E4702 Fall 2004 Midterm Solutions

# E4702 Fall 2004 Midterm Solutions - E4702 Midterm#2...

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Unformatted text preview: E4702 Midterm#2 solutions by Anmo Kim ([email protected]) 1. Consider the triangular spectrum shown below. 2 ( ) WP f 1.0 f W 2 2 (a) Does it satisfy Nyquist’s criterion for no intersymbol interference for the corresponding pulse of duration T b ? ( T b = 1 / 2 W ) Explain. (Solution) - 4pts The Nyquist criterion for the distortionless baseband transmission in the absence of noise is as follows: ∞ X n =-∞ P ( f- nR b ) = T b where R b = 1 /T b . The plotting of this signal can be done by locating the frequency shifted P ( f ) at every 2 W and summing all of them. 2 ( 2 ) W P f nW ¦ 1.0 f W 2 2 After the summation, it becomes a constant T b , which satisfies the Nyquist criterion. (b) What is the corresponding pulse? (Solution) 3pts The corresponding pulse is the inverse Fourier transform of the given spectrum. By taking the inverse Fourier transform of P ( f ) = 1 (2 W ) 2 (2 W- | f | ), if- 2 W ≤ f ≤ 2 W , we get 1 p ( t ) = Z ∞-∞ P ( f ) e j 2 πft df = 2 Z 2 W 1 (2 W ) 2 (2 W- f )cos(2 πft ) df = 1 2 W 2 " • (2 W- f ) sin(2 πft ) 2 πt ‚ 2 W + 1 2 πt Z 2 W sin(2 πft ) df # = 1 4 W 2 πt •- 1 2 πt cos(2 πft ) ‚ 2 W = 1- cos(4 πWt ) 8 W 2 π 2 t 2 = sin(2 πWt ) 2 πWt ¶ 2 = sinc 2 (2 Wt ) = sinc 2 ( t/T b ) (c) Compare the above pulse with that obtained from a rectangular spectrum and a full raised cosine spectrum. Give advantages and disadvantages of each. (Solution) 3pts The signals obtained from the rectangular spectrum and a full raised cosine spectrum are p rect = sinc(2 Wt ) p cosine = sinc(4 Wt ) 1- 16 W 2 t 2-3-2-1 1 2 3 0.2 0.4 0.6 0.8 1 f/W 2WP(f) raised consine triangular...
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## This note was uploaded on 08/05/2008 for the course ELEN E4702 taught by Professor Lazano during the Summer '08 term at Columbia.

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E4702 Fall 2004 Midterm Solutions - E4702 Midterm#2...

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