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132A_1_132A_1_2008SpringMidtermsolutions

# 132A_1_132A_1_2008SpringMidtermsolutions - EE132A Spring...

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EE132A, Spring 2008 Communication Systems Prof. John Villasenor Midterm Exam 1 Solutions Student Name: Student ID: Name of the student sitting to your left: Name of the student sitting to your right: QUESTION SCORE FULL SCORE 1 25 2 25 3 25 4 25 TOTAL 100 Note: Please be sure to clearly indicate your answers, and to show how you obtained them.

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EE132A, Spring 2008 Communication Systems Prof. John Villasenor Midterm Exam 2 1. (25 points) (a) (15 points) Let the Fourier transform of x(t) = sinc 3 (t) be denoted by X(f). Find X(f) in the range < f 2 1 . For full credit your answer must fully describe X(f) in this range. Please note that you are NOT being asked to find X(f) outside this range. Solution: ( ( ( ( 29 ( 29 3 2 ( ) sinc sinc sinc ( ) * x t t t t X f tri f rect f = = = ( f X will be zero for 2 3 f (no overlap between rect and tri). For 1 3 : 2 2 f ( 29 1 1 2 1 1 2 2 2 2 2 2 ( ) 1 1 2 1 1 1 1 2 2 1 1 2 2 2 2 1 1 1 1 3 9 1 2 2 8 2 2 8 f f u X f u du u f f f f f f f f f - - = - = - - - = - - + + = - + = - + - + = - + 2 1 - f 1 f 1 y f = - 2 1 + f 0 1 f
EE132A, Spring 2008 Communication Systems Prof. John Villasenor Midterm Exam 3 (b) (10 points) Find X 2 (f), the Fourier transform of -∞ = - n n t x 2 where x(t) = sinc 3 (t). Solution: This is a convolution in time with an impulse train, which corresponds to a multiplication in frequency by another impulse train: ( 29 -∞ = -∞ = - = - x x n t t x n t x 2 * 2 δ ( 29 ( 29 ( 29 2 2 2 n X f X f f n δ =-∞ = - Since ( f X is zero everywhere except from 2 3 - to 2 3 , there will be only one impulse (at the origin) that will fall where ( f X has nonzero value: ( ( ( 2 0 2 X f X f δ = 2 2 1 -1 f 3 2 3 2 - 3 4 2 0

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EE132A, Spring 2008 Communication Systems Prof. John Villasenor Midterm Exam 4 Now we need to compute ( 0 X .
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132A_1_132A_1_2008SpringMidtermsolutions - EE132A Spring...

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