1 IMAGING Seeing invisible things oRecap equations for wave
travel oExamine a range of digital images oExplain what is meant by
image resolution
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Pixel Pixels are the tiny building blocks from which a digital
image is built.
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Resolution Smallest discernible feature or smallest detectable
difference. Resolution = image dimension/number of pixels
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Current is kept constant so a record on the up and down motion
is a record of the surface. Scanning tunnelling microscope.
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Ultrasound imaging Explain principles of the technique
Calculate key parameters such as horizontal and vertical
resolution, minimum pulse duration, maximum pulse rate
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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
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Ultrasound pulse sequence showing two pulses being sent out by
the probe. The listening time is much longer than the duration of a
pulse.
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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.
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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?
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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?
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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
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Each Pixel is represented by a number
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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
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DecimalBinary 00 11 2 311 4 5101 Binary All 0s and 1s
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DecimalBinary 00 11 210 311 4100 5101 Binary All 0s and 1s
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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)
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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
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Each pixel is assigned a byte of info 00000001 = A 00000010 =
B. 10010010 = &
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Bits and Bytes N = number of alternatives l = number of bits N=
2 8 = 256 In general: N = 2 l log 2 N= l
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Amount of data in image = no. of pixels x bits per pixel
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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).
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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.
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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?
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Image processing Process a range of digital images to extract
the maximum amount of useful information Explain how the processing
algorithms work
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
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111333 111333 111333 111333 111333 111333 Smoothing out sharp
edges Replace each pixel by the mean of it and its eight
neighbours
Finding Edges Laplace Rule -1x +4x-1x Subtract the N, S, E and
W neighbours from 4x the value of each pixel.
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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.
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Finding edges with the Laplace Rule If there is no edge but a
gradient then the Laplace rule simply smooths the data
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Improving Contrast How might we make image brighter? How might
we improve the contrast? 444 424 444 Decimal numbers of image
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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.
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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.
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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
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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
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The King of all Imagers: The Eye. 100 million rods 5-10 photons
are required to trigger a response.
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Nerve Fibres
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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.
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Long Focal LengthShort Focal Length
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Fish eye lens
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How lenses shape light. v v focus
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A lens modifies the curvature of the wave by 1/f v v f
focus
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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
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v v If the object is close to the lens
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If we have a object near to the lens NEGATIVE POSITIVE
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1 = 1 + 1 v u f The lens changes the curvature of the incoming
wave by 1/f
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Question 1: What is the distance of the object from the lens? v
= 0.8m f = 0.31m
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Question 2: What is the focal length? v = 0.65m u = 0.45m
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Question 3: What is the distance of the image from the lens? f
= 0.21m u = 0.3m
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m = image height = image distance object height object distance
Magnification by a lens m = v/u
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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
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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.
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1/v = 1/u + 1/f When 1/u = 0 (object very far from lens), 1/v =
1/f
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Try these: Questions on Page 25. Homework Summary Questions on
page 27