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 isthe denominator. Correct answer: Output power & Input power What parameter of the intelligent signal causes or determinesthe instantaneousrate 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 signal Determine 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 linearsign-magnitude PCM code with a resolution of 0.02 V, determine the voltage range that would be converted to the PCM code 000 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 allsubcarriers & effectively usesthe 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 = 2.5 2.5 - 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 thissystem? 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 thissystem? 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() = 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 = 2000 N = 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 factorsthat 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°K What is the wavelength (in meters) of a 500 kHzsignal? Correct answer: 600m What is the wavelength (in meters) of a 750 kHzsignal? 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 meansthe 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 division Signal wavelength isinversely proportional to parabolic antenna gain. Correct answer: True is the product of antenna gain and the transmitted power minustransmission line loss. Corrective answer: Effective isotropic radiated power Which MIMO antenna use improvessystem 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 = 30dB What 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 signalsin a wireless system are transmitted via Correct answer: analog transmission carrier Carson’srule 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 refersto the rate of change of a digitalsignal. 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 hasthe 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 meansthere 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, orseries ofstages 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 ofshort 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 hasfrequency 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 theoreticalsampling rate is given asfollows: fs 2fm Where, fs represents the rate of sampling and, fm representsthe 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 =255 255 = 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 linearseven-bit sign magnitude PCM code. Correct answer: = 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 factorsthat 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 gain Long-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: True Attenuation is a decrease in atmospheric index of refraction with increasing altitude... Correct answer: False Multipath resultsin 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 refersto changes in signalstrength between a transmitter and receiver asthe 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 kHz Audio 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 agitation Thermal 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 elementsreceive 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 beamsto multiple simultaneous users