pcshw5_soln - THE STATE UNIVERSITY OF NEW JERSEY RUTGERS...

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THE STATE UNIVERSITY OF NEW JERSEY RUTGERS College of Engineering Department of Electrical and Computer Engineering 332:322 Principles of Communications Systems Spring 2004 Problem Set 5 Reading: Haykin 3.1–3.4 1. Nyquist 101: Specify the Nqyusit rate and Nyquist interval for each of the following signals. Note that sinc ( x ) sin( πx ) πx . (a) g ( t ) = sinc (200 t ) SOLUTION: This sinc pulse corresponds to a bandwidth of W = 100 Hz. Hence, the Nyquist rate is 200 Hz, and the Nyquist interval is 1 / 200 seconds. (b) g ( t ) = sinc 2 (200 t ) SOLUTION: This signal may be viewed as the product of the sinc pulse sinc (200 t ) with itself. Since multiplication in time domain corresponds to convolution in frequency domain, we find that the signal g ( t ) has a bandwidth equal to twice that of the sinc pulse sinc (200 t ) , that is 200 Hz. The Nyquist rate of g ( t ) is therefore 400 Hz, and the Nyquist interval is 1 / 400 seconds. (c) g ( t ) = sinc (200 t ) + sinc 2 (200 t ) SOLUTION: The bandwidth of g ( t ) is determined by the highest frequency content of either sinc (200 t ) or sinc 2 (200 t ) . From earlier parts, we know that sinc 2 (200 t ) has the higher bandwidth equal to 200 Hz. Correspondingly, the Nyquist rate is 400 Hz and the Nyquist interval is 1 / 400 seconds. 2.
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pcshw5_soln - THE STATE UNIVERSITY OF NEW JERSEY RUTGERS...

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