Post on 24-Feb-2016
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1 IMAGINGSeeing invisible things
o Recap equations for wave travel
o Examine a range of digital images
o Explain what is meant by image resolution
Pixel
Pixels are the tiny building blocks from which a digital image is
built.
Resolution
Smallest discernible feature or smallest detectable difference.
Resolution = image dimension/number of pixels
Current is kept constant so a record on the up and down motion is a record of the surface.Scanning
tunnelling microscope.
Ultrasound imaging
• Explain principles of the technique
• Calculate key parameters such as horizontal and vertical resolution, minimum pulse duration, maximum pulse rate
Principles of UltrasoundKey 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
Frequency, speed and wavelength
Time picture
Position picture: two different wave speeds compared (same frequency)
higher speed
lowerspeed
wavelengthshorter
wavelengthlonger
source
source
displacement
time T for1 oscillation
time
distance
frequency ftime T
T = 1f
Ultrasound pulses have a high frequency
speed vfrequency fwavelength
= vT = vT
v = f
Frequency, speed and wavelength
Time picture
Position picture: two different wave speeds compared (same frequency)
higher speed
lowerspeed
wavelengthshorter
wavelengthlonger
source
source
displacement
time T for1 oscillation
time
distance
frequency ftime T
T = 1f
Ultrasound pulses have a high frequency
speed vfrequency fwavelength
= vT = vT
v = f
Ultrasound pulse sequence showing two pulses being sent out by the probe. The “listening time” is much longer than the duration of a pulse.
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.
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?
There are only 10 types of people in the world…….
…..those who understand binary numbers
…..and those who don’t.
Can you explain this rather lame joke?
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
Each Pixel is represented by a number
0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 00 0 0 4 4 4 0 0 0 0 00 0 2 10 255 35 2 0 0 0 00 0 2 12 130 34 2 0 0 0 00 0 2 23 24 67 2 0 0 0 00 0 0 4 4 4 0 0 0 0 00 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0
Each pixel is assigned a byte of info = 28 = 256 alternatives = 256 levels of grey
Decimal Binary0 0
1 1
2
3 11
4
5 101
BinaryAll 0’s and 1’s
Decimal Binary0 0
1 1
2 10
3 11
4 100
5 101
BinaryAll 0’s and 1’s
Bits and BytesOne 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)
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 28 = 256 alternatives or 0-255
Number of alternatives = 2I
Where I is the number of bits.
e.g. 16 bit computer uses 216 = 65536
Each pixel is assigned a byte of info
00000001 = A
00000010 = B...
10010010 = &
Bits and BytesN = number of alternativesl = number of bits
N= 28 = 256In general: N = 2l
log2N= l
Amount of data in image = no. of pixels x bits per pixel
Questions: 120S – Logarithms and PowersThe 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 loge (written ln).
Questions: 120S – Logarithms and Powers4. 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.
Questions: 120S – Logarithms and Powers6. 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 (I2 / I1 )or more commonly the sound level in decibels (1 dB = 0.1 B) is given bynumber of decibels = 10 log10 (I2 / I1)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 I1. 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?
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
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”
1 1 1 3 3 31 1 1 3 3 31 1 1 3 3 31 1 1 3 3 31 1 1 3 3 31 1 1 3 3 3
Smoothing out sharp edges
Replace each pixel by the mean of it
and its eight neighbours
1 1 2 2 3 3
1 1 2 2 3 3
1 1 2 2 3 3
1 1 2 2 3 3
1 1 2 2 3 3
1 1 2 2 3 3
Resulting in....
Image before sharpening
Image after sharpening
3 3 3 3 3 33 3 6 3 3 33 3 3 3 3 33 3 3 1 3 33 4 3 3 3 33 3 3 3 3 3
Removing Noise
Replace each pixel by the median of its value and those of its neighbours
3 3 3 3 3 3
3 3 3 3 3 3
3 3 3 3 3 3
3 3 3 3 3 3
3 3 3 3 3 3
3 3 3 3 3 3
Resulting in...
Finding Edges
Laplace Rule
-1x
-1x +4x -1x
-1x
Subtract the N, S, E and W neighbours from 4x the value of each pixel.
Finding Edges
Laplace Rule
Result if there is an edge.
-1x
-1x +4x -1x
-1x
Subtract the N, S, E and W neighbours from 4x the value of each pixel.
Finding edges with the Laplace Rule
If there is no edge but a gradient then the Laplace rule simply smooths the data
Improving Contrast
How might we make image brighter?
How might we improve the contrast?
4 4 4
4 2 4
4 4 4
Decimal numbers of
image
Improving Contrast
4 4 4
4 2 4
4 4 4
8 8 8
8 4 8
8 8 8
8 8 8
8 6 8
8 8 8+4
x2
Adding fixed positive value makes image brighter
Multiplying by fixed value increases contrast and makes
brighter.
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………………….
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
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
The King of all Imagers: The Eye.
100 million rods
5-10 photons are required to trigger a response.
Nerve Fibres
Refractive indexRefractive index(n) =
Speed of light in vacuumSpeed 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.
Long Focal Length Short Focal Length
Fish eye lens
How lenses shape light.
v
v
focus
A lens modifies the curvature of the wave by 1/f
v
v
f
focus
v
v
f
f is the focal length
1 / f = the power of the lens measured in dioptree.g. If focal length f = 50mm
Power = 1 / 50 x10-3 = 20 dioptre
f
v
v
If the object is close to the lens
If we have a object near to the lens
NEGATIVE POSITIVE
1 = 1 + 1v u f
The lens changes the curvature of the incoming wave by 1/f
Question 1: What is the distance of the
object from the lens?
v = 0.8m
f = 0.31m
Question 2: What is the focal length?
v = 0.65mu = 0.45m
Question 3: What is the distance of the
image from the lens?
f = 0.21m
u = 0.3m
m = image height = image distance object height object distance
Magnification by a lens
m = v/u
What is the focal length of the lens?
u (m) v (m)-0.22 0.99-0.28 0.52-0.33 0.40-0.40 0.33-0.49 0.29-0.55 0.27
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.
-5 -4.5 -4 -3.5 -3 -2.5 -2 -1.50
0.5
1
1.5
2
2.5
3
3.5
4
f(x) = 0.987963442147401 x + 5.485155103524091/v (D)
1/u (D)
1/v = 1/u + 1/f When 1/u = 0 (object very far from lens), 1/v = 1/f
Try these:
Questions on Page 25.
Homework
Summary Questions on page 27