Comparing Simulink simulation with Laplace transform solution

jackalsniper · Aug 9, 2009

the question goes like this:

Typeset differential equation 5 d²y/dt² = 5 - 2 dy/dt - 5y(t), with initial conditions y(0) = 0 and y(0)' = 0.

the first part asks to utilize simulink to obtain the graph pattern:

Simulink block diagram: Constant 5 feeds a summing junction with signs, then Gain3 (1/5) into Integrator1 (1/s), then Integrator2 (1/s); Gain2 (2) feedback from Integrator2 output to the summing junction, Gain4 (5) feedback from Integrator2 output to the junction, with outputs to a Scope and To Workspace block (y).

Simulink scope plot of a step-response curve: output rises from 0 to a peak near 1.5 around t = 3, dips to about 0.7 near t = 6, then rises again to about 1.15 at t = 10, with axes running from 0 to 30 in time.

the second part requires us to make use of laplace transform and inverse laplace transform to solve the initial value problem:

MATLAB Command Window session: symbolic setup of the ODE 5y'' + 2y' + 5y - 5 with Laplace transform, solve for Y giving 5*(D(y)(0)*s+1)/s/(5*s^2+2*s+5), then ilaplace(Y,s,t) returning Y*dirac(t).

the third part asks us to compare both the results and comment, but how do i compare them

i'm not sure if the images could display or not, sorry
let me know if the images don't display, thanks

jackalsniper · Aug 9, 2009

bump, visible image now

trambolin · Aug 10, 2009

Compare means plot the second part's solution and compare it with the previous plot. And your solution does not seem correct with looking at it 0.5 seconds. So i didn't check.

Comparing Simulink simulation with Laplace transform solution

Discussion

Similar threads