COMMUNICATION SYSTEMS 1. Briefly define amplitude modulation (AM). - Amplitude modulation is a process of impressing a low frequency information-bearing signal onto the amplitude of a high-frequency carrier signal. Or Another suitable answer (inside lecture notes): - AM is the process of changing the amplitude of relatively high frequency carrier signal in proportional with the instantaneous value of modulating signal. 2. State one disadvantage of using conventional AM. - It is an inefficient method of modulation because for 100% modulation, two-thirds of the total transmitted power is being wasted in the carrier which carries no information. Or - Large power consumption, where carrier power constitutes more than 2/3 of the transmitted power. Or - Large bandwidth utilized. 3. Given an audio signal x(t ) = 2 cos1000 t + cos2000 t and a carrier of 10 cos105 t . Write down the expression for the upper sideband terms of the product signal for AM. Signal: x(t ) = 2 cos1000 t + cos2000 t Carrier: c (t) = 10 cos105 t The product signal: v(t ) = x(t )c(t ) = ( 2 cos1000 t + cos2000 t )( 10 cos105 t ) = 20 cos103 t cos105 t + 10 cos2000 t cos105 t = 10 (cos10.1 10 4 t + cos9.9 10 4 t ) + 5(cos10.2 10 4 t + cos9.8 10 4 t ) The upper sideband term: v( t ) = 10 cos10.1 10 4 t + 5 cos10.2 10 4 t 4. An AM modulated wave with the output wave changes of 7.5V p is represent by the following equation. Vam (t ) = 20 sin( 2500 103 )t − 3.75 cos(2510 103 )t + 3.75 cos(2490 103 )t Based on the information given, a) Calculate the modulation coefficient, m and percent modulation. 7.5 m= = 0.375 20 Percent modulation = 100 X 0.375 = 37.5% b) Calculate the peak amplitude of the modulated carrier, upper and lower side frequency voltages. Ec (mod ulated ) = Ec (un mod ulated ) = 20V p Eusf = Elsf = mEc (0.375)(20) = = 3.75V p 2 2 c) Calculate the maximum and minimum amplitude of the envelope. Vmax = Ec + Em = 20 + 7.5 = 27.5V p Vmin = Ec − Em = 20 − 7.5 = 12.5V p d) Draw and label the frequency spectrum. e) Sketch and label the output envelope. 5. A commercial AM station is broadcasting with an average transmitted power of 10 kW. The modulation index is set at 0.707 for a sinusoidal message signal. Find the following: a) The average power in the total sideband of the transmitted signal. Modulation index, m = 0.707 Average transmitted power, Pc = 10kW Psb = m2 0.707 2 Pc = (10 kW ) = 2.5kW 2 2 b) The transmission power efficiency. Total power transmits: Pt = Pc + Psb = 12.5kW P 2.5 The transmission power efficiency: = sb 100% = 100% = 20% Pt 12.5 6. What is the maximum modulating signal frequency that can be used for a conventional AM system with a 20 kHz bandwidth? Given BW = 20kHz BW = 2fm (max) Fm (max) = BW/2 = 10kHz 7. Draw a block diagram of a basic filter type system SSB transmitter. Briefly describe its operation. Antenna DSB signal Carrier oscillator Balanced modulator SSB signal Sideband filter Linear amplifier Microphone Filter response curve Audio amplifier Lower sidebands Upper sidebands Operations: - The carrier and information signal are applied to the balanced modulator, to eliminate the carrier. The DSB output of the balance modulator is then applied to the sideband filter. The filter is designed to pass the desired sideband and to block the unwanted sideband. Therefore, the SSB output is either the lower or upper sideband depending on the filter passband. The SSB output is feed through linear amplifier before being transmitted via antenna. 8. An antenna transmits a 10kW power at 95% modulation using conventional AM. Determine the amount of power saving if Single Side Band (SSB) transmission is used for the same intelligibility. Given Pt = 10kW, m = 0.95 Conventional AM: Pt = Pc + PUSB + PLSB m2 m2 m2 10 103 = Pc + Pc + Pc = Pc 1 + = 1.45125 Pc 4 4 2 Pc = 6890.61W SSBSC AM: m2 m2 0.95 2 PSSBSC = PSB = Pc = (6890.61) = 1554.69W 4 4 4 Power Saving = Pt − PSSBSC = 10 x10 3 − 1554 .69 = 8445 .31W % Power Saving = 8445.31 10x10 3 x100% = 84.45% 9. An 800 kHz sinusoidal carrier signal is amplitude-modulated by a 5 kHz audio signal to produce an AM envelope with maximum and minimum peaks voltages of 120Vp and 30Vp respectively. The transmitting antenna has an equivalent resistance of 75Ω. Based on the given information, a) Determine peak amplitude of the audio, carrier and sidebands voltages. V + Vmin Vc = max = 75Vp 2 Vm = Vmax − Vmin = 45Vp 2 Vusb / lsb = Vm = 22.5Vp 2 b) Calculate the bandwidth of the modulated signal. BW = 2fm (max) = 2 x 5 kHz = 10 kHz c) Calculate the power of the carrier signal and total sidebands power. 2 V Pc = c = 37.5W 2R Ptotalsb = m 2 Pc (0.62 )(37.5) = = 6.75W 2 2 d) Sketch and label the power spectrum of the modulated signal. 10. Give three significant importance of modulation in electronic communications. - To match the characteristic of channel - The low bandpass signal translated to higher frequency so that it will match with the passband characteristics of the channel. - Signal propagates with higher efficiency at higher frequency - Difficult or impractical to radiate low-frequency info-bearing signals through earth’s atmosphere in form of EM energy. E.g: design of antenna size will be enormous if radiated directly without modulation. - To accommodate for simultaneous transmission of signals from several source by means of multiplexing to avoid interference. 11. For an AM Double Side Band Full Carrier (DSBFC) wave with an unmodulated carrier of 25Vp and a load resistance of 50Ω, determine the unmodulated carrier power, modulated carrier power, upper and lower sideband powers and total transmitted power for a modulation index, m = 0.6. Given Vc = 25Vp and R = 50Ω 2 Vc 252 Pc(unmod) = Pc (mod) = = = 6.25W 2R 2 50 m2 0.62 Pc = (6.25) = 0.5625 W PUSB = PLSB = 4 4 Pt = Pc + PUSB + PLSB = 6.25 + 2(0.5625) = 7.375W 12. Differentiate between coherent and non-coherent type of AM receiver. - Coherent – the frequencies generated in the receiver and used for demodulation are synchronized to oscillator frequencies generated in the transmitter. - Non-coherent – receivers, either no frequencies are generated in the receiver or frequency used for demodulation are completely independent from the transmitter’s carrier frequency. 13. An AM superheterodyne receiver having no RF amplifier with a preselector Q of 100. Given the intermediate frequency (IF) is 455 kHz and the RF carrier is 1000 kHz respectively. Answer the following questions: (i) Determine the image frequency. fIM = fRF +2 fIF = 1000kHz +2(455kHz) = 1910kHz (ii) Calculate the Image Frequency Rejection Ratio, IFRR. ρ= (fIM/fRF)- (fRF/ fIM)=(1910 kHz/1000kHz)- (1000kHz/(1910kHz)=1.3864 IFRR = (1+ (Q)2(ρ) 2) ½ = (1+ (100)2(1.3864) 2) ½= 138.64 (iii) Find the preselector Q required to achieve the same IFRR as calculated in (ii) for an RF carrier of 600 kHz. fIM = fRF +2 fIF = 600kHz +2(455kHz) = 1510kHz ρ= (fIM/fRF)- (fRF/ fIM)=(1910 kHz/600kHz)- (600kHz/(1910kHz)=2.869 IFRR = (1+ (Q)2(ρ) 2) ½ Q= (IFRR2 - 1)/(ρ) 2) ½= (138.642 - 1)/(2.869) 2) ½=48.32 (iv) Conclude the importance of Q in AM reception circuitry based on the value of Q calculated in (iii). The higher Q the better the image frequency prevented from entering the receiver thus preventing an image frequency from interfering with the desired radio frequency. Image frequency is rejection is the primary purpose of the RF preselector.
