depend on the presence of [*both*]{} stimulus delta functions (i.e., the delta function $\delta (t)$ of strength $d_0$ and the delta function $\delta
(t-\tau )$ of strength $d_1$), we are interested in the component of $f_1^{(2)}$ that is proportional to the product $d_0d_1$ for $t>\tau $. We denote this component $f^{(2)}_{1,e}$, where the subscript $e$ stands for ‘echo’. From Eq. (23) we see that for $t>\tau $, the $f^{(2)}_{1,e}$ component of $f_1^{(2)}$ satisfied the following initial value problem $$\frac{\partial f_{1,e}^{(2)}}{\partial t}+(i\omega +\xi
)f^{(2)}_{1,e}-\frac{1}{2}KR_e^{(2)*}=0 \ ,$$ $$f^{(2)}_{1,e}(\omega ,\tau ^+)=id_1\sum
^{+\infty}_{n=-\infty}h_{-(n-1)}f^{(1)}_n(\omega ,\tau ^-) \ ,$$ $$R_e^{(2)*}(t)=\int ^{+\infty}_{-\infty} g(\omega
)f^{(2)}_{1,e}(\omega ,t)d\omega \ .$$ Since $f^{(1)}_n(\omega ,\tau^- )$ is proportional to $d_0$ (see Eqs. (19) and (22)), we see that the solution of Eqs. (24)–(26) for $f_{1,e}^{(2)}$ and $R_e^{(2)}$ will indeed be proportional to $d_0d_1$ as desired.

We solve Eqs. (24)–(26) by taking Laplace transforms, $$\hat f^{(2)}_{1,e}(\omega ,s)=\int ^\infty _\tau
e^{-st}f_{1,e}^{(2)}(\omega ,t)dt \ ,$$ $$\hat R^{(2)}_{e*}(s)\equiv \int ^\infty _\tau
e^{-st}R_e^{(2)*}(t)dt \ ,$$ in terms of which we obtain from Eq. (24) $$\hat f^{(2)}_{1,e}(\omega ,s)=\hat R^{(2)}_{e*}\frac{K/2}{s+\xi
+i\omega }+\frac{f^{(2)}_{1,e}(\omega ,\tau ^+)e^{-s\tau }}{s+\xi
+i\omega } \ .$$ Multiplying Eq. (29) by $g(\omega )d\omega $ and integrating from $\omega =-\infty$ to $\omega =+\infty$, then yields $$\hat R^{(2)}_{e*}(s)=\frac{e^{-s\tau }}{D(s)}\int
