Term 3

Question 1

Compare wire-pairs and coaxial cables

Answer 1

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

Question 2

Define AM and FM

Answer 2

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.

Question 3

Compare AM and FM

Answer 3

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

Question 4

What are the advantages of using a digital rather than analogue signal

Answer 4

• 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

Question 5

Define sampling rate

Answer 5

The number of samples made per second

Question 6

Describe features of a geostationary satellite

Answer 6

• 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.

Question 7

Compare geostationary and polar satellites

Answer 7

• 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

Question 8

State the formulas for attenuation

Answer 8

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)

Question 9

Define a photon

Answer 9

A ‘packet of energy’ or a quantum of electromagnetic energy.

Question 10

State the formulas to find the energy of a photon

Answer 10

E = hf

E = hc/λ

hf = E1 - E2

Question 11

Describe the photoelectric effect

Answer 11

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.

Question 12

Define threshold frequency and work function

Answer 12

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

Question 13

State work function formulas

Answer 13

Φ = hf0

hf = Φ + 1/2mv(max)^2

Question 14

Describe why intensity does not affect the max photoelectric effect

Answer 14

Greater intensity increases the number of photons released per second but not more energetic photons. This means intensity has no effect.

Question 15

Define the De Broglie wavelength

Answer 15

The wavelength of the waves associated with the movement of electrons.

λ = h/p

λ = h/mv

Question 16

Describe band theory in different materials

Answer 16

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.

Question 17

Describe how X-rays are produced

Answer 17

• 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.

Question 18

Describe how intensity and hardness of an X-ray beam is controlled

Answer 18

• 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.

Question 19

Describe how the X-ray image can be improved

Answer 19

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

Question 20

State formula for attenuation of X-ray

Answer 20

I = Io e^-μx