I am applying at a Faculty for Electrical Engineering, and I have an entrance exam in two days. I have few exams for exercise, from previous years and I keep getting stuck in one particular type of problem:

"A dynamicaI system is described by a differential equation of the following form: $$y''(t) + 5y'(t)+4y(t) = u(t)$$ where t is the independent variable of time, $u(t)$ is the input signal into the system, and $y(t)$ is the system response."

I need to prove that if the system is excited with stepped amplitude signal of $1$, the output of the system in the steady state will reach a constant value of $0.25$.

Can anyone please suggest how can I solve this problem. I think I should use Laplace Transformation to solve it, but I do not know to how transform the above differential equation in some type that can be solved with Laplace. Also, i do not get where does the amplitude signal of $1$ is used.

Any help or hints are welcomed and appreciated, I need all the help I can get. Thank you.

  • $\begingroup$ Maybe this would suit ELECTRICAL ENGINEERING.SE better? $\endgroup$ – Mr. Xcoder Aug 29 '17 at 9:21
  • $\begingroup$ Thank you for the suggestion, I did not know there is a page for Electrical Engineering. I will try there. $\endgroup$ – mchingoska Aug 29 '17 at 9:33
  • $\begingroup$ The question definitely belongs to math.SE. To solve this, why not use that every solution of the differential equation is $$y(t)=ae^{-t}+be^{-4t}+\frac13e^{-t}\int_0^tu(s)e^sds-\frac13e^{-4t}\int_0^tu(s)e^{4s}ds$$ for some constants $(a,b)$ depending on the initial conditions $(y(0),y'(0))$, and plug in the function $u(t)$ you are interested in, whatever it is? For instance, if $u(t)=1$ for every $t>0$, one gets a fully explicit formula for $y(t)$, from which the limit when $t\to\infty$ is direct... $\endgroup$ – Did Aug 29 '17 at 9:36
  • $\begingroup$ An electrical engineer would probably do Laplace tranformation $$(s^2+5s+4)Y=\frac{1}{s}$$ then do the partial fractions, solve the system and ensure that the limit is as expected. Or maybe use the final value theorem instead if allowed. $\endgroup$ – A.Γ. Aug 29 '17 at 10:05
  • $\begingroup$ I started solving it like that and ended with $$Y=\frac{1}{4}-\frac{1}{3}e^{-t}+\frac{1}{12}e^{-4t}$$ , and I am stuck again. $\endgroup$ – mchingoska Aug 29 '17 at 10:17

To answer your question you do not actually need to solve the differential equation. Indeed, in steady state conditions both $y´´$ and $y´$ will vanish: hence you are (asymptotically) left with the equation $$ 4 y(t) = 1$$ (as $u(t)$ is where the step input function comes into play) from which one concludes $$ y = 1/4 = 0.25$$

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    $\begingroup$ Assuming there is a steady state... $\endgroup$ – Did Aug 29 '17 at 9:40
  • $\begingroup$ @Did, needless to say you are right. I was just wondering if the existence of a steady state was in some way implied by the question formulation itself, which I believe the OP adapted from the exam sheet, "I need to prove that if the system is excited with stepped amplitude signal of 1 , the output of the system in the steady state will reach a constant value of 0.25" $\endgroup$ – An aedonist Aug 29 '17 at 9:42
  • $\begingroup$ Yes, the steady state was given in the question formulation. I did not know that in steady state the condition y'' and y' vanish. @An aedonist, your answer helped me a lot, thank you very much. $\endgroup$ – mchingoska Aug 29 '17 at 9:51
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    $\begingroup$ @KirylPesotski If you feed in the harmonic signal $e^{i\omega t}$ you will get your output $y_p(t)$ as a harmonic oscillation with no existing limit as $t\to\infty$. Your $y_s$ is not a steady state, but an amplitude (frequency response at $\omega$). I do not see your point. $\endgroup$ – A.Γ. Aug 29 '17 at 11:01
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    $\begingroup$ in the harmonic case, the steady state solution does not exist! $\endgroup$ – Kiryl Pesotski Aug 29 '17 at 11:19

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