Question 21.38: For a particular circuit it can be shown that the transfer f......
For a particular circuit it can be shown that the transfer function, G(s), is given by
G(s)=\frac{V_{\mathrm{o}}(s)}{V_{\mathrm{i}}(s)}=\frac{1}{s+2}where V_{\mathrm{i}}(s) \text { and } V_{\mathrm{o}}(s) are the Laplace transforms of the input and output voltages respectively.
(a) Find v_{\mathrm{o}}(t) \text { when } v_{\mathrm{i}}(t)=\delta(t).
(b) Use the convolution theorem to find v_0(t) when
v_{\mathrm{i}}(t)= \begin{cases}\mathrm{e}^{-t} & t \geqslant 0 \\ 0 & t<0\end{cases}(a) We are given the transfer function
G(s)=\frac{V_{\mathrm{o}}(s)}{V_{\mathrm{i}}(s)}=\frac{1}{s+2}When v_{\mathrm{i}}(t)=\delta(t), V_{\mathrm{i}}(s)=1 and hence
V_{\mathrm{o}}(s)=\frac{1}{s+2}and
v_{\mathrm{o}}(t)=\mathcal{L}^{-1}\left\{\frac{1}{s+2}\right\}=\mathrm{e}^{-2 t}This is known as the impulse response of the system, g(t). If the impulse response, g(t), is known then the response v_{\mathrm{o}}(t), \text { to any other input, } v_{\mathrm{i}}(t) can be obtained by convolution, that is v_0(t)=g(t) * v_{\mathrm{i}}(t).
(b) The response to an input, v_{\mathrm{i}}(t)=u(t) \mathrm{e}^{-t}, is given by
\begin{aligned} v_{\mathrm{o}}(t) & =g(t) * v_{\mathrm{i}}(t)=\int_0^t g(\lambda) v_{\mathrm{i}}(t-\lambda) \mathrm{d} \lambda \\ & =\int_0^t \mathrm{e}^{-2 \lambda} \mathrm{e}^{-(t-\lambda)} \mathrm{d} \lambda \\ & =\mathrm{e}^{-t} \int_0^t \mathrm{e}^{-\lambda} \mathrm{d} \lambda \\ & =\mathrm{e}^{-t}\left[\frac{\mathrm{e}^{-\lambda}}{-1}\right]_0^t \\ & =\mathrm{e}^{-t}\left(1-\mathrm{e}^{-t}\right) \\ & =\mathrm{e}^{-t}-\mathrm{e}^{-2 t} \end{aligned}