Section 1: Imaging and signalling

Question 1

What does a wave do?

Answer 1

A progressive (moving) wave carries energy and usually information from one place to another without transferring material.

Question 2

Frequency and period link

Answer 2

Frequency = 1/period

Question 3

Wavespeed, freq and wavelength equation

Answer 3

v=frequency x wavelength

Question 4

Intensity of waves

Answer 4

Intensity is a measure of how much energy each wave carries
measured in Watts per square metre (Wm^-2)
Intensity = Power/Area

Question 5

Give 5 examples of how waves can be used to generate images

Answer 5

Sight- visible light waves hitting retinas and trnsferring an image
Medical scanning- E.g Ultrasound scans build up an image by detecting reflected ultrasound waves
Heat cameras- sense infrared waves being emitted by the object being observed. also the same type of wave carries signal from tv remote to tv
Remote sensing- Satellites use sensors to detect waves from distant objects to do things such as keep track of vegetation, whether or make accurate elevation maps
Communications - mobile phones send and receive microwaves that carry signal containing information sent from your phone.

Question 6

Difference between a polarised and non-polarised wave.

Answer 6

Waves that are vibrating on more than one plane are referred to as non-polarised waves
A polarised wave is a wave that only oscillates in one direction or only vibrates in one plane.

Question 7

Electromagnetic polarisation

Answer 7

Electromagnetic radiation is made up of two types of transverse wave, the vibrations are in electric and magnetic fields.
Polarising filter acts like a fence, light that passes through will only be vibrating in one direction

Question 8

two polarising filters at right angles

Answer 8

No waves will pass through

Question 9

Which type of waves do polarisation filters work on

Answer 9

Polarisation filters only work on transverse waves.

This is one indication that light is a transverse wave as it can be polarised.

Question 10

Investigating the polarisation of light

Answer 10

• Align transmission axis of both filters so they are both vertical, keep position of 1st filter fixed
• light is always vertically polarised coming through the first filter
• when two filters are aligned all light passes through
  -as rotate the 2nd, the light that passes through varies
  (think of filters as vectors having vertical and horizontal component, the larger the vertical component the more vertical light)
• when full transmission axis are at 45 degrees to each other half the light will shine through. At right angles no light will pass through - the intensity is 0

Question 11

Polarising microwaves

Answer 11

Using a metal grille

• place grill between the microwave transmitter and receiver, microwave transmitters transmit polarised waves
• Intensity is a maximum when grill and direction of vibrations are at right angles to each other
• as grille is rotated, intensity decreases
• when lines and direction are aligned, no signal

Question 12

Which direction should the grill be when intensity is 0

polarisation of microwaves

Answer 12

The intensity drops to zero, when wires are aligned with the direction of polarisation of microwaves, because the grille is absorbing their energy

Question 13

How polarisation of microwaves work

Answer 13

• vibration electric field of microwaves excites electrons in the metal grille
• energy is absorbed by the grille and re-emitted in all directions
• only a few of the re-emitted waves are vibrating in the general direction of the microwave receiver
• the receiver only recieves waves in one plane, so even if waves travel towards it, they may not be picked up
• When wires and vibrations of waves are aligned, more electrons vibrate as they absorb vertical energy and have more room to more when grid is aligned
• when wires and vibrations are at right angles some electrons are still excited so there is a small drop in intensity

Question 14

What occurs when a wave reaches a boundary in materials

Answer 14

Refraction
If the wave (light) meets the boundary at an angle to the normal, transmitted ray is bent or refracted as it travals at a different speed in different medium
the amount of refraction depends on the wave length of light,
(so focal length will change depending on wavelength)

Question 15

What do converging lenses do?

Answer 15

Converging lenses change the curvature of wave-fronts by refraction.

Question 16

How do converging lenses do what they do?

Answer 16

A converging lens adds curvature to the wave as it passes through.
They curve the wave fronts by slowing down the light travelling through the centre of the lens more than the light entering at the edges.
All points on the wave take an equal amount of time to get to the focus point.

Question 17

Power and thickness relationship

Answer 17

The more powerful the lens is the thicker it is, the more it will curve the wave-fronts that travel through.
shorter focal length.

Question 18

Lens equation

Answer 18

1/v = 1/u +1/f
(u is almost always negative)
curvature after = curvature before + curvature added by lens

Question 19

power, and curvature equations

Answer 19

P= 1/f    (P measured in Dioptres, D)
Curvature = 1/f

Question 20

Lens makers equation

Answer 20

1/f = (n-1)[1/R’ -1/R]
f is focal length
n is index of refraction of lens
R’ (after lens) and R (before lens) are the radii of the two sides of the two sides of lens.

Question 21

what happens when 1/u =0

Answer 21

the lens will add a curvature of 1/f and the focus will be at the focal point

Question 22

What happens if the object is at the focus point of the lens ( on the u side)

Answer 22

The lens will add curvature (-ve) and the waves will leave parallel to the axis of the lens. There will be no focus point/it will be at infinity

Question 23

Lens magnification equation

Answer 23

linear magnification = image height / object height

m =v/u

Question 24

Define focal length of a lens

Answer 24

the distance between the centre of a lens and it’s focus

Question 25

Describe The experiment to determine the focal length of a thin converging lens

Answer 25

- set up a bulb, a lens and a screen all on the same axis will a ruler to measure the distances between them. - place the bulb an exact distance from the lens, i.e 0.2 m so u= 0.2 move the screen until you can see a clear picture of the filament on the screen. - measure the distance between the lens and the screen with image. this is v -Repeat - use equation to find 1/f and therefore power of lens -find average of your values for 1/f and divide by 1 to give focal length f for your lens

Question 26

What is a bit

Answer 26

A single binary digit is called a bit

Question 27

What is a byte

Answer 27

A byte is a group of eight bits

Question 28

How does a computer store data

Answer 28

As a string of bits, the number of bits in a string determines how many alternatives the string can represent.

Question 29

Number of alternatives equation

Answer 29

Number of alternatives = 2^(number of bits) N=2^b Number of bits = Log base 2 (number of alternatives)

Question 30

How are images stored

Answer 30

As an array of binary numbers each pixel is represented by a binary number, with 1 byte and 8 bits (for b&w images) binary numbers are stored as an array, so the location of each number refers to the location of the pixel in the photo. The value of the number refers to the colour of the pixel

Question 31

How are coloured images stored

Answer 31

Each pixel is described as three binary numbers. 1 byte for each primary colour ( red, blue and GREEN) 24 bits per pixel Value of number gives colour of corresponding pixel

Question 32

Image resolution definition

Answer 32

Image resolution is the detail an image holds. usually refers tothe length represented by each pixel. If an object of width 1m is represented by 200 pixels in an image then 1/200 = resolution = 0.005 smaller resolution is better RESOLUTION CAN MEAN DIFFERENT THINGS

Question 33

What does amount of info depend on? b/w and coloured

Answer 33

Bits per pixel pixels range from 0-225 or 2^0 -- 2^8 with 0 being black and 255 being white colours are made up of three numbers each providing 256 possible shades of red green and blue this means there are (2^8)^3 = 2^24 possible different colour combinations

Question 34

information in image equation

Answer 34

total amount of info = number of pixels x bits per pixel

Question 35

Manipulate brightness

Answer 35

Pixels are stored as a number, changing the brightness means changing the value of each pixel adding a fixed positive value to the image makes it lighter (will shift the histogram to the right by fixed value)

Question 36

Manipulating contrast

Answer 36

mnipulating by a fixed value greater than 1 makes the image brighter and increases the contrast (e.g. multiply every pixel by 2) However this is flawed as the image will become significantly brighter If all pixels were in range 50--> 150 scaling all pixels to have a range from 0-->255 will change the contrast without editing the brightness significantly so: 50-> 0 and 150->255

Question 37

Manipulate to highlight feature

Answer 37

Adding false colour. important features could be made v bright colours by mapping numbers to colours. used for remote sensing of data e.g thermal imaging showing heat as colour nebula images are not colourised. colourised to represent the elements that are known to be present

Question 38

Manipulating for noise reduction

Answer 38

Noise is unwanted interference, usually appears as bright or dark spots in images. In the process of sending images some pixel values can be lost two ways of reducing noise - Take the median of surrounding pixels --- lose fine detail, requires more processing power, slower but a much better method, better at preserving fine edges - take the mean of the surrounding pixels called smoothing - -- Spreads damage out, blurring of image, less sharp but preserves fine detail both of these are done to every pixel in the image not just the noisy pixels

Question 39

Edge detection

Answer 39

Laplace rule- multiply the pixel value by 4 and subtract the value of the north, south, east and west values OR take an average of the neighbouring pixels and subtract it from the centre pixel where the light intensity changes sharply there is an edge

Question 40

Analogue and Digital signals

Answer 40

Digital signals are represented by binary numbers, number of values a digital signal can take depend on number of bits. Analogue signals are not limited in the values they can take, they vary continuously over a range of volume and frequency

Question 41

Noise effect on different signals

Answer 41

Digital signals are resistant to effects of noise, pick up disturbance from electrical signals think of the diagram you've seen since day 1

Question 42

Digitising analogue signals

Answer 42

Take the value of the signal at regular time intervals, then find the nearest digital value. Each digital value is represented by a binary number, so you can convert the analogue values to binary numbers. Will not be exact but close enough

Question 43

How well a digitised signal depends on what two things?

Answer 43

The difference between the possible digital values (resolution) and the time from one sample to the next. higher resolution i.e more possible digital values there are, the more closely the digitised signal will match the original high resolution is needed for CDs / DVDs low resolution is used for telephone lines (need to be audible but not completely accurate)

Question 44

Digital advantages/Disadvantages

Answer 44

``` Advantages Sent received and reproduced more easily can be compressed to reduce size, manipulated easily for artistic effect Represent different types of info can be easily produced from computers ``` Disadvantages Digital never reproduce analogue exactly info can be lost signals are much more easily intercepted or hacked without owners knowledge or consent

Question 45

Analogue ad/disad vantages

Answer 45

Advantages Can not be hacked as easily wide range of info Disadvantages noise is easily picked up info is lost very easily

Question 46

The musical note example of signals madeup of lots of different freqs

Answer 46

Waves can contain multiple frequency and they will superimpose over each other amplitude of final signal is the sum of amplitudes of all other signals in that instant fundamental frequency is the lowest frequency wave that makes up the sound wave find the shortest repeating part, calculate the inverse of it's period

Question 47

How does noise limit sampling?

Answer 47

If the original signal contains noise, fine resolution will reproduce small vibrations caused by noise resolution is limited by the ratio of the total variation in the signal to variation caused by noise

Question 48

number of bits for sampling equation

Answer 48

Maximum number of bits = log base 2 (total variation/noise variation) b=log2(Vtotal/Vnoise)

Question 49

minimum sampling range?

Answer 49

Twice the maximum frequency to ensure all with in spectrum are transmitted accurately music needs high sampling rate to make sure high freq details aren't lost (44100 Hz) DVDs (192000 Hz)

Question 50

Flaw of low sampling rate

Answer 50

Can create low frequency signals called aliases that were not in the original signal at all

Question 51

Avoiding aliases equation

Answer 51

minimum rate of sampling > 2 x maximum frequency of signal

Question 52

Rate of transmissions equation

Answer 52

Rate of transmission = samples per second x bits per sample

Question 53

Bits per sample

Answer 53

High enough that the transmitted signal closely matches the original, but not so high it is affected bu noise. (size of bits should include freq of noise, check your notes you have really good notes on this yay)

Question 54

Time taken to transmit signal equation

Answer 54

time taken to transmit a signal = number of bits to transmit/rate of transmission