Problem Set Solutions

Problem Set Solutions - ECE 301 Detailed Solutions - Set 8...

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Unformatted text preview: ECE 301 Detailed Solutions - Set 8 : Signal Spectrum 1 8-1. Several signals s t are given below. For each signal use the Fourier transform to compute the mathematical form for the spectrum S f . Also, plot each s t and j S f j . (a) s t = 2 h u t : 001 u t : 001 i , (b) s t = 5 h u t 5 10 6 u t 5 10 6 i DETAILED SOLUTION: Let A be the amplitude of a time-domain rectangular pulse, and let the pulse extend in time from t = T = 2 to t = T = 2. Using this notation, parts (a) and (b) of this problem can be worked using the following Fourier transform pair: A h u t T 2 u t T 2 i AT sinc f T a In equation a note that T is the duration (in sec) of the time-domain pulse. This will be very helpful in working the problems. To compute the desired transforms, substitute the appropriate T and A into a and plot the resulting waveforms. Notice that you do not have to start from the integral definition of the Fourier transform. (a) s t = 2 h u t : 001 u t : 001 i : If you plot this s t you will see that the duration of the pulse is T = : 002 sec. and that the amplitude is A = 2. Substituting these values into equation a then gives 2 h u t : 002 2 u t : 002 2 i 2 ¢ : 002 sinc f ¢ : 002 b The transform S f , given by the term on the right side of b , may be simplified to give the useful form S f = : 004sinc f 500 c Equation c is useful for plotting the sinc because the zero-crossings of the sinc occur at integer multiples of the sinc argument. That is, with n equal to any integer, the sinc in c is zero when f 500 = n f = 500 n d Some of these values from d are f = 0 (when n = 0), f = ¦ 500 (when n = ¦ 1), f = ¦ 1000 (when n = ¦ 2), and so on, at intervals of 500 Hz. ECE 301 Detailed Solutions - Set 8 : Signal Spectrum 2 8-1. (cont.) (b) s t = 5 h u t 5 10 6 u t 5 10 6 i : If you plot this s t you will see that the duration of the pulse is T = 10 5 sec. and the amplitude is A = 5. Substituting these values into equation a then gives 5 h u t 10 5 2 u t 10 5 2 i 5 ¢ 10 5 sinc f ¢ 10 5 e The transform S f , given by the term on the right side of e , may be simplified to give the useful form S f = 5 10 5 sinc f 10 5 f Equation f is useful for plotting the sinc because the zero-crossings of the sinc occur at integer multiples of the sinc argument. That is, with n equal to any integer, the sinc in f is zero when f 10 5 = n f = n ¢ 10 5 g Some of these values from g are f = 0 (when n = 0), f = ¦ 10 5 (when n = ¦ 1), f = ¦ 2 10 5 (when n = ¦ 2), and so on, at intervals of 10 5 Hz = 100 KHz. ECE 301 Detailed Solutions - Set 8 : Signal Spectrum 3 8-2. Several signals x t are given below. For each, use the Fourier transform to compute the mathematical form for the spectrum X f . Also, plot each x t and j X f j ....
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This note was uploaded on 01/25/2010 for the course ECE 301 taught by Professor Hoozturk during the Spring '08 term at N.C. State.

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Problem Set Solutions - ECE 301 Detailed Solutions - Set 8...

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