ECE-350 Midterm: ECE 350 Boise State Exam3 sample Solutions s14

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Spring 2014: (a)what is the equation describing x(j )? jx jx jx. 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. See o&w chapter 8. 6 (b) c1(j ) comes from fourier series table #6 with w=5, x0=1, t0 = 20 table 3. 1 #2 (time shift) and. Table 4. 2#1. a0=(5/20)e-j(0)(2 /20)(-2. 5)= 5/20 = 1/4, a1=(5/20)sinc(5(1)(2 /20/2)) e-j(1)(2 /20)(-2. 5) =(1/4)sinc( /4)ej /4, 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), Y(j ) = (1/2 )x(j )*c1(j ), so it will be the x(j ) spectrum repeated at intervals of s=2 /t with amplitudes (1/2 )(2 ak)(2) (x(j ) has a max amplitude of 2) 4 /20 (c) this is time division multiplexing because we are modulating with a square pulse train.

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