HW05f09_soln

# HW05f09_soln - SOLUTIONS TO ECE 2025 FALL 2009 PROBLEM...

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1 S OLUTIONS TO ECE 2025 F ALL 2009 P ROBLEM S ET #5 PROBLEM 5.1*: Approach: (a) The equation describes one period; replicating this segment gives the periodic signal (b) The given signal is a scaled and stretched version of a signal whose Fourier series coefficients are already known; we will use the scaling property to find the Fourier series of the given signal. (c) Integrate the signal using the formula for the area of a triangle (d) The spectrum of a periodic signal has lines at integer multiples of the fundamental frequency, with weights given by the Fourier series coefficients. (e) Use Euler to combine the positive- & negative-frequency exponentials into a single cosine The above triangle wave x ( t ) has the same shape as that of Fig. 3-18 on page 55 of the book, let’s call it s ( t ) , but with a different period and amplitude. This means we can write x ( t ) = cs ( dt ) for some constants c and d , namely c = 10 and d = 0.1 . From this we can see that the Fourier series coefficients for x ( t ) are simply related to those of s ( t ) , for if s ( t ) = k a k e jk π t / T 0 , then x ( t ) = cs ( dt ) = c k a k e jk π ( dt )/ T 0 = k a k ´ e jk π t / T 0 ´ has Fourier series coefficients a k ´ = ca k and period T 0 ´ = T 0 / d . We conclude that stretching the signal horizontally (along the time axis) has no impact on the Fourier series coefficients; it merely changes the fundamental period. And scaling the amplitude of a periodic signal scales the Fourier 2 4 6 4 2 6 0 t x ( t ) 10

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2 series coefficients by the same amount. Scaling the Fourier series coefficients from (3.39) by c = 10 yields: a k ´ = . a 0 = x ( t ) dt = ( area under triangle ) = (0.5)( base )( height ) = (0.5)(4)(10) = 5. x ( t ) = k = a k ´ e jk π t / T 0 ´ = a ´ e j π t / T 0 ´ + a ´ e j π t / T 0 ´ + a 0 ´ + a ´ e j π t / T 0 ´ + a ´ e j π t / T 0 ´ = a 0 ´ + a ´ ( e j
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