Question

Model for Evaluation The model used for evaluation is the single degree of freedom lumped mass model defined by second order differential equation with constant coefficients. This model is shown in Figure 1. x(t)m m f(t) Figure 1 - Single Degree of Freedom Model The equation of motion describing this system can easily be shown to be md-x + cdx + kx = f(t) dt dt where m is the mass, c is the damping and k is the stiffness with the displacement, velocity and acceleration and the forcing function as shown. Parameters such as the natural frequency, damped natural frequency, damping factor, critical damping, and techniques for estimating damping are important terms that have been defined as part of the development of class notes. Given Information Mass (m)- lkg Damping constant-2 N.s/m Stiffness 5 N/m Percentage overshoot-20.78 % f(t) unit step x(0) = 0 m x(0) 0 m/s 4 Perform the following tasks in the order specified as follows: 1. Compute On, ,Xss, Xp, tp, tr, ta, tss and identify system characteristics (overdamped, 2. 3. Solve the second order differential equation due to initial conditions using Laplace approach. underdamped, etc). Solve the second order differential equation due to initial conditions using classical ODE approach.

Answer 1

Characteristics of the system:

RiseTime: 0.6903 SettlingTime: 3.7352 SettlingMin: 0.1830 SettlingMax: 0.2416 Overshoot: 20.7866 Undershoot: 0 Peak: 0.2416 P

Manual ODE and Laplace Approach Solution

Solutfon Given Dala- IL 劜(o)-0 Deining the states let, ti- and 1 억 2

χ2. 겟2- Splutiono oDE - Soling χη The solcaltonto his uo be 2eplale x and p and p

Rnd cIan leswe the Lrftial (onditions (9-0 and χ 10%。 wher (x (o), 10 1 0 solofng ustng (aplale approac

Taietng LT 5
Matlab code to get displacement using the above equations :

%% Manual solution us ing ode t0:0.1:10 Xt- (-1/5).*exp (-t).*cos (2.*t) + (-1/10).*exp (-t).*sin (2.*t) + (1/5); -

%% Manual solution using laplace approach % To find inverse laplace transform syms stsym x = 1/ (s*3 + 2*s*2 + 5*s); ilaplace (x) displacement(1/5) ((exp (-t)(cos (2.*t) + (sin (2. t)/2)))/5):
Matlab Solution to above problem:

'eom' is a function defined as follows:

%% Initialisation global ABCD k m c %% Defining the state-space variables D-0: %% Identifying system characteristics sys = ss (A, B, C, D) ; stepinfo (sys) %% Using ode45 to solve the system equations x0 = [0 0]; [time, x dot]ode45 ('eom', [0 10], x0) %% Using laplace approach to solve the system equations [disp, t l] step (sys, 10);

-function X dot = eom (t, x0) global A BCDk m ft=1; gStep input end

Graphical solution comparing all the above solution

As illustrated in the graph above, all the methods present the same solution.

0.25 Manual ODE Solution Manual solution using laplace approach ODE45 solution using MATLAB aplace approach solution using lapiace approach 0.2 0.15 O 0.1 0.05 10 time(s)