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In a simple linear system with feedback (figure 1), the closed-loop transfer function $H(s)$ can be written as $$ H(s)=\frac{X_o(s)}{X_i(s)} = \frac{G(s)}{1+G(s)F(s)} $$ by solving the equations $$ \begin{align} X_o(s)&=G(s)X_e\\ X_e(s)&=X_i(s)-X_f(s)\\ X_f(s)&=X_o(s)F(s) \end{align} $$ with respect to $X_o(s)$ and dividing with $X_i(s)$ on both sides as the last step. (Substitute $X_f(s)$ in the second equation by the third equation, and then substitute $X_e(s)$ in the first equation by the second equation).

Closed-loop system in the $s$-domain

It is then possible to compute the impulse response $h(t)$ and the unit step response $h_u(t)$ by the inverse Laplace transform: $$ \begin{align} h(t) &= \mathcal{L}^{-1}\{H(s)\}\\ h_u(t) &= \mathcal{L}^{-1}\left\{\frac{1}{s} H(s)\right\} \end{align} $$ I would like to do the same in the time domain (figure 2). Suppose $g(t)$ and $f(t)$ are known impulse responses for the transfer functions $G(s)$ and $F(s)$ respectively, I want to get an expression for $h(t)$ (the impulse response of the closed-loop transfer function $H(s)$) without going to the $s$-domain and back to the $t$-domain.

Closed-loop system in the time domain

I suppose one could do that by the convolution theorem $$ \mathcal{L}^{-1}\{F(s)G(s)\} = f(t) * g(t) $$ but I have trouble writing an expression for $h(t)$ in terms of $f(t)$, $g(t)$ and the convolution operator: $$ h(t) = \mathcal{L}^{-1}\left\{\frac{G(s)}{1+G(s)F(s)} \right\} = \ldots ? $$ I have found no book or website explaining how to do this so I guess it is not possible? If that is true, I would like to know why. Also I am unsure if $h(t)=x_o(t)/x_i(t)$. Actually I am a bit confused by the Laplace transform in general.

Thanks in advance :-)

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First of all, the impulse response is definitely not equal to $x_o(t)/x_i(t)$. In the time domain, the output signal $x_o(t)$ can be expressed by an integral equation. With

$$x_e(t)=x_i(t)-(x_o\star f)(t)\tag{1}$$

the output signal is given by

$$x_o(t)=(x_e\star g)(t)=(x_i\star g)(t)-(x_o\star f\star g)(t)\tag{2}$$

where $\star$ denotes convolution. With $a(t)=(f\star g)(t)$, Eq. $(2)$ can be written as

$$x_o(t)+\int_{-\infty}^{\infty}x_o(\tau)a(t-\tau)d\tau=\int_{-\infty}^{\infty}x_i(\tau)g(t-\tau)d\tau\tag{3}$$

An integral equation for the impulse response $h(t)$ is obtained from $(3)$ for $x_i(t)=\delta(t)$:

$$h(t)+\int_{-\infty}^{\infty}h(\tau)a(t-\tau)d\tau=g(t)\tag{4}$$

However, in the time domain there is generally no way to get an explicit expression for $h(t)$ (or $x_o(t)$ in general), that's why the Laplace transform is so useful because it transforms integral or differential equations into algebraic ones.

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  • $\begingroup$ Thank you very much for your response, Matt L. I do have one question, though: In case that the one-sided Laplace transform is used, is it possible to compute the convolution by integrating from 0 to t instead of +/- infinity? $\endgroup$
    – pvh1987
    Jan 11 '16 at 13:39
  • $\begingroup$ @pvh1987: Yes, if all signals are zero for $t<0$ the convolution integral actually becomes $$\int_0^tf(\tau)g(t-\tau)d\tau$$ $\endgroup$
    – Matt L.
    Jan 11 '16 at 15:07
  • $\begingroup$ OK - thanks again :-) $\endgroup$
    – pvh1987
    Jan 12 '16 at 21:54

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