ASU IFT 372 Fall 2025 Helm Weeks 1-7 Quizzes + Midterm Multisource

ASU IFT 372 Fall 2025 Helm Weeks 1-7 Quizzes + Midterm Multisource 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 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 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 = 2.5 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 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 = 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