Quiz1sol_FA08 - UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN...

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Unformatted text preview: UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN Department of Electrical and Computer Engineering ECE 410 Digital Signal Processing Quiz Number 1 Thursday, September 4, 2008 Student Name: Solutions Section: Prof. Bresler / Prof. Jones NOTE: You may not use any calculators, cell phones, earphones (or other forms of electronic media) on this quiz. You may use one side of one sheet of handwritten notes. Problem 1 (10 points) Given that 1 0 0 0 plot the following discrete-time signal. Label the horizontal and vertical axes for full credit. 4 1 We have 4 and 1 From the unit circle, we know that Thus, 1 1 0 4 1 0 1. 1 0 1 1 4 0 0 4 Problem 2 (30 points) Find the Continuous Time Fourier Transform of the pulsed cosine: cos Ω 1, where . You may do it directly using the CT Fourier Transform formula, or 0, use the convolution property in the frequency domain. Using the CT Fourier Transform formula directly, we have: / Ω / 1 2 / 1 2 1 2 / / 1 2 Ω sin Ω 2 /2 Ω / 1 2Ω /2 Ω Ω 2Ω /2 Ω Ω Ω 2 Ω sin 1 Ω Ω 2 /2 Ω Ω 2 2 Ω Else, one can realize that since we know: cos Ω Ω 2 and multiplication in the time domain is equivalent to convolution in the frequency domain, we get: Ω 2 2 Ω Ω 2 Ω Ω Ω Ω which intuitively corresponds to two shifted (and scaled) ‘sinc’ functions in the frequency domain. Also, since the sinc() function is symmetric, this is also a valid answer: Ω 2 Ω 2 Ω 2 Ω Problem 3 (a) (10 points) Compute the Continuous Time Fourier Transform of 1, 0, using the CT Fourier Transform formula. (b) (10 points) Find the time-domain function, f(t), from Ω 1, |Ω| 0, |Ω| Ω Ω using the inverse CT Fourier Transform formula. (a) Using the CT Fourier Transform formula, we have: Ω 12 Ω2 1 2 Ω 1 Ω 2 sin Ω Ω Ω Ω 1 Ω 2 Ω 2 (b) Using the inverse CT Fourier Transform formula, we have: 1 2 1 2 1 2 sin Ω 1 2 Ω 1 Ω Ω Ω 1 Ω Ω Ω Ω Problem 4 (20 points) Find the Continuous Time Fourier Transform of where K is a constant. From the continuous time Fourier transform formula, we have (using the sifting property of impulse functions): Ω Problem 5 (20 points) Simplify the following expression 1 6 3 √3 Give your solution both in rectangular and polar form. Note that 1 6 Thus, 1 6 and 1 3 2 6 1 3 6 √2 2 3 Therefore, the two first terms cancel out, and we’re left with: √3 2 ...
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