PracticeProblemsMidterm_solu

T b where 1 is the input frequency and a and b are the

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Unformatted text preview: ! t + b), y2 (t) = 4a sin(! t + b), where ! = 1 is the input frequency and a and b are the gain and phase of P (s) at frequency ! , that is, where and a and are the gain and phase of b are the gain and phase is, where ! = 1 is the input frequency! = 1 is thebinput frequency and a and P (s) at frequency ! , thatof P (s) at frequency p 3j jb p P (j ! ) = ae or p = e j (⇡/6) . 3j 3 2j jb P = ! ) j (⇡aejb or (j e = /6) . = e j (⇡/6) . P (j ! ) = ae or 2 2 Hence, the overall response is Hence, the overall response Hence, the overall response is is s.s. 10 y (t) = y1 (t) + y2 (t) = p + 4 sin(t ⇡ /6) s.s. 10 s.s. 10 3 y (t) = y1 (t) + y2 (t) = p y (t) sin(t (t)⇡ /6)2 (t) = p + 4 sin(t ⇡ /6) + 4 = y1 + y 3 3 10. Sketch y (t) in the previous problem. Solution: The amplitude Sketch y frequency is ! = 1, and the period is T = 2⇡ /! = p⇡ . The 2 10. is 8, the p 10. Sketch y (t) in the previous problem. (t) in the previous problem. whole curve is shifted up Solution: The amplitude. is 8, fu...
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This note was uploaded on 01/27/2014 for the course MAE 171A taught by Professor Idan during the Winter '09 term at UCLA.

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