ECE405HW6Su2010Sol

ECE405HW6Su2010Sol - ECE 405 HOMEWORK#6 Solutions SU 2010 DR JAMES S KANG 1(a x(t = 2 cos(210t 2 1 x(t 0-1-2-0.2-0.15-0.1-0.05 0 t 0.05 0.1 0.15

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ECE 405 HOMEWORK #6 Solutions SU 2010 ©DR. JAMES S. KANG 1 (a) x(t) = 2 cos(2 π ×10t) -0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.2 -2 -1 0 1 2 t x(t) (b) X(f) = δ (f 10) + δ (f + 10) -100 -80 -60 -40 -20 0 20 40 60 80 100 0 0.2 0.4 0.6 0.8 1 f X(f) (c) x(k) = 2 cos(2 π ×10×T s ×k) -0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.2 0 1 2 t xs(t) (d) () 1 Ss n s X ff T δ =−∞ =− n f 1
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-100 -80 -60 -40 -20 0 20 40 60 80 100 0 10 20 30 40 50 f Xs(f) (e) The ideal lowpass filter has a bandwidth of f s /2 Hz and gain of T s . -100 -80 -60 -40 -20 0 20 40 60 80 100 0 0.2 0.4 0.6 0.8 1 f Y(f) (f) y(t) = x(t) -0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.2 -2 -1 0 1 2 t y(t) 2 T s = 1/15 (a) x(t) = 2 cos(2 π ×10t) 2
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-0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.2 -2 -1 0 1 2 t x(t) (b) X(f) = δ (f 10) + δ (f + 10) -100 -80 -60 -40 -20 0 20 40 60 80 100 0 0.2 0.4 0.6 0.8 1 f X(f) (c) x(k) = 2 cos(2 π ×10×T s ×k) -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 0 1 2 t xs(t) (d) () 1 Ss n s X ff T δ =−∞ =− n f -30 -20 -10 0 10 20 30 0 5 10 15 f Xs(f) 3
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(e) The ideal lowpass filter has a bandwidth of f s /2 Hz and gain of T s . -30 -20 -10 0 10 20 30 0 0.2 0.4 0.6 0.8 1 f Y(f) (f) y(t) = 2 cos(2 π ×5×t) -0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.2 -2 -1 0 1 2 t y(t) 3 T s = 1/10 (a) x(t) = 2 cos(2 π ×10t) -0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.2 0 1 2 t x(t) (b) X(f) = δ (f 10) + δ (f + 10) 4
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-100 -80 -60 -40 -20 0 20 40 60 80 100 0 0.2 0.4 0.6 0.8 1 f X(f) (c) x(k) = 2 cos(2 π ×10×T s ×k) -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 0 0.5 1 1.5 2 t xs(t) (d) () 1 Ss n s X ff T δ =−∞ =− n f -20 -15 -10 -5 0 5 10 15 20 0 5 10 15 20 f Xs(f) (e) The ideal lowpass filter has a bandwidth of 25 Hz and gain of T s . 5
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-20 -15 -10 -5 0 5 10 15 20 0 0.5 1 1.5 2 f Y(f) (f) y(t) = 2 -0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.2 1 1.5 2 2.5 3 t y(t) 4 x(t) = 4 cos(2 π ×20t) + 2 cos(2 π ×10t), T s = 1/50.
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This note was uploaded on 08/09/2010 for the course ECE ECE 405 taught by Professor Jameskang during the Summer '10 term at Cal Poly Pomona.

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ECE405HW6Su2010Sol - ECE 405 HOMEWORK#6 Solutions SU 2010 DR JAMES S KANG 1(a x(t = 2 cos(210t 2 1 x(t 0-1-2-0.2-0.15-0.1-0.05 0 t 0.05 0.1 0.15

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