ASU IFT 372 Fall 2023 Helm Weeks 1-7 Quizzes +Midterm Multisource with complete solution

ASU IFT 372 Fall 2023 Helm Weeks 1-7 Quizzes +Midterm Multisource with complete solution What is a logarithm? Correct answer: A logarithm is an exponent based mathematical representation of a numerical value. What logarithmic base is used in digital communications? Correct answer: Two How do you find a log2 value on a calculator that only has log10 functionality? (Hint: what is the formula?) Correct answer: log2(a) = [log10(a) / log10(2)] What is the definition of a decibel? Correct answer: a = 10lob(b), where a is the decibel value and b is the decimal value Correct answer: A decibel is a mathematical representation of a power level on a logarithmic scale. You cannot mix decimal and decibel values in an equation (i.e., you either have to work with decimal values or decibel values) Correct answer: True There can be more than one power value in a decibel (dB) equation. Correct answer: False When you add two decibel values, you are multiplying those values in decimal (I.e. 3 dB + 6 dB = 9 dB which is equivalent to 2 * 4 = 8 in decimal) Correct answer: True is a decibel value referenced to watts and is a decibel value referenced to milliwatts. Correct answer: dBW & dBm How do you convert dBW to dBm? Correct answer: Add 30 dB to dBW to obtain dBm value The net gain or (loss) of a transmission system is a radio between the and where the is the numerator and the is the denominator. Correct answer: Output power & Input power What parameter of the intelligent signal causes or determines the instantaneous rate of carrier frequency? Correct answer: Amplitude of the modulating signal What parameter of the intelligent signal causes frequency deviation of the carrier? Correct answer: Amplitude of the modulating signalDetermine the noise power (in dBm) at 27° C in a 500 kHz bandwidth.Correct answer: PN = 10 log (1.38*10^-23*(27+273)*500kHz) = -146.8 dBw = -116.8 dBm Determine the noise power (in dBm) at 27° C in a 20 MHz bandwidth system. Correct answer: Pn=kTB; k=1.38*(10^-23)W°K-Hz; T(°K)=27°C+273=300°K; B=20*10^6Hz =10log(1.38*(10^-23)*300*20*10^6) =10log(8.28*(10^-14)) =-130.82 dBW =-130.82 dBW +30 dB =-100.8 dBm Determine the noise power (in dBm) at 27°C in a 20MHz bandwidth system. Correct answer: Pn=kTB; k=1.38*10^-23W°K-Hz; T(°K)=27°C+273=300°K; B=20*10^6Hz =10log(1.38*10^-23*300*20*10^6) =10log(8.28*10^-14) =-130.82 dBW =-130.82 dBW + 30dB = -100.8dBm Determine the noise power (in dBM) at 27°C in a 20 MHz bandwidth system. k = 1.38 * 10^-23 T = 27 + 273 = 300 B = 20 * 10^6 PN = 10log(kTB) PN = 10log((1.38 * 10^-23)(300)(20 * 10^6)) = -130.819 = -130.82 = -130.82 + 30 dB = -100.8 dBm Determine the noise power (in dBm) at 23°C in a 500 kHz bandwidth. k = 1.38 * 10^23 T = 27 + 273 = 300 B = 5e5 PN = 10log(kTB) = 10log(1.38 * 10^23 * 300 * 5e5) = -146.8 = -146.8 + 30 dB = -116.8 dBm In a data transmission system, the is the physical path between the transmitter and receiver Transmission path and channel For a 12-bit linear sign-magnitude PCM code with a resolution of 0.02 V, determine the voltage range that would be converted to the PCM code Correct answer: 1024 * .02 = 20.47V to 20.49V An FM signal has a deviation of 75 kHz and a modulating frequency of 15 kHz. The carrier frequency is 1600 MHz. Using Carson's rule, what is the required bandwidth for this system? Correct answer: BSCR = 2 * (75kHz + 15kHz) = 180kHz Which of the following are strengths of OFDM? Correct answer: Frequency selective fading does not adversely affect all subcarriers & effectively uses the wireless spectrum. Given a three stage amplifier system with stage 1 having a gain of 16 dB and a NF=3.5 dB, Stage 2 gain of 19 dB and a NF = 4.8 dB, Stage 3 having a gain of 20 dB and a NF = 4.17 dB, calculate the system noise figure. Correct answer: 16 dB = 40; 3.5 dB = 2.24; 19 dB = 80; 4.8 dB = 3.02; 20 dB = 100; 6.2 dB = 4.17 F = 2.24 + ((3.02-1)/40) + ((4.17-1)/(40*80)) = 2.29 = 2.6 dB Consider an OFDM implementation that uses 15 kHz subcarriers and use an OFDM symbol of 2048 subcarriers. The nominal cyclic prefix accounts for a 7% guard time. The extended cyclic prefix can use up to 25%. 1600 subcarriers can be used for data transmission. The rest are needed for pilot and null subcarriers. The transmission bandwidth is 25 MHz using 8 QAM modulation. What is the data rate for the nominal CP? Correct answer: R = 25MHz * (1600 / [2048 + 205])*5=88.8 Mbps What is the effective Temperature (Te in °K) for a system with a 29 dB input SNR and a 23 dB output SNR? Correct answer: Noise Factor (F) = 10^2.9 / 10^2.3 = 4 Effective Temp (Te) = (F – 1) * 290 = (4 – 1) * 290 = 3 * 290 = 870 OR F = (1 + Te/290) 4 = (1 + Te/290) 4 - 1 = Te/290 870°K = Te What is the effective Temperature (Te in °K) for a system with a 28 dB input SNR and a 24 dB ouptut SNR? F = (1 + Te/290) NF = 4 dB F = 2.52.5 - 1 = Te/290 438°K = Te Signal exist in either the time domain or the frequency domain, but not both. Correct answer: False Given the decimal equation: y = a/b. Find the value of y (in decimal), given that a = 16 dB and b = 7 dB. Correct answer: 8 Two multiplexing techniques used in telecommunications are and . Correct answer: time division multiplexing (TDM); frequency division multiplexing (FDM) Given the decimal equation: y = a + b. Find the value of y (in dB), given that a = 20 (decimal) and b = 40 dB (decibel)? Correct answer: 40 dB A transmission channel has a bandwidth of 8 kHz. Neglecting the effects of noise determine the channel for (a) 64- level encoding (b) If the SNR is 38 dB, what is... (c) Which factor is limiting in this system? Correct answer: a. 96 kHz b. 101 kHz c. Levels C = 2B * log2 (128) = 2 * 16kHz * 7 = 224 kHz C = B log2 (1 + SNR) = 16 kHz 8 log2 (10001) = 212.6 kHz A transmission channel has a bandwidth of 16 kHz. Neglecting the effects of noise, determine the channel capacity for a. 128-level encoding b. if the SNR is 40 dB , what is the maximum data rate? c. Which factor is limiting in this system? a. 224 kHz b. 212.6 kHz c. SNR C = 2B * log2(128) = 2 * 16kHz * 7 = 224 kHz C = Blog2(1 + SNR) = 16 kHz * 8log2(100001) = 212.6 kHz A VOIP network is converting the analog voice into digital data. The full range of the converter is 4 volts and the rms quantization error (Qe) is 1 millivolt. a. How many bits are required for the converter, including a sign bit? b. What is the dynamic range of the converter (in dB)? Correct answer: a. 2^N - 1 = 4V/.002 = 2000N = log(2001)/log(2) = 10.97 11 Magnitude bits +1 sign bit 12 bits total b. 20 log(2^11 - 1) = 66.2 dB A VoIP network is converting the analog voice into digital data. The full range of converter is 3 volts and the rms quantization error (Qe) is 0.5 millivolt a. How many bits are required for the converter, including a sign bit? b. What is the dynamic range of the converter (in dB)? a. 2^N - 1 = 3V/0.001 = 3000 2^N - 1 = 3000 2^N = 3001 N = log(3001)/log(2) N = 11.55 = 12 Magnitude bits + 1 sign bit = 13 bits b. 20log(2^12-1) = 72.2 dB Given that a repeater is 15 km from a transmitter sending data at a frequency of 1900 MHz, what is the path loss between the transmitter and repeater? Correct answer: LdB = 92.4 + 20log10(F GHz) + 20log10(Dkm) LdB = 92.4 dB + 20log(1.9) + 20log(15) LdB = 121.5 dB A digital communications system is being designed with the goal of transmitting 64 kbits/s over a channel having a bandwidth of 10kHz. Determine... a. The number of encoding levels required b. The minimum signal-to-noise ratio required in the channel. Correct answer: a. 64 kbps = 2 * 10 khz * log2 (M) = (64 kbps/20 kHz) = log2 (M) = 9.18 levels = round up to 16 b. 6.4 = log2(1 + SNR) = 2^6.4 = 1 + SNR = 84.4 - 1 = SNR = 83.3 = 19.2 dB C = B log2(1 + SNR) = (64 kbps/10 kHz) = log2(1 + SNR) = 6.4 C = log2(1 + SNR) = 2^6.4 = 1 + SNR = 84.4 -1 = SNR = 19.2 dB Two different factors affecting path loss are frequency and wavelength Correct answer: False Two different factors that influence path loss are distance and amplitude. Correct answer: False What are the (a) noise figure, (b) noise factor, and (c) effective temperature (Te) for a system with a 16 dB input SNR and a 12 dB output SNR? Correct answer: a. 4dB b. 2.51 c. 437.9°K F = (1 + Te/290) = 290(2.51 - 1) = 437.9°KWhat is the wavelength (in meters) of a 500 kHz signal? Correct answer: 600m What is the wavelength (in meters) of a 750 kHz signal? Correct answer: W = (3.0*10^8)/(750*10^6) = 0.4 m Use the following information for this question. Pr = 10^-12 W; T=17°C; fb=2Mbps; B=1MHz; Te=296°K (Hint: Don’t confuse dBW and dBm) a. What is the carrier-to-noise ratio (SNR) at the input to the amplifier? b. What is the energy per bit to noise density ratio (Eb/N0) for the receiver system (including the amplifier)? Correct answer: a. C = -120 dBW; N = -144 dBW C/N = 24 dB b. Eb = -120 dBw - 63 dB = -183 dBJ; No = -204 + 3.05 dB = -200.95 Eb / No = -183 dBj -(-200.95) = -17.95 dB Use the following information for this question. Pr = 10^-13 W; T=23°C; fb=1Mbps; B=1MHz; Te=296°K (Hint: Don’t confuse dBW and dBm) a. What is the carrier-to-noise ratio (SNR) at the input to the amplifier? b. What is the energy per bit to noise density ratio (Eb/N0) for the receiver system (including the amplifier)? Correct answer: a. C = -130 dBW; PN = -143.9 dBW C/N = -130 dBW – (-143.9) dBW = 13.9 b. Eb = -130 - 60 dB = -190 dBJ; No = -204 + 3.05 dB = -200.95 Eb / No = -190 - ( -200.95) = 10.95 dB Parasitic elements that are shorter than its associated driven elements are called . Correct answer: Directors Which of the following are feed mechanisms used in a parabolic antenna? Correct answer: Center feed & cassegrain feed means the direction of the electric field. Correct answer: Polarization The directional patter of an antenna is the same when the antenna is used to receive electromagnetic energy as it is when the antenna is used to transmit electromagnetic energy. Correct answer: True Energy is radiated into space when an alternating current is passed through a wire. Correct answer: True Which MIMO antenna use improves capacity? Correct answer: Spacial divisionSignal wavelength is inversely proportional to parabolic antenna gain. Correct answer: True is the product of antenna gain and the transmitted power minus transmission line loss. Corrective answer: Effective isotropic radiated power Which MIMO antenna use improves system performance? Correct answer: Diversity Given the decimal equation: y = a * b. Find the value of y (in decimal), given that a=13dB and b=7dB. Correct answer: y = 13dB + 7dB = 20dB = 10^2 = 100 OR y = 10^1.3 * 10^.7 y = 20 * 5 y = 100 Given the decimal equation: y = a * b. Find the value of y (in decimal), given that a=16dB and b=7dB. Correct answer: 16dB + 7dB = 23dB = 10^2.3 = 199.5 Round to 200 Given the decimal equation: y = a / b. Find the value of y (in decimal), given that a=16dB and b=7dB. Correct answer: 10^(16/10) = 40 10^(7/10) = 5 y = 40/5 y = 8 Given the decimal equation: y = a / b. Find the value of y (in decimal), given that a=13dB and b=7dB. Correct answer: 13dB - 7dB = 6dB = 10^.6 = 3.98 Round to 4 Given the decimal equation: y = a + b. Find the value of y (in dB), given that a=15 and b=69dB. Correct answer: 60dB Given the two amplifier gains in a chain: x=23dB and y=7dB, how much would the decimal gain value be for this amplifier chain? Correct answer: 23dB + 7dB = 30dB = 10^(30/10) = 1000 Given the two amplifier gains in a chain: x=20dB and y=13dB, how much would the decimal gain value be for this amplifier chain? Correct answer: decimal gain = 2000 Given the two amplifier gains in a chain: x=33dB and y=4dB, how much would the decimal gain value be for this amplifier chain? Correct answer: decimal gain = 5000 Given x=23dB and y=7dB, how much is xdB + ydB? Correct answer: 23dB + 7dB = 30dBWhat is the Base 2 logarithmic value for the following equation? log2 (3096)? Correct answer: log2(956) = log10(956) / log10(2) log2 (956) = 9.9 What is the Base 2 logarithmic value for the following equation? log2 (3096)? Correct answer: log2 (3096) = 11.6 Both analog and digital radio systems use analog carriers to transport the information through the transmission medium. Correct answer: True Signals can exist in the time domain or the frequency domain, but not both. Correct answer: False Modulation is a linear process. Correct answer: False All modulated signals in a wireless system are transmitted via Correct answer: analog transmission carrier Carson’s rule is an approximation for signal bandwidth Correct answer: Frequency modulation (FM) Information capacity represents the number of independent symbols that can be carried through a system in a given unit of time. Correct answer: True What is the Nyquist equation that identifies the maximum allowable data rate with multilevel digital signaling? Correct answer: C=2B log2(M) rate refers to the rate of change of a digital signal. Correct answer: Bit rate refers to the rate of change of a signal on the transmission medium after encoding and modulation. Correct answer: baud Quadrature Amplitude Modulation (QAM) is a form of digital modulation where the information is contained in both the amplitude and the phase of the transmitted carrier. Correct answer: True What is carrier recovery? Correct answer: The process of extracting a phase-coherent reference carrier from a receiver signal. Which carrier recovery method has the fastest acquisition time of the three methods? Correct answer: Remodulator P9e) and BER are often used interchangeably because they both have the exact same meaning.Correct answer: False Error rates on a wireless link can be high, resulting in a large number of retransmissions. Correct answer: True Error rate is proportional to Eb/No. Correct answer: Inversely Both analog and digital signals use analog carrier signals. Correct answer: True Given an AM frequency of 20 kHz, what is the required bandwidth? Options: 20kHz, 30kHz, 50kHz, 10kHz, none of these Correct answer: None of these For an RF bandwidth of 500 kHz, determine the maximum information rate (in bps) that can be transmitted in 8PSK? Correct answer: C = 2B log2 (M) C = 2 * 500 kHz * log2 (8) C = 1000 kHz *3 C = 3000 kHz = 3 MHz Given an RF bandwidth of 500 kHz, determine the maximum information rate (in bps) that can be transmitted if the SNR is 26 dB. Correct answer:I = B log2(1+SNR); SNR = 26 = dB = 400 I = 500 kHz * log2(1 + 400) I = 500 kHz * log2(401) I = 4324 kHz or 4.324 MHz An FM signal has a deviation of 15 kHz and a modulating frequency of 5 kHz. The carrier frequency is 1600 kHz. Using Carson’s Rule, what is the required bandwidth for this system? Correct answer: BSCR = 2 * (15kHz + 5kHz) = 40 kHz Given an AM frequency of 30 kHz, what is the required bandwidth? Given: 30 kHz, 40 kHz, 50 kHz, 15 kHz, None of the above Correct answer: None of the above Audio signals are in the time-domain, but radio signals are in the frequency domain. Correct answer: False 16-QAM means there are 5 bits represented by each symbol Correct answer: False Given an AM frequency of 20 kHz, what is the required bandwidth? Given: 20 kHz, 30 kHz, 50 kHz, 10 kHz, None of the above Correct answer: None of the above.M-ary PSK and QAM are digital modulation techniques used to increase data rate and limit overall bandwidth. Correct answer: True An FM signal has a deviation of 15 kHz and a modulating frequency of 5 kHz. The carrier frequency is 1600 kHz. Using Carson's Rue, what is the required bandwidth for this system? Correct answer: 40kHz Given an RF bandwidth of 500 kHz, determine the maximum information rate (in bps) that can be transmitted if the SNR is 14 dB. Correct answer: I = B log2 (1 + SNR); SNR = 14dB = 25 I = 500 kHz * log2 (1 + 25) I = 500 kHz * log2 (26) I = 2350kHz or 2.35MHz Mathematically, modulation involves addition of the carrier to the information signal. Correct answer: False Given a binary signal with a bit rate of 50kHz, what is the bandwidth? Correct answer: 25 KHz is a quadrature process that uses two parallel tracking loops (I & Q) simultaneously to derive and recover carrier information. Correct answer: Costas loop The energy per bit (Eb) is equal to the divided by the bit rate. Correct answer: Received signal power (Pr) What is the noise figure (NF) for a system with a 36dB input SNR and a 26dB output SNR? Convert your answer into noise factor (F) Correct answer: SNRin(dB)-SNRout(dB) = 36dB - 26dB = 10dB noise produces signals at a frequency that is the sum or difference of the two original frequency or multiples of those frequencies. Correct answer: Intermodulation The thermal noise level of a receiver is a function of the receiver noise figure and its bandwidth. Correct answer: True What is the definition of noise? Correct answer: Noise is defined as any unwanted interference that completes with the signal of interest and diminishes the signal of interest’s ability to successfully communicate the information. Name two places where noise originates. Correct answer: Channel and comm equipment Correct answer: Internal within equipment and external in the channel Noise power is measured in .Correct answer: Watts Internal noise power can be calculated. What is the formula – Pn = ? Correct answer: Pn = kTB Signal-to-noise ratio is a dimensionless value with power as the numerator and power as the denominator. Correct answer: Signal & noise SNR is one factor that must be considered when calculating the upper bound on achievable data ratge. Correct answer: True What is the noise power density (No) of the system (in dBm) including the amplifier given Te=500°K? (Hint use No dB formula) Correct answer: No = 10log(kTF) = -204dBW/Hz + NFdB = NF = 10log(1 + Te/290) = 4.4 dB = -199.6 dBW + 30dB = -169.6 dBm The receiver on a digital line of sight microwave link has a noise figure of 5 dB. What is the No for this receiver in dBM? Correct answer: -169 dBm The receiver on a digital line of sight microwave link has a noise figure of 7 dB. What is the No for this receiver in dBM? Correct answer: -197 dBm CCITT recommends a bit error rate (BER) of 1 * 10^-6 for error free communication. Correct answer: True CCITT recommends a bit error rate (BER) of 1 * 10^-5 for error free communication. Correct answer: False CCITT recommends a bit error rate (BER) of 1 * 10^-4 for error free communication. Correct answer: False Noise Figure is a figure of merit, indicating how much a component, stage, or series of stages degrades the signal-to-noise ratio of a system. Correct answer: True The thermal noise level of a receiver is a function of the receiver noise figure and its bandwidth. Correct answer: True Thermal noise is not present in all transmission media including all communication equipment and passive devices. Correct answer: False The bit error rate (or bit error probability) for digital data in a digital amplifier chain is a function of Eb/No Correct answer: True Eb/No is directly proportional to bit error rate.Correct answer: False Dynamic Range is the ratio of the largest possible magnitude to the smallest possible magnitude. Correct answer: True noise is noncontiguous consisting of irregular pulses or noise spikes of short duration and of relatively high amplitude. Correct answer: Impulse The energy per bit (Eb) is equal to the divided by the bit rate. Given: Noise power, noise spectral density, effective isotropic radiated power, isotropic receive level, none of the above. Correct answer: None of the above. A baseband signal has frequency components from dc to 6.2kHz. Determine the Nyquist theoretical minimum sampling rate. Correct answer: 12.4kHz A baseband signal has frequency components from dc to 15 kHz. Determine the Nyquist theoretical minimum sampling rate. The signal frequency components are as follows: 0 - 7.5 Hz In this, 0 is the dc frequency. The Nyquist theoretical sampling rate is given as follows: fs 2fm Where, fs represents the rate of sampling and, fm represents the component of maximum frequency. Therefore, fs for the given component is as follows: fs 2 * 15 kHz fs 30 kHz The upper bound on achievable data rate is the maximum data rate between the max data rate using the SNR bound and the max data rate. Correct answer: False Determine the minimum number of magnitude bits plus an additional sign bit required in a PCM code for a dynamic range of 60dB. Correct answer: 60dB = 20log(2^N-1) = 10(60/20) = 2^N - 1 = 1001 = 2^N = log(1001)/log(2) = 10 = N = 10 Magnitude bits required - 11 bits required when sign bit added Determine the minimum number of magnitude bits plus an additional sign bit required in a PCM code for a dynamic range of 70dB. Correct answer: 70dB = 20log(2^N-1) = 10(70/20) = 2^N - 1 = 3163 = 2^N = log(3163)/log(2) = 11.6 = N = 12 Magnitude bits required - 13 bits required when sign bit added Determine the resolution and quantization error for a nine-bit linear sign magnitude PCM code for a maximum decoded voltage of 2.55V. Correct answer: DR = 2^N - 1 = 2^8 - 1 =255255 = 2.55/Vres = Vres = 2.55/255 = .01; Qe=.01/2=.005 For a resolution of 0.02 V, determine the voltage and ranged for the following linear seven-bit sign magnitude PCM code. 1001011 Correct answer: 1001011 = 11 * .02 = 0.22 +- .01 = 0.21V to 0.23V Linear prediction coding (LPC) and are two main approaches used in vocoders. Correct answer: CELP Which source of delay is a function of the link bandwidth and the packet length? Correct answer transmission A communications network is a collection of switching nodes. Correct answer: True Nyquist states that in order to reconstruct a band-limited signal from periodic samples, the sampling rate must be at least twice the frequency of the highest frequency component. Correct answer: True Nyquist states that in order to reconstruct a band-limited signal from periodic samples, the sampling rate must be at least three times the frequency of the highest frequency component. Correct answer: False What are the two factors that affect wave propagation? Correct answer: Frequency and distance results in waves arriving at the destination at different times and appears as noise interference. Correct answer: Multipath Identify four propagation factors. Correct answer: Anomalous; propagation; diffraction; attenuation; and environmental noise Free space propagation path loss is [a] / octave or [b] / decade where an octave means doubling the distance and a decade means a period of ten. Correct answer: a = 6dB; b = 20dB What is the dB related free space path loss formula given frequency in GHz and distance in km? Correct answer: LdB = 92.4 + 20log10(FGHz) = 20log10(Dkm) It takes 22dB to launch a wave just 1 wavelength distance from an antenna. Correct answer: True The path loss exponent does not change even if the environment changes. Correct answer: False What are the three values used to calculate the effective isotropic radiated power (EIRP)? Correct answer: Transmitter power, transmission line loss, antenna gainLong-term fading is the average or envelope of the fading signal. Correct answer: True Short-term fading is caused mainly by absorption of a transmitted wave by local obstacles such a houses, buildings and other human-built structures or by natural obstacles, such as forests. Correct answer: False The isotropic receive level (IRL) is can be calculated is dB by subtracting the path loss (in dB) from the EIRP (in dBW or dBm). Correct answer: True The power entering the first active stage of the receiver is called the isotropic receive level (IRL) Correct answer: False Okumura and Hata created models derived from empirical measurements that can be used to calculate path loss in various types environments. Correct answers: True What are the three commonly identified probability density functions? Correct answer: log normal, Rayleigh, Rician In a dispersive medium, what are the two kinds of spread? Correct answer: Doppler and multipath spread The resolution of a pulse-coded modulated signal is determined by dividing the by the number of positive (or negative) non-zero PCM codes. Correct answer: maximum positive (or negative) voltage Which type of coding and error detection techniques are used to request transmission from the receiver to the transmitter in the presence of a transmission error? Correct answer: ARQ Packet switching is the most commonly used switching technique in the internet. Correct answer: True Anomalous Propagation is a decrease in atmospheric index of refraction with increasing altitude. Correct answer: True Affects unequally the different spectral components of a radio signal Correct answer: Selective fading The common cause of fading is Correct answer: Multipath Received Signal Level (RSL) is the power level entering the first active stage of the receiver. Correct answer: TrueAttenuation is a decrease in atmospheric index of refraction with increasing altitude... Correct answer: False Multipath results in waves arriving at the destination at different times; however, it has no effect on the reconstructed output. Correct answer: False Multipath results in waves arriving at the destination at different frequencies and appears as noise interference. Correct answer: False Is spreading in frequency in a dispersive channel. Correct answer: Doppler Attenuation occurs when the strength of a signal falls off with distance over the transmission medium. Correct answer: True In a dispersive medium, what are the two kinds of spread? Correct answer: Doppler and Multipath Fades vary in Correct answer: depth, duration, frequency is spreading in frequency in a dispersive channel. Correct answer: Doppler Received Signal Level (RSL) is the RF power impinging on the receive antenna. Correct answer: False refers to changes in signal strength between a transmitter and receiver as the distance between the two changes by a small distance of about one-half a wavelength. Correct answer: Fast fading For an RF bandwidth of 500 kHz, determine the maximum information rate (in bps) that can be transmitted in QPSK (4PSK) C = 2 * B log2(m) C = 2 * 500 kHz * log2(4) C = 1000 kHz * 2 = 2000 kHz = 2 MHz Given an AM frequency of 25 kHz, what is the required bandwidth? 50 kHzAudio signals are in the time-domain, but radio signals are in the frequency domain. False The receiver on a digital line of sight microwave link has a noise figure of 5 dB. What is the No for this receiver in dBm? -169 What is the noise power density (No) of the system (in dBm) including the amplifier given Te = 450 degrees K? (Hint: Use No dB formula) k = 1.38 * 10^-23 Te = T = 450°C F = 1 + [Te / 290] N(x) = 10log(x) No = N(kTF) = N(kT) + N(F) = 10log((1.38 * 10^-23)(450)) + N(F) = 10log((1.38 * 10^-23)(450)) + 4.1 dB = -202.07 dBW/Hz + 4.07 dB = -199.9 dBW = -169.9 dBm An FM signal has a deviation of 75 kHz and a modulating frequency of 15 kHz. The carrier frequency is 1600 MHz. Using Carson's Rule, what is the required bandwidth for this system? 180 kHz Given an RF bandwidth of 500 kHz, determine the maximum information rate (in bps) that can be transmitted if the SNR is 14 dB. SNR = 14 dB = 25 I = B log2 (1 + SNR) I = 500 kHz * log2(1 + 25) I = 500 kHz * log2(26) = 2350 kHz = 2.35 MHz M-Ary PSK and QAM are digital modulation techniques used to increase data rate and limit overall bandwidth True The process of extracting a phase-coherent signal reference carrier from a received signal is called Carrier recovery Given a binary signal with a bit rate of 50 kHz, what is the bandwidth? 25 kHz noise is due to thermal agitationThermal SNR is not a factor to consider when calculating the upper bound on achievable data rate. False Thermal noise is not present in all transmission media including all communication equipment and passive devices False Energy per bit (Eb) is a function of the received signal power (Pr) divided by bit rate The receiver on a digital line of sight microwave link has a noise figure of 4 dB. What is the No for this receiver in dBm? -170 dBm What is the noise figure (NF) for a system with a 36 dB input SNR and a 26 dB output SNR? Convert your answer into noise factor (F) Noise figure = SNR(in)dB - SNR(out)dB = 36dB - 26dB Noise factor (F) = 10^(10/10) = 10 Given that Pr = 10^-10 W; T = 30°C; fb = 3 Mbps; B = 400 kHZ; Te = 285°K. What is the energy per bit (Eb), in dB, in the amplifier? Pr = 1e-10 fb = 3e6 Eb = (Pr/fb) = (1e-10/3e6) = 3.3e-17 = 10log(5e-16) = -164.8 dBJ Anomalous Propagation is a decrease in atmospheric index of refraction with increasing altitude. True Given that Pr = 10^-11 W; T = 30°C; fb = 2Mbps; B = 500 kHZ; Te = 295°K. What is the received noise power (in dBW) at the input to the amplifier chain? Correct answer: Pr = 1 * 10e-11; Pn = kTB = 1.38e-23 W/Hz-deg K * (273 + 30) deg K * 500e3 Hz = 2.09 * e-15 = -146.8 dBW Pr/Pn = Pr(dBW) – Pn (dBW) = -110 dBW – ( - 146.8 dBW ) = 36.8 dB Reciprocity in an antenna is defined as the directional pattern of an antenna being the same when the antenna is used to receive electromagnetic energy as it is when the antenna is used to transmit electromagnetic energy. Correct answer: True Radiation fields close to an antenna are the same as those at a distance.Correct answer: False When current flows in an antenna, there is no energy loss because the energy is dissipated into electromagnetic energy. Correct answer: False The radio of the power radiated to the total input power is called . Correct answer: Antenna efficiency is a relationship of radiated power in a direction referenced to a spherical radiator. Correct answer: Directivity What are the two type of antenna elements? Correct answer: Driven and parasitic What type element is directly connected to the transmission line? Correct answer: Driven element elements receive energy through mutual inductance. Correct answer: Parasitic At a given frequency, the gain of a parabolic antenna is a function of its effective area. Correct answer: True The frequency of a signal does not influence the gain of a parabolic reflector. Correct answer: False What is the direction of the electric field called? Correct answer: Polarization Antenna arrays are used for: Correct answer: Diversity, Parallel data streams, beamforming, directional beams to multiple simultaneous users List four uses of MIMO Correct answer: Diversity for better system performance, beam-forming, spatial diversity, increased data rates A typical parabolic antenna gain has n antenna efficiency. Given answers: 80%, 90%, 75%, 60%, None of the above Correct answer: None of the above Which of the following are feed mechanisms used in a parabolic antenna? Correct answer: Center feed and yagi feed means the direction of the electric field. Correct answer: Polarization The directional pattern of an antenna is the same when the antenna is used to receive electromagnetic energy as it is when the antenna is used to transmit electromagnetic energy.Correct answer: True Given that the radiation resistance of an antenna is 300 ohm and the elemental resistance is 275 ohm, what is the efficiency of this antenna? Correct answer: E = Rr / (Rr + Re) E = 300 / (300 + 275) E = 0.522 or 52.2% Energy is radiated into space when an alternating current is passed through a wire. Correct answer: True Given that the radiation resistance for a short dipole antenna is 300 ohm and the length is 150mm, at what frequency would one expect this antenna to radiate? Correct answer: Rr = 200 * (I/A)^2; Given Rr = 300ohm; I = 150mm; A = c/f 300 = 200 * ((150 * 10^-3)^2 / (3 * 10^8 / f)^2 f^2 = 300 * (3 * 10^8)^2) / (200 * (150 * 10^-3)^2) f^2 = 6 * 10^18 f = 2.5 GHz OFDMA uses a combination of by following users to use a subset of the subcarriers at different times. FDMA and TDMA Orthogonal Frequency Division Multiple Access (OFDMA) is an OFDM MAC mechanism where multiple uses share a channel by using different OFDM subcarriers. True Why does the OFDM system implement an IFFT function on the transmit side and FFT on the receive side? Ensure carriers do not interfere with each other The cyclic prefix addresses the problem of inter symbol inter symbol interference caused by multipath delay spread True What is(are) the reason(s) for using SC-FDMA? None of the above For a resolution of 0.02V, determine the voltages and ranges for the following linear seven-bit sign magnitude PCM code 1001011 = 1001011 = -1 (2^0 + 2^1 + 2^3) * 0.02 Voltage = -1 (11) * 0.02 = -0.22 Range = (-0.21, -0.23) A packet switched network does not use switching nodes.False Nyquist states that in order to reconstruct a band-limited signal from periodic sampling, the sampling rate must be at least three times the frequency of the highest frequency component. False CCITT recommends a bit error rate (BER) of 1*10^-6 for error free communication True Thermal noise is not present in all transmission media including all communication equipment and passive devices False Energy per bit (Eb) is a function of the received signal power (Pr) divided by the bit rate The receiver on a digital line of sight microwave link has a noise figure of 4 dB; What is the No for this receiver in dBm? -170 dBm Which source of delay depends on the congestion level of the router? Queueing Which source of delay is due to checking for bit errors and identifying the output link? Correct answer: Nodal The of a pulse-coded modulation signal is determined bu dividing the maximum positive (or negative) voltage by the number of positive (or negative) non-zero PCM codes. Correct answer: resolution Determine the minimum number of magnitude bits plus an additional sign bit required in a PCM code for a dynamic range of 65 dB DR(db) = 20 log(Vmax/Vmin) = 20 log(2^n - 1) = 6.02n 6.02n = 65 n = 10.797 n - 11 bits 11 bits + 1 sign bit = 12 bits Determine the resolution and quantization error for a nine-bit linear sign magnitude PCM code for a maximum decoded voltage of 2.55 V = 2.55/(2^8 - 1)= 2.55/(255) = 0.01 Qe = 0.01/2 = 0.005 Determine the resolution and quantization error for a nine-bit linear sign magnitude PCM code for a maximum decoded voltage of 5.10 V Correct answer: 255= 5.10/Vres = Vres=5.10/255 = .02; Qe = .02/2 = .01 Which type of coding and error detection techniques are used to detect and correct the presence of a transmission error? FEC Linear prediction coding (LPC) and are two main approaches used in vocoders. CELP Fades vary in Depth, Duration, Frequency OFDM signals have a higher peak-to-average power ratio because of the following Correct answer: The peak value of the signal is substantially larger than the average value. Explain the difference between an OFDM "symbol" and a QAM "symbol" Correct answer: A QAM symbol represents multiple bits in a digital system; i.e., 4 bits can be represented by 16 levels. An OFDM symbol is much more complex. An 802.11a OFDM carrier signal (burst type) is the sum of one or more OFDM symbols each comprised of 52 orthogonal subcarriers, with baseband data on each subcarrier being independently modulated using quadrature amplitude modulation (available formats: BPSK, QPSK, 16-QAM, or 64-QAM). So an OFDM symbol is really a multiple of QAM symbols. If the received power at 1 km is 3 dBm, determine the received power at 18 km. Correct answer: 3 dBm - 18 dB = -15 dBm = .032 mW If the received power at 1 km is 2 dBm, determine the received power at 4 km. 2 dBm - 12 dB = -10 dBm is spreading in frequency in a dispersive channel. Correct answer: Doppler Receive Signal Level (RSL) is the RF power impinging on the receive antenna. Correct answer: False affects unequally the different spectral components of a radio signal. Selective fading What is the noise power for a system with 30°C noise temperature in a 250 kHz bandwidth? Correct answer: -148.3 dBwSingle-carrier FDMA (SC-FDMA) is a variant of OFDM that performs extra DFT operations at the transmitter and receiver True For a 45 Mbps LTE data stream with a symbol time of 66.67 us, how many subcarriers are created? T = 1/66.67 = 15 kHz = 45 * 10^6 / 15 * 10^3 = 3000 us The of a pulse-coded modulated signal is determined by dividing the maximum positive (or negative) voltage by the number of positive (or negative) nonzero PCM codes Resolution In a dispersive medium, what are the two kinds of spread? Doppler and Multipath Multipath results in waves arriving at the destination at different times, however, it has no effect on the reconstructed output. False What is the receive signal level (in dBm) of a system that has an EIRP of 57 dBW, a path loss of 130 dB, a receive antenna gain of 24 dB and receive transmission line loss of 0.6 dB? RSL = RSL(dBW) = IRL(dBW) + Rx Antenna Gain (dB) - Rx Transmission Line Loss (dB) IRL(dBw) = EIRP(dBW) - Path Loss(dB) Given: EIRP = 57 dBW Path Loss = 130 dB Rx Antenna = 24 dB IRL = 57 dBW - 130 dB = -73 dBW RSL = -73 dBW + 24 - 0.6 = -49.6 dBW = -19.6 dBm This is a constellation diagram. *Description* 16 points plotted in a circle on x y axis. 4 points in each quadrant. Correct answer: 16-PSK Given the decimal equation: y=a*b. Find the value of y (in decimal), given that a = 13dB and b = 7 dB Correct answer: y=100 Given the decimal equation: y=a/b. Find the value of y (in decimal), given that a=16dB and b=7dB Correct answer: y=8 Given the decimal equation: y=a+b. Find the value of y (in dB), given that a=15 and b=60 dB Correct answer: y=60dBGiven the two amplifier gains in a chain: x=20dB and y=13dB, how much would the decimal gain value be for this amplifier chain? Correct answer: decimal gain = 2000 Given x=20dB and y=13dB, how much is x dB + y dB? Correct answer: xdB+ydB=33dB What is the Base 2 logarithmic value for the following equation? Log2(956) Correct answer: 9.9 For an RF bandwidth of 500 kHz, determine the maximum information rate (in bps) that can be transmitted in 8PSK? Correct answer: C=1000kHz*3 = 3000kHz or 3MHz is a quadrature process that uses two parallel tracing loops (I & Q) simultaneously to derive and recover carrier information. Correct answer: Costas loop Given an RF bandwidth of 500 kHz, determine the maximum information rate (in bps) that can be transmitted if the SNR is 26dB. Correct answer: I=B log2(1+SNR); SNR=26 db 400 I=500kHz*log2(1+400) I=500kHz*log2(401) = 4324 kHz or 4.324 MHz An FM signal has a deviation of 15 kHz and a modulating frequency of 5 kHz. The carrier frequency is 1600kHz. Using Carson’s Rule, what is the required bandwidth for this system? Correct answer: 40kHz Given an AM frequency of 30kHz, what is the required bandwidth? Given: 30kHz, 40kHz, 50kHz, 15 kHz, None of the above Correct answer: None of the above Audio signals are in the time-domain, but radio signals are in the frequency domain. Correct answer: False 16-QAM means there are 5 bits represented by each symbol. Correct answer False Given an AM frequency of 20kHz, what is the required bandwidth? Given: 20kHz, 30kHz, 50kHz, 10 kHz, None of the above Correct answer: None of the above Analog signals can be transmitted digitally Correct answer: True This is a constellation diagram *Description* 8 points on an x y axis in the shape of an X. 2 points in each quadrant.Correct answer: 8-QAM noise is noncontiguous consisting of irregular pulses or noise spikes of short duration and of relatively high amplitude. Correct answer: Impulse The energy per bit (Eb) is equal to the divided by the bit rate. Given: Noise power, noise spectral density, effective isotropic radiated power, isotropic receive level, none of the above Correct answer: None of the above What is the noise power density (No) of the system (in dBm) including the amplifier given Te=450°K? (Hint: Use No dB formula) Correct answer: No=10log(kTF)=-204dBW/Hz+NFdb=10log(1+Te/290)=4.1dB =-199.9dBw = -169.9dBm What is the noise figure (NF) for a system with a 27 dB input SNR and a 24dB output SNR? Convert your answer into noise factor (F) Correct answer: SNRin(dB)-SNRout(dB)=27dB-24dB=3dB Noise Figure SNR is not a factor to consider when calculating the upper bound on achievable data rate. Correct answer: False The receiver on a digital line of sight microwave link has a noise figure of 5dB. What is the No for this receiver in dBm? Correct answer: -169dBm CCITT recommends a bit error rate (BER) of 1*10^-6 for error free communication. Correct answer: True Thermal noise is not present in all transmission media including all communication equipment and passive devices. Correct answer: False A communication network is a collection of switching nodes. Correct answer: True Circuit switching is the most efficient switching architecture. Correct answer: False In a packet-switched network, two stations with different data rates can exchange unformation. Correct answer: True Which network blocks calls when traffic heavy? Correct answer: Circuit-switched In a circuit-switched network, each packet is treated independently, without reference to previous packets. Correct answer: FalseWhich of the following are sources of packet delay? Correct answer: Nodal processing, queueing, transmission delay, propagation delay Throughput is the rate at which bits are transferred between a sender and a receiver. Correct answer: True Sampling alone at the Nyquist rate in not a digital technique. Correct answer: True What is the most commonly used digital modulation scheme? Correct answer: Pulse-code Modulation What is the ratio of the strongest possible signal that can be transmitted and the weakest discernable signal? Correct answer: dynamic range The practice of assigning a binary sequence to each voltage sample is called Correct answer: quantization The error from the quantum value to the true value is quantizing distortion. Correct answer: True Single-carrier FDMA (SC-FDMA) is a variant of OFDM that performs extra DFT operations at the transmitter and receiver. Correct answer: True The OFDM process converts the parallel input to frequency domain output in the receiver. NOT Correct answer: IFFT The cyclic prefix addresses the problem of intersymbol interference caused by multipath delay spread Correct answer: True Why does the OFDM system implement and IFFT function on the transmit side and FFT on the receiver side? Correct answer: Ensure carriers don’t interfere What is (are) the reasons for using SC-FDMA? Correct answer: Reduce the PAPR Problem For a 45 Mbps LTE data stream with a symbol time of 66.67 us, how many subcarriers are created? Needs Grading: 3000 subcarriers created Orthogonal frequency division multiplexing (OFDM) is a scheme that divides a broadband signal into multiple, Correct answer: True OFDMA uses a combination of by allowing users to use a subset of the subcarriers at different times.Possible answer: FDMA & TDMA Why does the OFDM system implement and IFFT function on the transmit side and FFT on the receiver side? Correct answer: Ensure carriers do not interfere with each other OFDM signals have a higher peak-to-average power ratio because of the following Correct answer: The peak value of the signal is substantially larger than the average value. One main technical problems with OFDM is the intercarrier interference caused by subcarriers that are not in frequency synchronization. Correct answer: True The OFDM process uses to compute the frequency domain output in the receiver. Correct answer: FFT For a 45 Mbps LTE data stream with a symbol time of 66.67 us, how many subcarriers are created? Needs grading: 45*10^6 / 15*10^3 = 3000 subcarriers created For a 12 Mbps LTE data stream with a symbol time of 66.67 us, how many subcarriers are created? Needs grading: 1/66.67 = hz (12*10^6)/(15*10^3) 800 OFDM does not efficiently use the wireless spectrum. Correct answer: False Why does the term "symbol" take on different meaning in an OFDM system? Needs grading: Good luck Single-carrier FDMA (SC-FDMA) is a variant of OFDM that performs extra DFT operations at the transmitter and receiver. Correct answer: True What is the effective Temperature (Te in °K) for a system with 27 dB input SNR and a 24 dB output SNR? Correct answer: Noise Factor (F) = [10^(27/10)] / [10^(24/10)] = 2; F = (1+Te/290) 2 – 1 = Te/290) Te = 290 OR Effective Temp (Te) = (F – 1) * 290 = (2 – 1) * 290 = 1 * 290 = 290A baseline signal has frequency components from 7.5 kHz to 20 kHz. Determine the Nyquist***INCOMPLETE PIC*** Correct Answer: 1/(2*[20-7.5])=.04 = 40kHz All radio signals exist in both the time and frequency domains simultaneously. Correct answer: True Given that the repeater is 15 km from a transmitter sending data at a frequency of 1900 MHz, w***INCOMPLETE PIC*** Correct answer: Ldb = 92.4 + 20 log(FGHz) + 20 log(Dkm) = 92.4 + 20 log(1.9) + 20 log(15) = 121.5 dBw Given the two amplifier gains in a chain: x = 23 dB and y = 7 dB, how much would the decimal gain value be for this amplifier chain? Voltage gain in dB A = B + P = 23 dB + 7 dB = 30 dB For decimal gain A(gain in dB)/20 A(dec) = 10 = 10 ^ (30/20) = 10 ^ 1.5 = 31.6 What is the wavelength (in meters) of a 750 MHz signal? wvl = c/f = (3 * 10^8 m/s)/(750 * 10^6) = 0.4 m/s Given that a repeater is 45 km from a transmitter sending data at a frequency of 3100 MHz, what is the path loss between the transmitter and repeater? L(dB) = 92.4 + 20log10(F(GHZ)) + 20log10(D(km)) = 92.4 + 20log10(3.1) + 20log(45) dB = 135.33 Two multiplexing techniques used in telecommunications are and . Frequency division multiplexing (FDM) Time division multiplexing (TDM) All radio signals exist in both the time and frequency domains simultaneously. True Consider an OFDM implementation that uses 15 kHz subcarriers and use an OFDM symbol of 2048 subcarriers. The nominal cyclic prefix accounts for a 10% guard time. The extended cyclic prefix can use up to 25%. 1600 subcarriers can be used to for data transmission. The rest are needed for pilot and nullsubcarriers. The Transmission bandwidth is 25 MHz using 32 QAM modulation. What is the data rate for the extended CP? R = 25 MHz * (1600 / [2048 + 512]) * 5 = 78.1 Mbps What parameter of the intelligence signal causes or determines the instantaneous rate of carrier frequency? The frequency of intelligence signal determines the rate of carrier frequency deviation Given the decimal equation: y = a + b. Find the value of y (in dB), given that a = 10 (decimal) and b = 13 dB (decibel) a = 10 b = 13 dB b(dec) = 10 ^(b(dB)/20) = 10^(13/20) = 10^0.65 = 4.47 Since decimal equation y = a + b = 10 + 4.47 = 14.47 Convert to dB y(dB) = 20log(y) = 20log(14.47) = 23.2 Use the following information for this question. Pr = 10^-12W; T = 17°C; fb = 2 Mb ps; B = 1 MHz; Te = 296°K (Hint: Don't confuse dBW and dBm) a. What is the carrier-to-noise ratio (SNR) at the input to the amplifier? b. What is the energy per bit to noise density ratio (Eb/No) for the recei ver system (including the amplifier) a. C = -120 dBW; N = -144 dBW; C/N = 24 dB b. Eb = -120 dBW - 63 dB = -183 dBJ No = -204 + 3.05 dB = -200.95 Eb/No = -183 dBJ - (-200.95) = 17.95 dB Two different factors affecting path loss are distance and frequency. True A digital communications system is being designed with the goal of transmitting 32 kbits/s over a channel having a bandwidth of 8 kHz. Determine (a) the number of encoding levels required and (b) the minimum signal to noise ratio required i n the channel. a. 32 kbps = 2 * 8 kHz * log2(M) 32 kbps = 16 kHz * log2(M) log2(M) = 32 kbps/16 kHZ log2(M) = 2 = 2 * log(2) = 0.602= 10^0.602 = 3.9994475 = 4 b. C = Blog2(1 + SNR) = (32 kbps/8kHz) = log2(1 + SNR) = 4 = log2(1 + SNR) = 2^4 = 1 + SNR = 16 - 1 = SNR = 11.8 dB A modulated carrier with a digital information signal typically occupies multiple frequencies True A certain transmission channel has a bandwidth of 2 MHz, neglecting the effects of noise, what is the channel capacity for 32 level encoding? 2MHz is 2000kHz C = 2B * Logbase2(32) = 2 * 2000kHz * 5 = 20000 kHz or 20MHz Determine the minimum number of bits required in a PCM code with a sign bit for a dynamic range of 80 dB. What is the coding efficiency? DR(db) = 20 log(Vmax/Vmin) 20 log(2^n - 1) 6.02n = 80dB n = 13.3 14 bits + 1 sign bit = 15 bits (round it up and add one extra bit) Determine the total radiated power from an elemental antenna 118.5 mm long if the uniform rms current is 260 mA and the frequency is 1900 MHz. P = 789 * [l/λ]^2 (i^2) = RMS P = 789/2 * [l/λ]^2 (i^2) = Peak P = 789 * [l/λ]^2 (i^2) λ = (3*10^8 / f) = 0.158m = 789* [ (118.5* (10) ^(-3))/(0.158) ]^2 * (300* ) (10^-3 )^2 = 16 dBW When current flows in an antenna, there is no energy loss. TrueDigital communication is to environmental changes? Less sensitive Advantages of digital communication are Easy multiplexing, Easy processing, Reliable What is necessary for digital communication? Precision timing, Frame synchronization, Character synchronization What are the disadvantages of digital communication? Needs more bandwidth & Is more complex Which system uses digital transmission? ISDN & LANs The attenuation level in bounded power spectral density is 35 & 50 The interval of frequencies outside which the spectrum is zero is called as absolute bandwidth Synchronization available in digital communication are Symbol synchronization, Frame synchronization, Carrier synchronization