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Unformatted text preview: ECE 101 – Linear Systems Fundamentals Problem Set # 5 Solutions November 19, 2009 (solutions by Nick Spiliakos. send questions/comments to [email protected]) Problem 1. (a) x(t) is bandlimited to 20 kHz, so its spectrum is zero for f > 20000 Hz. In radial frequency, that just means the spectrum is zero for  ω  > 20000*2 π = 40000 π rad/s. In Fourier Transform problems, it’s often helpful to draw a picture to see what’s going on, so we’ll do that here. Suppose X(j ω ) looked like the shape shown below (A): (A) (B) The signal then gets corrupted with noise in the range 15kHz < f < 20kHz ↔ 30000 π <  ω  < 40000 π (plot (B) above). Recall that when we sample the signal at rate ω s , the sampled signal’s FT is just the original spectrum, scaled and copied every ω s rad/s. To avoid overlap (i.e. aliasing), we need to ensure, as the Sampling Theorem stipulates, that ω s is fast enough. The only part of the spectrum we’re interested in is the usable (uncorrupted) part, from 30000 π to 30000 π rad/s (15 to 15 kHz). When we sample the signal, as long as we don’t get overlapping onto the usable part, we’re fine. We can permit aliasing on the noise part, because we don’t care about it and are eventually going to filter it out anyway. The smallest sampling rate ω s that gives us copies of the spectrum which don’t overlap with the usable part from 15 to 15 kHz is ω s = 70000 π rad/s , or in Hz, f s = 35kHz . To see this, draw the copies of the spectrum you get, centered every ω s = 70000 π rad/s (every 35 kHz): The copies of the noise overlap each other, which is fine. But the uncorrupted spectrum from 30000...
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 Spring '09
 Digital Signal Processing, Signal Processing, copies, Lowpass filter, sampling rate, rad/s

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