IFT 372 WIRELESS COMMUNICATION - HELM - QUIZ 1-7 + MIDTERM + LECTURE ACTIVITIES EXAM QUESTIONS AND ANSWERS | LATEST VERSION 2026

IFT 372 WIRELESS COMMUNICATION - HELM - QUIZ 1-7 + MIDTERM + LECTURE ACTIVITIES EXAM QUESTIONS AND ANSWERS | LATEST VERSION 2026 What is a logarithm? - answer-A logarithm is an exponent based mathematical representation of a numerical value. What logarithmic base is used in digital communications? - answer-Two How do you find a log2 value on a calculator that only has log10 functionality? (Hint: what is the formula?) - answer-log2(a) = [log10(a) / log10(2)] What is the definition of a decibel? - answer-A decibel is a mathematical representation of a power level on a logarithmic scale. a = 10lob(b), where a is the decibel value and b is the decimal value. You cannot mix decimal and decibel values in an equation (i.e., you either have to work with decimal values or decibel values) - answer-True There can be more than one power value in a decibel (dB) equation. - 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) - answer-True__________ is a decibel value referenced to watts and ______ is a decibel value referenced to milliwatts. - answer-dBW & dBm How do you convert dBW to dBm? - 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. - answer-Output power & Input power What parameter of the intelligent signal causes or determines the instantaneous rate of carrier frequency? - answer-Amplitude of the modulating signal What parameter of the intelligent signal causes frequency deviation of the carrier? - answer-Amplitude of the modulating signal A modulated carrier occupies a single frequency. - answer-False Determine the noise power (in dBm) at 27° C in a 500 kHz bandwidth. - 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. - 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 dBmDetermine the noise power (in dBm) at 23°C in a 500 kHz bandwidth. - answer-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