solnsmid1_501

solnsmid1_501 - ECE-501 Midterm #1 Introduction to Analog...

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ECE-501 Introduction to Analog and Digital Communications Winter 2008 Midterm #1 Feb. 6, 2008 MIDTERM #1 SOLUTIONS 1. (a) For a path length of d and propagation speed c , the propagation delay equals τ = d/c . So, the direct path delay is τ d = 3 × 10 3 m 3 × 10 8 m/sec = 1 × 10 - 5 sec = 10 μ sec , and the indirect path delay is τ i = 6 × 10 3 m 3 × 10 8 m/sec = 2 × 10 - 5 sec = 20 μ sec . (b) Here we model the channel that produces x ( t ) from s ( t ) as a linear system with im- pulse response h ( t ). In other words, x ( t ) = h ( t ) * s ( t ). Because x ( t ) is a superposition of two delayed versions of s ( t ), one scaled by 1 2 , we can write x ( t ) = s ( t - τ d ) + 1 2 s ( t - τ i ) . Now, if s ( t ) had been the impulse δ ( t ), then the output would have been δ ( t - τ d ) + 1 2 δ ( t - τ i ). Thus, the “impulse response” equals h ( t ) = δ ( t - τ d ) + 1 2 δ ( t - τ i ) . (c) Notice that, for sampling rate 1 T s = 100 kHz, the sampling interval is T s = 1 × 10 - 5 sec = 10 μ sec. Thus, τ d corresponds to 1 sample of delay, while τ i corresponds to 2 samples of delay. The causal sampled impulse response vector that we would pass to plottf would then be h = b 0 , 1 T s , 0 . 5 T s B = [0 , 100000 , 50000].
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solnsmid1_501 - ECE-501 Midterm #1 Introduction to Analog...

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