Term 3 Flashcards
Compare wire-pairs and coaxial cables
Wire Pairs:
- cheap and convenient
- strongly attenuate a signal
- have a low bandwidth
- pick up some noise and interference
- suffer from cross-talk
- have low security
Coaxial cable
- is more expensive
- less attenuating
- higher bandwidth
- less electrical interference and noise
- has little crosstalk
- is more secure
Define AM and FM
Amplitude modulation: The amplitude of the modulated wave changes in synchrony with the amplitude of the signal.
Frequency modulation: The frequency of the modulated wave changes in synchrony with the amplitude of the signal.
Compare AM and FM
FM:
- less electrical interference and noise
- greater bandwidth produces a better quality of sound
AM:
- greater area covered by one transmitter
- smaller bandwidth means more stations available in any frequency range.
- cheaper radio sets
What are the advantages of using a digital rather than analogue signal
- less noise compared to analogue as regeneration removes noise.
- digital signals are compatible with modern technology and can be stored and processed more easily.
- more reliable and easier to design and build systems
- errors in reception are noticed and parts of the signal can be sent again
Define sampling rate
The number of samples made per second
Describe features of a geostationary satellite
- satellite rotates with same period as the Earth
- satellite is in orbit above the equator with a period of 1 day
- satellite is not moved as it is always above the same point above the equator.
Compare geostationary and polar satellites
- polar satellites have a shorter period of orbit as they travel from pole to pole
- at a smaller height above the earth and can detect objects of smaller detail
- not always in same position relative to Earth and so dishes must be moved
- has smaller delay times
State the formulas for attenuation
dB = 10 lg (P1/P2)
signal to noise ratio = 10 lg (signal power/noise power)
attenuation per unit length = attenuation (dB)/ length of cable (km)
Define a photon
A ‘packet of energy’ or a quantum of electromagnetic energy.
State the formulas to find the energy of a photon
E = hf
E = hc/λ
hf = E1 - E2
Describe the photoelectric effect
As light shines on a metal surface electrons are released from it as light is a wave that carries energy which is used to release the electrons.
Define threshold frequency and work function
Threshold frequency: minimum frequency required to release electrons from surface of metal
Work Function: Minimum amount of energy required by an electron to escape its surface
State work function formulas
Φ = hf0
hf = Φ + 1/2mv(max)^2
Describe why intensity does not affect the max photoelectric effect
Greater intensity increases the number of photons released per second but not more energetic photons. This means intensity has no effect.
Define the De Broglie wavelength
The wavelength of the waves associated with the movement of electrons.
λ = h/p
λ = h/mv
Describe band theory in different materials
Metals:
- conduction band is partially filled meaning they are free electrons and so give the metal its conductivity.
Insulator:
- conduction band is unoccupied. Gap between valence band and conduction band is large.
Semiconductor:
- conduction band is unoccupied but gap between valence and conduction band is much smaller and so some electrons can jump up into conduction band.
Describe how X-rays are produced
- produced when high speed electrons are slowed down causing their kinetic energy to be transformed into photons of EM radiation.
- an X-ray tube consists of a cathode and anode with an external power supply of 200 kV
- when electrons strike the anodes at high speed they lose some Ek in the form of X-ray photons.
Describe how intensity and hardness of an X-ray beam is controlled
- intensity is increased by increasing current as this means more electrons will be released per second and more electrons will hit the anode.
- hardness is the measure of the beam’s energy. Hard X-rays penetrate more easily. Increasing hardness is done by increasing voltage to produce higher energy X-rays.
Describe how the X-ray image can be improved
Improving sharpness:
- determined by width of X-ray beam, narrower beam of parallel X-rays leads to better image.
- width is determined by width of electron beam and target, size of aperture at exit window, and collimation of the beam to ensure it does not fan out.
Improving contrast:
- Clear difference between different types of tissue
- Hardness of X-ray beam is a factor
- Contrast media such as barium meal used
State formula for attenuation of X-ray
I = Io e^-μx
