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I'm trying to get a transfer function of F=ma in the Laplace domain. This should be simple, but yet I'm confused. The transfer function is displacement over force. So, I have two approaches.

  1. First approach: integrate w.r.t. time twice, and laplace transform of that. $$\ddot{x}=F \\ \dot{x}=Ft+c_1 \\ x=\dfrac{1}{2}Ft^2+c_1t+c_2$$ which makes for a transfer function with $c_1$ and $c_2$ zero: $$h(t)=\frac{x}{F}=\frac{1}{2}t^2$$ The right hand side has a Laplace transform of $$H(s)=\frac{1}{s^3}$$

  2. Laplace transform of double integration, which is simply $$H(s)=\frac{1}{s^2}$$

I have heard that the second one is correct, but why is the first one not correct?

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First, I don't think the second one is correct. I'm going to assume that the problem is that of finding the solution to $\ddot{x}=F$ with $F$ constant with time and $x(0)=c_2,\, \dot{x}(0)=c_1$ known. Then, denoting $\mathcal{L}(x)(s)\equiv X(s)$ the LT of $x(t)$ and using standard identities, $$ \mathcal{L}(\ddot{x})(s)=s\mathcal{L}(\dot{x})(s)-\dot{x}(0)=s^2 X(s)-sx(0)-\dot{x}(0). $$ and, by linearity $$ \mathcal{L}(F)=F\mathcal{L}(1)=\frac{F}{s}, $$ so upon acting by LT on both sides of $\ddot{x}=F$, it becomes $$ s^2 X(s)-sc_2-c_1 =\frac{F}{s}. $$ Hence $$ X(s)=\frac{F}{s^3}+\frac{c_1}{s^2}+\frac{c_2}{s}. $$ Thus, for $c_{1,2}=0$, we get $$ H(s)=\frac{X(s)}{F}=\frac{1}{s^3}, $$ and hence $h(t)=t^2/2$, as it should be (recall that $\mathcal{L}(t^n)(s)=n!/s^{n+1}$ for $n$ non-negative integer).

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  • $\begingroup$ Thanks! I was just wondering, was my first option also a correct derivation? Or is it incorrect, or correct but frowned upon? $\endgroup$ – Sanchises Jul 2 '14 at 18:53
  • $\begingroup$ The first method seems correct to me (just straightforward application of Fundamental Theorem of Calculus). In the second method, I believe you mean "LT of second derivative" rather than "LT of double integration". Anyway, there must be a problem with your calculation in 2. - the two approaches must yield same answers, as they are both correct and the answer is obviously unique. We may find the problem if you give more details as to how you obtained $H(s)=1/s^2$ in 2. $\endgroup$ – Kubav Jul 2 '14 at 19:09
  • $\begingroup$ Well, I didn't; my tutor (a student) did, presumably using the identity $\int(x)=>\frac{1}{s}X(s)$, so $\int\int(x)=>\frac{1}{s^2}X(s)$ by their logic... $\endgroup$ – Sanchises Jul 2 '14 at 19:52
  • $\begingroup$ Thanks! That explains a lot, because I couldn't find the fault in that reasoning either. $\endgroup$ – Sanchises Jul 2 '14 at 20:07
  • $\begingroup$ Oh, I see. Well, this also works. More precisely, you have $\mathcal{L}\left(\int_0^t f(\tau)\,d\tau\right)=\frac{1}{s}\mathcal{L}(f)(s)$ so, in sloppy notation (here we tacitly use $c_{1,2}=0$), $X(s)=\mathcal{L}\left(\int\int F\,d t\right)=\frac{1}{s}\mathcal{L}\left(\int F\,d t\right)=\frac{1}{s^2}\mathcal{L}(F)=\frac{F}{s^3}$, which is what we want. $\endgroup$ – Kubav Jul 2 '14 at 20:12

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