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 🙂
Best Answer
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.