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Unformatted text preview: Solution of ECE 316 Test 2 Su08 1. The signal x t ( 29 = 5tri( t 1) 2 t ( 29 is sampled at a rate of 8 samples/second with the first sample (sample number 1) occurring at time t = . (a) What is the numerical value of sample number 6? The signal is a periodic repetition of a triangle function of basewidth 2 shifted to the right by one. The period is 2. f s = 8 T s = 1 / 8 . Sample number 6 occurs at time t = 5 T s = 5 / 8 . This time is within the original triangle so the value is x 5 / 8 ( 29 = 5tri(5 / 8 1) = 5tri 3 / 8 ( 29 = 5 5 / 8 = 25 / 8 = 3.125 (b) What is the numerical value of sample number 63? The samples repeat periodically with period 16 (2 seconds multiplied by 8 samples/second). So the numerical value of sample number 63 occurs at time t = 62 T s = 62 / 8 and that same numerical value occurs at time t = 62 16 3 ( 29 T s = 14 T s = 7 / 4 which is within the original triangle. The numerical value of the sample at t = 7 / 4 is x 7 / 4 ( 29 = 5tri(7 / 4 1) = 5tri 3 / 4 ( 29 = 5 1 / 4 = 5 / 4 = 1.25 2. A signal x t ( 29 is sampled 4 times to produce the signal x n and the sample values are x 0 ,x 1 ,x 2 ,x 3 { } = 7,3, 4, a { } . This set of 4 numbers is the set of input data to the DFT which returns the set X 0 ,X 1 ,X 2 ,X 3 { } . (Be sure to notice that some xs are lower case and some Xs are upper case.) (a) What numerical value of a makes X 1 a purely real number? X k = x n e j 2 kn / N F n = N F 1 Since the DFT is periodic with period 4, X 1 = X 3 = x n e j 3 n /2 n = 3 = 7 + j 3 + 4 ja Therefore a = 3 makes X 1 purely real. (b) Let a = 9 . What is the numerical value of X 29 ? X 29 = X 29 7 4 = X 1 = x n e j n /2 n = 3 = 7 j 3 + 4 + j 9 = 11 + j 6 (c) If X 15 = 9 j 2 , what is the numerical value of X 1 ? X 15 = X 1 = 9 j 2 X 1 = X * 1 = 9 + j 2 3. Find the numerical Nyquist rates of the following signals. (If a signal is not bandlimited write infinity.) (a) x t ( 29 = 11tri t / 4 ( 29 sinc t / 2 ( 29 Time limited signal. Therefore Nyquist rate is infinite. (b) x t ( 29 = 11cos 100 t ( 29 sin 300 t ( 29 X f ( 29 = 11/ 2 ( 29 f 50 ( 29 + f + 50 ( 29 j / 2 ( 29 f + 150 ( 29 f 150 ( 29 X f ( 29 = j 11/ 4 ( 29 f + 100 ( 29 + f + 200 ( 29 f 200 ( 29 f 100 ( 29 The maximum frequency in the signal is 200. Therefore the Nyquist rate is 400....
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This note was uploaded on 09/18/2008 for the course ECE 316 taught by Professor Roberts during the Summer '08 term at University of Tennessee.
 Summer '08
 ROBERTS

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