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Response of First Order Systems to Sinusoidal InputFirst Order System - Sinusoidal InputSlide 3Slide 4Slide 5Plotting Frequency responseSlide 7Slide 8Adding terms of Frequency Response2/3/2006 BAE 5413 1Response of First Order Systems to Sinusoidal Input2/3/2006 BAE 5413 2First Order System - Sinusoidal InputConsider the following first order system:11)()()(ssIsOsGDetermine the response of the system if input is a sinusoidal:)sin()( tAtIWhich may be transformed to:22)(sAsIthe system response, O(s), is then:iscisbsassAsO1))(1()(22Solving then for a,b, and c:2/3/2006 BAE 5413 3First Order System - Sinusoidal InputTo solve for a, multiply by s + 1/ , and let s = -1/ 222211AaATo solve for b, multiply by s + i, and let s = -iiiAiiAiiiA2222))(1(2221)(2222iAiiiAb2/3/2006 BAE 5413 4First Order System - Sinusoidal InputUsing the solution in the S&C for a sinusoidal function, the solution becomes: teCBirsiCBirsiCBLrtsin2221Where:BC1tan21122AC2122AB,r = 0  teAtsin1211202222  1tanc will be the complex conjugate of b and is:1)(222iAc2/3/2006 BAE 5413 5First Order System - Sinusoidal InputIt is important to note that the solution is made-up of a transient (the first term) and a non-transient part (the second term).  tAeAtOtsin11)(2222The complete solution is then:Consider simply substuting for s in the original transfer function and solving for)(iG)(iG11111))(Re())(Im()(22222222iGiGiG11)(iiG11111111)(2222222iiiiiiG  11tan))(Re())(Im(tan)(iGiGiGNote, this gives us the non-transient solution2/3/2006 BAE 5413 6Plotting Frequency responseG(i) 20log(G(i))j. /0 011 0 .0 5729 . /1 0 .5 7106 /0.7071 -3 45  /0.1 -20 .84 289   /0.01 -40   .    1tan)(iG11)(22iG2/3/2006 BAE 5413 7Plotting Frequency response-40-35-30-25-20-15-10-500.01 0.1 1 10 100Frequency/ (/)Magnitude Ratio (db)2/3/2006 BAE 5413 8Plotting Frequency response-90-80-70-60-50-40-30-20-1000.01 0.1 1 10 100Frequency/ (/)Phase Lag (degrees)2/3/2006 BAE 5413 9Adding terms of Frequency Response        iGiGiGiG2121log20log20log20        iGiGiGiG2121 iG1 iG2We can simply add terms on the Bode Magnitude plotand on the Bode Phase plot to get total


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O-K-State BAE 5413 - Lecture Notes

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