I am following the MIT open course you can find here. My question is about one of the exercises given as homework in the latter and more specifically I think I am missing something on the concept of "system initially at rest".

  • Homework [exercise 8]: HW2
  • Solutions: SOL2

In the proposed solution he is solving the problem using differential equations. What I am trying to do is to use functionals (as this is new to me). I think I end up missing an integration constant due to the hypothesis of "system initially at rest". My solution goes as follows:

System differential equation: $$ \ddot{y}(t) = x(t)-y(t)$$

We seek the response to a step function i.e. $x(t) = u(t)$ where $u(t)$ is zero for $t<0$ and one elsewhere. Now the system functional is:

$$\frac{Y}{X} = \frac{A^2}{1+A^2} = \frac{i}{2}\left( \frac{A}{1+iA}+ \frac{A}{1-iA} \right)$$

where $Af(t) = \int_{-\infty}^{t} f(\tau) d\tau $ is the accumulator operator as defined in the course. Given the impulse function(al) $\delta(t)$, knowing that $u(t) = A\delta(t)$ and using the commutative property of operators:

$$\frac{Y}{X}u(t) = \frac{iA}{2}\left( \frac{A}{1+iA}+ \frac{A}{1-iA} \right)\delta(t)$$

The fundamental mode being $e^{pt}$ where $p$ is a pole:

$$\frac{Y}{X}u(t) = \frac{iA}{2} \left( e^{-it} - e^{it} \right)u(t) = \frac{i}{2}\left( \frac{e^{-it}}{-i} - \frac{e^{it}}{i}\right) = -cos(t)u(t)$$

While the correct solution is (physically obvious): $\left(1-cos(t)\right) u(t)$. I don't understand where I am doing wrong except for the initial hypothesis "system at rest". Can you tell me where the mistake is and, if the hypothesis is wrong, tell me exactly in what conditions it holds?


1 Answer 1


The mistake is in the final integration of the term


because you seem to ignore the step function $u(t)$. Integrating $(1)$ gives


  • $\begingroup$ Oh yes, I was looking for my mistake in the wrong place! Thank you! Is the concept if system initially at rest in signal processing the same as in physics? $\endgroup$
    – Worldsheep
    Nov 4, 2018 at 11:14
  • $\begingroup$ @Worldsheep: In discrete-time it means that all memory elements are initialized with zeros. In continuous time it means that all capacitor voltages and inductor currents are initially zero. $\endgroup$
    – Matt L.
    Nov 4, 2018 at 11:36
  • $\begingroup$ Ok all potentials at zero and all currents at zeros, hence in the spring model, no potential energy in the spring and no momentum on the mass. Sounds good, thank you! $\endgroup$
    – Worldsheep
    Nov 4, 2018 at 11:49
  • $\begingroup$ @Worldsheep: Yes, that's it. $\endgroup$
    – Matt L.
    Nov 4, 2018 at 12:02

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