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Problem Set Solutions

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

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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 S f . (a) s t 2 u t 001 u t 001 , (b) s t 5 u t 5 10 6 u t 5 10 6 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 u t T 2 u t T 2 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 u t 001 u t 001 : If you plot this s t you will see that the duration of the pulse is T 0 002 sec. and that the amplitude is A = 2. Substituting these values into equation a then gives 2 u t 0 002 2 u t 0 002 2 2 0 002 sinc f 0 002 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 0 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.
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ECE 301 Detailed Solutions - Set 8 : Signal Spectrum 2 8-1. (cont.) (b) s t 5 u t 5 10 6 u t 5 10 6 : 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 u t 10 5 2 u t 10 5 2 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.
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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 X f . (a) x t 2 u t 001 u t 003 , (b) x t 5 u t 5 10 6 u t 15 10 6 DETAILED SOLUTION: Each time signal in parts (a) and (b) is a rectangular pulse shifted from t 0. We can compute the transforms of these signals using the following property: If g t G f then g t d e j 2 π f d G f a Therefore, we can find the Fourier transform of the shifted pulse by first finding the transform of the un- shifted pulse and then multiplying the result by the exponential phase term.
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