ECE 101 lab_solution4

ECE 101 lab_solution4 - ECE 101 Linear Systems...

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ECE 101 – Linear Systems Fundamentals, Winter 2008 Lab # 4 Solutions February 12, 2008 (send questions/comments to psiegel@ucsd.edu) In this lab assignment, you’re given a signal prototype for “dashes” and “dots” in the Morse Code system, and your goal is to decode an encrypted message by using modulation techniques. The message you’re given by Agent 007 is: x(t) = m 1 (t)cos(2 π f 1 t) + m 2 (t)sin(2 π f 2 t) + m 3 (t)sin(2 π f 1 t) m 1 (t), m 2 (t), and m 3 (t) are signals which represent 3 letters in Morse Code. We would like to extract these signals from the total signal x(t). Also at our disposal is a lowpass filter (with polynomial coefficients bf and af). Agent 007 told you, Agent 008, that “The future of technology lies in…”. The rest of the agent’s message is [m 1 (t) m 2 (t) m 3 (t)] in Morse Code. It’s up to you, Agent 008, to decode this message, given the total signal x(t)! (a) Using the Morse Code table, the letter Z is constructed as [ dash dash dot dot ]. This is plotted below: (b) The frequency response of the low-pass filter is (on a log scale):
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(c) When we apply the above low-pass filter to the signals dot and dash , we note that the shape and amplitude of the signals remains essentially intact, with a slight time delay: Because the frequency spectrum of the signals lies mostly within the passband of the lowpass filter, the shape of the signals are mostly unmodified. (d) When the signal is modulated by a cosine, recall from Problem Set 5 (or eq. 4.16 in the Matlab text) that this modifies the spectrum by shifting the spectrum to a copy at plus/minus the frequency of the cosine. That is, if X(j ω ) is the FT of the signal x(t), then modulating x(t) by cos(2 π f 1 t) results in: x(t)cos(2 π f 1 t) Å FT Æ ½X(j( ω – 2 π f 1 )) + ½X(j( ω + 2 π f 1 )) Here, f 1 = 200, so as we see from part (b), shifting the spectrum to ±2 π *200 will put it in the stopband of the filter. Hence, we should expect that filtering dash (t)cos(2 π f 1 t) should essentially wipe out the signal. The signal dash (t)cos(2 π f 1 t) and the output after applying the filter are plotted below, which verifies our expectation (note that the second plot’s vertical axis is scaled by 10 -5 , i.e. the magnitude is small ):
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(e) From Equation (4.16) in the Matlab text and also simply from the Multiplication/Convolution Property of the FT, we know that modulating by a cosine leads to shifted, scaled copies of the original FT: FT{ m(t)cos(2 π f 1 t) } = (1/2 π ) FT{ m(t) } FT{ cos(2 π f 1 t) } (Multiplication Property) = (1/2 π )M(j ω ) ∗π [ δ ( ω – 2 π f 1 ) + δ ( ω – 2 π f 1 )] = ½M(j( ω – 2 π f 1 )) + ½M(j( ω + 2 π f 1 )) Now let’s examine what happens if we modulate again , i.e. multiply m(t) by another cosine or sine.
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ECE 101 lab_solution4 - ECE 101 Linear Systems...

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