# HW1s - AE 353 HW 1 Solutions 1.1 We consider an...

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AE 353: HW 1 Solutions 1.1 We consider an air-conditioning system installed in a building. Physical input would be temperature setting desired by the users. Output would be the actual temperature of the room. A thermometer could be used as a sensor. The sensor function H ( s ) would be a ff ected by the building geometry and location of the sen- sor. Controller could be any device which controls volume rate of cool air pumped into the room depending on input it receives. The plant is the building itself–the dynamics of air blown in, shape of the room would all be represented in G ( s ). External disturbances such as opening of doors by people entering or leaving the room would be represented in D ( s ). The following link contains some information on attitude control: http://en.wikipedia.org/wiki/Attitude_control 1.2 Using the superposition property of linear systems we can compute the transfer functions of output due to each input individually and then obtain the overall transfer function by adding the individual contributions. In what follows we do not indicate the functional dependence on s , sometimes, for the sake of brevity i.e. we write R for R ( s ) etc. From C ( s ) to R ( s ): Set D = N = 0. Then C = KGE = KG ( R - B ) = KGR - KGHC . On rearranging we have C ( s ) = K ( s ) G ( s ) 1 + K ( s ) G ( s ) H ( s ) R ( s ). From C ( s ) to D ( s ): Set R = N = 0. Then C = ( D + EK ) G = DG - CHKG . On rearranging we have C ( s ) = G ( s ) 1 + K ( s ) G ( s ) H ( s ) D ( s ). From C ( s ) to N ( s ): Set R = D = 0. Then C = GKE = - GKH ( C + N ) = - GKHC - GKHN . On rearranging we have C ( s ) = - G ( s ) K ( s ) H ( s ) 1 + K ( s ) G ( s ) H ( s ) N ( s ). Overall transfer function is then C ( s ) = K ( s ) G ( s ) 1 + K ( s ) G ( s ) H ( s ) R ( s ) + G ( s ) 1 + K ( s ) G ( s ) H ( s ) D ( s ) + - G ( s ) K ( s ) H ( s ) 1 + K ( s ) G ( s ) H ( s ) N ( s ) 1.3 N = D = 0. H ( s ) = 1. Now, E ( s ) = R ( s ) - B ( s ) = R ( s ) - C ( s ) = R ( s ) - KG ( s ) 1 + KG ( s ) R ( s ) = 1 1 + KG ( s ) R ( s ) Page 1 of 6

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AE 353: HW 1 Solutions 1.4 The steady state error is denoted by e ss . We use the final value theorem. e ss = lim t →∞ e ( t ) = lim s 0 sE ( s ) = lim s 0 s 1 1 + KG ( s ) 1 s = lim s 0 1 1 + KG ( s ) Then e ss 1 = lim s 0 1 1 + K 1 s 2 + 2 = 2 2 + K ***The above is not a correct application of FVT theorem. Assume K > 0. Consider E ( s ). E ( s ) = R ( s ) 1 + KG 1 ( s ) = 1 1 + K 1 s 2 + 2 1 s = 1 K + 2 2 s + K 2 1 s + j 2 + K + 1 s - j 2 + K On taking the inverse Laplace Transform we find e 1 ( t ) = 2 K + 2 + K K + 2 cos( 2 + Kt ) Note that the poles were on the imagainary axis. Therefore the error associated with these terms still persists and as such there is no steady state error. One should apply FVT only if all the other poles are in the LHP. Why? We know that if for a
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