Q1S - ECE 402 Communications Engineering Quiz 1 Fall 2009...

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ECE 402 Communications Engineering Quiz 1 Fall 2009 KEY NAME: STUDENT NUMBER: INSTRUCTIONS Write your name and student number in the space provided above. This quiz is closed book and closed notes. One handwritten 8.5” × 11” reference sheet may be used (both sides). Conventional calculators may be used (but no computers, PDAs or other wireless devices) Give step-by-step solutions that include all details. Assume the graders know only the basics. Box your final answer . GRADES 1 2 3 4 Total 1
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Problem 1 (25 pts): A signal is digitally recorded using sampling and uniform quantization. Assuming the highest frequency tone to be recorded is 8,000 Hz, what is the minimum sampling rate that can be used? How many bits would be required to store one minute of this signal at a signal- to-quantization noise ratio of at least 80 dB? Assume that the signal is peak-limited to | x ( t ) | ≤ 1 V and has normalized power P x = 0 . 25 W. Solution: The sampling rate must be greater than the Nyquist rate: f s > 2 W = 16 , 000 samples/sec . To achieve an SNR 80 dB, the number of quantizer bits must satisfy 10log 10 ± P x V 2 + 4 . 8 + 6 k 80 dB = 10log 10 ± 1 4 + 6 k 75 . 2 dB = 6 k 81 . 2 dB , so k = 14 is required. To encode this signal requires a data rate of R = kf s = 224 kb/s. Thus for one minute, we require 60 × 224 , 000 = 13 , 440 , 000 bits, or 13 . 44 Mb . 2
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Find the Fourier Transform of x ( t ) = 1 1 + t 2 . Solution: From the table, we have F n e -| t | /T o = 2 T 1+(2 πfT ) 2 , which reduces to F n e - 2 π | t | o = 1 π (1+ f 2 ) for T = 1 / 2 π . From duality, we then obtain F 1 π (1 + t 2 ) ± = e - 2 π | f | = F { x ( t ) } = πe - 2 π | f | . 3
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Q1S - ECE 402 Communications Engineering Quiz 1 Fall 2009...

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