20072ee132A_1_ca4_sol

20072ee132A_1_ca4_so - EE 132A Spring 2007 Prof John Villasenor TA Choo Chin(Jeffrey Tan Communication Systems Handout#23 Computer Assignment 4

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EE 132A, Spring 2007 Communication Systems Prof. John Villasenor Handout #23 TA: Choo Chin (Jeffrey) Tan Computer Assignment 4 Solutions Part 1 a) Figure 1 shows the PAM sent and received signals. Figure 1 b) Figure 2 shows an eye diagram of the PAM received signal. The signal clearly contains ISI, which reduces the aperture of the “eye” by a factor of 2. This is a caused by the spreading introduced by the channel impulse response, which extends beyond one symbol time. Figure 2
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Part 2 a) Figure 3a shows the real part of the four subcarriers in a block interval. These waveforms are orthogonal in the interval 2<t<6. b) Figure 3b shows the magnitude of the DFTs of the four subcarriers. c) Figure 3c shows the magnitude of the DFTs of the four subcarriers considering only the interval 2<t<6. We can see that in figure 3b the carriers are not orthogonal and interfere with each other at the peak frequencies. Hence the signal is not free from Inter-Carrier Interference (ICI). In figure 3c, the carriers are orthogonal, and there is no ICI. This case corresponds to OFDM without a cyclic prefix. Hence, the cyclic prefix is introducing ICI, but in general we can make this interference small by choosing T cp <<T. Figure 3 d) The PAR obtained in this case is PAR=4 (a value of 4.0336 is obtained if using a discrete sample time of 0.1). A high PAR is undesirable because large signal amplitudes may saturate the power amplifier at the transmitter, distorting the signal.
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Part 3 a) Figure 4 shows the transmitted and received signals.
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This note was uploaded on 11/07/2009 for the course EE 132A taught by Professor Walker during the Spring '08 term at UCLA.

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20072ee132A_1_ca4_so - EE 132A Spring 2007 Prof John Villasenor TA Choo Chin(Jeffrey Tan Communication Systems Handout#23 Computer Assignment 4

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