imaging + signalling

Question 1

formation of real image by thin converging lens

Answer 1

lens changes curvature of incident wavefront

curvature = 1 / r

Question 2

what is power of converging lens?

Answer 2

power is curvature added to wavefronts as they pass through converging lens

p = 1 / f

Question 3

lens equation:

Answer 3

1/v = 1/u + 1/f

(u is negative)

Question 4

where focus is, based on object / source position:

Answer 4

1) light from object closer to lens than fl: focus beyond fp
2) light from distant object (parallel incoming rays): focus at fp
3) light from near object but beyond fl (at distance u): focus at distance v
4) light from object at fl: focus at long distance (rays are made parallel)

Question 5

linear magnification

Answer 5

m = image height / object height

on magnification diagram, triangles formed by paths of light rays before and after lens are congruent

Question 6

magnification values:

Answer 6

1) negative: image is inverted
2) m < 1: image is smaller + closer to lens than object (source)
3) m = 1: image is same size + distance from lens to object (source)
4) m > 1 = image is larger + further from lens than object (source)

Question 7

wave speed equation:

Answer 7

v = f x λ

Question 8

frequency equation:

Answer 8

f = 1 / T

T: period of wave

Question 9

characteristics of em waves (polarising):

Answer 9

1) electromagnetic waves are transverse + when unpolarised they oscillate in randomly changing plane
2) when polarised they oscillate in 1 fixed plane

Question 10

how to polarise microwaves:

Answer 10

1) using metal grate + microwave transmitter (+ detector), if wave oscillates in plane parallel to grate alignment, wave is absorbed as electrons in grate can move length of grate so can absorb higher energy photons
2) if grate is rotated 90° so wave oscillates perpendicular to grate alignment, wave passes through as electrons in metal grate can only move width of each bar of grate - so, it cannot absorb photons of microwave as their energy is too large

Question 11

pixel def:

Answer 11

pixel: single ‘picture element’ created by light sensitive detectors

Question 12

bit def:

Answer 12

bit: smallest unit of digital information

Question 13

byte

Answer 13

1 byte = 8 bits = 256 alternatives

Question 14

resolution (imaging + signalling):

Answer 14

resolution: scale of smallest detail that can be distinguished

r = width of object in image / number of pixels across it

Question 15

image processing:

Answer 15

1) changing brightness: add (+ / -) value on each pixel (to inc brightness, inc value on each pixel until brightest is coded at 255)
2) smoothing (removing noise): each pixel is replaced with mean of surrounding pixels
3) noise reduction (removing noise): each pixel is replaced with median of surrounding pixels
4) edge detection: mean of surrounding 8 pixels is subtracted from each pixel
5) changing contrast: to change contrast, multiply each pixel by (+ / -) value (where|value| < 1). to improve contrast, stretch range of pixels in image to full range (256) (values tend to 0 or 255 depending on which they are closer to) (image with little contrast does not use full range of pixels)

Question 16

signal

Answer 16

signal transfers information from one place to another

(can be coded into binary digits to form digital signal)

Question 17

analogue signal + cons:

Answer 17

analogue signal: continuously varying signal

1) can be amplified when distorted by noise, however this amplifies noise too
2) noise can be filtered out, but this loses signal clarity

Question 18

digital signal + pros + cons:

Answer 18

digital signal: signal that only contains two values / ‘modes’, 0 or 1 / on or off

pros:
1) signal doesn’t lose detail when noise is filtered out
2) easy to detect, only takes values of 0 or 1 so can be regenerated perfectly
3) travels faster
4) carries more information

cons:
1) digital signals / numbers can be changed or scrambled, threats to online banking
2) digitally enhanced images on films - achieve level of reality we can’t achieve

Question 19

amount of information in image? [equation]

Answer 19

info in image amount = num of pixels x bits per pixel

Question 20

equations for quantisation levels + num of bits per sample needed to store them?

Answer 20

N = 2^b
b = log2[N]

where:
N = number of quantisation levels
b = bits per sample

Question 21

def for noise

Answer 21

noise is random variation on a signal

Question 22

what is sampling?

Answer 22

sampling is the process where the displacement of a continuous [analogue] signal is measured at small Δt and turned into a string of binary numbers (samples)

Question 23

what is sampling rate / sampling frequency?

Answer 23

sampling rate / frequency is the number of samples per second

Question 24

how to sample varying signal accurately?

Answer 24

• to sample a varying signal accurately, the time between samples must be shorter than the time between when important changes in the signal occur (eg troughs, peaks)
• with a larger Δt, details of the original signal is lost => can cause aliasing

Question 25

quantisation levels description:

Answer 25

quantisation levels description:
• the signal being quantised can be represented graphically, with voltage (amplitude) on the y-axis and time on the x-axis (levels are kind of like the notches on the y-axis)
• number of levels = 2^b

Question 26

what happens when you increase the number of quantisation levels?

Answer 26

increasing the number of quantisation levels causes a better signal match, however it also increases the demand on storage and transmission

Question 27

what happens if the spacing between subsequent [/ consecutive] quantisation levels is smaller than the size of the noise variation?

Answer 27

if the spacing between subsequent [/ consecutive] quantisation levels is smaller than the size of the noise variation, then digitising the signal is useless as it would detail the noise [/ sample the noise to a higher detail], not reduce it

Question 28

equation for max number of quantisation / alternative levels?

Answer 28

max number of levels (2^b) = V total / V noise

Question 29

equation for max number of bits per sample?

Answer 29

max number of bits per sample (b) = log2

Question 30

what to do if b (number of bits per sample) is decimal?

Answer 30

• if b is a decimal, round up!
• this ensures there are enough bits to provide the number of levels required

Question 31

what is a quantisation error?

Answer 31

a quantisation error is the difference between the signal value and the quantisation level value

Question 32

what is resolution (in signalling context)?

Answer 32

resolution is the smallest change in pd that can be determined

Question 33

what is the equation for resolution (in signalling context)?

Answer 33

resolution =
pd range of signal / number of quantisation levels

Question 34

what are the two conditions for sampling?

Answer 34

the two conditions for sampling are:
• the signal cannot contain frequencies above a certain maximum
• the min sampling rate must be at least 2x the max frequency of the signal (the Nyquist rate / Nyquist’s rule)

Question 35

what is the standard sampling frequency for music?

Answer 35

the standard sampling frequency for music is 44.1kHz

(the limit of human hearing is 20 kHz, so the sampling frequency must be greater than 40 kHz)

Question 36

what happens when frequencies above a certain maximum are sampled?

Answer 36

if frequencies above a certain maximum are sampled, the signal won’t be regenerated accurately and aliases will be generated

Question 37

what happens if the sampling frequency used to sample a signal is too low [/ slow]?

Answer 37

if sampling is too slow, high frequencies will be missed and false, low frequency aliases will be created

Question 38

what are aliases?

Answer 38

aliases are low frequency values that were not in the original signal

Question 39

what is bit rate?

Answer 39

bit rate is the rate of transmission of digital information (units: Hz or bits^-1)

Question 40

what is the question for bit rate?

Answer 40

the equation for bit rate is:

bit rate = samples per second x bits per sample

Note:
samples per second = sampling rate or sampling frequency

Question 41

what is the equation for signal duration?

Answer 41

the equation for signal duration is:

signal duration =
number of bits per signal / bit rate