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pretty new to this concept, so do bear with me.

A linear dynamic system is described by the following differential equation:enter image description here

Transfer function H(s) is calculated to be = enter image description here enter image description here

I've already found the transfer function by assuming that the initial conditions are set to 0, getting the calculated stuff as shown above. I will then inverse Laplace it in order to find h(t), which turns out to be (exp^(-4t)t)

I will then use the formula, which is : enter image description here

This is what I got, and I was wondering if it is correct:enter image description here

If so, why did the questions give me the initial conditions? What should I do with them? Also, how do I determine whether the system is stable? Unclear of this.

Thanks :)

As per EDIT:

I've converted them into differential equations where by: enter image description here

In this case, how do I find the TF? I'm supposed to divide by X(s) in both sides, but the equation on the right will have a problem won't it?

EDIT 2:

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You can't compute the output by convolving the input with the system's impulse response if the initial conditions are non-zero. The reason is simple: there is no information about the initial conditions in the impulse response or, equivalently, in the system's transfer function.

What you can do is use the following properties of the unilateral Laplace transform:

$$\mathcal{L}\{y'(t)\}=sY(s)-y(0^+)\tag{1}$$

$$\mathcal{L}\{y''(t)\}=s^2Y(s)-sy(0^+)-y'(0^+)\tag{2}$$

If you use $(1)$ and $(2)$ when you transform the differential equation, you'll get an equation for the Laplace transform of $y(t)$ which takes into account the given initial conditions. The inverse Laplace transform of that expression gives you the desired system output for non-zero initial conditions.

As for stability, causal systems with a rational transfer function are stable if all poles (i.e., the zeros of the transfer function's denominator) are inside the left half-plane.

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  • $\begingroup$ Hi there! I edited as per your comment and I'm still stuck at a part when dividing X(S). Thanks :) $\endgroup$ – Jisbon May 30 at 1:46
  • $\begingroup$ @Jisbon: Your second attempt looks good. As for stability, I'm not sure where to start. Didn't you learn anything about system stability, about how transfer function poles relate to stability, etc.? $\endgroup$ – Matt L. May 30 at 8:09
  • $\begingroup$ Unfortunately I don't think so. How do I relate transfer function to stability? $\endgroup$ – Jisbon May 30 at 9:44
  • $\begingroup$ @Jisbon: I've added some information about stability to my answer. Note how the location of the poles is reflected in the exponents of the time domain output. Stability implies decaying transients, and decaying transients occur if the poles are in the left half-plane, i.e., if they have a negative real part. $\endgroup$ – Matt L. May 30 at 10:30
  • $\begingroup$ Hi :) For the transfer function, which equation should I be taking for the transfer function? Should it be the one without initial conditions or the one with the initial condition? Should if it is the one with the initial conditions, how do I get the transfer function? I understand that it is Y(s)/X(s), but it doesn't work in this case does it? $\endgroup$ – Jisbon May 31 at 3:32

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