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- Title: ee752hw1sol
- Type: Notes
- School: Ohio State
- Course: EE 752
- Term: Fall
Coursehero >> Ohio >> Ohio State >> EE 752
Path: Ohio State >> EE >> 752 Fall, 2009
Path: Ohio State >> EE >> 752 Fall, 2009
Path: Ohio State >> EE >> 752 Fall, 2009
Path: Ohio State >> EE >> 752 Fall, 2009
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Homework#1 EE752 Solution 2-1 & With h0 is an equilibrium point for the system with input 0 , ( h(t ) = f (h0 , u0 ) = 0 ). Around this equilibrium point, linearize the system with respect to the first order taylor s series. f f & & h(t ) = h(t ) f (h0 , u0 ) + |u0 u (t ) + |h0 h(t ) u h V0 f 1 f |u = and |h = , substitute into equation above and apply Laplace where u 0 A0 h 0 2 A0 h0 Transform, we will obtain, h( s ) e s = . G(s) = u ( s ) A ( s + V0 ) 0 2 A0 h0 & u (t ) v(t ) The system dynamics h = A(h(t )) v(t ) is known, define u '(t ) = u (t ) v(t ) . Since we need h(t ) hd (t ) as t , the & designed dynamics could be h(t ) + h(t ) = hd (t ) , > 0 ( h(t ) hd (t ) with 0 , what if < 0 ?). A(h(t )) ( h(t ) + hd (t )) with small To achieve the designed dynamics, choose u '(t ) = 2-2 (i) = 0 (why?). (ii) 11 10.5 10 9.5 9 8.5 8 K=27 K=50 7.5 0 20 40 60 80 100 120 140 160 180 Matlab code: clear all; % Given constants A0=300; A1=12; A2=120; Tao=12; % the time delay v0=24; h_d=10; h_0=8; % Feedback gain to be designed K_min=27; % lower bound for K K_max=50; % upper bound for K % Initialize the simulation Ts=0.04; t=0:Ts:180; N=length(t); % number of simulation steps step_of_delay=Tao/Ts; % Using Euler's first order method: X(k+1)=X(k)+Ts*f(k) h_1=zeros(size(t)); % Initialize vector h for K_min case h_1(1)=h_0; h_2=zeros(size(t)); % Initialize h vector for K_max case h_2(1)=h_0; for i=2:N, if i<=step_of_delay u1=0; % control for K_min case u2=0; % control for K_max case else u1=K_min*(h_d-h_1(i-step_of_delay))+v0; % control for K_min case u2=K_max*(h_d-h_2(i-step_of_delay))+v0; % control for K_max case end h_1(i)=h_1(i-1)+Ts*(u1-(v0+20*sin(0.5*t(i-1))))/(A0+A1*h_1(i-... 1)+A2*sqrt(h_1(i-1))); h_2(i)=h_2(i-1)+Ts*(u2-(v0+20*sin(0.5*t(i-1))))/(A0+A1*h_2(i-... 1)+A2*sqrt(h_2(i-1))); end % Plot the results: figure(1) plot(t, h_1, 'b-', t, h_2, 'r--');grid on;zoom on; legend('K=27','K=50', 4); You can compare the performance of different Ks by using the code above. 2-5 P( s) . 1 + P( s )C ( s ) Because this is a linear system, the output is the superposition from the input and the disturbance. Now consider the part of the output from the disturbance. P( s) P( s) 3 yv ( s ) = v( s) = (*) [because v(t ) = sin(3t ) ]. 2 1 + P( s )C ( s ) 1 + P( s )C ( s ) s + 9 From y (t ) 0 as t , we have yv (t ) 0 as t . The transfer function between disturbance v(t ) and output y (t ) is Thus, in (*), s 2 + 9 must be eliminated from the denominator. Otherwise, after the partial expansion, we know there will always be some sinusoidal components in the output. To cancel s 2 + 9 , it is clear that the controller C ( s ) must have s 2 + 9 term in the denominator.
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Description: Autonomous Helicopter Landing A Nonlinear Output Regulation Perspective Andrea Serrani Department of Electrical and Computer Engineering Collaborative Center for Control Sciences The Ohio State University Applications of Nonlinear Output Regulation...
Problem24.pdfPath: Ohio State >> MATH >> 151 Spring, 2008
Description: Problem #24, Section 2.6 Here x , so x is passing through negative values When x < 0 , x 2 = x (remember this while simplifying the denominator) x + x 2 + 2x x x 2 + 2x x + x 2 + 2x x x 2 + 2x lim = lim x 1 1 x x 2 + 2 x x x x 2 + 2x x + ...
solutionHW3.pdfPath: Ohio State >> MATH >> 151 Spring, 2008
Description: Section 5.3, Problem #5 Figure 1 Consider a rectangle with length 11 inches and width 9 inches. Imagine 11 small squares of sides 1 inch placed along the length of the rectangle and 9 small squares of sides 1 inch placed along the width of the recta...
syllabus151.pdfPath: Ohio State >> MATH >> 151 Spring, 2008
Description: http:/www.newark.osu.edu/gnair/ Page 1 of 4 The Ohio State University MATH 151 (5 credits) Summer Quarter 2007 Instructor: Girija Nair-Hart Meeting Time: MTWR, 11:30 am 12:30 pm Office: FH 84 Office Hours: MTWR 12:30 1:30 by appointment Email: n...
hw2.pdfPath: Ohio State >> EE >> 341 Fall, 2009
Description: The Ohio State University Department of Electrical Engineering EE 341 Energy Conversion Home work Set # 2 Print Your Name _ The Last Four Digits of Your OSU I.D. number: _ 1 Problem 1: Consider a 3- distribution system as shown below: Source bus...