Lecture 14-15 MEEN 364: Transient Response of First and Second Order Systems - Prof. Alexa, Study notes of Mechanical Engineering

Download Lecture 14-15 MEEN 364: Transient Response of First and Second Order Systems - Prof. Alexa and more Study notes Mechanical Engineering in PDF only on Docsity! MEEN 364 Parasuram Lecture 14,15 August 22, 2001 1 HANDOUT E.15 - EXAMPLES ON TRANSIENT RESPONSE OF FIRST AND SECOND ORDER SYSTEMS, SYSTEM DAMPING AND NATURAL FREQUENCY Example 1 In the system shown below, x(t) is the input displacement and θ(t) is the output angular displacement. Assume all masses involved are negligibly small and that all motions are restricted to be small. Obtain the response of the system for a unit step input. Assume zero initial conditions. Writing the force balance equation for the above system, we get ,)( .. θθ kLLxb =− . .. xL b kL =+ θθ (1) Equation (1) represents the governing differential equation of motion. Taking the Laplace transforms of equation (1), we get ),()( ssXsL b kLs =Θ     + . )( 1 )( )( b ks s LsX s + =Θ⇒ (2) MEEN 364 Parasuram Lecture 14,15 August 22, 2001 2 Equation (2) represents the transfer function of the system shown above. For a unit step input ssX 1)( = , therefore the output Θ(s) becomes, . )( 11)( b ksL s + =Θ (4) Taking the inverse Laplace transform of equation (4), we get tb k e L t )(1)( −=θ . (5) Equation (5) represents the response of the system for a given step input. The MATLAB sequence to obtain the step response for a given ‘L’, ‘k’ and ‘b’ is given below. L = 2; k = 100; b = 20; num = 1/L; den = [1 k/b]; sys = tf(num,den); step(sys) xlabel('Time') ylabel('Angular Displacement, Theta') Title('Step response of a first order system') The response of the system is as shown below. MEEN 364 Parasuram Lecture 14,15 August 22, 2001 5 Example 3 When the system as shown in Figure (a) is subjected to a unit step input, the system output responds as shown in Figure (b). Determine the values of ‘K’ and ‘T’ from the response curve. The maximum overshoot from the response curve is 25.4%. Therefore .4.0 ,254.0 ,254.0 21 =⇒ =⇒ = − − ζ ζ ζπ e M p MEEN 364 Parasuram Lecture 14,15 August 22, 2001 6 From the response curve we have .14.1 ,3 )4.0(11 ,3 22 =⇒ = − = − ==⇒ = n nnd p p t t ω ω π ζω π ω π From the block diagram, the closed loop transfer function is . )( )( 2 KsTs K sR sC ++ = Hence .12, TT K nn == ωζω Therefore the values of ‘K’ and ‘T’ can be determined as .42.109.1)14.1( ,09.1 14.14.02 1 2 1 22 =×== = ×× == TK T n n ω ωζ Example 4 Figure (a) shows a mechanical vibratory system. When a 2 N force (step input) is applied to the system, the mass oscillates as shown in Figure (b). Determine ‘m’, ‘b’, and ‘k’ of the system from the response curve. The displacement ‘x’ is measured from the equilibrium position. MEEN 364 Parasuram Lecture 14,15 August 22, 2001 7 The transfer function of the system is .1 )( )( 2 kbsmssP sX ++ = Since for step input of 2 lb, ,2)( ssP = we obtain . )( 2)( 2 kbsmss sX ++ = From the response curve, the steady state value of x is 0.1, hence from the final value theorem, we have ,1.022lim)(lim)( 200 ==++ ==∞ >−>− kkbsms ssXx ss .20 m Nk = From the response curve, the maximum overshoot is 0.0095, hence applying the formula for the maximum overshoot, we get .6.0 ,0095.0 ,0095.0 21 =⇒ =⇒ = − − ζ ζ ζπ e M p MEEN 364 Parasuram Lecture 14,15 August 22, 2001 10 Example 6 Derive the governing differential equation of motion of a swinging bar supported at its ends by a cord. Solve the derived differential equation and plot the initial response of the system for the following initial conditions: a) Arbitrary initial condition Choose l1 = 1m; l2 = 1m; m2 = 5Kg φ l1 l2 , m2 θ The differential equations of motion for the above system when represented in a matrix form is                 −+− −+− =                     − − 2 )sin(sin 2 )sin(sin 32 )cos( 2 )cos( 2. 21222 . 2. 22 2 .. .. 2 22212 22 12 φθφθ φθθφ θ φ φθ φθ llmlw lmw lmllm lmlm Initial response The second order differential equation has to be converted into a first order differential equation. Let Substituting the above relations in the original nonlinear differential equation, we get the following nonlinear first order differential equation, which when represented in matrix form is );4( );3( );2( );1( . . y y y y = = = = θ θ φ φ MEEN 364 Parasuram Lecture 14,15 August 22, 2001 11 MATLAB Code In this type of a problem where the inertia matrix (mass matrix) is a function of the states or the variables, a separate M-file has to be written which incorporates a switch/case programming with a flag case of ‘mass’. For example if the differential equation is of the form, M (t, y) *y’ (t)=F (t, y), then the right hand side of the above equation has to be stored in a separate m-file called ‘F.m’. Similarly the inertia matrix (mass matrix) should also be stored in a separate m-file named ‘M.m’. So, when the flag is set to the default, the function ‘F.m’ is called and later when the flag is set to ‘mass’ the function ‘M.m’ is called. The code with the switch/case is given below. Note that it is a function file and should be saved as ‘indmot_ode.m’ in the current directory. function varargout=indmot_ode(t,y,flag) switch flag case '' %no input flag varargout{1}=FF(t,y); case 'mass' %flag of mass calls MM.m varargout{1}=MM(t,y); otherwise error(['unknown flag ''' flag '''.']); end To store the right hand side of the original matrix form of differential equation, a separate function file must be written as shown below. Note that the name of the function is ‘FF’, so this file must be saved as ‘FF.m’. %the following function contains the right hand side of the %differential equation of the form %M(t,y)*y'=F(t,y) %i.e. it contains F(t,y).it is also stored in a separate file named, FF.m. function yp=FF(t,y) l1=1; l2=1;                 −+− −+− =                                 − − 2 ))1()3(sin()2()3(sin )4( 2 ))1()3(sin()4( )1(sin )2( )4( )3( )2( )1( 3 0 2 ))1()3(cos( 0 0100 2 ))1()3(cos( 00 0001 2 21222 2 22 2 . . . . 2 22212 22 12 yyyllmylw y yyylmyw y y y y y lmyyllm yylmlm MEEN 364 Parasuram Lecture 14,15 August 22, 2001 12 m2=5; g=9.81; w2=m2*g; yp=zeros(4,1); yp(1)=y(2); yp(2)=-w2*sin(y(1))+ (m2*l2/2)*(y(4)^2)*sin(y(3)-y(1)); yp(3)=y(4); yp(4)=(-w2*l2*sin(y(3)))/2+(m2*l1*l2/2)*(y(2)^2)*sin(y(3)-y(1)); Similarly, to store the mass matrix a separate function file is written which is stored as ‘MM.m’. % the following function contains the mass matrix. %it is separately stored in a file named, MM.m function n = MM(t,y) l1=1; l2=1; m2=5; g=9.81; w2=m2*g; n1=[1 0 0 0]; n2=[0 m2*l1 0 (m2*l2/2)*cos(y(3)-y(1))]; n3=[0 0 1 0]; n4=[0 (m2*l1*l2/2)*cos(y(3)-y(1)) 0 m2*l2*l2/3]; n=[n1;n2;n3;n4]; To plot the response, the main file should call the function ‘indmot_ode.m’, which has the switch/case programming which in turn calls the corresponding functions depending on the value of the flag. For the main file to recognize the inertia matrix, the MATLAB command ODESET is used to set the mass to ‘M (t, y)’. tspan=[0 30] y0=[0.5233;0;1.0467;0] % Arbitrary Initial condition options=odeset('mass','M(t,y)') [t,y]=ode113('indmot_ode',tspan,y0,options) subplot(2,1,1) plot(t,y(:,1)) grid xlabel('Time') ylabel('phi') subplot(2,1,2) plot(t,y(:,3)) grid xlabel('Time') ylabel('Theta') The above code plots the values of ‘theta’ and ‘phi’ with respect to time for the arbitrary initial condition case.