.$$ According to our previous assumptions, we require $K<K_c\equiv
2(\Delta +\xi )$ so that the $R=0$ state is stable, and $(\Delta
-K/2)\tau - \xi \tau \gg 1$ so that we can neglect contributions from the pole at the root $D(s)=0$ in our approximation of (31) by (32). Using Eq. (36) and the $n=-1$ contribution to $f^{(2)}_{1,e}$ (i.e., the first term in (33)) in Eq. (32) we obtain for the echo term at $t=2\tau $ (denoted $R^{(2)*}_{2\tau
}(\epsilon )$), $$R^{(2)*}_{2\tau }(t)=2ih^*_1h_2d_0d_1\Delta
\int^{+\infty}_{-\infty} \frac{d\omega }{2\pi i}\cdot
\frac{\exp[-i\omega (t-2\tau )-\xi t]}{[(\omega -\bar \omega
)-i(\Delta -\frac{K}{2})][(\omega -\bar \omega )+i( \Delta
-\frac{K}{2})]}\ .$$ For $t>2\tau $ $(t<2\tau )$ the integrand exponentially approaches zero as $Im(\omega )\rightarrow -\infty $ $(Im(\omega )\rightarrow
+\infty)$, and we can therefore close the integration path with a large semicircle in the lower half $\omega $-plane (upper half $\omega $-plane). Thus the integral (38) is evaluated from the pole enclosed by the resulting path \[i.e., the pole $\omega
=\omega _0-i(\Delta -\frac{K}{2})$ for $t>2\tau $, and the pole $\omega =\omega _0+i(\Delta -\frac{K}{2})$ for $t<2\tau $\], $$R^{(2)*}_{2\tau}(t)=\frac{h^*_1h_2d_0d_1\Delta }{\Delta
-(K/2)}e^{-i\bar \omega (t-2\tau )-\xi t}e^{-(\Delta
-\frac{K}{2})|t-2\tau |} \ .$$ From Eq. (39) we see that we obtain an echo that is approximately symmetric in shape about 