hmw8_sol_s08 - i ECE220 Homework Assignment#8 Solutions...

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i ECE220 Homework Assignment #8 Solutions Problem 9.11 Without using any integrals, determine the Fourier series co- efficients, a k , b k and α k , of the following functions. Note the frequency that corresponds to each index k . The time variable is measured in seconds. You will need to determine the period, T , over which to determine the coefficients. Use the shortest possible period. (a) s 1 ( t ) = 0 . 45 cos(7 . 1 t ) (b) s 2 ( t ) = 3 cos(4 . 3 t ) + 2 sin(12 . 9 t ) (c) s 3 ( t ) = 3 . 0 cos(2 π 30 t ) - 1 . 7 cos(2 π 100 t ) (d) s 4 ( t ) = 3 . 0 cos(2 π 30 t ) - 1 . 7 sin(2 π 30 t ) (e) s 5 ( t ) = 20 cos(2 π 30 t ) - 17 cos(2 π 100 t - 5 π/ 6) + 8 cos(2 π 200 t + π/ 4) Solution (a) s 1 ( t ) = . 45 cos(7 . 1 t ) = . 45 cos ( 2 π 2 π/ 7 . 1 ) t The period is = T = 2 π 7 . 1 = 0 . 8850 seconds. We have a k = . 45 , k = 1 0 , o.w. b k = 0 , k = 1 , 2 , ... From the conversion equations (9.7)-(9.9), page 445, we have α k = . 45 2 , k = ± 1 0 , o.w. A plot of the α k coefficients is given in Figure 0.1. It was produced using the problem9 11a.m file in the class locker.
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ii -15 -10 -5 0 5 10 15 0 0.05 0.1 0.15 0.2 0.25 frequency (Hz) |S(n)| Problem 9.11a: The magnitude spectrum -15 -10 -5 0 5 10 15 -1 -0.5 0 0.5 1 frequency (Hz) θ (n) Problem 9.11a: The phase spectrum Figure 0.1. The spectrum in Problem 9.11a. (b) s 2 ( t ) = 3 cos(4 . 3 t ) + 2 sin(12 . 9 t ) = 3 cos( 2 π 2 π/ 4 . 3 t ) + 2 sin( 2 π 2 π/ (3 · 4 . 3) t ) = 3 cos( 2 π · 1 2 π/ 4 . 3 t ) + 2 sin( 2 π · 3 2 π/ 4 . 3 t ) T = 2 π 4 . 3 = 1 . 4612 sec a k = 3 , k = 1 0 , o.w. b k = 2 , k = 3 0 , o.w. From the conversion equations (9.7)-(9.9), page 445, we have α k = 3 2 , k = ± 1 - 2 j 2 = - j , k = 3 j , k = - 3 0 , o.w.
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iii A plot of the α k coefficients is given in Figure 0.2. It was produced using the problem9 11b.m file in the class locker. -8 -6 -4 -2 0 2 4 6 8 0 0.5 1 1.5 frequency (Hz) |S(n)| Problem 9.11b: The magnitude spectrum -8 -6 -4 -2 0 2 4 6 8 -2 -1 0 1 2 frequency (Hz) θ (n) Problem 9.11b: The phase spectrum Figure 0.2. The spectrum in Problem 9.11b. (c) s 3 ( t ) = 3 . 0 cos(2 π 30 t ) - 1 . 7 cos(2 π 100 t ) = 3 . 0 cos( 2 π · 3 1 / 10 t ) - 1 . 7 cos( 2 π · 10 1 / 10 t ) T = 1 10 sec a k = 3 . 0 , k = 3 - 1 . 7 , k = 10 0 , o.w. b k = 0 , 2200 k = 1 , 2 , ... From the conversion equations (9.7)-(9.9), page 445, we have
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iv α k = 3 2 , k = ± 3 - 1 . 7 2 , k = ± 10 0 , o.w. A plot of the α k coefficients is given in Figure 0.3. It was produced using the problem9 11c.m file in the class locker.
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