Solution of ECE 342 Test 3 S12

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1 Solution of ECE 34 Test 3 S1 1 A random power signal has a mean of three and a standard deviation of five Find its numerical total average signal power Signal Power P = = 34 A random energy signal has an autocorrelation R x τ signal energy Λ x ( ) = 1 x, x < 1 Signal Energy 0, otherwise P = R x 0 ( ) = Λ ( ) = Λ τ 1 + Λ = 3 / / 4 = Λ τ Find its numerical

2 3 A random signal with power spectral density G x ( f ) = 10Λ( f / 50) is the input signal to a filter with frequency response H f signal from the filter Π x ( ) = 5 Π f 0 10 ( ) = 1, x < 1 / Output Signal Power 0, x > 1 / P out = G x ( f ) H( f ) df = 5 10 Λ( f / 50)df = Π f Find the numerical signal power of the output = 3000

3 4 An AM signal with carrier amplitude A c = 10 is modulated by x( t) = cos( 1000πt) + 4sin( 4000πt) with a modulation index µ = 1 The modulated carrier is transmitted through a channel with a loss of 10 db It is received by an AM receiver using synchronous detection and DC blocking with noise temperature T = 4000K The predection bandpass filter in the AM receiver is practically ideal and just wide enough to pass the AM signal The last lowpass filter (after the synchronous detection) is just wide enough to pass the demodulated signal (a) Find the numerical signal power of the transmitted signal { ( ) + 4sin( 4000πt) cos ( ω ct) } = cos 1000πt = 100( ) 1 / = = 550 (b) Find the numerical signal power of the received signal S R S R S R = /10 1 = (c) Find the numerical destination signal power S D S D This answer depends on the assumptions about the synchronous detector Do we multiply by cos ω c t ( ) and take the low-frequency part, yielding a factor of 1/ in amplitude and 1/4 in power as ( ), yielding a factor of 1 in amplitude and in power in my slides? Or do we multiply by cos ω c t as in the book? The signal-to noise ratio is the same either way and I accepted either answer if the signal power and noise power were computed consistently with either assumption S D = 1 4 A c L S = 1 10 x = OR S D = A c L S = 10 x 10 = (d) Find the numerical destination noise power in the receiver's bandwidth N D N D σ n = N 0 B T = kt 4000 = = N D = σ n / 4 = / 4 = OR N D = σ n = (e) Find the numerical signal-to-noise ratio at the destination in db = / = or 6656 db OR = 10 9 / = or 6656 db (f) In db, how much better could the signal-to-noise ratio be made by using a bandpass filter after the detection process instead of a lowpass filter?

4 Improvement in db We could use a bandpass filter between 500 Hz and khz instead of a lowpass filter between 0 and khz That would reduce the noise by a factor of 075, realizing an improvement in signal-tonoise ratio of about 149 db 5 A signal with signal power P in = 100 mw is the input signal to a transmission cable with loss of α = 15 db/km The cable is 100 km in length Repeaters are used to avoid the signal's getting lost in noise The signal power during transmission cannot be allowed to fall below 1 mw at any point on the cable and each repeater has a power gain of 0 db (a) What is the minimum number of repeaters necessary? Minimum number of repeaters is The minimum number of repeaters occurs when the signal is allowed to fall to 1 mw at the input of all the repeaters The cable length at which the signal power falls from 100 mw to 1mW is km 100 km/133333km is 75 So we need a minimum of 7 repeaters (b) Using the minimum number of repeaters what is the output signal power? Output signal power is The seventh repeater is at the km point Its input power is 1 mw and its output power is 100 mw The distance to the cable end is 6667 km The signal loses 10 db in that distance So the output signal power at the end of the cable is 10 mw 6 What is the advantage of using an envelope detector compared with using a synchronous detector? It allows for a very simple and economical demodulation of the signal compared to synchronous techniques 7 What is the purpose of guard bands in frequency-division multiplexing? Guard bands are used to minimize crosstalk between channels when multiple signals are frequency multiplexed onto to one signal Crosstalk occurs because real filters do not have vertical sides and their stopbands do not have infinite rejection

5 8 Given that the full range of AM carrier frequencies permitted by law is from 530 khz to 1600 khz in 10 khz steps, and that the intermediate frequency used is 455 khz, identify all the legal AM carrier frequencies for which an image frequency exists in the AM range You may do this by simply listing all these frequencies or by specifying with <, >, and/or the ranges of frequencies The answer to this question depends on whether one assumes high-side injection or low-side injection of the local oscillator With high-side injection, all carrier frequencies for which the frequency minus 910 khz are in the AM range are image frequencies Those are 1440, 1450, 1460, 1470, 1480, 1490, 1500, 1510, 150, 1530, 1540, 1550, 1560, 1570, 1580, 1590, 1600 or 1440 f c 1600 With low-side injection, all carrier frequencies for which the frequency plus 910 khz are in the AM range are image frequencies Those are 530, 540, 550, 560, 570, 580, 590, 600, 610, 60, 630, 640, 650, 660, 670, 680, 690 or 530 f c If a phase locked loop VCO free runs at a frequency f 0 and the incoming frequency to be locked to is at f c and if f 0 = f c and if the loop is locked, describe the exact phase relationship between the VCO output signal and the incoming signal (What is the phase difference and which signal is leading or lagging?) The phase difference is exactly 90 with the VCO output leading the incoming signal If the incoming signal increases in size by a factor of two, what happens to this phase relationship? It is momentarily perturbed but then returns back exactly to the same phase relationship 10 What is the effect on a demodulated USSB signal of a local oscillator at a frequency that is too high? (Be specific) The lower edge of the USSB signal, instead of being shifted to zero, will be shifted to a frequency below zero So all the frequencies in the demodulated signal will be too low by the error in the local oscillator frequency and the two replicas of the USSB signal will overlap causing distortion

6 Solution of ECE 34 Test 3 S1 1 A random power signal has a mean of two and a standard deviation of eight Find its numerical total average signal power Signal Power P = + 8 = 68 A random energy signal has an autocorrelation R x τ signal energy Λ x ( ) = 1 x, x < 1 Signal Energy P = R x 0 0, otherwise ( ) = Λ ( ) = Λ τ Λ τ Find its numerical + Λ = 5 / / 6 = 5 / 3 or 16667

7 3 A random signal with power spectral density G x ( f ) = 4Λ( f / 50) is the input signal to a filter with frequency response H f signal from the filter Π x ( ) = 5 Π f 5 10 ( ) = 1, x < 1 / Output Signal Power 0, x > 1 / P out = G x ( f ) H( f ) df = 5 4 Λ( f / 50)df = Π f Find the numerical signal power of the output = 1000

8 4 An AM signal with carrier amplitude A c = 8 is modulated by x( t) = cos( 000πt) + 4sin( 4000πt) with a modulation index µ = 1 The modulated carrier is transmitted through a channel with a loss of 10 db It is received by an AM receiver using synchronous detection and DC blocking with noise temperature T = 6000K The predection bandpass filter in the AM receiver is practically ideal and just wide enough to pass the AM signal The last lowpass filter (after the synchronous detection) is just wide enough to pass the demodulated signal (a) Find the numerical signal power of the transmitted signal { ( ) + 4 sin( 4000πt) cos ( ω ct) } = cos 000πt = 64( ) 1 / = 3 11 = 35 (b) Find the numerical signal power of the received signal S R S R S R = /10 1 = (c) Find the numerical destination signal power S D S D This answer depends on the assumptions about the synchronous detector Do we multiply by cos ω c t ( ) and take the low-frequency part, yielding a factor of 1/ in amplitude and 1/4 in power as ( ), yielding a factor of 1 in amplitude and in power in my slides? Or do we multiply by cos ω c t as in the book? The signal-to noise ratio is the same either way and I accepted either answer if the signal power and noise power were computed consistently with either assumption S D = 1 4 A c 8 L S = 1 x = OR S D = A c L S = 8 x 10 = (d) Find the numerical destination noise power in the receiver's bandwidth N D N D σ n = N 0 B T = kt 4000 = = N D = σ n / 4 = / 4 = OR N D = σ n = (e) Find the numerical signal-to-noise ratio at the destination in db = / = or 6861 db OR = / = or 6861 db

9 (f) In db, how much better could the signal-to-noise ratio be made by using a bandpass filter after the detection process instead of a lowpass filter? Improvement in db We could use a bandpass filter between 1 khz and khz instead of a lowpass filter between 0 and khz That would reduce the noise by a factor of 05, realizing an improvement in signal-tonoise ratio of about 3 db 5 A signal with signal power P in = 00 mw is the input signal to a transmission cable with loss of α = 15 db/km The cable is 100 km in length Repeaters are used to avoid the signal's getting lost in noise The signal power during transmission cannot be allowed to fall below 1 mw at any point on the cable and each repeater has a power gain of 3 db (a) What is the minimum number of repeaters necessary? Minimum number of repeaters is The minimum number of repeaters occurs when the signal is allowed to fall to 1 mw at the input of all the repeaters The cable length at which the signal power falls from 00 mw to 1mW is km 100 km/153333km is 6517 So we need a minimum of 6 repeaters (b) Using the minimum number of repeaters what is the output signal power? Output signal power is The sixth repeater is at the 9 km point Its input power is 1 mw and its output power is 00 mw The distance to the cable end is 8 km The signal loses 1 db in that distance So the output signal power at the end of the cable is mw 6 What is the effect on a demodulated USSB signal of a local oscillator at a frequency that is too high? (Be specific) The lower edge of the USSB signal, instead of being shifted to zero, will be shifted to a frequency below zero So all the frequencies in the demodulated signal will be too low by the error in the local oscillator frequency and the two replicas of the USSB signal will overlap causing distortion 7 What is the advantage of using an envelope detector compared with using a synchronous detector? It allows for a very simple and economical demodulation of the signal compared to synchronous techniques

10 8 What is the purpose of guard bands in frequency-division multiplexing? Guard bands are used to minimize crosstalk between channels when multiple signals are frequency multiplexed onto to one signal Crosstalk occurs because real filters do not have vertical sides and their stopbands do not have infinite rejection 9 Given that the full range of AM carrier frequencies permitted by law is from 530 khz to 1600 khz in 10 khz steps, and that the intermediate frequency used is 455 khz, identify all the legal AM carrier frequencies for which an image frequency exists in the AM range You may do this by simply listing all these frequencies or by specifying with <, >, and/or the ranges of frequencies The answer to this question depends on whether one assumes high-side injection or low-side injection of the local oscillator With high-side injection, all carrier frequencies for which the frequency minus 910 khz are in the AM range are image frequencies Those are 1440, 1450, 1460, 1470, 1480, 1490, 1500, 1510, 150, 1530, 1540, 1550, 1560, 1570, 1580, 1590, 1600 or 1440 f c 1600 With low-side injection, all carrier frequencies for which the frequency plus 910 khz are in the AM range are image frequencies Those are 530, 540, 550, 560, 570, 580, 590, 600, 610, 60, 630, 640, 650, 660, 670, 680, 690 or 530 f c If a phase locked loop VCO free runs at a frequency f 0 and the incoming frequency to be locked to is at f c and if f 0 = f c and if the loop is locked, describe the exact phase relationship between the VCO output signal and the incoming signal (What is the phase difference and which signal is leading or lagging?) The phase difference is exactly 90 with the VCO output leading the incoming signal If the incoming signal increases in size by a factor of two, what happens to this phase relationship? It is momentarily perturbed but then returns back exactly to the same phase relationship

11 Solution of ECE 34 Test 3 S1 1 A random power signal has a mean of four and a standard deviation of twelve Find its numerical total average signal power Signal Power P = = 160 A random energy signal has an autocorrelation R x τ signal energy Λ x ( ) = 1 x, x < 1 Signal Energy P = R x 0 0, otherwise ( ) = Λ ( ) = Λ τ Λ τ Find its numerical + Λ = 7 / / 8 = 7 / 4 or 175

12 3 A random signal with power spectral density G x ( f ) = 6Λ( f / 50) is the input signal to a filter with frequency response H f signal from the filter Π x ( ) = 5 Π f 0 0 ( ) = 1, x < 1 / Output Signal Power 0, x > 1 / P out = G x ( f ) H( f ) df = 5 6 Λ( f / 50)df = Π f Find the numerical signal power of the output = 3600

13 4 An AM signal with carrier amplitude A c = 1 is modulated by x( t) = cos( 3000πt) + 4sin( 4000πt) with a modulation index µ = 1 The modulated carrier is transmitted through a channel with a loss of 10 db It is received by an AM receiver using synchronous detection and DC blocking with noise temperature T = 5000K The predection bandpass filter in the AM receiver is practically ideal and just wide enough to pass the AM signal The last lowpass filter (after the synchronous detection) is just wide enough to pass the demodulated signal (a) Find the numerical signal power of the transmitted signal { ( ) + 4sin( 4000πt) cos ( ω ct) } = cos 3000πt = 144( ) 1 / = 7 11 = 79 (b) Find the numerical signal power of the received signal S R S R S R = /10 1 = (c) Find the numerical destination signal power S D S D This answer depends on the assumptions about the synchronous detector Do we multiply by cos ω c t ( ) and take the low-frequency part, yielding a factor of 1/ in amplitude and 1/4 in power as ( ), yielding a factor of 1 in amplitude and in power in my slides? Or do we multiply by cos ω c t as in the book? The signal-to noise ratio is the same either way and I accepted either answer if the signal power and noise power were computed consistently with either assumption S D = 1 4 A c L S = 1 1 x 10 = OR S D = A c L S = 1 x 10 = (d) Find the numerical destination noise power in the receiver's bandwidth N D N D σ n = N 0 B T = kt 4000 = = N D = σ n / 4 = / 4 = OR N D = σ n = (e) Find the numerical signal-to-noise ratio at the destination in db = / = or db OR = / = or db

14 (f) In db, how much better could the signal-to-noise ratio be made by using a bandpass filter after the detection process instead of a lowpass filter? Improvement in db We could use a bandpass filter between 15 Hz and khz instead of a lowpass filter between 0 and khz That would reduce the noise by a factor of 05, realizing an improvement in signal-tonoise ratio of about 6 db 5 A signal with signal power P in = 400 mw is the input signal to a transmission cable with loss of α = 15 db/km The cable is 100 km in length Repeaters are used to avoid the signal's getting lost in noise The signal power during transmission cannot be allowed to fall below 1 mw at any point on the cable and each repeater has a power gain of 6 db (a) What is the minimum number of repeaters necessary? Minimum number of repeaters is The minimum number of repeaters occurs when the signal is allowed to fall to 1 mw at the input of all the repeaters The cable length at which the signal power falls from 400 mw to 1mW is km 100 km/173333km is 5769 So we need a minimum of 5 repeaters (b) Using the minimum number of repeaters what is the output signal power? Output signal power is The fifth repeater is at the km point Its input power is 1 mw and its output power is 400 mw The distance to the cable end is km The signal loses 0 db in that distance So the output signal power at the end of the cable is 4 mw 6 What is the purpose of guard bands in frequency-division multiplexing? Guard bands are used to minimize crosstalk between channels when multiple signals are frequency multiplexed onto to one signal Crosstalk occurs because real filters do not have vertical sides and their stopbands do not have infinite rejection 7 Given that the full range of AM carrier frequencies permitted by law is from 530 khz to 1600 khz in 10 khz steps, and that the intermediate frequency used is 455 khz, identify all the legal AM carrier frequencies for which an image frequency exists in the AM range You may do this by simply listing all these frequencies or by specifying with <, >, and/or the ranges of frequencies The answer to this question depends on whether one assumes high-side injection or low-side injection of the local oscillator With high-side injection, all carrier frequencies for which the frequency minus 910 khz are in the AM range are image frequencies Those are 1440, 1450, 1460, 1470, 1480, 1490, 1500, 1510, 150, 1530, 1540, 1550, 1560, 1570, 1580, 1590, 1600 or 1440 f c 1600 With low-side injection, all carrier frequencies for which the frequency plus 910 khz are in the AM range are image frequencies Those are 530, 540, 550, 560, 570, 580, 590, 600, 610, 60, 630, 640, 650, 660, 670, 680, 690 or 530 f c 690

15 8 If a phase locked loop VCO free runs at a frequency f 0 and the incoming frequency to be locked to is at f c and if f 0 = f c and if the loop is locked, describe the exact phase relationship between the VCO output signal and the incoming signal (What is the phase difference and which signal is leading or lagging?) The phase difference is exactly 90 with the VCO output leading the incoming signal If the incoming signal increases in size by a factor of two, what happens to this phase relationship? It is momentarily perturbed but then returns back exactly to the same phase relationship 9 What is the effect on a demodulated USSB signal of a local oscillator at a frequency that is too high? (Be specific) The lower edge of the USSB signal, instead of being shifted to zero, will be shifted to a frequency below zero So all the frequencies in the demodulated signal will be too low by the error in the local oscillator frequency and the two replicas of the USSB signal will overlap causing distortion 10 What is the advantage of using an envelope detector compared with using a synchronous detector? It allows for a very simple and economical demodulation of the signal compared to synchronous techniques

( ) D. An information signal x( t) = 5cos( 1000πt) LSSB modulates a carrier with amplitude A c

( ) D. An information signal x( t) = 5cos( 1000πt) LSSB modulates a carrier with amplitude A c An inormation signal x( t) 5cos( 1000πt) LSSB modulates a carrier with amplitude A c 1. This signal is transmitted through a channel with 30 db loss. It is demodulated using a synchronous demodulator.