IFT 372 Wireless Communication (all answers correct)

What is a logarithm? correct answers A logarithm is an exponent based mathematical representation of a numerical value. What logarithmic base is used in digital communications? correct answers Two How do you find a log2 value on a calculator that only has log10 functionality? (Hint: what is the formula?) correct answers log2(a) = [log10(a) / log10(2)] What is the definition of a decibel? correct answers 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) correct answers True There can be more than one power value in a decibel (dB) equation. correct answers 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 answers True __________ is a decibel value referenced to watts and ______ is a decibel value referenced to milliwatts. correct answers dBW & dBm How do you convert dBW to dBm? correct answers 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 answers Output power & Input power What parameter of the intelligent signal causes or determines the instantaneous rate of carrier frequency? correct answers Amplitude of the modulating signal What parameter of the intelligent signal causes frequency deviation of the carrier? correct answers Amplitude of the modulating signal A modulated carrier occupies a single frequency. correct answers False Determine the noise power (in dBm) at 27° C in a 500 kHz bandwidth. correct answers 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 answers Pn=kTB; k=1.38*(10^-23)W°K-Hz; T(°K)=27°C+273=300°K;