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## DYNAMIC SYST AND CONTROL

by: Orrin Weissnat

69

0

3

# DYNAMIC SYST AND CONTROL MEEN 364

Orrin Weissnat
Texas A&M
GPA 3.58

Alexander Parlos

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COURSE
PROF.
Alexander Parlos
TYPE
Class Notes
PAGES
3
WORDS
KARMA
25 ?

## Popular in Mechanical Engineering

This 3 page Class Notes was uploaded by Orrin Weissnat on Wednesday October 21, 2015. The Class Notes belongs to MEEN 364 at Texas A&M University taught by Alexander Parlos in Fall. Since its upload, it has received 69 views. For similar materials see /class/225991/meen-364-texas-a-m-university in Mechanical Engineering at Texas A&M University.

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Date Created: 10/21/15
MEEN 364 Recitation 2 September 12 2007 Recitation 2 Due September 12 2007 1 Compute the Laplace transform of the following functions a ft t coswt SOLUTION First find the Laplace transform of coswt in 1 in jax ijax J Ocosmte dt Ej oe e e dt 2640713960 eitsja7dt 2 jg 1647 e xHw s 10 0 sm 0 1 1 I 1 1sjas ju 2s jw 39 sjw 2 sz02 s szwz Using this result we know that Ltftl Keeping this in mind apply a derivative in the sdomain dFS ds F dFs 32 aZ 2s2 ds 32a222 2 Compute the time function given the Laplace transform F S SOLUTION l ss 22 Expand using partial fractions MEEN 364 Recitation 2 September 12 2007 FSZ i2 ss2 3 32 32 1 sFmH Z A s 22Fs572 C d 2 i l il Es2 Fss2ds3 2 B 24s 4s2 2s22 1 1722 1 722 t e te f 4 4 2 3 Consider the analog signal xa t 20 cos600721 12sin300721 7cos100m a Determine the minimum sampling rate for xa I to avoid aliasing b Suppose the signal is sampled at F5250 samples per second What is the discrete time signal obtained after sampling a The frequency components in the above signal are f1 300 HZ 27 f2 300 150 Hz 27 1007r 50Hz f3 27 The maximum frequency component in the signal is fmax 300HZ According to sampling theorem the minimum samplingrate required to avoid aliasing should be fs gt 2me 600 Hz Thus fs 600 Hz Ans b MEEN 364 Recitation 2 September 12 2007 The discrete time signal obtained by sampling at 250 samples per second is xn 20 cos 600 n 12 sin 300 n 7cos 100 n 250 250 250 3 xn 20 cos 27r n 12 sin 27r n 7 cos 27r n 1 2 1 3 xn 20 cos27r1 12 s1n27r1 7 cos27r g n 1 2 1 3 xn 20 cos27r g n 12 s1n27r 7 cos27r g 71 Finally we obtain xn 13 COS27L YI 12sin2 n

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