Electromagnetic spectrum — Paper 4 · A Level Physics 9702

9 Different frequencies and wavelengths are used in different channels of communication. Suggest why (a) infra-red radiation rather than visible light is usually used with optic fibres, … … … [2] (b) the base stations in mobile phone networks operate on UHF, … … … [2] (c) for satellite communication, frequencies of the order of GHz are used, with the uplink having a different frequency to the downlink. … … … [2]

6 marks

Mark scheme: 9 (a) IR has less attenuation (per unit length) B1 fewer (repeater) amplifiers / longer uninterrupted length B1 [2] (b) either limited range B1 (so) cells do not overlap (appreciably) B1 [2] or short wavelength (B1) so convenient length aerial (on mobile phone) (B1) (c) large bandwidth / large information carrying capacity B1 different so that uplink signal not swamped by downlink B1 [2]

This question in 9702/41 Oct/Nov 2008

13 (a) Fig. 13.1 is a block diagram illustrating part of a mobile phone handset used for receiving For a signal from a base station. Examiner’s Use switch D A C Fig. 13.1 Complete Fig. 13.1 by labelling each of the blocks. [4] (b) Explain the role of the base station and the cellular exchange when a mobile phone is switched on and before a call is made or received. … … … … … … …

4 marks

Mark scheme: 13 (a) switch; tuning cct; (r.f.) amplifier; demodulator; serial-to-parallel converter; DAC; (a.f.) amplifier mark as 2 sets of 2 marks each 5 blocks identified correctly B2 (each error or omission, deduct 1 mark) 5 blocks in correct order B2 [4] (4 or 3 blocks in correct order, allow 1 mark) (b) phone transmits signal (to identify itself) (1) signal received by (several) base stations (1) transferred to cellular exchange (1) computer selects base station with strongest signal (1) assigns a (carrier) frequency (1) (any four, 1 each, max 4) B4 [4]

This question in 9702/41 May/June 2009

11 The variation with time of the signal transmitted from an aerial is shown in Fig. 11.1. For Examiner’s Use ls / time 200 190 180 170 160 150 140 130 120 110 11.1 100 Fig. 90 80 70 60 50 40 30 20 10 0 (a) State the name of this type of modulated transmission. For Examiner’s … [1] Use (b) Use Fig. 11.1 to determine the frequency of (i) the carrier wave, frequency = … Hz [2] (ii) the information signal. frequency = … Hz [1] (c) (i) On the axes of Fig. 11.2, draw the frequency spectrum (the variation with frequency of the signal voltage) of the signal from the aerial. Mark relevant values on the frequency axis. signal voltage frequency Fig. 11.2 [3] (ii) Determine the bandwidth of the signal. bandwidth = … Hz [1]

8 marks

Mark scheme: 11 (a) amplitude modulation ……(allow AM) … B1 [1] (b) (i) frequency = 1 / period … C1 = 100 kHz … A1 [2] (ii) frequency = 10 kHz … A1 [1] (c) (i) vertical line at 100 kHz … B1 vertical lines at 90 kHz and 110 kHz … B1 lines at 90 kHz and 110 kHz same length and shorter than at 100 kHz … B1 [3] (ii) 20 kHz … B1 [1] [Total: 8]

This question in 9702/42 Oct/Nov 2009

11 Many radio stations now broadcast on FM rather than on AM. In general, FM is broadcast at For much higher frequencies than AM. Examiner’s Use (a) Explain what is meant by FM (frequency modulation). … … … … [2] (b) State two advantages and two disadvantages of FM transmissions when compared with AM transmissions. advantages of FM transmissions 1. … … 2. … … disadvantages of FM transmissions 1. … … 2. … … [4]

6 marks

Mark scheme: 11 (a) frequency of carrier wave varies (in synchrony) with signal M1 (in synchrony) with displacement of signal A1 [2] (b) advantages e.g. less noise / less interference greater bandwidth / better quality (1 each, max 2) disadvantages e.g. short range / more transmitters / line of sight more complex circuitry greater expense (1 each, max 2) B4 [4]

This question in 9702/41 May/June 2010

12 A ground station on Earth transmits a signal of frequency 14 GHz and power 18 kW towards For a communications satellite orbiting the Earth, as illustrated in Fig. 12.1. Examiner’s Use ground station, signal power 18 kW frequency signal 14 GHz satellite Earth Fig. 12.1 The loss in signal power between the ground station and the satellite is 190 dB. (a) Calculate the power of the signal received by the satellite. power = … W [3] (b) The signal received by the satellite is amplified and transmitted back to Earth. (i) Suggest a frequency for the signal that is sent back to Earth. frequency = … GHz [1] (ii) Give a reason for your answer in (i). … … [1]

5 marks

Mark scheme: 12 (a) gain / loss/dB = 10 lg(P1/P2) C1 190 = 10 lg(18 × 103 / P2) or –190 = 10 lg P2 / 18 × 103) C1 power = 1.8 × 10–15 W A1 [3] (b) (i) 11 GHz / 12 GHz B1 [1] (ii) e.g. so that input signal to satellite will not be ‘swamped’ to avoid interference of uplink with / by downlink B1 [1]

This question in 9702/41 May/June 2010

11 (a) Describe what is meant by frequency modulation (FM ). For Examiner’s … Use … … [2] (b) A sinusoidal carrier wave has a frequency of 600 kHz and an amplitude of 5.0 V. The carrier wave is frequency modulated by a sinusoidal wave of frequency 7.0 kHz and amplitude 2.0 V. The frequency deviation of the carrier wave is 20 kHz V–1. Determine, for the modulated carrier wave, (i) the amplitude, amplitude = … V [1] (ii) the maximum frequency, maximum frequency = … Hz [1] (iii) the minimum frequency, minimum frequency = … Hz [1] (iv) the number of times per second that the frequency changes from maximum to minimum and then back to maximum. number = … [1]

6 marks

Mark scheme: 11 (a) frequency of carrier wave varies M1 (in synchrony) with the displacement of the information signal A1 [2] (b) (i) 5.0 V A1 [1] (ii) 640 kHz A1 [1] (iii) 560 kHz A1 [1] (iv) 7000 (condone unit) A1 [1]

This question in 9702/41 May/June 2011

12 Many television receivers are connected to an aerial using a coaxial cable. Such a cable is For illustrated in Fig. 12.1. Examiner’s Use copper wire polythene plastic insulator covering copper braid Fig. 12.1 (a) State two functions of the copper braid. 1. … … 2. … … [2] (b) Suggest two reasons why a coaxial cable is used, rather than a wire pair, to connect the aerial to the receiver. 1. … … 2. … … [2] (c) A coaxial cable has an attenuation per unit length of 200 dB km–1. The length of the co-axial cable between an aerial and the receiver is 12 m. Calculate the ratio input signal power to coaxial cable . output signal power from coaxial cable

4 marks

Mark scheme: 12 (a) e.g. acts as ‘return’ for the signal shields inner core from noise / interference / cross-talk (any two sensible answers, 1 each, max 2) B2 [2] (b) e.g. greater bandwidth less attenuation (per unit length) less noise / interference (any two sensible answers, 1 each, max 2) B2 [2] (c) attenuation is 2.4 dB C1 attenuation = 10 lg(P1 / P2) C1 ratio = 1.7 A1 [3]

This question in 9702/41 May/June 2011

11 The use of ionospheric reflection of radio waves for long-distance communication has, to a For great extent, been replaced by satellite communication. Examiner’s Use (a) State and explain two reasons why this change has occurred. 1. … … … 2. … … … [4] (b) The radio link between a geostationary satellite and Earth may be attenuated by as much as 190 dB. Suggest why, as a result of this attenuation, the uplink and downlink frequencies must be different. … … … … [2]

6 marks

Mark scheme: 11 (a) e.g. unreliable communication (M1) because ion layers vary in height / density (A1) e.g. cannot carry all information required (M1) bandwidth too narrow (A1) e.g. coverage limited (M1) reception poor in hilly areas (A1) (any two sensible suggestions, M1 & A1 for each, max 4) [4] (b) signal must be amplified (greatly) before transmission back to Earth B1 uplink signal would be swamped by downlink signal B1 [2] GCE AS/A LEVEL – May/June 2011 9702 42

This question in 9702/42 May/June 2011

11 The use of ionospheric reflection of radio waves for long-distance communication has, to a For great extent, been replaced by satellite communication. Examiner’s Use (a) State and explain two reasons why this change has occurred. 1. … … … 2. … … … [4] (b) The radio link between a geostationary satellite and Earth may be attenuated by as much as 190 dB. Suggest why, as a result of this attenuation, the uplink and downlink frequencies must be different. … … … … [2]

6 marks

Mark scheme: 11 (a) e.g. unreliable communication (M1) because ion layers vary in height / density (A1) e.g. cannot carry all information required (M1) bandwidth too narrow (A1) e.g. coverage limited (M1) reception poor in hilly areas (A1) (any two sensible suggestions, M1 & A1 for each, max 4) [4] (b) signal must be amplified (greatly) before transmission back to Earth B1 uplink signal would be swamped by downlink signal B1 [2] GCE AS/A LEVEL – May/June 2011 9702 43

This question in 9702/43 May/June 2011

10 (a) Cable television uses optic fibres for the transmission of signals. For Suggest four advantages of optic fibres over coaxial cables for the transmission of data. Examiner’s Use 1. … … 2. … … 3. … … 4. … … [4] (b) Electromagnetic radiation of wavelength 1310 nm is frequently used for optic fibre communication, rather than visible light. (i) State the region of the electromagnetic spectrum in which radiation of wavelength 1310 nm is found. … [1] (ii) Suggest why this radiation is used, rather than visible light. … … [1] (c) An optic fibre has an attenuation per unit length of 0.2 dB km–1. For A signal is transmitted along the optic fibre of length 30 km to a receiver. The noise Examiner’s power at the receiver is 9.3 μW. Use The minimum acceptable signal-to-noise ratio at the receiver is 26 dB. Calculate (i) the minimum signal power at the receiver, power = … W [2] (ii) the minimum input signal power to the optic fibre. power = … W [2]

10 marks

Mark scheme: 10 (a) e.g. large bandwidth/carries more information low attenuation of signal low cost smaller diameter, easier handling, easier storage, less weight high security/no crosstalk low noise/no EM interference (allow any four sensible suggestions, 1 each, max 4) B4 [4] (b) (i) infra-red B1 [1] (ii) lower attenuation than for visible light B1 [1] (c) (i) gain/dB = 10 lg(P2/P1) C1 26 = 10 lg(P2/9.3 × 10–6) P2 = 3.7 × 10–3 W A1 [2] (ii) power loss along fibre = 30 × 0.2 = 6.0 dB C1 either 6 = 10 lg(P/3.7 × 10–3) or 6 dB = 4 × 3.7 × 10–3 or 32 = 10 lg(P/9.3 × 10–6) input power = 1.5 × 10–2 W A1 [2] GCE AS/A LEVEL – October/November 2011 9702 41

This question in 9702/41 Oct/Nov 2011

11 A simplified block diagram of a mobile phone handset is shown in Fig. 11.1. For Examiner’s aerial Use A B amplifier r.f. amplifier modulator oscillator demodulator parallel-to- serial-to- serial parallel C DAC a.f. D amplifier microphone loudspeaker Fig. 11.1 (a) Name and state the function of (i) block A, … … … [2] (ii) block B, … … … [2] (iii) block C, For Examiner’s … Use … … [2] (iv) block D. … … … [2] (b) Give two reasons why communication between a mobile phone handset and the base station is conducted using UHF. 1. … … 2. … … [2]

10 marks

Mark scheme: 11 (a) (i) switch M1 so that one aerial can be used for transmission and reception A1 [2] (ii) tuning circuit M1 to select (one) carrier frequency (and reject others) A1 [2] (iii) analogue-to-digital converter/ADC M1 converts microphone output to a digital signal A1 [2] (iv) (a.f.) amplifier (not r.f. amplifier) M1 to increase (power of) signal to drive the loudspeaker A1 [2] (b) e.g. short aerial so easy to handle short range so less interference between base stations larger waveband so more carrier frequencies (any two sensible suggestions, 1 each, max 2) B2 [2]

This question in 9702/41 Oct/Nov 2011

10 (a) Cable television uses optic fibres for the transmission of signals. For Suggest four advantages of optic fibres over coaxial cables for the transmission of data. Examiner’s Use 1. … … 2. … … 3. … … 4. … … [4] (b) Electromagnetic radiation of wavelength 1310 nm is frequently used for optic fibre communication, rather than visible light. (i) State the region of the electromagnetic spectrum in which radiation of wavelength 1310 nm is found. … [1] (ii) Suggest why this radiation is used, rather than visible light. … … [1] (c) An optic fibre has an attenuation per unit length of 0.2 dB km–1. For A signal is transmitted along the optic fibre of length 30 km to a receiver. The noise Examiner’s power at the receiver is 9.3 μW. Use The minimum acceptable signal-to-noise ratio at the receiver is 26 dB. Calculate (i) the minimum signal power at the receiver, power = … W [2] (ii) the minimum input signal power to the optic fibre. power = … W [2]

10 marks

Mark scheme: 10 (a) e.g. large bandwidth/carries more information low attenuation of signal low cost smaller diameter, easier handling, easier storage, less weight high security/no crosstalk low noise/no EM interference (allow any four sensible suggestions, 1 each, max 4) B4 [4] (b) (i) infra-red B1 [1] (ii) lower attenuation than for visible light B1 [1] (c) (i) gain/dB = 10 lg(P2/P1) C1 26 = 10 lg(P2/9.3 × 10–6) P2 = 3.7 × 10–3 W A1 [2] (ii) power loss along fibre = 30 × 0.2 = 6.0 dB C1 either 6 = 10 lg(P/3.7 × 10–3) or 6 dB = 4 × 3.7 × 10–3 or 32 = 10 lg(P/9.3 × 10–6) input power = 1.5 × 10–2 W A1 [2] GCE AS/A LEVEL – October/November 2011 9702 42

This question in 9702/42 Oct/Nov 2011

11 A simplified block diagram of a mobile phone handset is shown in Fig. 11.1. For Examiner’s aerial Use A B amplifier r.f. amplifier modulator oscillator demodulator parallel-to- serial-to- serial parallel C DAC a.f. D amplifier microphone loudspeaker Fig. 11.1 (a) Name and state the function of (i) block A, … … … [2] (ii) block B, … … … [2] (iii) block C, For Examiner’s … Use … … [2] (iv) block D. … … … [2] (b) Give two reasons why communication between a mobile phone handset and the base station is conducted using UHF. 1. … … 2. … … [2]

10 marks

Mark scheme: 11 (a) (i) switch M1 so that one aerial can be used for transmission and reception A1 [2] (ii) tuning circuit M1 to select (one) carrier frequency (and reject others) A1 [2] (iii) analogue-to-digital converter/ADC M1 converts microphone output to a digital signal A1 [2] (iv) (a.f.) amplifier (not r.f. amplifier) M1 to increase (power of) signal to drive the loudspeaker A1 [2] (b) e.g. short aerial so easy to handle short range so less interference between base stations larger waveband so more carrier frequencies (any two sensible suggestions, 1 each, max 2) B2 [2]

This question in 9702/42 Oct/Nov 2011

13 (a) In a mobile phone system, the area covered by the system is divided into a number of For cells. Examiner’s For this system, explain why Use (i) neighbouring cells use different carrier frequencies, … … [1] (ii) each cell has a limited area, even in sparsely populated regions. … … [1] (b) A mobile phone handset is left switched on. Explain why, although a call is not being made, the computer at the cellular exchange is still operating for this phone. … … … … [3]

5 marks

Mark scheme: 13 (a) (i) no interference (between signals) near boundaries (of cells) B1 [1] (ii) for large area, signal strength would have to be greater and this could be hazardous to health B1 [1] (b) mobile phone is sending out an (identifying) signal M1 computer/cellular exchange continuously selects cell/base station with strongest signal A1 computer/cellular exchange allocates (carrier) frequency (and slot) A1 [3]

This question in 9702/41 May/June 2012

13 (a) In a mobile phone system, the area covered by the system is divided into a number of For cells. Examiner’s For this system, explain why Use (i) neighbouring cells use different carrier frequencies, … … [1] (ii) each cell has a limited area, even in sparsely populated regions. … … [1] (b) A mobile phone handset is left switched on. Explain why, although a call is not being made, the computer at the cellular exchange is still operating for this phone. … … … … [3]

5 marks

Mark scheme: 13 (a) (i) no interference (between signals) near boundaries (of cells) B1 [1] (ii) for large area, signal strength would have to be greater and this could be hazardous to health B1 [1] (b) mobile phone is sending out an (identifying) signal M1 computer/cellular exchange continuously selects cell/base station with strongest signal A1 computer/cellular exchange allocates (carrier) frequency (and slot) A1 [3]

This question in 9702/43 May/June 2012

11 In commercial radio, transmissions are made by means of carrier waves that are modulated For by the audio signals. Examiner’s Use (a) State what is meant by a modulated carrier wave. … … … … [3] (b) State three reasons why modulated carrier waves are used, rather than the direct transmission of electromagnetic waves having audio frequencies. 1. … … 2. … … 3. … … [3]

6 marks

Mark scheme: 11 (a) high frequency wave B1 the amplitude or the frequency is varied M1 the variation represents the information signal / in synchrony with (the displacement of) the information signal. A1 [3] GCE AS/A LEVEL – October/November 2012 9702 43 (b) e.g. shorter aerial required longer transmission range / lower transmitter power / less attenuation allows more than one station in a region less distortion (allow any three sensible suggestions, 1 mark each) B3 [3]

This question in 9702/43 Oct/Nov 2012

11 A radio station emits an amplitude-modulated wave for the transmission of music. For Examiner’s (a) (i) State what is meant by an amplitude-modulated (AM) wave. Use … … … [2] (ii) Give two reasons why the transmitted wave is modulated, rather than transmitting the information signal directly as a radio wave. 1. … … 2. … … [2] (b) The variation with frequency f of the amplitude A of the transmitted wave is shown in For Fig. 11.1. Examiner’s Use A 0 900 909 918 f / kHz Fig. 11.1 For this transmission, determine (i) the wavelength of the carrier wave, wavelength = … m [2] (ii) the bandwidth, bandwidth = … kHz [1] (iii) the maximum frequency, in Hz, of the transmitted audio signal. frequency = … Hz [1]

8 marks

Mark scheme: 11 (a) (i) amplitude of the carrier wave varies M1 (in synchrony) with the displacement of the information signal A1 [2] (ii) e.g. more than one radio station can operate in same region/less interference enables shorter aerial increased range/less power required/less attenuation less distortion (any two sensible answers, 1 each) B2 [2] (b) (i) frequency = 909 kHz C1 wavelength = (3.0 × 108) / (909 × 103) = 330 m A1 [2] (ii) bandwidth = 18 kHz A1 [1] (iii) frequency = 9000 Hz A1 [1]

This question in 9702/41 May/June 2013

11 A radio station emits an amplitude-modulated wave for the transmission of music. For Examiner’s (a) (i) State what is meant by an amplitude-modulated (AM) wave. Use … … … [2] (ii) Give two reasons why the transmitted wave is modulated, rather than transmitting the information signal directly as a radio wave. 1. … … 2. … … [2] (b) The variation with frequency f of the amplitude A of the transmitted wave is shown in For Fig. 11.1. Examiner’s Use A 0 900 909 918 f / kHz Fig. 11.1 For this transmission, determine (i) the wavelength of the carrier wave, wavelength = … m [2] (ii) the bandwidth, bandwidth = … kHz [1] (iii) the maximum frequency, in Hz, of the transmitted audio signal. frequency = … Hz [1]

8 marks

Mark scheme: 11 (a) (i) amplitude of the carrier wave varies M1 (in synchrony) with the displacement of the information signal A1 [2] (ii) e.g. more than one radio station can operate in same region/less interference enables shorter aerial increased range/less power required/less attenuation less distortion (any two sensible answers, 1 each) B2 [2] (b) (i) frequency = 909 kHz C1 wavelength = (3.0 × 108) / (909 × 103) = 330 m A1 [2] (ii) bandwidth = 18 kHz A1 [1] (iii) frequency = 9000 Hz A1 [1]

This question in 9702/43 May/June 2013

12 (a) Suggest why attenuation of a signal in channels of communication is usually measured For on a logarithmic rather than a linear scale. Examiner’s Use … … [1] (b) For a particular channel of communication having low attenuation, the input power is 6.5 mW and the attenuation per unit length is 1.8 dB km–1. (i) Suggest the name of this channel of communication. … [1] (ii) Calculate the distance over which the power of the signal is reduced to 1.5 × 10–15 W. distance = … km [3]

5 marks

Mark scheme: 12 (a) e.g. logarithm provides a smaller number e.g. gain of amplifiers is series found by addition, (not multiplication) (any sensible suggestion) B1 [1] (b) (i) optic fibre B1 [1] (ii) attenuation/dB = 10 lg(P2/P1) C1 attenuation/dB = 10 lg({6.5 × 10–3}/{1.5 × 10–15}) C1 attenuation/dB = 126 length = 126 / 1.8 length = 70 km A1 [3]

This question in 9702/41 Oct/Nov 2013

12 (a) Suggest why attenuation of a signal in channels of communication is usually measured For on a logarithmic rather than a linear scale. Examiner’s Use … … [1] (b) For a particular channel of communication having low attenuation, the input power is 6.5 mW and the attenuation per unit length is 1.8 dB km–1. (i) Suggest the name of this channel of communication. … [1] (ii) Calculate the distance over which the power of the signal is reduced to 1.5 × 10–15 W. distance = … km [3]

5 marks

Mark scheme: 12 (a) e.g. logarithm provides a smaller number e.g. gain of amplifiers is series found by addition, (not multiplication) (any sensible suggestion) B1 [1] (b) (i) optic fibre B1 [1] (ii) attenuation/dB = 10 lg(P2/P1) C1 attenuation/dB = 10 lg({6.5 × 10–3}/{1.5 × 10–15}) C1 attenuation/dB = 126 length = 126 / 1.8 length = 70 km A1 [3]

This question in 9702/42 Oct/Nov 2013

12 (a) State two reasons why frequencies in the gigahertz (GHz) range are used in satellite For communication. Examiner’s Use 1. … … 2. … … [2] (b) In one particular satellite communication system, the frequency of the signal transmitted from Earth to the satellite (the up-link) is 6 GHz. The frequency of the signal transmitted back to Earth from the satellite (the down-link) is 4 GHz. Explain why the two signals are transmitted at different frequencies. … … … [2] (c) A signal transmitted from Earth has a power of 3.1 kW. This signal, received by a satellite, has been attenuated by 185 dB. Calculate the power of the signal received by the satellite. power = … W [3]

7 marks

Mark scheme: 12 (a) e.g. no / little ionospheric reflection large information carrying capacity (any two sensible suggestions, 1 each) B2 [2] (b) prevents (very) low power signal received at satellite M1 being swamped by high-power transmitted signal A1 [2] (c) attenuation / dB = 10 lg(P2/P1) C1 185 = 10 lg({3.1 × 103}/P) C1 P = 9.8 × 10–16 W A1 [3]

This question in 9702/43 Oct/Nov 2013

14 (a) State what is meant by the attenuation of a signal. … … [1] (b) A transmission cable has a length of 30 km. The attenuation per unit length of the cable is 2.4 dB km−1. Calculate, for a signal being transmitted along the cable, (i) the total attenuation, in dB, attenuation = … dB [1] (ii) the ratio input power of signal . output power of signal ratio = … [3] (c) By reference to your answers in (b), suggest why the attenuation of transmitted signals is usually expressed in dB. … … [1]

6 marks

Mark scheme: 14 (a) reduction in power (allow intensity/amplitude) B1 [1] (b) (i) attenuation = 2.4 × 30 = 72 dB A1 [1] (ii) gain/attenuation/dB = 10 lg(P2/P1) C1 72 = 10 lg(PIN/POUT) or –72 = 10 lg(POUT /PIN) C1 ratio = 1.6 × 107 A1 [3] (c) e.g. enables smaller/more manageable numbers to be used e.g. gains in dB for series amplifiers are added, not multiplied B1 [1]

This question in 9702/41 May/June 2014

3 A microwave cooker uses electromagnetic waves of frequency 2450 MHz. The microwaves warm the food in the cooker by causing water molecules in the food to oscillate with a large amplitude at the frequency of the microwaves. (a) State the name given to this phenomenon. … [1] (b) The effective microwave power of the cooker is 750 W. The temperature of a mass of 280 g of water rises from 25 °C to 98 °C in a time of 2.0 minutes. Calculate a value for the specific heat capacity of the water. specific heat capacity = … J kg−1 K−1 [3] (c) The value of the specific heat capacity determined from the data in (b) is greater than the accepted value. A student gives as the reason for this difference: ‘heat lost to the surroundings’. Suggest, in more detail than that given by the student, a possible reason for the difference. … … [1]

5 marks

Mark scheme: 3 (a) resonance B1 [1] (b) Pt R mc ∆θ C1 750 × 2 × 60 R 0.28 × c × (98 – 25) C1 c R 4400 J kg–1 K–1 A1 [3] (use of ∆θ R 73 N 273 max. 1 / 3) (use of t R 2 s not 120 s max. 2 / 3) GCE A LEVEL – May/June 2014 9702 42 (c) e.g. some microwave leakage from the cooker e.g. container for the water is also heated (any sensible suggestion) B1 [1] 2

This question in 9702/42 May/June 2014

12 Two people, living in different regions of the Earth, communicate either using a link provided by a geostationary satellite or using optic fibres. (a) (i) Explain what is meant by a geostationary satellite. … … … … … [3] (ii) The uplink frequency for communication with the satellite is 6 GHz and the downlink has a frequency of 4 GHz. Explain why the frequencies are different. … … … … [2] (b) Comment on the time delays experienced by the two people when communicating either using geostationary satellites or using optic fibres. Explain your answer. … … … … … [3]

8 marks

Mark scheme: 12 (a) (i) satellite is in equatorial orbit B1 travelling from west to east B1 period of 24 hours / 1 day B1 [3] GCE A LEVEL – May/June 2014 9702 42 (ii) either uplink signal is highly attenuated or signal is highly amplified (before transmission) as downlink signal B1 prevents downlink signal swamping the uplink signal B1 [2] (b) speed of signal is same order of magnitude in both systems B1 optic fibre link (much) shorter than via satellite M1 time delay using optic fibre is less A1 [3]

This question in 9702/42 May/June 2014

14 (a) State what is meant by the attenuation of a signal. … … [1] (b) A transmission cable has a length of 30 km. The attenuation per unit length of the cable is 2.4 dB km−1. Calculate, for a signal being transmitted along the cable, (i) the total attenuation, in dB, attenuation = … dB [1] (ii) the ratio input power of signal . output power of signal ratio = … [3] (c) By reference to your answers in (b), suggest why the attenuation of transmitted signals is usually expressed in dB. … … [1]

6 marks

Mark scheme: 14 (a) reduction in power (allow intensity/amplitude) B1 [1] (b) (i) attenuation = 2.4 × 30 = 72 dB A1 [1] (ii) gain/attenuation/dB = 10 lg(P2/P1) C1 72 = 10 lg(PIN/POUT) or –72 = 10 lg(POUT /PIN) C1 ratio = 1.6 × 107 A1 [3] (c) e.g. enables smaller/more manageable numbers to be used e.g. gains in dB for series amplifiers are added, not multiplied B1 [1]

This question in 9702/43 May/June 2014

12 (a) Information may be carried by different channels of communication. State one application, in each case, where information is carried using (i) microwaves, … … [1] (ii) coaxial cables, … … [1] (iii) wire pairs. … … [1] (b) A station on Earth transmits a signal of initial power 3.1 kW to a geostationary satellite. The attenuation of the signal received by the satellite is 190 dB. (i) Calculate the power of the signal received by the satellite. power = … kW [2] (ii) By reference to your answer in (i), state and explain the changes made to the signal before transmission back to Earth. … … … … … [3]

8 marks

Mark scheme: 12 (a) (i) e.g. satellite communication, mobile phones, line of sight communication, wifi B1 [1] (ii) e.g. connection of TV to aerial, loudspeaker, microphone (if clearly identified) B1 [1] (iii) e.g. a.f. amplifier to loudspeaker, landline for phone B1 [1] (b) (i) attenuation / dB = 10 lg (P2 / P1) C1 –190 = 10 lg (P2 / 3.1) P2 = 3.1 × 10–19 kW A1 [2] (ii) signal is amplified M1 frequency is changed M1 to prevent swamping of up-link signal by down-link (signal) A1 [3]

This question in 9702/41 Oct/Nov 2014

12 (a) Information may be carried by different channels of communication. State one application, in each case, where information is carried using (i) microwaves, … … [1] (ii) coaxial cables, … … [1] (iii) wire pairs. … … [1] (b) A station on Earth transmits a signal of initial power 3.1 kW to a geostationary satellite. The attenuation of the signal received by the satellite is 190 dB. (i) Calculate the power of the signal received by the satellite. power = … kW [2] (ii) By reference to your answer in (i), state and explain the changes made to the signal before transmission back to Earth. … … … … … [3]

8 marks

Mark scheme: 12 (a) (i) e.g. satellite communication, mobile phones, line of sight communication, wifi B1 [1] (ii) e.g. connection of TV to aerial, loudspeaker, microphone (if clearly identified) B1 [1] (iii) e.g. a.f. amplifier to loudspeaker, landline for phone B1 [1] (b) (i) attenuation / dB = 10 lg (P2 / P1) C1 –190 = 10 lg (P2 / 3.1) P2 = 3.1 × 10–19 kW A1 [2] (ii) signal is amplified M1 frequency is changed M1 to prevent swamping of up-link signal by down-link (signal) A1 [3]

This question in 9702/42 Oct/Nov 2014

13 In a mobile phone system, the country is divided into a number of cells, each with its own base station. State and explain (a) why the country is divided into cells, … … … [2] (b) two reasons why the base stations operate on UHF frequencies. 1. … … … 2. … … … [4]

6 marks

Mark scheme: 13 (a) same carrier frequencies can be re-used M1 but not in neighbouring cells / possible to use more handsets A1 [2] (b) e.g. wavelength is short (M1) so aerial on mobile phone conveniently short (A1) e.g. limited range (M1) so low power / less interference between cells (A1) e.g. large number of channels / greater bandwidth (M1) so more simultaneous callers (A1) [4]

This question in 9702/43 Oct/Nov 2014

12 (a) Information may be carried by means of various channels of communication. Name examples, one in each case, of devices where information is carried to the device using (i) a wire pair, … [1] (ii) a coaxial cable, … [1] (iii) microwaves. … [1] (b) State two advantages of optic fibres as compared with coaxial cables for long-range communication. 1. … 2. … [2] (c) An optic fibre has length 62 km and an attenuation per unit length of 0.21 dB km−1. The input power to the fibre is P. At the receiver, the noise power is 9.2 μW. The signal-to-noise ratio at the receiver is 25 dB. (i) Calculate the ratio, in dB, of the input power P to the noise power at the receiver. ratio = … dB [2] (ii) Use your answer in (i) to determine the input power P. P = … W [2]

9 marks

Mark scheme: 12 (a) (i) loudspeaker/doorbell/telephone etc. B1 [1] (ii) television set/audio amplifier etc. B1 [1] (iii) satellite/satellite dish/mobile phone etc. B1 [1] (b) e.g. lower attenuation/fewer repeaters more secure less prone to noise/interference physically smaller/less weight lower cost greater bandwidth (any two sensible suggestions, 1 each) B2 [2] (c) (i) ratio = 25 + (62 × 0.21) C1 = 38 dB A1 [2] (ii) ratio / dB = 10 lg (P2 / P1) C1 38 = 10 lg (P / {9.2 × 10–6}) P = 58 mW or 5.8 × 10–2 W A1 [2] (allow 1/2 for missing 10 in equation)

This question in 9702/41 May/June 2015

12 A mobile phone handset is, at its simplest, a radio transmitter and receiver. Outline the role of base stations and the cellular exchange when a mobile phone is switched on and before a call is made. … … … … … … … [4]

4 marks

Mark scheme: 12 handset transmits (identification) signal to number of base stations B1 base stations transfers (signal) to cellular exchange B1 (idea of stations needed at least once in first two marking points) computer at cellular exchange selects base station with strongest signal B1 computer at cellular exchange selects a carrier frequency for mobile phone B1 [4] (idea of computer needed at least once in these two marking points)

This question in 9702/42 May/June 2015

12 (a) Information may be carried by means of various channels of communication. Name examples, one in each case, of devices where information is carried to the device using (i) a wire pair, … [1] (ii) a coaxial cable, … [1] (iii) microwaves. … [1] (b) State two advantages of optic fibres as compared with coaxial cables for long-range communication. 1. … 2. … [2] (c) An optic fibre has length 62 km and an attenuation per unit length of 0.21 dB km−1. The input power to the fibre is P. At the receiver, the noise power is 9.2 μW. The signal-to-noise ratio at the receiver is 25 dB. (i) Calculate the ratio, in dB, of the input power P to the noise power at the receiver. ratio = … dB [2] (ii) Use your answer in (i) to determine the input power P. P = … W [2]

9 marks

Mark scheme: 12 (a) (i) loudspeaker/doorbell/telephone etc. B1 [1] (ii) television set/audio amplifier etc. B1 [1] (iii) satellite/satellite dish/mobile phone etc. B1 [1] (b) e.g. lower attenuation/fewer repeaters more secure less prone to noise/interference physically smaller/less weight lower cost greater bandwidth (any two sensible suggestions, 1 each) B2 [2] (c) (i) ratio = 25 + (62 × 0.21) C1 = 38 dB A1 [2] (ii) ratio / dB = 10 lg (P2 / P1) C1 38 = 10 lg (P / {9.2 × 10–6}) P = 58 mW or 5.8 × 10–2 W A1 [2] (allow 1/2 for missing 10 in equation)

This question in 9702/43 May/June 2015

13 Polar orbiting satellites have orbits over the poles of the Earth. Geostationary satellites are in equatorial orbits. Both are used as part of communication channels. (a) State one advantage and one disadvantage of the use of a polar orbiting satellite as compared with a geostationary satellite. advantage: … … disadvantage: … … [2] (b) A geostationary satellite is known to operate on the 6/4 GHz band. Explain (i) what is meant by the 6/4 GHz band, … … … [2] (ii) why two different frequencies are necessary. … … … … [2]

6 marks

Mark scheme: 13 (a) advantage: e.g. shorter time delay greater coverage over a long time B1 disadvantage: e.g. satellite needs to be tracked more satellites for (continuous) coverage/communication (any sensible suggestions) B1 [2] (b) (i) frequencies linking Earth with satellite B1 6 GHz is uplink frequency } 4 GHz is downlink frequency } (allow vice versa) B1 [2] (ii) either signal from Earth to satellite is attenuated greatly or downlink must be amplified greatly before transmission B1 downlink would swamp uplink unless frequencies are different B1 [2]

This question in 9702/43 Oct/Nov 2015

5 (a) In radio communication, the radio wave is usually modulated. State what is meant by amplitude modulation (AM ). … … … [2] (b) A sinusoidal radio carrier wave has a frequency of 900 kHz and an unmodulated amplitude measured to be 4.0 V. The carrier wave is amplitude modulated by a signal of frequency 5.0 kHz. For the amplitude modulated wave, (i) determine the wavelength, wavelength = … m [1] (ii) describe the amplitude variation, … … … [2] (iii) state the bandwidth. bandwidth = … Hz [1] (c) Communication is sometimes made using satellites in geostationary orbits that have a period of rotation about the Earth of 24 hours. (i) State two other features, apart from the period, of a geostationary orbit. 1. … … 2. … … [2] (ii) Suggest why 1. frequencies of the order of gigahertz are used for satellite communication, … … [1] 2. the uplink frequency to the satellite is different from the downlink frequency. … … … [2] [Total: 11]

11 marks

Mark scheme: 5(a) amplitude of carrier (wave) varies B1 variation in synchrony with displacement of information signal B1 5(b)(i) wavelength = (3.0 × 108) / (900 × 103) = 3.3 × 102 m A1 5(b)(ii) amplitude varies (continuously) between a maximum and a minimum B1 variations repeat 5000 times each second or variations repeat every 0.2 ms or variations above and below 4.0 V B1 5(b)(iii) 10000 Hz A1 5(c)(i) Any two from: • (orbit is) above the Equator • (orbit is) from west to east/same direction as Earth’s rotation • orbit is circular/orbit has a particular radius B2 5(c)(ii) 1. minimal reflection/absorption/attenuation by atmosphere or maximum penetration of/transmission through atmosphere B1 2. uplink signal is greatly attenuated/must be greatly amplified B1 prevents downlink signal swamping the uplink signal B1

This question in 9702/42 Oct/Nov 2018

4 (a) During the transmission of a signal, attenuation occurs and noise is picked up. State what is meant by: (i) attenuation … … [1] (ii) noise. … … … [2] (b) By reference to (a)(ii), explain the advantage of the transmission of the signal in digital form rather than in analogue form. … … [1] (c) Part of an analogue signal is shown in Fig. 4.1. 10 signal voltage / mV 8 6 4 2 0 0 1 2 3 4 5 6 7 time t / ms Fig. 4.1 The signal is to be transmitted in digital form. The analogue signal is sampled at a frequency of 1.0 × 103 Hz using an analogue-to-digital converter (ADC). The ADC produces 4-bit numbers. The times t at which the analogue signal is sampled are shown in Fig. 4.2. time t / ms 0 1.0 2.0 3.0 4.0 5.0 6.0 digital number 0010 0110 0100 0101 ……… ……… ……… Fig. 4.2 On Fig. 4.2: (i) for the digital number at time t = 3.0 ms, underline the least significant bit (LSB) [1] (ii) state the digital numbers corresponding to the sampling times between time t = 4.0 ms and time t = 6.0 ms. [2] (d) The transmitted digital signal is converted back to an analogue signal using a digital-to- analogue converter (DAC). On Fig. 4.3, show the variation with time t of the output levels of the DAC for time t = 0 to time t = 4.0 ms. Assume that there is negligible time delay in the transmission line. 8 output level 6 4 2 0 0 1 2 3 4 time t / ms Fig. 4.3 [3] [Total: 10]

10 marks

Mark scheme: 4(a)(i) loss of (signal) power/amplitude/intensity B1 4(a)(ii) unwanted/random signal B1 superposed on (transmitted) signal B1 4(b) noise can be eliminated (from digital signals) or signal can be regenerated (from digital signals) B1 4(c)(i) 0101 A1 4(c)(ii) 1000 at t = 4.0 ms B1 0110 at t = 5.0 ms and 0100 at t = 6.0 ms B1 4(d) series of equally-spaced steps of width 1 ms B1 each step in correct time interval (0–1 ms, 1–2 ms, 2–3 ms, 3–4 ms) B1 correct step heights (2, 6, 4 and 5) B1

This question in 9702/41 May/June 2019

5 (a) For a signal transmitted along an optic fibre, state what is meant by: (i) attenuation … … [1] (ii) noise. … … … [2] (b) The initial section of the transmission line for a signal from a telephone exchange is illustrated in Fig. 5.1. 52 km exchange amplifier gain 115 dB Fig. 5.1 At the exchange, the input signal to the transmission line has a power of 2.5 × 10–3 W. After the signal has travelled a distance of 52 km along the transmission line, the power of the signal is 7.8 × 10–16 W. The signal is then amplified. (i) Calculate the attenuation per unit length, in dB km–1, in the transmission line. attenuation per unit length = … dB km–1 [3] (ii) The gain of the amplifier is 115 dB. Calculate the power of the signal at the output of the amplifier. power = … W [2] [Total: 8]

8 marks

Mark scheme: 5(a)(i) loss of (signal) power/amplitude/intensity B1 5(a)(ii) unwanted/random signal B1 superposed on (transmitted) signal B1 5(b)(i) attenuation = 10 lg(P2 / P1) C1 attenuation per unit length = (1 / L) × 10 lg(P2 / P1) = (1 / 52) × 10 lg [(2.5 × 10–3) / (7.8 × 10–16)] C1 = 2.4 dB km–1 A1 5(b)(ii) gain / dB = 10 lg(P2 / P1) 115 = 10 lg [P / (7.8 × 10–16)] C1 P = 2.5 × 10–4 W A1

This question in 9702/42 May/June 2019

4 (a) During the transmission of a signal, attenuation occurs and noise is picked up. State what is meant by: (i) attenuation … … [1] (ii) noise. … … … [2] (b) By reference to (a)(ii), explain the advantage of the transmission of the signal in digital form rather than in analogue form. … … [1] (c) Part of an analogue signal is shown in Fig. 4.1. 10 signal voltage / mV 8 6 4 2 0 0 1 2 3 4 5 6 7 time t / ms Fig. 4.1 The signal is to be transmitted in digital form. The analogue signal is sampled at a frequency of 1.0 × 103 Hz using an analogue-to-digital converter (ADC). The ADC produces 4-bit numbers. The times t at which the analogue signal is sampled are shown in Fig. 4.2. time t / ms 0 1.0 2.0 3.0 4.0 5.0 6.0 digital number 0010 0110 0100 0101 ……… ……… ……… Fig. 4.2 On Fig. 4.2: (i) for the digital number at time t = 3.0 ms, underline the least significant bit (LSB) [1] (ii) state the digital numbers corresponding to the sampling times between time t = 4.0 ms and time t = 6.0 ms. [2] (d) The transmitted digital signal is converted back to an analogue signal using a digital-to- analogue converter (DAC). On Fig. 4.3, show the variation with time t of the output levels of the DAC for time t = 0 to time t = 4.0 ms. Assume that there is negligible time delay in the transmission line. 8 output level 6 4 2 0 0 1 2 3 4 time t / ms Fig. 4.3 [3] [Total: 10]

10 marks

Mark scheme: 4(a)(i) loss of (signal) power/amplitude/intensity B1 4(a)(ii) unwanted/random signal B1 superposed on (transmitted) signal B1 4(b) noise can be eliminated (from digital signals) or signal can be regenerated (from digital signals) B1 4(c)(i) 0101 A1 4(c)(ii) 1000 at t = 4.0 ms B1 0110 at t = 5.0 ms and 0100 at t = 6.0 ms B1 4(d) series of equally-spaced steps of width 1 ms B1 each step in correct time interval (0–1 ms, 1–2 ms, 2–3 ms, 3–4 ms) B1 correct step heights (2, 6, 4 and 5) B1

This question in 9702/43 May/June 2019

5 (a) State two advantages of the transmission of data in digital form, rather than analogue form. 1. … … 2. … … [2] (b) Optic fibres are used for the transmission of data. (i) A signal in an optic fibre is carried by an electromagnetic wave of frequency 1.36 × 1014 Hz. The speed of the wave in the fibre is 2.07 × 108 m s−1. For this electromagnetic wave, determine the ratio: wavelength in free space . wavelength in fibre ratio = … [2] (ii) The attenuation per unit length of the signal in the fibre is 0.40 dB km−1. The input power is 1.5 mW and the output power is 0.060 mW. Calculate the length of the fibre. length = … km [3] [Total: 7]

7 marks

Mark scheme: 5(a) Any 2 from: • noise can be filtered out / noise can be removed / signal can be regenerated • can carry more information per unit time / greater rate of transmission of data • can have extra bits of data to check for errors • can be encrypted B2 5(b)(i) v ∝ λ C1 ratio = vair / vfibre = 3.00 × 108 / 2.07 × 108 = 1.45 A1 5(b)(ii) attenuation = 10 log (P2/P1) C1 0.40 × L = 10 log (1.5 / 0.06) 0.40 × L = 13.979 C1 L = 35 km A1

This question in 9702/42 Feb/March 2020

6 (a) The transmission of signals using optic fibres has, to a great extent, replaced the use of coaxial cables. Advantages of optic fibres include greater bandwidth and very little crosslinking. (i) Suggest an advantage of greater bandwidth. … … [1] (ii) State what is meant by crosslinking. … … … [2] (b) In telecommunications, a signal power of 1.0 mW is used as a reference power. Signal powers relative to this reference power and expressed in dB are said to be measured in ‘dBm’. Show that a signal power of 13 dBm is equivalent to 20 mW. [2] (c) A signal of input power 20 mW is transmitted along an optic fibre for an uninterrupted distance of 45 km. The optic fibre has an attenuation per unit length of 0.18 dB km–1. Calculate the output power P from the optic fibre. P = … mW [2] [Total: 7]

7 marks

Mark scheme: 6(a)(i) greater information carrying capacity B1 6(a)(ii) power/energy is radiated B1 signal picked up by adjacent fibre/wire B1 6(b) ratio / dB = 10 lg(P2 / P1) C1 13 = 10 lg [P / (1.0 × 10–3)] and so P = 20 mW A1 6(c) 45 × 0.18 = 10 lg (20 / P) C1 P = 3.1 mW A1

This question in 9702/41 May/June 2020

6 (a) Telephone signals may be transmitted either by means of an optic fibre or by means of a wire pair. State three advantages of the use of an optic fibre rather than a wire pair. 1. … … 2. … … 3. … … [3] (b) It is proposed to transmit a signal over a distance of 4.5 × 103 km by means of an optic fibre. The input signal has a power of 9.8 mW. The minimum signal that can be detected at the output has a power of 6.3 × 10–17 W. For this signal power, the signal‑to‑noise ratio is 21 dB. Calculate: (i) the power of the background noise power = … W [2] (ii) the maximum attenuation per unit length of the optic fibre that allows for uninterrupted transmission of the signal. attenuation per unit length = … dB km–1 [2] [Total: 7]

7 marks

Mark scheme: 6(a) • greater bandwidth • less noise • less attenuation or fewer repeaters • less crosslinking or greater security Any three points, 1 mark each 6(b)(i) ratio / dB = 10 lg(P1 / P2) C1 21 = 10 lg [(6.3 × 10–17) / P] P = 5.0 × 10–19 W A1 6(b)(ii) attenuation per unit length = (1 / 4.5 × 103) × 10 lg [(9.8 × 10–3) / (6.3 × 10–17)] C1 = 0.032 dB km–1 A1

This question in 9702/42 May/June 2020

6 (a) The transmission of signals using optic fibres has, to a great extent, replaced the use of coaxial cables. Advantages of optic fibres include greater bandwidth and very little crosslinking. (i) Suggest an advantage of greater bandwidth. … … [1] (ii) State what is meant by crosslinking. … … … [2] (b) In telecommunications, a signal power of 1.0 mW is used as a reference power. Signal powers relative to this reference power and expressed in dB are said to be measured in ‘dBm’. Show that a signal power of 13 dBm is equivalent to 20 mW. [2] (c) A signal of input power 20 mW is transmitted along an optic fibre for an uninterrupted distance of 45 km. The optic fibre has an attenuation per unit length of 0.18 dB km–1. Calculate the output power P from the optic fibre. P = … mW [2] [Total: 7]

7 marks

Mark scheme: 6(a)(i) greater information carrying capacity B1 6(a)(ii) power/energy is radiated B1 signal picked up by adjacent fibre/wire B1 6(b) ratio / dB = 10 lg(P2 / P1) C1 13 = 10 lg [P / (1.0 × 10–3)] and so P = 20 mW A1 6(c) 45 × 0.18 = 10 lg (20 / P) C1 P = 3.1 mW A1

This question in 9702/43 May/June 2020

5 (a) (i) State what is meant by the amplitude modulation (AM) of a radio wave. … … … [2] (ii) State two advantages of AM transmissions when compared with frequency modulation (FM) transmissions. 1. … … 2. … … [2] (b) The variation with frequency f of the amplitude A of a transmitted radio wave after amplitude modulation by an audio signal is shown in Fig. 5.1. A 0 1490 1500 1510 f / kHz Fig. 5.1 For this transmission, determine: (i) the wavelength of the carrier wave wavelength = … m [1] (ii) the maximum frequency of the transmitted audio signal. frequency = … kHz [1] (c) Another audio signal with the same maximum frequency is transmitted using a different carrier wave frequency. The lowest frequency of this modulated wave is equal to the highest frequency of the modulated wave in (b). Determine the frequency of this carrier wave. frequency = … kHz [1] [Total: 7]

7 marks

Mark scheme: 5(a)(i) amplitude of the carrier wave varies M1 in synchrony with the displacement of the (information) signal A1 5(a)(ii) Any 2 from: • fewer transmitters needed / each transmitter can cover a greater distance • more stations can share waveband • transmitters and receivers are cheaper B2 Question Answer Marks 5(b)(i) v f λ = 8 6 3.0 10 200 1.5 10 m × = = × A1 5(b)(ii) 10 kHz B1 5(c) 1520 kHz B1

This question in 9702/42 Feb/March 2021

5 (a) State what is meant by the amplitude modulation (AM) of a radio wave. … … … [2] (b) A radio wave is modulated by an audio signal. The variation with frequency f of the amplitude of the modulated wave is shown in Fig. 5.1. amplitude 0 292 300 308 f / kHz Fig. 5.1 Determine: (i) the wavelength of the carrier wave wavelength = … m [1] (ii) the bandwidth of the modulated wave bandwidth = … kHz [1] (iii) the maximum frequency of the audio signal. maximum frequency = … kHz [1] (c) The power of a radio signal at a transmitter is PT. At a receiver, the received power PR is given by the expression 0.082 PT PR = x2 where x is the distance, in metres, between the transmitter and the receiver. For the transmission of this signal, the attenuation is 73 dB. Determine the distance x. x = … m [3] [Total: 8]

8 marks

Mark scheme: 5(a) amplitude of the carrier wave varies M1 in synchrony with the displacement of the (information) signal A1 5(b)(i) wavelength = (3.0 × 108) / (300 × 103) = 1000 m A1 5(b)(ii) bandwidth = 16 kHz A1 5(b)(iii) frequency = 8 kHz A1 5(c) attenuation = 10 lg (P1 / P2) C1 73 = 10 lg (PT / PR) 73 = 10 lg (PT x2 / 0.082 PT) or x2 / 0.082 = 107.3 C1 x = 1300 m A1

This question in 9702/41 May/June 2021

4 (a) A sinusoidal carrier wave has a constant amplitude and a frequency of 1.2 MHz. The carrier wave is modulated by a signal wave such that a 1.0 V displacement of the signal wave causes a change in frequency of 25 kHz. The signal wave has frequency 8.0 kHz and amplitude 2.0 V. (i) State the name of this type of modulation of the carrier wave. … [1] (ii) For this modulated carrier wave, determine the variation, if any, in: 1. its amplitude … … 2. its frequency. … … … [3] (b) An audio signal is transmitted by means of a modulated radio wave. The variation with frequency of the amplitude of the radio wave is shown in Fig. 4.1. amplitude 0 225 240 255 frequency / kHz Fig. 4.1 For this transmission, determine: (i) the wavelength, in km, of the carrier wave wavelength = … km [2] (ii) the bandwidth bandwidth = … kHz [1] (iii) the frequency of the audio signal. frequency = … kHz [1] [Total: 8]

8 marks

Mark scheme: 4(a)(i) frequency (modulation) B1 4(a)(ii) 1. zero B1 2. frequency (of 1.2 MHz) varies by ±50 kHz B1 frequency varies (by ±50 kHz) at a rate of 8000 times per second B1 4(b)(i) wavelength = (3.00 × 108) / (240 × 103) C1 (= 1250 m) = 1.25 km A1 4(b)(ii) bandwidth = 30 kHz A1 4(b)(iii) frequency = 15 kHz A1

This question in 9702/42 May/June 2021

5 (a) State what is meant by the amplitude modulation (AM) of a radio wave. … … … [2] (b) A radio wave is modulated by an audio signal. The variation with frequency f of the amplitude of the modulated wave is shown in Fig. 5.1. amplitude 0 292 300 308 f / kHz Fig. 5.1 Determine: (i) the wavelength of the carrier wave wavelength = … m [1] (ii) the bandwidth of the modulated wave bandwidth = … kHz [1] (iii) the maximum frequency of the audio signal. maximum frequency = … kHz [1] (c) The power of a radio signal at a transmitter is PT. At a receiver, the received power PR is given by the expression 0.082 PT PR = x2 where x is the distance, in metres, between the transmitter and the receiver. For the transmission of this signal, the attenuation is 73 dB. Determine the distance x. x = … m [3] [Total: 8]

8 marks

Mark scheme: 5(a) amplitude of the carrier wave varies M1 in synchrony with the displacement of the (information) signal A1 5(b)(i) wavelength = (3.0 × 108) / (300 × 103) = 1000 m A1 5(b)(ii) bandwidth = 16 kHz A1 5(b)(iii) frequency = 8 kHz A1 5(c) attenuation = 10 lg (P1 / P2) C1 73 = 10 lg (PT / PR) 73 = 10 lg (PT x2 / 0.082 PT) or x2 / 0.082 = 107.3 C1 x = 1300 m A1

This question in 9702/43 May/June 2021

5 (a) (i) When audio signals are transmitted over long distances, modulation of radio waves is used. Suggest a reason why modulation is used. … … [1] (ii) State a technical advantage and a technical disadvantage of using frequency modulation rather than amplitude modulation. advantage: … … disadvantage: … … [2] (b) An audio signal of amplitude 2.0 μV and frequency 4.2 kHz is to be transmitted using a carrier wave of amplitude 10.0 mV and frequency 100 kHz. Either amplitude modulation or frequency modulation may be used. The amplitude modulation is at a rate of 1 mV μV–1. The frequency modulation is at a rate of 5 kHz μV–1. Complete Table 5.1 to show the maximum and minimum values of the amplitude and of the frequency of the modulated wave for each type of modulation. Table 5.1 amplitude / mV frequency / kHz minimum maximum minimum maximum amplitude modulation frequency modulation [4] (c) For the amplitude modulated wave in (b), determine the bandwidth. bandwidth = … kHz [1] [Total: 8]

8 marks

Mark scheme: 5(a)(i) unmodulated (radio) waves would interfere with each other or not modulating would require aerials too long (to be practical) B1 5(a)(ii) advantage: • can transmit higher frequencies • higher quality reproduction • less prone to interference • same frequency can be used in different areas (any one point) B1 disadvantage: • takes up greater bandwidth • shorter range of transmission • requires a greater number of transmitting aerials (any one point) B1 5(b) AM amplitude: min. 8 mV and max. 12 mV B1 AM frequency: min. 100 kHz and max. 100 kHz B1 FM amplitude: min. 10 mV and max. 10 mV B1 FM frequency: min. 90 kHz and max. 110 kHz B1 5(c) 8.4 kHz A1

This question in 9702/42 Oct/Nov 2021