ECE-350 Midterm: ECE 350 Boise State Exam3 sample Solutions f04

• School: Boise State University
• Department: Electrical and Computer Engineering
• Course: ECE-350
• Professor: All

Document Summary

20 (g) from transform #9 in table 4. 2, tb. )20 t (h) this is frequency division multiplexing because two signals share different frequency bands: (a) this is pulse amplitude modulation. 4 /20 (b) where a0=(5/20)e-j(0)(2 /20)(-2. 5)= (2. 5/20), a1=(5/20)sinc(5(1)(2 /20/2)) e-j(1)(2 /20)(-2. 5), a-1=(5/20)sinc(5(-1)(2 /20/2)) e-j(-1)(2 /20)(-2. 5), a2=(5/20)sinc(5(2)(2 /20/2)) e-j(2)(2 /20)(-2. 5), ak=(5/20)sinc(5(k)(2 /20/2)) e-j(k)(2 /20)(-2. 5), (c) this is time division multiplexing because we are modulating with a square pulse train. This will allow another signal to fill in the gaps of time not used by this signal: for the transfer function sh s. Re{s}-1. c & d) sh. Re{s} -1 h1(t)= e-tu(t) so. Re{s}>-1 : for the transfer function h(t)= - e-3tu(-t) - e-tu(-t), h(t)= e-3tu(t) - e-tu(-t) h(t)= e-3tu(t)+ e-tu(t) not stable not stable stable not causal not causal causal. 1: sketch the pole-zero plot for this transfer function. Because of the infinite amplitudes, i would accept either stop band.