1 IMAGING Seeing invisible things oRecap equations for wave travel oExamine a range of digital...
Preview:
Citation preview
- Slide 1
- 1 IMAGING Seeing invisible things oRecap equations for wave
travel oExamine a range of digital images oExplain what is meant by
image resolution
- Slide 2
- Slide 3
- Pixel Pixels are the tiny building blocks from which a digital
image is built.
- Slide 4
- Resolution Smallest discernible feature or smallest detectable
difference. Resolution = image dimension/number of pixels
- Slide 5
- Current is kept constant so a record on the up and down motion
is a record of the surface. Scanning tunnelling microscope.
- Slide 6
- Slide 7
- Ultrasound imaging Explain principles of the technique
Calculate key parameters such as horizontal and vertical
resolution, minimum pulse duration, maximum pulse rate
- Slide 8
- Principles of Ultrasound Key things to know about: How is it
generated? Why the need for short pulses? Why the need for high
frequency? What information do we gain from: The pulse-echo times
The reflected intensity
- Slide 9
- Slide 10
- Slide 11
- Ultrasound pulse sequence showing two pulses being sent out by
the probe. The listening time is much longer than the duration of a
pulse.
- Slide 12
- 4-D ultrasound is similar, except that the images are
constructed rapidly to give almost real-time information. 3-D
ultrasound sends sound pulses in at different angles. A computer
algorithm constructs a highly detailed image from the
reflections.
- Slide 13
- Ultrasound is also used to detect cracks in objects such as
aeroplane components and rails. How does it work? Why is it
preferable to using x rays?
- Slide 14
- There are only 10 types of people in the world. ..those who
understand binary numbers ..and those who dont. Can you explain
this rather lame joke?
- Slide 15
- Information in digital images Explain how information is stored
in digital images Use binary arithmetic to work out the values
stored for each pixel Compute the amount of information in an image
in bits and bytes
- Slide 16
- Each Pixel is represented by a number
- Slide 17
- 00000000000 00000000000 00000000000 00000000000 00044400000
002102553520000 002121303420000 00223246720000 00044400000
00000000000 00000000000 Each pixel is assigned a byte of info = 2 8
= 256 alternatives = 256 levels of grey
- Slide 18
- DecimalBinary 00 11 2 311 4 5101 Binary All 0s and 1s
- Slide 19
- DecimalBinary 00 11 210 311 4100 5101 Binary All 0s and 1s
- Slide 20
- Bits and Bytes One Bit of information = 0 or 1 (two
possibilities) Eight Bits of information = 256 possibilities. (8
bits = 1 byte) HOW? There are 256 alternative arrangements of 8
bits. (each bit is either 1 or 0)
- Slide 21
- There is 0/1 alternative for each position One 0/1 choice = One
bit In 8 bit data there are eight 0/1 alternatives Eight bits =
eight 0/1 choices = One byte There are 2 8 = 256 alternatives or
0-255 Number of alternatives = 2 I Where I is the number of bits.
e.g. 16 bit computer uses 2 16 = 65536
- Slide 22
- Each pixel is assigned a byte of info 00000001 = A 00000010 =
B. 10010010 = &
- Slide 23
- Bits and Bytes N = number of alternatives l = number of bits N=
2 8 = 256 In general: N = 2 l log 2 N= l
- Slide 24
- Amount of data in image = no. of pixels x bits per pixel
- Slide 25
- Slide 26
- Slide 27
- Slide 28
- Questions: 120S Logarithms and Powers The response of the eye
(and ear) to light (and sound) intensity is logarithmic not linear.
A logarithm is just another name for the power (or exponent or
index lots of different names for the same idea!) that the constant
ratio is raised to. 1.We can choose convenient ratios to consider:
take a constant ratio of x 10. Write out a series of intensity
values starting with 1 with this constant ratio property. 2. If you
haven't done so, repeat question 1, writing out the series using
scientific notation and powers of ten. 3. What is happening to the
power of ten (its index), or logarithm to base 10? The base is just
the initial constant ratio we chose to work with; any number will
do. (Base 2 gives binary, base 10 gives log10, base e (e =
2.718...) gives the natural log e (written ln).
- Slide 29
- Questions: 120S Logarithms and Powers 4. Using your calculator,
record log10 of the series 1, 10, 100, 1000, 10000 etc. What do you
notice? 5. Sketch a graph of the series of intensities plotted on
the y-axis, against the powers of ten or log10 plotted on the
x-axis. This graph is logarithmic in shape (we say it grows
exponentially). Now plot the graph on a log scale (non-linear),
i.e. log10 (series) on y-axis against the powers of ten on the
x-axis. What has the log scale done to the exponentially growing
data? You will use log scales for representing quantities that vary
enormously, and for testing for logarithmic or exponential
variations.
- Slide 30
- Questions: 120S Logarithms and Powers 6. If you have followed
these steps so far, you have in fact learnt to master the scale for
the measurement of sound intensity ratios, the bel scale (after
Alexander Graham Bell) where: number of bels = log10 (I 2 / I 1 )
or more commonly the sound level in decibels (1 dB = 0.1 B) is
given by number of decibels = 10 log10 (I 2 / I 1 ) A sound that is
on the threshold of audible intensity is 10 -12 W m -2 (like
hearing a pin drop). This is taken as the baseline intensity I 1. A
painful sound (like a jet taking off) has an intensity of about 10
W m -2. What is the sound level of the jet in bels and dB?
- Slide 31
- Image processing Process a range of digital images to extract
the maximum amount of useful information Explain how the processing
algorithms work
- Slide 32
- Image processing algorithms Smoothing Noise removal Edge
detection Change brightness Change contrast
- Slide 33
- Image processing algorithms Smoothing Noise removal Edge
detection Change brightness Change contrast For each of the image
processing algorithms: 1.Describe its effect on an image. 2.Explain
how the algorithm operates on the data. 3.Describe a situation
where it might be useful. 4.Discuss any drawbacks of the algorithm.
The Mars Face
- Slide 34
- 111333 111333 111333 111333 111333 111333 Smoothing out sharp
edges Replace each pixel by the mean of it and its eight
neighbours
- Slide 35
- 112233 112233 112233 112233 112233 112233 Resulting in....
- Slide 36
- Image before sharpening
- Slide 37
- Image after sharpening
- Slide 38
- 333333 336333 333333 333133 343333 333333 Removing Noise
Replace each pixel by the median of its value and those of its
neighbours
- Slide 39
- 333333 333333 333333 333333 333333 333333 Resulting in...
- Slide 40
- Finding Edges Laplace Rule -1x +4x-1x Subtract the N, S, E and
W neighbours from 4x the value of each pixel.
- Slide 41
- Finding Edges Laplace Rule Result if there is an edge. -1x
+4x-1x Subtract the N, S, E and W neighbours from 4x the value of
each pixel.
- Slide 42
- Finding edges with the Laplace Rule If there is no edge but a
gradient then the Laplace rule simply smooths the data
- Slide 43
- Slide 44
- Improving Contrast How might we make image brighter? How might
we improve the contrast? 444 424 444 Decimal numbers of image
- Slide 45
- Improving Contrast 444 424 444 888 848 888 888 868 888 +4 x2
Adding fixed positive value makes image brighter Multiplying by
fixed value increases contrast and makes brighter.
- Slide 46
- A histogram analysis showing how many pixels there are in an
image that have a particular number of pixel values is very useful
in explaining how the contrast and brightness algorithms work.
- Slide 47
- Starter: Match up the image processing technique with the
algorithm that describes how it is applied to the data A Smoothing
B Noise removal C Edge detection D Change brightness E Change
contrast 1.Subtract N,S,E,W neighbours from 4 times value of pixel.
2.Multiply all pixel values by a constant factor. 3.Add a fixed
value to all pixel values. 4.Replace each pixel value by the median
of its value and those of its neighbours 5.Replace each pixel value
by the mean of its value and those of its neighbours
- Slide 48
- Making images using lenses Recap relationship between light
rays and waves Investigate properties of converging lenses Solve
problems using 1/v = 1/u+1/f
- Slide 49
- The King of all Imagers: The Eye. 100 million rods 5-10 photons
are required to trigger a response.
- Slide 50
- Slide 51
- Nerve Fibres
- Slide 52
- Slide 53
- Slide 54
- Slide 55
- Slide 56
- Slide 57
- Slide 58
- Slide 59
- Slide 60
- Slide 61
- Slide 62
- Slide 63
- Slide 64
- Refractive index Refractive index(n) = Speed of light in vacuum
Speed of light in material The refractive index of glass is 1.5 Q1.
What happens to the speed, frequency and wavelength of light waves
when they pass from air into glass? Hint: use the equation above,
the equation speed = frequency x wavelength and if you are stuck,
see page 21 of the textbook Q2. The larger the refractive index of
a material, the more light will bend when it enters the material at
an angle. Use this fact to explain why the makers of spectacle
lenses prefer to use glass with a higher refractive index.
- Slide 65
- Long Focal LengthShort Focal Length
- Slide 66
- Fish eye lens
- Slide 67
- How lenses shape light. v v focus
- Slide 68
- A lens modifies the curvature of the wave by 1/f v v f
focus
- Slide 69
- v v f f is the focal length 1 / f = the power of the lens
measured in dioptre e.g. If focal length f = 50mm Power = 1 / 50
x10 -3 = 20 dioptre f
- Slide 70
- Slide 71
- v v If the object is close to the lens
- Slide 72
- If we have a object near to the lens NEGATIVE POSITIVE
- Slide 73
- 1 = 1 + 1 v u f The lens changes the curvature of the incoming
wave by 1/f
- Slide 74
- Question 1: What is the distance of the object from the lens? v
= 0.8m f = 0.31m
- Slide 75
- Question 2: What is the focal length? v = 0.65m u = 0.45m
- Slide 76
- Question 3: What is the distance of the image from the lens? f
= 0.21m u = 0.3m
- Slide 77
- m = image height = image distance object height object distance
Magnification by a lens m = v/u
- Slide 78
- What is the focal length of the lens? u (m) v (m) -0.220.99
-0.280.52 -0.330.40 -0.400.33 -0.490.29 -0.550.27
- Slide 79
- You could just substitute individual values of u and v into the
equation 1/v = 1/u + 1/f, but there is a better way: Plot 1/v
against 1/u, and then note where the line cuts the y- axis. This
corresponds to 1/u = 0, in other words the source is very far from
the lens. The line cuts the y-axis where 1/v = 1/f, so f can be
determined.
- Slide 80
- 1/v = 1/u + 1/f When 1/u = 0 (object very far from lens), 1/v =
1/f
- Slide 81
- Try these: Questions on Page 25. Homework Summary Questions on
page 27