos(a) / sin(a) = 1 / tan(a). Just like in 4.11 we can look geometric shapes of the same shape but different dimensions, like

 z / b = f / (fcot(a) - u).

In this expression z is the distance to the target, b is the distance betweem the projecting laser and the lens, f is the focal length of the lens, a is the angle of the projecting laser and u is the image coordinate defined such that it increases in the same direction as x. The definition of direction of u in Fig 4.15c therefore seems to be wrong. The figure below shows a close up of the triangles "in the camera" which has teh same shape as the triangle defined by the laser, lens and target point (x,y).

The second equation to work with is

x/b = -u/(fcot(a)-u)

(notice that there seems to be a minus sign missing here)

Q: Equations 4.13?

We can rewrite the right equation in 4.12 to

 u = -b*f/z + fcot(a)

Taking the derivative w.r.t. z gives

 du/dz = x*f/(z*z)

Q: Equations 4.19?

This is of the same form as the thin lens equation that states that

 1 / f = 1 / d + 1 / i

where f is the focal length, d is the distance from the lens to the object and i is the distance from the lens to the image plane (e in 4.19). We can see that thi sis true by looking at the figure below

Q: Equation 4.20?

Normally you put the image plane in the focal plane, ie delta=0. If we move the image plane out from the focal plane the rays will no longer meet in a point. The will spread into a circle roughly. If you look at rays that hit the outmost part of the lens, up and down, these will be focused in the focal plane and then cross each other. The distance between them in the image plane will be 2R. If we again use similar shapes we can get

 2R / delta = L / e

which can be rewritten as

 R = L * delta / (2*e)

This is illustrated in the figure below

Q: Equations 4.23-24

Given that the radius of the blurred cirlce from equation 4.20 above is R, the light is distributed over a circle of area (pi*R*R) which means that the denisty is 1 / (pi*R*R). The density is this only inside the circle and 0 outside, this is what the two different cases are in equation 4.23.
The second equation, 4.24 just sums the contribution from all pixels.

Q: Equations 4.25-28

Basically an application of similar triangles as above