ECE
Exam 3.pdf

# Detailed solution the analog signal x t 10 e 50 t

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DETAILED SOLUTION: The analog signal x ( t ) = 10 e - 50 t cos(2 π 50 t ), the sampling waveform s δ ( t ), and the resulting impulse-sampled signal x δ ( t ) are shown below for f s = 250 Hz Note: With f s = 250 we have a sampling interval of T s = 1 / 250 = 0.004 sec. Additionally, only the impulses in x δ ( t ) in the range 0 t 15 msec have been shown: t, msec t, msec ( ) s t δ t, msec ( ) x t δ ( ) x t 0 . . . 4 . . . 1 - 5.42 -6 . . . 5 T s = 0.004 sec 8 16 12 2.52 - 4.43 - 5.42 2.52 10 - 4.43 10 The values of x ( t ) at the sample times in the figure are given by x [ m ] = x ( t ) vextendsingle vextendsingle vextendsingle t = m T s = 10 e - 50 m T s cos parenleftBig 2 π 50 mT s parenrightBig ( a ) Substituting T s = 0 . 004 into ( a ) then gives x [ m ] = 10 e - 50 m (0 . 004) cos parenleftBig 2 π 50 m (0 . 004) parenrightBig = 10 e - 0 . 2 m cos parenleftBig 0 . 4 π m parenrightBig ( b )

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ECE 301-001 Spring 2018 Test 3 w/Answers 7 Problem 3 (cont.) Evaluating ( b ) for m = 0 , 1 , 2 , 3 then gives the x [ m ] in the range 0 t 5 msec: x [0] = 10 e - 0 . 2(0) cos parenleftBig 0 . 4 π (0) parenrightBig = 10 e (0) cos(0) = 10 . 00 x [1] = 10 e - 0 . 2(1) cos parenleftBig 0 . 4 π (1) parenrightBig = 10 e - 0 . 2 cos(0 . 4 π ) = 2 . 52 x [2] = 10 e - 0 . 2(2) cos parenleftBig 0 . 4 π (2) parenrightBig = 10 e - 0 . 4 cos(0 . 8 π ) = - 5 . 42 x [3] = 10 e - 0 . 2(3) cos parenleftBig 0 . 4 π (3) parenrightBig = 10 e - 0 . 6 cos(0 . 8 π ) = - 4 . 43 ( c ) The impulse response of the filter h ( t ) is a rectangle lasting 0 . 5 T s = 0.002 sec. This h ( t ) is shown in the figure below: 0.5 T s = 0.002 t = 0 . h(t) 1 t Therefore, the output p ( t ) is the convolution of the above h ( t ) with the x δ ( t ) shown on the pre- ceeding page. The result is shown below: 0 t = t, ms 2 - 4.43 - 5.42 12 10 2.52 ( ) p t 6 4
ECE 301-001 Spring 2018 Test 3 w/Answers 8 4. Let time-domain signal x ( t ) have frequency-domain spectrum X ( f ) given by X ( f ) = 10 parenleftbigg | f | 1000 parenrightbigg , | f | < 600 0 , else The signal x ( t ) is sampled at times t = m T s to produce the digital signal x [ m ] = x ( m T s ). Compute the largest T s value such that x ( t ) can be reconstructed exactly from the x [ m ]. DETAILED SOLUTION: First, plot the spectrum X ( f ) given in the problem statement: ( ) X f Hz f, 6 0 600 - 600 From this figure you can see that the maximum frequency in X ( f ) is f max = 600 Hz. Therefore, the sampling frequency f s necessary to reconstruct x ( t ) is f s = 2 f max ( a ) Substituting f max = 600 into ( a ) then gives the necessary sampling frequency f s : f s = 2 (600) = 1200 ( b ) The sampling interval T s is the inverse of the sampling frequency: T s = 1 f s ( c ) Substituting ( b ) into ( c ) then gives the desired answer: T s = 1 1200 = 0.833 msec ( d )
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