After read and made yourself familiar with the following topics, please answer /discuss the following 15 questions. Your answers should have detailed explanations. Topics: frequency division multiplexing, time division multiplexing, electromagnetic waves, radio-wave propagation, TDM in telephony, FDMA, TDMA access, spread spectrum, frequency-hopping systems, pseudo-random noise sequence, direct-sequence systems, attenuation, transmitting antenna gain, receiving antenna gain.Questions:1) What do these terms stand for? TDM, FDM, CDMA, FDMA, TDMA.2) Why is FDMA (Frequency Division Multiple Access) using radio more difficult to coordinate than FDM (Frequency Division Multiplexing) on a single cable?3) A frequency-hopping spread-spectrum system hops to each of 300 frequencies every 15 seconds. How long does it spend on each frequency?4) How random is pseudo-random noise in a frequency-hopping system?5) How many methods are there for doing spread-spectrum?6) Can TDM (Time-division multiplexing) be used for analog signals?7) Is this a correct statement? With CDMA, all stations can be transmitting on the same band at the same time.8) Why is high frequency radio wave propagation over long distances better during peaks in the sunspot cycle?9) When do we use TDMA instead of TDM?10) What is the difference between Processing gain and Spreading gain?11) What do TDR, HF, SWR, VSWR, EIRP stand for?12) Name at least 3 forms of ElectroMagnetic radiation.13) At what frequencies are the Ground waves most effective?14) Why can AM (Amplitude Modulation) radio stations often be received at distances beyond the horizon?15) What is the input impedance to a lossless antenna at resonance?
Obtain the general solution of the following DEs: i. y′′′ + y′′ − 4y′ + 2y = 0 ii. y(4) + 4y(2) = 0 iii. x(x − 2)y′′ + 2(x − 1)y′ − 2y = 0; use y1 = (1 − x) iv. y′′ − 4y = sin2(x) v. y′′ − 4y′ + 3y = x ; use y1 = e3x vi. y′′ + 5y′ + 6y = e2xcos(x) vii. y′′ + y = sec(x) tan(x)
Obtain the general solution of the following DEs: i. y′′′ + y′′ − 4y′ + 2y = 0 ii. y(4) + 4y(2) = 0 iii. x(x − 2)y′′ + 2(x − 1)y′ − 2y = 0; use y1 = (1 − x) iv. y′′ − 4y = sin2(x) v. y′′